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Tap A can fill a tank in 36 minutes, Tap B can fill the same

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Tap A can fill a tank in 36 minutes, Tap B can fill the same  [#permalink]

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New post Updated on: 05 Dec 2016, 09:25
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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 (12-40).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.

Originally posted by conty911 on 27 Sep 2012, 09:21.
Last edited by bb on 05 Dec 2016, 09:25, edited 1 time in total.
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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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New post 27 Sep 2012, 09:25
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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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New post 14 Nov 2012, 03:42
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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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New post 27 Sep 2012, 15:28
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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 (12-40).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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New post 25 Oct 2012, 06:32
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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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New post 02 Nov 2012, 02:10
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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 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.
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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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New post 21 Nov 2016, 08:02
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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.


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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New post 27 Sep 2012, 09:35
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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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Tap A can fill a tank in 36 minutes, Tap B can fill the same  [#permalink]

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New post 05 Jun 2018, 02:05
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rajsinghudr wrote:
grandtheman wrote:
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 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.


Your calculation is wrong at the first step:

(1/36+1/48) = 6/144
this is not right.
(1/36+1/48) = 7/144

So eventually you will also get 12 minutes as answer.
"Approaches can be different in maths, but the answer will only be one."


Hey rajsinghudr

Appreciate the effort but grandtheman posted that post in 2012 :)

Btw, Welcome to GMATClub. All the best with your GMAT journey!
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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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New post 25 Oct 2012, 17:59
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,
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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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New post 12 Feb 2018, 20:21
[quote="conty911"]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




Ans. Let the volume be = 144

Efficiency of A = 4
Efficiency of B = 3

Tap A will run for 27 minutes so 27*4 = 108 units done
Remaining units will be done by B that is 36/3 = 12 minutes
Tap B will work for 12 minutes.
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Tap A can fill a tank in 36 minutes, Tap B can fill the same  [#permalink]

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New post Updated on: 05 Jun 2018, 02:09
grandtheman wrote:
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 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.


Your calculation is wrong at the first step:

(1/36+1/48) = 6/144
this is not right.
(1/36+1/48) = 7/144

So eventually you will also get 12 minutes as answer.
"Approaches can be different in maths, but the answer remains same."

Originally posted by rajsinghudr on 05 Jun 2018, 01:09.
Last edited by rajsinghudr on 05 Jun 2018, 02:09, edited 1 time in total.
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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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New post 30 Jan 2020, 15:53
1=(1/36 + 1/48)t1 + (1/36)t2
1= 1/36(t1+t2) + (1/48)t1
1= 1/36(27) + (1/48)t1
1= 27/48 + t1/48
1-27/48=t1/48
1/4 = t1/48
12 = t1

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Re: Tap A can fill a tank in 36 minutes, Tap B can fill the same   [#permalink] 30 Jan 2020, 15:53
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