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Tap A can fill the water cooler in 20 minutes and Tap B can fill water cooler in 1 hour. Tap A is alone opened for 10 minutes. After 10 minutes, tap B is alone opened for 20 min. After this both taps are opened for 2 minutes, then tap B is closed. In what time remaining water cooler is completely filled by tap A?

Tap A can fill the water cooler in 20 minutes and Tap B can fill water cooler in 1 hour. Tap A is alone opened for 10 minutes. After 10 minutes, tap B is alone opened for 20 min. After this both taps are opened for 2 minutes, then tap B is closed. In what time remaining water cooler is completely filled by tap A? Correct Answer 40 seconds

Given,

Tap A's 1 min work = 1/20

Tap B's 1 min work = 1/60

Tap(A + B)’s 1 min work = 1/20 + 1/60 = 1/15

Tap A's 10 min work = 10/20 = 1/2

Tap B's 20 min work = 20/60 = 1/3

Tap (A + B)’s 2 min work = 2/15

Tank filled is = 1/2 + 1/3 + 2/15 = 29/30

Part of water cooler that is empty = 1 – 29/30 = 1/30

Tap A will fill 1/30th part of water cooler in =

= 1/30 × 20

= 2/3 minutes

= 2/3 × 60 seconds

= 40 seconds

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