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?

120 hours

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.