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There is a leak in the bottom of the tank. This leak can empty the tank in 15 hours. Tap A alone can fill the tank in 6 hours. Tap A opens for 4 hours after that remaining part of the tank is filled by tap B in 18 hours. When both pipes are opened simultaneously, then in what time tank is completely filled?

There is a leak in the bottom of the tank. This leak can empty the tank in 15 hours. Tap A alone can fill the tank in 6 hours. Tap A opens for 4 hours after that remaining part of the tank is filled by tap B in 18 hours. When both pipes are opened simultaneously, then in what time tank is completely filled? Correct Answer 5 hours

Given:

Leakage 1 hour's work = 1/15

Tap A's 1 hour's work = 1/6

Calculation:

(Tap A + leakage)’s 1 hour's work = 1/6 – 1/15 = 1/10

(Tap A + leakage)’s 4 hours work = 4 × 1/10 = 4/10 = 2/5

⇒ Remaining work = 1 – 2/5 = 3/5

Let Tap B's 1 hours of work be 1/B.

(Tap B + leakage)’s 1 hour's work = 1/B – 1/15 = (15 – B)/15B

Then,

⇒ 3/5 × = 18

⇒ B = 30 – 2B

⇒ B = 10

Then,

(Tap A + tap B + leakage)’ 1 hours work = 1/6 + 1/10 – 1/15 = 1/5

∴ When two pipes are opened then the tank is filled in 5 hours. 

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