Question : Two pipes can fill a cistern in 12 hours and 16 hours, respectively. The pipes are opened simultaneously and it is found that due to leakage at the bottom, it takes 90 minutes more to fill the cistern. How much time will the leakage take to empty the filled tank?
Option 1: $36 \frac{29}{49} \mathrm{~h}$
Option 2: $38 \frac{10}{49} \mathrm{~h}$
Option 3: $39 \frac{13}{49} \mathrm{~h}$
Option 4: $37 \frac{15}{49} \mathrm{~h}$
Correct Answer: $38 \frac{10}{49} \mathrm{~h}$
Solution :
Works done by two pipes in 1 hours = $\left(\frac{1}{12}+\frac{1}{16}\right)=\frac{\left(4+3\right)}{48}=\frac{7}{48}$
Time taken by the pipes to fill the tank = $\frac{48}{7}$ hr = $6\frac{6}{7}$ hr
Due to leakage time taken = $6\frac{6}{7}$ hr + 90 min = 8$\frac{5}{14}$ hr
Work done by (two pipes + leak) in 1 hours = $\frac{14}{117}$ hr
⇒ Work done by the leak in 1 hour = ($\frac{7}{48} - \frac{14}{117}$) = $\frac{49}{1872}$
⇒ Leak will empty the full as a term in = $\frac{1872}{49}$ hours
Hence, the correct answer is $39\frac{10}{49} \text{h}$.
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