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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 opened at the same time to fill the empty tank and leakage point was repaired after 1 hour, then in how much time the empty tank will be completely filled?

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 opened at the same time to fill the empty tank and leakage point was repaired after 1 hour, then in how much time the empty tank will be completely filled? Correct Answer 7 hours 24 minutes

Given:

Time taken by first water tap to fill the empty tank = 12 hours

Time taken by second water tap to fill the empty tank = 18 hours

Time taken by leakage to empty the filled tank = 36 hours

Formula used:

Work = Efficiency × Time

Calculation:

LCM of 12, 18 and 36 is 36

Efficiency of first water tap = 36/12 = 3 units/hr

Efficiency of second water tap = 36/18 = 2 units/hr

Efficiency of leakage = 36/36 = 1 unit/hr

Efficiency of these 2 taps and leakage together = (3 + 2 – 1) = 4 units/hr

Remaining units = (36 – 4) = 32 units

Work done by two tap water together to fill the empty tank = 32/(3 + 2) hr

⇒ 32/5 hr

⇒ 6.4 hours = 6 hours + 0.4 hours (0.4 hours = 24 minutes)

⇒ 6 hr 24 min

Total time taken to fill the empty tank = (6 hr 24 min + 1 hr)

⇒ 7 hr 24 min

∴ The empty tank will completely filled in 7 hr 24 minutes

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For the first one hour = (3 + 2 - 1) × 1 = 4 units

Time to fill remaning tank = (36 - 4) ÷ (3 + 2) = 6.4 hours

Total time taken = (6.4 + 1) = 7.4 hours = 7 hr 24 min

∴ The empty tank will completely filled in 7 hr 24 minutes

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