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There are two water taps in a tank which can fill the empty tank in 12 hours and 18 hours respectively. It is seen that there is a
leakage point at the bottom of the tank which can empty the completely filled tank in 36 hours. If both the water taps are
openedat the same timetofill the empty tank and the leakage point was repaired after 1 hour, then in how muchtime the empty
tank will be completely filled?
7 hours
8 hours 24 minutes
7 hours 24 minutes
7 hours 12 minutes
- Calculate the filling rates:
- First tap fills \(\frac{1}{12}\) of the tank per hour.
- Second tap fills \(\frac{1}{18}\) of the tank per hour.
- Together, they fill \(\frac{1}{12} + \frac{1}{18} = \frac{5}{36}\) of the tank per hour.
- Calculate the leakage rate:
- The leak empties \(\frac{1}{36}\) of the tank per hour.
- Combined effect in the first hour:
- \(\frac{5}{36} - \frac{1}{36} = \frac{4}{36} = \frac{1}{9}\) of the tank is filled in 1 hour.
- Remaining to fill after 1 hour:
- \(\frac{8}{9}\) of the tank left to fill.
- Rate without leakage:
- Combined rate of filling without leakage = \(\frac{5}{36}\).
- Time to fill remaining \(\frac{8}{9}\) of the tank:
- Time = \(\frac{8}{9} \div \frac{5}{36} = \frac{8}{9} \times \frac{36}{5} = \frac{32}{5} = 6.4\) hours = 6 hours 24 minutes.
- Total time = 1 hour (first hour) + 6 hours 24 minutes = 7 hours 24 minutes.
- Correct Answer: Option 3 - 7 hours 24 minutes
By: santosh ProfileResourcesReport error
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