Two pipes A and B can fill a tank in 8 hours and 10 hours respectively. Both pipes are opened together, but 2 hours before the tank is full, B is closed. How much time does it take to fill the tank?

The correct answer is 5 hours 20 minutes.

Step-by-step Explanation:

1. Work done per hour by each pipe:
Pipe A can fill the tank in 8 hours. Therefore, the work done by Pipe A in 1 hour is:
$\frac{1}{8}$
Pipe B can fill the tank in 10 hours. Therefore, the work done by Pipe B in 1 hour is:
$\frac{1}{10}$

2. Combined work done per hour when both pipes are open:
When both Pipe A and Pipe B are open together, the work done in 1 hour is:
$\frac{1}{8}+\frac{1}{10}=\frac{5+4}{40}=\frac{9}{40}$

3. Work done in the last 2 hours:
Pipe B is closed 2 hours before the tank is full, which means Pipe A works alone during the final 2 hours.
Work done by Pipe A alone in 2 hours is:
$2\times \frac{1}{8}=\frac{2}{8}=\frac{1}{4}$

4. Remaining work done by both pipes together:
The total work to fill the tank is 1.
Remaining work completed by both pipes A and B working together is:
$1-\frac{1}{4}=\frac{3}{4}$

5. Time taken for both pipes working together:
Time taken by A and B together to complete $\frac{3}{4}$ of the work is:
$\frac{\frac{3}{4}}{\frac{9}{40}}=\frac{3}{4}\times \frac{40}{9}=\frac{10}{3}\text{hours}$

6. Total time taken to fill the tank:
Total time = (Time A and B worked together) + (Time A worked alone)
$\text{Total time}=\frac{10}{3}+2=\frac{16}{3}\text{hours}$

Converting $\frac{16}{3}$ hours into hours and minutes:
$\frac{16}{3}\text{hours}=5\frac{1}{3}\text{hours}$
Since $\frac{1}{3}\text{hour}=\frac{1}{3}\times 60\text{minutes}=20\text{minutes}$, the total time taken is 5 hours 20 minutes.