There are two water taps in tank which can fill the empty tank in 12 hours and 18 hours respectively. It is seen that there is a leakage point at the bottom of the tank which can empty the completely filled tank in 36 hours. If both the water taps are opened at the same time to fill the empty tank and the leakage point was repaired after 1 hour, then in how much time the empty tank will be completely filled?
Show answer & explanation
Correct answer: C
Explanation
The correct answer is Option 3 i.e 7 hours 24 minutes.
P1 can fill the empty tank in 12 hours.
P2 can fill the empty tank in 18 hours.
P3 can empty the full tank in 36 hours.
Total work = LCM of ( 12 , 18 , 36 ) = 36 units.
Efficiency of P1 = 36/12 = 3 units/hours.
Efficiency of P2 = 36/18 = 2 units/hours.
Efficiency of P3 = 36/36 = 1 units/hours.
In the first hour, amount of water fill in the tank = 3 + 2 - 1 = 4 units.
Remaining work = 36 - 4 = 32 units.
Time is taken by pipes P1 and P2 to fill 32 units of water in a tank = 32 / 5 = 6 hours 24 minutes.
Total time taken to fill the tank = 7 hours 24 minutes.