Three pipe P1, P2 and P3 fill the tank in 15 hours, 10 hours and 20 hours respectively. If all the three pipes opened together and pipe P2 & P1 are closed after three hours, then pipe P3 fill the remaining tank. Find the total time to fill whole tank.

The correct option is 10 hours.

Let us solve the problem step-by-step by determining the individual work rates of the pipes and tracking the total work completed.

Step 1: Determine the rate of work done by each pipe in 1 hour.
Let the total capacity of the tank be 1 unit (or 100%).

Work done by pipe P1 in 1 hour = $\frac{1}{15}$

Work done by pipe P2 in 1 hour = $\frac{1}{10}$

Work done by pipe P3 in 1 hour = $\frac{1}{20}$

Step 2: Calculate the combined work done by all three pipes in the first 3 hours.
In 1 hour, all three pipes operating together fill:

Combined 1-hour rate = $\frac{1}{15}+\frac{1}{10}+\frac{1}{20}$

To add these fractions, find the Least Common Multiple (LCM) of denominators 15, 10, and 20, which is 60:

Combined 1-hour rate = $\frac{4+6+3}{60}=\frac{13}{60}$

In 3 hours, the tank filled by P1, P2, and P3 together is:

Work done in 3 hours = $3\times \frac{13}{60}=\frac{13}{20}$

Step 3: Calculate the remaining part of the tank to be filled.
Remaining work = $1-\frac{13}{20}=\frac{7}{20}$

Step 4: Find the time taken by pipe P3 to fill the remaining tank alone.
Since P1 and P2 are closed, only P3 continues to fill the remaining $\frac{7}{20}$ of the tank.

Time taken by P3 = $\frac{Remaining\; Work}{\text{P3's 1-hour rate}}=\frac{7/20}{1/20}=7\text{hours}$

Step 5: Calculate the total time taken to fill the entire tank.
Total time = (Time when all three pipes were open) + (Time when only P3 was open)

Total time = $3\text{hours}+7\text{hours}=10\text{hours}$

Thus, the total time required to fill the whole tank is 10 hours.