Two pipes A and B can fill a tank in 18 minutes and 24 minutes respectively. If both the pipes are opened simultaneously, then after how much time should pipe B be closed so that the tank is full in 12 minutes?

Options

A. 6 minutes
B. 9 minutes
C. 8 minutes
D. 5 minutes

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the efficiencies of pipes A and B.
- Pipe A can fill the tank in 18 minutes, so its efficiency is:
\[ \text{Efficiency of A} = \frac{1 \text{ tank}}{18 \text{ minutes}} = \frac{1}{18} \text{ tanks per minute} \]
- Pipe B can fill the tank in 24 minutes, so its efficiency is:
\[ \text{Efficiency of B} = \frac{1 \text{ tank}}{24 \text{ minutes}} = \frac{1}{24} \text{ tanks per minute} \]

### Step 2: Find the combined efficiency of pipes A and B when both are opened.
- The combined efficiency when both pipes are open is:
\[ \text{Combined Efficiency} = \text{Efficiency of A} + \text{Efficiency of B} = \frac{1}{18} + \frac{1}{24} \]
- To add these fractions, we find a common denominator (which is 72):
\[ \frac{1}{18} = \frac{4}{72}, \quad \frac{1}{24} = \frac{3}{72} \]
- Therefore, the combined efficiency is:
\[ \text{Combined Efficiency} = \frac{4}{72} + \frac{3}{72} = \frac{7}{72} \text{ tanks per minute} \]

### Step 3: Calculate the total work done in 12 minutes.
- If both pipes are open for 12 minutes, the total work done is:
\[ \text{Total Work} = \text{Combined Efficiency} \times \text{Time} = \frac{7}{72} \times 12 = \frac{7 \times 12}{72} = \frac{84}{72} = \frac{7}{6} \text{ tanks} \]
- However, since we need to fill only 1 tank, we will focus on how much work is done by each pipe.

### Step 4: Determine how long pipe B should remain open.
- Let \( t \) be the time (in minutes) that both pipes A and B are open together, and after that, only pipe A continues to work for the remaining time \( (12 - t) \).
- The work done by both pipes in \( t \) minutes is:
\[ \text{Work by A and B} = \left(\frac{7}{72}\right) t \]
- The work done by pipe A alone for the remaining \( (12 - t) \) minutes is:
\[ \text{Work by A} = \left(\frac{1}{18}\right)(12 - t) \]
- The total work done should equal 1 tank:
\[ \left(\frac{7}{72}\right) t + \left(\frac{1}{18}\right)(12 - t) = 1 \]

### Step 5: Solve the equation.
- Substitute \(\frac{1}{18}\) with \(\frac{4}{72}\) to have a common denominator:
\[ \left(\frac{7}{72}\right) t + \left(\frac{4}{72}\right)(12 - t) = 1 \]
- Multiply through by 72 to eliminate the fraction:
\[ 7t + 4(12 - t) = 72 \]
- Expand and simplify:
\[ 7t + 48 - 4t = 72 \]
\[ 3t + 48 = 72 \]
\[ 3t = 72 - 48 \]
\[ 3t = 24 \]
\[ t = 8 \]

### Step 6: Conclusion
- Pipe B should be closed after 8 minutes.

### Final Answer:
Pipe B should be closed after **8 minutes**.