Two pipes `P_1 and P_2` can fill a tank in 8 h and 5 h, respectively. They are opened together, after an hour, pipe `P_2` is closed, in how many more hours will the tank be filled?

Options

A. `27/5h`
B. `27/2h`
C. `23/5h`
D. `28/5h`

Answer

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

1. **Determine the rate of each pipe:**
- Pipe P1 can fill the tank in 8 hours, so in 1 hour, it fills \( \frac{1}{8} \) of the tank.
- Pipe P2 can fill the tank in 5 hours, so in 1 hour, it fills \( \frac{1}{5} \) of the tank.

2. **Calculate the combined rate when both pipes are open:**
- When both pipes are open together for 1 hour, the total volume filled is:
\[ \text{Volume filled in 1 hour} = \frac{1}{8} + \frac{1}{5} \]
- To add these fractions, we need a common denominator, which is 40:
\[ \frac{1}{8} = \frac{5}{40}, \quad \frac{1}{5} = \frac{8}{40} \]
- Therefore,
\[ \text{Total volume filled in 1 hour} = \frac{5}{40} + \frac{8}{40} = \frac{13}{40} \]

3. **Calculate the volume filled after 1 hour:**
- After 1 hour, the volume filled by both pipes is \( \frac{13}{40} \) of the tank.

4. **Determine the remaining volume to be filled:**
- The total volume of the tank is considered as 1 (whole tank), so the remaining volume after 1 hour is:
\[ \text{Remaining volume} = 1 - \frac{13}{40} = \frac{40}{40} - \frac{13}{40} = \frac{27}{40} \]

5. **Calculate the time taken by Pipe P1 to fill the remaining volume:**
- After 1 hour, Pipe P2 is closed, and only Pipe P1 is filling the tank. The rate of Pipe P1 is \( \frac{1}{8} \) of the tank per hour.
- Let \( T \) be the time in hours that Pipe P1 takes to fill the remaining volume:
\[ T \times \frac{1}{8} = \frac{27}{40} \]
- Rearranging gives:
\[ T = \frac{27}{40} \times 8 = \frac{27 \times 8}{40} = \frac{216}{40} = \frac{27}{5} = 5.4 \text{ hours} \]

6. **Conclusion:**
- Therefore, after Pipe P2 is closed, Pipe P1 will take an additional 5.4 hours to fill the tank.