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Tap A can fill a tank in 36 minutes, Tap B can fill the same [#permalink]
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27 Sep 2012, 10:21
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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. A. 8 B. 12 C. 20 D. 30 E. 36 This question was asked by one of the tutor who was giving presentation on a local GMAT prep course.I cannot remember the options exactly, but all options were integers. I guess b/w (1240).If the question sounds awkward or has other flaws please feel free to provide inputs/correct it.I am looking for its solution and similar rate problems in which 2 entities work together for some portion and then one of 'em leaves.
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Re: Tap A can fill a tank in 36 minutes, Tap B can fill the same [#permalink]
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27 Sep 2012, 10:25
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Tap A can fill a tank in 36 minutes which means that 1/36th of the tank will be filled up in a minute Tap B can fill a tank in 48minutes which means that 1/48th of the tank will be filled up in a minute and Given that A is open for 27 min > 27/36 or 3/4 of the tank will be filled and B has to fill the remaing 1/4th of the tank. B will fill the enitre tank in 48 mins and for it to fill 1/4 th of the tank it will take (48/4)12 mins.
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Re: Tap A can fill a tank in 36 minutes, Tap B can fill the same [#permalink]
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27 Sep 2012, 10:35
abhishekkpv wrote: Tap A can fill a tank in 36 minutes which means that 1/36th of the tank will be filled up in a minute Tap B can fill a tank in 48minutes which means that 1/48th of the tank will be filled up in a minute
and Given that A is open for 27 min > 27/36 or 3/4 of the tank will be filled and B has to fill the remaing 1/4th of the tank.
B will fill the enitre tank in 48 mins and for it to fill 1/4 th of the tank it will take (48/4)12 mins. Thanks that helps kudos to you .
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Re: Tap A can fill a tank in 36 minutes, Tap B can fill the same [#permalink]
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27 Sep 2012, 16:28
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conty911 wrote: 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. This question was asked by one of the tutor who was giving presentation on a local GMAT prep course.I cannot remember the options exactly, but all options were integers. I guess b/w (1240).If the question sounds awkward or has other flaws please feel free to provide inputs/correct it.I am looking for its solution and similar rate problems in which 2 entities work together for some portion and then one of 'em leaves. You can also use two equations provided: (1/36 + 1/48)t1 + (1/36)t2 =1 t1 + t2 = 27 > t2 = 27  t1 (1/36 + 1/48)t1 + (1/36)(27  t1) = 1 t1 = 12



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Re: Tap A can fill a tank in 36 minutes, Tap B can fill the same [#permalink]
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25 Oct 2012, 07:32
A to be opened for 27 mins that is 27/36 =3/4 of the tank is filled.. so remaining 1/4th of tank to be filled by B 48 /4 = 12 mins to fill 1/4 of the remaining...(logic is 1/4 =x/48) !!
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Re: Tap A can fill a tank in 36 minutes, Tap B can fill the same [#permalink]
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25 Oct 2012, 18:59
abhishekkpv wrote: Tap A can fill a tank in 36 minutes which means that 1/36th of the tank will be filled up in a minute Tap B can fill a tank in 48minutes which means that 1/48th of the tank will be filled up in a minute
and Given that A is open for 27 min > 27/36 or 3/4 of the tank will be filled and B has to fill the remaing 1/4th of the tank.
B will fill the enitre tank in 48 mins and for it to fill 1/4 th of the tank it will take (48/4)12 mins. Very good way of thinking. +1 Cheers, CJ
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Re: Tap A can fill a tank in 36 minutes, Tap B can fill the same [#permalink]
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02 Nov 2012, 03:10
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conty911 wrote: 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 27T (since total time is 27 minutes) = 1 (filling up the tank) the equation looks like this : (1/24)T + (1/36)(27T) = 1 >>> 3T/72 +(542T)/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.



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Re: Tap A can fill a tank in 36 minutes, Tap B can fill the same [#permalink]
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Rate of B x time + Rate of A x 27 minutes = 1 (combined output of Tap A and B) \(\frac{1}{48} t + \frac{27}{36}= 1\) \(\frac{1}{48} t = 1  \frac{3}{4}\) \(\frac{1}{48} t = \frac{1}{4}\) t = 12 minutes
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Re: Tap A can fill a tank in 36 minutes, Tap B can fill the same [#permalink]
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Re: Tap A can fill a tank in 36 minutes, Tap B can fill the same [#permalink]
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21 Nov 2016, 09:02
conty911 wrote: 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. Let the volume be = 144 Efficiency of A = 4 Efficiency of B = 3 Combinedd efficiency of A & B = 7 7*( 27  t ) + 4t = 144 189 7t + 4t = 144 Or, 3t = 45 So, t = 15 Thus tap B must be stopped after 12 ( ie, 27  15) minutes
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Re: Tap A can fill a tank in 36 minutes, Tap B can fill the same [#permalink]
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