There are two water taps in a 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?

7 hours 24 minutes

A+ = 12 hrs, B+ = 18 hrs, C- = 36 hours,

⇒ All of the were opened for 1 hours together = (3 + 2 - 1) x 1 = 4 x 1 = 4 units,

⇒ Remaining tank = 36 - 4 = 32 units,

⇒ Time required by A and B to fill the remaining tank = \(\frac{32}{5}\) = \( {6 }_{5 }^{2 } \) hours,

⇒ Total time to fill the tank completely = 1 + \( {6 }_{5 }^{2 } \) = \( {7}_{5 }^{2 } \) = 7 hours and 24 minutes.