Two pipes can fill a cistern in 8 and 12 hours respectively. The pipes are opened simultaneously, and it takes 12 minutes more to fill the cistern due to leakage. If the cistern is full, what will be the time taken by the leakage to empty it?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 120 hours
Rates of filling
First pipe $=\frac{1}{8}$ tank per hour
Second pipe $=\frac{1}{12}$ tank per hour
Combined filling rate without leakage
$=\frac{1}{8}+\frac{1}{12}=\frac{3+2}{24}=\frac{5}{24}$ tank per hour
Time without leakage
$=\frac{24}{5}=4.8$ hours $=4$ hours $48$ minutes
Due to leakage, extra time $=12$ minutes $=\frac{1}{5}$ hour
Actual time with leakage
$=4.8+\frac{1}{5}=5$ hours
Net filling rate with leakage
$=\frac{1}{5}$ tank per hour
Leakage rate
$=\frac{5}{24}-\frac{1}{5}$
$=\frac{25-24}{120}$
$=\frac{1}{120}$ tank per hour
Time taken by leakage to empty the cistern
$=120$ hours
The leakage alone will empty the cistern in $120$ hours.