Question : 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 and 40 minutes more to fill the cistern. When the cistern is full, in how much time will the leak empty the cistern?
Option 1: 45 hours
Option 2: 42 hours
Option 3: 48 hours
Option 4: 50 hours
Correct Answer: 50 hours
Solution :
Since pipe A can fill a cistern in $12 \frac{1}{2}$ hours
So, work done by Pipe A in 1 hr = $\frac{2}{25}$
Similarly work done by B in 1 hr = $\frac{1}{25}$
⇒ Work done by both A and B together in 1 hr = $\frac{2}{25}$ + $\frac{1}{25}$ = $\frac{3}{25}$
So, time taken by both A and B together to finish the work = $\frac{25}{3}$ hr = 8 hr 20 min
With the leak total time taken = 8 hr 20 min + 1 hr 40 min = 10 hr
1 hr work of leak = 1 hr work of A and B together without leak - 1 hr work of A and B together with leak
= $\frac{3}{25}$ – $\frac{1}{10}$
= $\frac{30 – 25}{250}$
= $\frac{1}{50}$
So, the leak can empty the filled cistern in 50 hours.
Hence, the correct answer is 50 hours.
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