Tap A can fill a tank in 36 minutes, Tap B can fill the same tank in 48 minutes. If they are opened simultaneously, at what time tap B (keep tap A running) can be stopped so that the tank can be filled completely filled in 27 minutes.
I would approach this question in some other way, since it says that the taps are opened simultaneously, and the question asks what time tap B can be stopped at interval "after both taps have been opened for some time" so that the entire tank can be filled in 27 minutes (not
running tap A for addition 27 minutes).
1. find the combined rate of two taps (since they are opened simultaneously): work/time>>> (1/36+1/48) = 6/144, meaning that combined rate is 144/6 = 24 minutes (both taps can fill up the tank in 24 minutes if with no pause on either tap).
2. let T be the time that tap B can be stopped. we can form the work formula letting a+b run together for T + tap A alone continues running for 27-T (since total time is 27 minutes) = 1 (filling up the tank)
the equation looks like this : (1/24)T + (1/36)(27-T) = 1
>>> 3T/72 +(54-2T)/72 = 1
solve this equation you will get T = 18 minutes, the time that tap B can be stopped so that tank can be filled in 27 minutes, which is the answer.
in this case, tap A alone will continue running for 9 minutes to fill up the tank.