Two pipes A and B can fill a cistern in 12 min and 16 min respectively. Both the pipes are opened together for a certain time but due to some obstruction, the flow of water was restricted to `7/8` of the full flow in pipe A and `5/6` of the full in pipe
B. The obstruction, is removed after some time and the tank is now filled in 3 min from that moment. For how many minutes was the obstruction there?

Options

A. 8 min
B. 3 min
C. 5.6 min
D. 4.5 min

Answer

To solve the problem step by step, we will break it down as follows:

### Step 1: Determine the rates of filling for each pipe
- Pipe A can fill the cistern in 12 minutes, so its rate of filling is:
\[ \text{Rate of A} = \frac{1}{12} \text{ cisterns per minute} \]
- Pipe B can fill the cistern in 16 minutes, so its rate of filling is:
\[ \text{Rate of B} = \frac{1}{16} \text{ cisterns per minute} \]

### Step 2: Calculate the effective rates due to obstruction
- Due to obstruction, the flow of water is restricted to \( \frac{7}{8} \) of A's full flow:
\[ \text{Effective rate of A} = \frac{7}{8} \times \frac{1}{12} = \frac{7}{96} \text{ cisterns per minute} \]
- Similarly, for pipe B, the flow is restricted to \( \frac{5}{6} \) of its full flow:
\[ \text{Effective rate of B} = \frac{5}{6} \times \frac{1}{16} = \frac{5}{96} \text{ cisterns per minute} \]

### Step 3: Combine the rates of both pipes
- When both pipes are open together with the obstruction, their combined effective rate is:
\[ \text{Combined rate} = \frac{7}{96} + \frac{5}{96} = \frac{12}{96} = \frac{1}{8} \text{ cisterns per minute} \]

### Step 4: Let the obstruction last for \( x \) minutes
- In \( x \) minutes, the part of the cistern filled with obstruction is:
\[ \text{Part filled with obstruction} = \left(\frac{1}{8}\right) \times x = \frac{x}{8} \]

### Step 5: Calculate the part filled after the obstruction is removed
- After the obstruction is removed, the pipes fill the cistern at their full rates for 3 minutes:
- The combined rate of both pipes without obstruction:
\[ \text{Combined rate without obstruction} = \frac{1}{12} + \frac{1}{16} \]
- Finding a common denominator (which is 48):
\[ \frac{1}{12} = \frac{4}{48}, \quad \frac{1}{16} = \frac{3}{48} \]
- Therefore, the combined rate is:
\[ \text{Combined rate} = \frac{4}{48} + \frac{3}{48} = \frac{7}{48} \text{ cisterns per minute} \]
- In 3 minutes, the part filled is:
\[ \text{Part filled in 3 minutes} = 3 \times \frac{7}{48} = \frac{21}{48} \text{ cisterns} \]

### Step 6: Set up the equation for the total filling
- The total part filled in the cistern must equal 1 (the whole cistern):
\[ \frac{x}{8} + \frac{21}{48} = 1 \]

### Step 7: Solve for \( x \)
- To solve for \( x \), first convert \( \frac{x}{8} \) to have a common denominator with \( \frac{21}{48} \):
\[ \frac{x}{8} = \frac{6x}{48} \]
- Now, substituting back into the equation:
\[ \frac{6x}{48} + \frac{21}{48} = 1 \]
- Combine the fractions:
\[ \frac{6x + 21}{48} = 1 \]
- Multiply both sides by 48:
\[ 6x + 21 = 48 \]
- Rearranging gives:
\[ 6x = 48 - 21 = 27 \]
- Finally, divide by 6:
\[ x = \frac{27}{6} = 4.5 \text{ minutes} \]

### Final Answer
The obstruction lasted for **4.5 minutes**.