Two taps A and B can fill a tank in 5 hours and 20 hours respectively. If both the taps are open then due to a leakage, it took 40 minutes more to fill the tank. If the tank is full, how long will it take for the leakage alone to empty the tank?
Part filled by pipe A in 1 hour $=\dfrac{1}{5}$
Part filled by pipe B in 1 hour =$\dfrac{1}{20}$
Part filled by pipe A and B in 1 hour $=\dfrac{1}{5}+\dfrac{1}{20}$=$\dfrac{1}{4}$
i.e., A and B together can fill the tank in 4 hours.
Due to leakage, it took 40 minutes more to fill the tank.
i.e., due to the leakage, the tank got filled in
4$\dfrac{40}{60}$ hour =4$\dfrac{2}{3}$ hour =$\dfrac{14}{3}$ hour.
Net part filled by pipe A, pipe B and the leak in 1 hour=$\dfrac{3}{14}$
Therefore, part emptied by the leak in 1 hour=$\dfrac{1}{4}-\dfrac{3}{14}=\dfrac{1}{28}$
i.e., the leak can empty the tank in 28 hours.