Two pipes S1 and S2 alone can fill an empty tank in 15 hours and 20 hours respectively. Pipe S3 alone can empty that completely filled tank in 40 hours. Firstly both pipes S1 and S2 are opened and after 2 hour pipe S3 is also opened. In how much time tank will be completely filled after S3 is opened?
Show answer & explanation
Correct answer: D
Explanation
The correct answer is option 4 i.e. 92/11 hours.
Let the total capacity of the tank be LCM (15, 20, 40) = 120 units
The efficiency of pipe S1 to fill the tank = 120/15 = 8
The efficiency of pipe S2 to fill the tank = 120/20 = 6
The efficiency of pipe S3 to empty the tank = 120/40 = 3
The tank are filled in 2 hours by S1 and S2 = (8 + 6) × 2 = 28 units
Remaining capacity of the tank = 120 - 28 = 92 units
Hence, all pipes together filled the remaining tank in = 92/(8 + 6 - 3) = 92/11 hours