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There is a leakage at the bottom of tank which can empty the tank in 24 hours. Two taps A and B alone can fill the tank in 12 hours and 18 hours respectively. If tap A alone opened for first two hours while tap B opened for next two hours. After 4 hours, both taps are opened simultaneously. In what time tank gets completely filled?

There is a leakage at the bottom of tank which can empty the tank in 24 hours. Two taps A and B alone can fill the tank in 12 hours and 18 hours respectively. If tap A alone opened for first two hours while tap B opened for next two hours. After 4 hours, both taps are opened simultaneously. In what time tank gets completely filled? Correct Answer 92/7 hours

Given:

Leakage's 1 hour's work = 1/24

Tap A's 1 hour's work = 1/12

Tap B's 1 hour's work = 1/18

(Tap A + Leakage)'s 2 hours work = 2 × (1/12 - 1/24) = 1/12

(Tap B + Leakage)'s 2 hours work = 2 × (1/18 - 1/24) = 1/36

Remaining tank after 4 hours = 1 - (1/12 + 1/36) = 1 - 1/9 = 8/9

(Tap A + Tap B + Leakage)'s 1 hours work = 1/12 + 1/18 - 1/24 = 7/72

Time taken two taps and leakage to fill remaining tank = 8/9 × 72/7 = 64/7 hours

∴ Total time required to fill the tank = 4 + 64/7 = 92/7 hours

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