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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