Two pipes A and B can fill a cistern in $$12\frac{1}{2}$$ hours and 25 hours, respectively. The pipes are opened simultaneously and it is found that due to a leakage in the bottom, it took 1 hour 40 minutes more to fill the cistern. When the cistern is full, in how much time will the leak empty the cistern?

Solution (By Examveda Team)

Given:
A can fill the cistern 12.5 hr
B can fill the cistern 25 hr
Formula Used:
Total work = Efficiency × Time
Calculation:
Let the capacity of the cistern be 25 units (LCM of 12.5 and 25)
⇒ Efficiency of A = $$\frac{{25}}{{12.5}}$$ = 2 units/hr
⇒ Efficiency of B = $$\frac{{25}}{{25}}$$ = 1 units/hr
⇒ Combined efficiency of A and B = 2 + 1 = 3 units/hr
Time taken by A and B to fill the cistern without leakage ⇒ $$\frac{{25}}{3}$$ = $$8\frac{1}{3}$$ hr = 8 hr 20 min
Time taken by A and B to fill the cistern with leakage ⇒ 8 hr 20 min + 1 hr 40 min = 10 hr
The combined efficiency of A and B with leakage ⇒ $$\frac{{25}}{{10}}$$ = 2.5 units/hr
⇒ Efficiency of leakage = 3 - 2.5 = 0.5 units/hr
Time required by the leak to empty the full cistern ⇒ $$\frac{{25}}{{0.5}}$$ = 50 hr
∴ The leak can empty the full cistern in 50 hours.