[28 hr] Two taps A and B can fill a tank in 5 hours and 20 hours respectively. If both the taps are open the

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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?

28 hr

16 hr

22 hr

32 hr

Explanation:

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.

Additional Questions

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