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Pipe A that can fill a tank in an hour and pipe B that can fill the [#permalink]
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Updated on: 29 Aug 2014, 09:18
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Pipe A that can fill a tank in an hour and pipe B that can fill the tank in half an hour are opened simultaneously when the tank is empty. Pipe B is shut 15 minutes before the tank overflows. When will the tank overflow? A) 30 mins B) 35 mins C) 40 mins D) 32 mins E) 36 mins Source : 4gmat
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Originally posted by alphonsa on 29 Aug 2014, 09:14.
Last edited by Bunuel on 29 Aug 2014, 09:18, edited 1 time in total.
Edited the question and the tags.



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Re: Pipe A that can fill a tank in an hour and pipe B that can fill the [#permalink]
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29 Aug 2014, 09:24
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alphonsa wrote: Pipe A that can fill a tank in an hour and pipe B that can fill the tank in half an hour are opened simultaneously when the tank is empty. Pipe B is shut 15 minutes before the tank overflows. When will the tank overflow?
A) 30 mins B) 35 mins C) 40 mins D) 32 mins E) 36 mins
Source : 4gmat The last 15 minutes only pipe A was open. Since it needs 1 hour to fill the tank, then in 15 minutes it fills 1/4th of the tank, thus 3/4 of the tank is filled with both pipes open. The combined rate of two pipes is 1 + 2 = 3 tanks/hour, therefore to fill 3/4th of the tank they need (time) = (work)/(rate) = (3/4)/3 = 1/4 hours = 15 minutes. Total time = 15 + 15 = 30 minutes. Answer: A.
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Pipe A that can fill a tank in an hour and pipe B that can fill the [#permalink]
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29 Aug 2014, 23:23
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1/60+1/30=1/20t/min working together
Let 's try backsolving:
better to start from B, i.e. 35 min. 3515=20, that is 1/20*20=1 meaning that tank was full when pipes worked together, it contradicts condition, eliminate B,C,E
go D, i.e. 32 min. 3215=17, that is 1/20*17=17/20 working together means 3/20 working A alone. Check, 1/60*15=15/60=1/4>3/20, so eliminate D
So, must be A
we can check 3015=15*1/20=15/20=3/4. 15/60=1/4. It is correct



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Re: Pipe A that can fill a tank in an hour and pipe B that can fill the [#permalink]
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14 Sep 2014, 04:29
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I think I might be able to give a more elaborate reply:
Rate for Pipe A: 1/60 Rate for Pipe B: 1/30 Working together, their rate is 1/60 + 1/30 = 1/20
Pipe A and B work together 15 minutes before Pipe B quits. So, Pipe A is practically running solo for the last 15 minutes.
Therefore, for Pipe A to work alone for the last 15 minutes: Rate x Time = Work 1/60 x 15 = 1/4  Means Pipe A has to fill up 1/4th of the tank alone.
Remaining capacity = 1  1/4 = 3/4
Now, it's obvious that Pipe A and B worked together and completed 3/4th of the task. So Rate x Time = Work 1/20 x Time = 3/4; Time = 3/4 x 20 = 15 minutes
Result = Time taken for Pipe A and B to work together + Time taken for Pipe A to work solo = 15 minutes + 15 minutes = 30 minutes.



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Re: Pipe A that can fill a tank in an hour and pipe B that can fill the [#permalink]
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14 Sep 2014, 22:10
alphonsa wrote: Pipe A that can fill a tank in an hour and pipe B that can fill the tank in half an hour are opened simultaneously when the tank is empty. Pipe B is shut 15 minutes before the tank overflows. When will the tank overflow?
A) 30 mins B) 35 mins C) 40 mins D) 32 mins E) 36 mins
Source : 4gmat Let, Work done by Pipe A in 1 min = 1/60 Work done by pipe B in 1 min = 1/30 Work done by both Pipe A & Pipe B in 1 min = 1/60 + 1/30 = 1/20 Pipe B is shut down for the last 15 Mins. So, the work in those fifteen minutes is done by Pipe A = 15 * (1/60) = 1/4 Remaining 3/4th part of the work was completed by Both Pipe A & Pipe B Setting up proportion in 1 min Both A & B do 1/20 th part of work Therefore to complete 3/4th part of the work time required = 15 min So, total time required for the tank to overflow = Time for which Pipe A was running alone + Time for which both Pipe A & Pipe B were running together = 15 min + 15 min = 30 Min Answer A



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Re: Pipe A that can fill a tank in an hour and pipe B that can fill the [#permalink]
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07 Oct 2015, 13:13
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alphonsa wrote: Pipe A that can fill a tank in an hour and pipe B that can fill the tank in half an hour are opened simultaneously when the tank is empty. Pipe B is shut 15 minutes before the tank overflows. When will the tank overflow?
A) 30 mins B) 35 mins C) 40 mins D) 32 mins E) 36 mins
Source : 4gmat Work= Rate * Time Let's say Work = 1 Task Rate of Pipe A= \(\frac{1}{60}\)\(\frac{task}{minute}\)Rate of Pipe B= \(\frac{1}{30}\)\(\frac{task}{minute}\)Pipe B is shut down 15 minutes before Pipe A, so if Pipe B worked for t15 minutes, then Pipe A worked for t minutes.So, Combined work = Pipe A work + Pipe B work > 1 Task = \(\frac{t15}{30}\)+\(\frac{t}{60}\)At this point, you can either solve the quadratic or plug in answer choices. It would be quickest to plug in answer choices. Choice A works.



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Re: Pipe A that can fill a tank in an hour and pipe B that can fill the [#permalink]
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16 Nov 2015, 04:32
Time needs to fill tanks: x A runs all the time so in x hour it fills x portion of the tank. B doesn't run the last 15 mins (=1/4 hour) so it runs (x1/4) hour. During this time it fills 2(x  1/4) of the tank. Together they fill the tank. So x + 2(x1/4) = 1 Solve for x we have 1/2 = 30 mins. So A.



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Pipe A that can fill a tank in an hour and pipe B that can fill the [#permalink]
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16 Nov 2015, 06:30
The question states B can fill the Tank when empty in 30 Minutes while A can do it in 60 Minutes. also B is turned off 15 minutes prior to overflow which means B was on for 15 minutes now since B can fill the tank in 30 minutes , it would have filled 50% of the tank in 15 minutes while A would have filled 25% in 15 minutes after this point only A is kept open and we know that , A can fill 25% in 15 minutes which also happens to be the remaining space left in the tank. Thus Tank will overflow after 15 minutes. Total time taken =30 minutes



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Re: Pipe A that can fill a tank in an hour and pipe B that can fill the [#permalink]
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31 Mar 2017, 07:10
To avoid fractions and percentages  which might look scary for some of us let work with Smart numbers. Let asume that the tank is 60 liters. (Of course, 120 or other numbers will lead to the same result) A fills the tank in 1 hour, A work at 1 liter per min. B fills the tank in 30 min, B work at 2 liters per min. A&B together work at 3 liters per min. In the last 15 min, A works alone to complete filling the tank with the remaining 15 liters. Therefore, before B was shut down, A&B had worked together to fill the tank with 45 liters (6015). This takes 15 min. Answer is 30 min.
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Pipe A that can fill a tank in an hour and pipe B that can fill the [#permalink]
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03 Apr 2018, 21:54
alphonsa wrote: Pipe A that can fill a tank in an hour and pipe B that can fill the tank in half an hour are opened simultaneously when the tank is empty. Pipe B is shut 15 minutes before the tank overflows. When will the tank overflow?
A) 30 mins B) 35 mins C) 40 mins D) 32 mins E) 36 mins let t=time to fill tank (t1/4)*3+(1/4)*1=1 t=1/2 hour=30 minutes A



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Re: Pipe A that can fill a tank in an hour and pipe B that can fill the [#permalink]
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06 Apr 2018, 08:50
alphonsa wrote: Pipe A that can fill a tank in an hour and pipe B that can fill the tank in half an hour are opened simultaneously when the tank is empty. Pipe B is shut 15 minutes before the tank overflows. When will the tank overflow?
A) 30 mins B) 35 mins C) 40 mins D) 32 mins E) 36 mins We see that the rate of pipe A = 1/1 = 1, and the rate of pipe B = 1/(½) = 2. We can let the time used to fill the tank by pipe A = n hours; thus, the time used to fill the tank by pipe B = (n  1/4), and create the equation: 1(n) + 2(n  1/4) = 1 n + 2n  1/2 = 1 3n = 3/2 n = (3/2)/3 = 1/2 hour = 30 minutes. Answer: A
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