Two pipes A and B can fill a cistern in 12½hours and 25 hours, respectively. The pipes were opened simultaneously, and it was found that, due to leakage in the bottom, it took one hour 40 minutes more to fill the cistern. if the cistern is full, in how much time(in hours) will the leak alone empty 70% of the cistern?
Show answer & explanation
Correct answer: D
Explanation
The correct answer is option 4 i.e. 35
A fill the tank in 25/2 hours.
B fill the tank in 25 hours.
Total work = LCM of (25/2, 25)
Total work = 625.
The efficiency of Pipe A = 50 units/ hour
The efficiency of Pipe B = 25 units/hour
Both A and B can fill the tank in 625/75hours = 8 hours 20 minutes.
Due to the leakage, the work is now completed in 8 hours 20 minutes + 1 hour 40 minutes.
Total time taken = 10 hours.
Let the efficiency of leakage be x
(50 + 25 - x) × 10 = 625
x = 12.5 units/hour.
70% of 625 = 437.5
Time taken by leakage to empty 437.5 units of water = 437.5/12.5 = 35 hours.