Question : Two pipes A and B can fill a cistern in $12 \frac{1}{2}$ hours 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?
Option 1: 40
Option 2: 35
Option 3: 30
Option 4: 50
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Correct Answer: 35
Solution : LCM of 12.5 and 25 = 25 Capacity of the cistern = 25 units Efficiency of A = $\frac{25}{12.5}$ = 2 units/hour Efficiency of B = $\frac{25}{25}$ = 1 unit/hour Combined efficiency of A and B = 2 + 1 = 3 units/hour Time taken by A and B to fill the cistern without leakage = $\frac{25}{3}$ = $8\frac{1}{3}$ = 8 hours 20 mins Time taken by A and B to fill the cistern with leakage = 8 hours 20 mins + 1 hour 40 mins = 10 hours The combined efficiency of A and B with leakage = $\frac{25}{10}$ = 2.5 units/hour Efficiency of leakage = 3 – 2.5 = 0.5 units/hour Time required by the leak to empty the full cistern = $\frac{25}{0.5}$ = 50 hours Time required to empty 70% of the tank = $\frac{70}{100}\times50$ = 35 hours Hence, the correct answer is 35.
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