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Pipe A can fill a tank in 3 hours. Pipe B can empty the same

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Pipe A can fill a tank in 3 hours. Pipe B can empty the same [#permalink] New post 27 May 2013, 02:16
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Pipe A can fill a tank in 3 hours. Pipe B can empty the same tank in 2 hours. If the tank is initially two thirds full then in how many hours will the tank be completely empty in if both the pipes are turned on simultaneously?

(A) 2 hours
(B) 3 hours
(C) 4 hours
(D) 5 hours
(E) 9 hours
[Reveal] Spoiler: OA

Last edited by gmatquant25 on 27 May 2013, 02:30, edited 1 time in total.
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Re: Pipe A can fill a tank in 3 hours. Pipe B can empty the same [#permalink] New post 27 May 2013, 02:21
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gmatquant25 wrote:
Pipe A can fill a tank in 3 hours. Pipe B can empty the same tank in 2 hours. If the tank is initially two thirds full then in how many hours will the tank be completely empty in if both the pipes are turned on simultaneously?
(A)2 hours
(B)3 hours
(C)4 hours
(D)5 hours


Lets say tank contains 6 liters
Pipe A works at \frac{6}{3}= 2 liters every hour (added)=+2
Pibe B works ar \frac{6}{2}= 3 liters every hour (subtracted)=-3

The tank is 2/3 full so contains 4 liters, and the pipes combined give us a -3+2=-1 liter every hour.(minus one liter every hour)
So they would take 4 hours to empty it.
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Re: Pipe A can fill a tank in 3 hours. Pipe B can empty the same [#permalink] New post 27 May 2013, 02:32
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gmatquant25 wrote:
Pipe A can fill a tank in 3 hours. Pipe B can empty the same tank in 2 hours. If the tank is initially two thirds full then in how many hours will the tank be completely empty in if both the pipes are turned on simultaneously?

(A) 2 hours
(B) 3 hours
(C) 4 hours
(D) 5 hours


The 2/3 of the tank will be emptied in (time)=\frac{(job)}{(rate)}=\frac{\frac{2}{3}}{\frac{1}{2}-\frac{1}{3}}=4 hours (1/2-1/3 is the rate at which the tank is emptied).

Answer: C.

Hope it's clear.
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Re: Pipe A can fill a tank in 3 hours. Pipe B can empty the same [#permalink] New post 27 May 2013, 02:55
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Rate of filling the tank by A =1/3
Rate of emptying the tank by B =1/2
Combined rate of emptying the tank = 1/2-1/3 =1/6
2/3rd of tank needs to be emptied.
R=w/t
1/6=(2/3)*(1/t)
On simplifying , t= 4 hrs

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Re: Pipe A can fill a tank in 3 hours. Pipe B can empty the same [#permalink] New post 27 Jun 2014, 05:57
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Re: Pipe A can fill a tank in 3 hours. Pipe B can empty the same   [#permalink] 27 Jun 2014, 05:57
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