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 cistren?

50 hours

A = $12\frac{1}{2}$ hours 

B = 25 hours,

⇒ Time taken by A + B to fill the cistern = \(\frac{25}{2+1}\) = \(\frac{25}{3}\) = \( {8 }_{3 }^{1 } \) hrs,

⇒ Due to leakage it takes 1 hr 40 min more to fill the tank, therefore,

⇒ A + B + C = 10 hrs,

⇒ A + B + C = 5,  (Efficiency)

⇒ 4 + 2 + C = 5,

⇒ 6 + C = 5,

⇒ C = -1,

Time required by C to completely empty the tank = \(\frac{50}{1}\) = 50 hrs.