Pipes A and B can fill a tank in 12 min and 16 min respectively. Both are kept open for 'n' min and then B is closed and A fills the rest of the tank in 5 min. The time 'n' after which B was closed is:

Options

A. 3 min
B. 2 min
C. 5 min
D. 4 min

Answer

To solve the problem, we need to determine the time 'n' after which pipe B is closed. Here’s the step-by-step solution:

### Step 1: Determine the rates of pipes A and B
- Pipe A can fill the tank in 12 minutes, so its rate of work is:
\[ \text{Rate of A} = \frac{1}{12} \text{ tank/min} \]
- Pipe B can fill the tank in 16 minutes, so its rate of work is:
\[ \text{Rate of B} = \frac{1}{16} \text{ tank/min} \]

### Step 2: Find the combined rate when both pipes are open
- When both pipes A and B are open, their combined rate is:
\[ \text{Combined Rate} = \text{Rate of A} + \text{Rate of B} = \frac{1}{12} + \frac{1}{16} \]
- To add these fractions, find a common denominator (which is 48):
\[ \text{Combined Rate} = \frac{4}{48} + \frac{3}{48} = \frac{7}{48} \text{ tank/min} \]

### Step 3: Calculate the amount of tank filled in 'n' minutes
- If both pipes are open for 'n' minutes, the amount of tank filled in that time is:
\[ \text{Amount filled in n minutes} = \text{Combined Rate} \times n = \frac{7}{48} n \text{ tanks} \]

### Step 4: Determine the amount filled by pipe A in 5 minutes
- After 'n' minutes, pipe B is closed, and pipe A continues to fill the tank for an additional 5 minutes. The amount filled by pipe A in 5 minutes is:
\[ \text{Amount filled by A in 5 minutes} = \text{Rate of A} \times 5 = \frac{1}{12} \times 5 = \frac{5}{12} \text{ tanks} \]

### Step 5: Set up the equation for the total tank filled
- The total amount of the tank filled by both pipes must equal 1 (the whole tank):
\[ \frac{7}{48} n + \frac{5}{12} = 1 \]

### Step 6: Solve the equation
- First, convert \(\frac{5}{12}\) to a fraction with a denominator of 48:
\[ \frac{5}{12} = \frac{20}{48} \]
- Substitute this back into the equation:
\[ \frac{7}{48} n + \frac{20}{48} = 1 \]
- Multiply through by 48 to eliminate the denominator:
\[ 7n + 20 = 48 \]
- Rearranging gives:
\[ 7n = 48 - 20 = 28 \]
- Finally, divide by 7:
\[ n = \frac{28}{7} = 4 \]

### Final Answer
The time 'n' after which pipe B was closed is:
\[ \boxed{4} \text{ minutes} \]