Two inlet pipes, P1 and P2, can fill a cistern in 20 hours and 30 hours, respectively. They were opened at the same time, but pipe P1 had to be closed 5 hours before the cistern was full. How many hours in total did it take for the two pipes to fill the cistern?

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Let the total time taken to fill the cistern be T hours.

Pipe P1 fills the cistern in 20 hours, so its rate = 1/20 of the cistern per hour.
Pipe P2 fills the cistern in 30 hours, so its rate = 1/30 of the cistern per hour.

Since P1 was closed 5 hours before the cistern was full, P1 worked for (T - 5) hours.
P2 worked for the entire T hours.

Total work done = Work by P1 + Work by P2 = 1 (full cistern)

=> (T - 5) * (1/20) + T * (1/30) = 1

Multiply both sides by 60 (LCM of 20 and 30) to clear denominators:

=> 60 * (T - 5) / 20 + 60 * T / 30 = 60
=> 3(T - 5) + 2T = 60
=> 3T - 15 + 2T = 60
=> 5T - 15 = 60
=> 5T = 75
=> T = 15 hours

Therefore, the total time taken to fill the cistern is 15 hours.

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