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

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.

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

Thanks that helps kudos to you .

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

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) !!

Very good way of thinking. +1

Cheers,
CJ

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.

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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[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.

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."

Hey rajsinghudr

Appreciate the effort but grandtheman posted that post in 2012

Btw, Welcome to GMATClub. All the best with your GMAT journey!

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

B

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