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A pool which was 2/3 full to begin with, was filled at a [#permalink]
02 Jun 2012, 22:51

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Question Stats:

83% (02:35) correct
17% (02:32) wrong based on 232 sessions

A pool which was 2/3 full to begin with, was filled at a constant rate for 5/3 hours until it was until it was 6/7 full. At this rate, how much time would it take to completely fill this pool if it was empty to begin with?

A. 8 hrs 45 mins. B. 9 hrs. C. 9 hrs 30 mins. D. 11 hrs 40 mins. E. 15 hrs 30 mins .

Re: A pool which was 2/3 full to begin with, was filled at a [#permalink]
03 Jun 2012, 02:57

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macjas wrote:

A pool which was 2/3 full to begin with, was filled at a constant rate for 5/3 hours until it was until it was 6/7 full. At this rate, how much time would it take to completely fill this pool if it was empty to begin with?

A. 8 hrs 45 mins. B. 9 hrs. C. 9 hrs 30 mins. D. 11 hrs 40 mins. E. 15 hrs 30 mins .

\(\frac{6}{7}-\frac{2}{3}=\frac{4}{21}\) of the pool wad filled in \(\frac{5}{3}\) hours.

So, to fill the pool completely at this rate it would take \(\frac{(\frac{5}{3})}{(\frac{4}{21})}=8\frac{3}{4}\) hours or 8 hours and 45 minutes.

Re: A pool which was 2/3 full to begin with, was filled at a [#permalink]
24 Jan 2014, 10:45

Bunuel wrote:

macjas wrote:

A pool which was 2/3 full to begin with, was filled at a constant rate for 5/3 hours until it was until it was 6/7 full. At this rate, how much time would it take to completely fill this pool if it was empty to begin with?

A. 8 hrs 45 mins. B. 9 hrs. C. 9 hrs 30 mins. D. 11 hrs 40 mins. E. 15 hrs 30 mins .

\(\frac{6}{7}-\frac{2}{3}=\frac{4}{21}\) of the pool wad filled in \(\frac{5}{3}\) hours.

So, to fill the pool completely at this rate it would take \(\frac{(\frac{5}{3})}{(\frac{4}{21})}=8\frac{3}{4}\) hours or 8 hours and 45 minutes.

Answer: A.

Bunuel, this problem is fairly simple but what has confused me a little bit was your approach. Are we always able to use Total Time \(\frac{5}{3}\) divided by Total Work \(\frac{4}{21}\) as a formula to give us the amount of Time it will take to complete the entire task? I thought that Total Time \(\frac{5}{3}\) divided by Total Work \(\frac{4}{21}\) gave us \(1/Rate\)

Thank you for all your great posts. They are helping me get my quant score up significantly. Work/Rate problems sometimes stump me for some reason. _________________

Re: A pool which was 2/3 full to begin with, was filled at a [#permalink]
28 Jan 2014, 07:04

I have gone in a little traditional way...

As its about fraction I have taken 21 liter is capacity. Now as given Initially it was 21*2/3=14 L full Then it was filled @ constant rate till 6/7 means= 21*6/7=18 L It means @ 5/3 hr it was filled 18-14=4 ltr Now R*5/3=4 so R=12/5 So to find Time(hours) to completely fill @ constant rate we have 12/5*T=21(full capacity) T=8 hr 45 min(A)

Re: A pool which was 2/3 full to begin with, was filled at a [#permalink]
04 May 2014, 06:02

PeterHAllen wrote:

Bunuel wrote:

macjas wrote:

A pool which was 2/3 full to begin with, was filled at a constant rate for 5/3 hours until it was until it was 6/7 full. At this rate, how much time would it take to completely fill this pool if it was empty to begin with?

A. 8 hrs 45 mins. B. 9 hrs. C. 9 hrs 30 mins. D. 11 hrs 40 mins. E. 15 hrs 30 mins .

\(\frac{6}{7}-\frac{2}{3}=\frac{4}{21}\) of the pool wad filled in \(\frac{5}{3}\) hours.

So, to fill the pool completely at this rate it would take \(\frac{(\frac{5}{3})}{(\frac{4}{21})}=8\frac{3}{4}\) hours or 8 hours and 45 minutes.

Answer: A.

Bunuel, this problem is fairly simple but what has confused me a little bit was your approach.

I am a little confused too. Shouldn't it be Rate = Time x Work ?

Re: A pool which was 2/3 full to begin with, was filled at a [#permalink]
04 May 2014, 07:10

Expert's post

pretzel wrote:

PeterHAllen wrote:

Bunuel wrote:

A pool which was 2/3 full to begin with, was filled at a constant rate for 5/3 hours until it was until it was 6/7 full. At this rate, how much time would it take to completely fill this pool if it was empty to begin with?

A. 8 hrs 45 mins. B. 9 hrs. C. 9 hrs 30 mins. D. 11 hrs 40 mins. E. 15 hrs 30 mins.

\(\frac{6}{7}-\frac{2}{3}=\frac{4}{21}\) of the pool wad filled in \(\frac{5}{3}\) hours.

So, to fill the pool completely at this rate it would take \(\frac{(\frac{5}{3})}{(\frac{4}{21})}=8\frac{3}{4}\) hours or 8 hours and 45 minutes.

Answer: A.

Bunuel, this problem is fairly simple but what has confused me a little bit was your approach.

I am a little confused too. Shouldn't it be Rate = Time x Work ?

No, it's Time * Rate = Work.

We have that \(\frac{4}{21}\) of the pool (job done) was filled in \(\frac{5}{3}\) hours (time). How much time would it take to completely fill the pool?

\(\frac{5}{3}*Rate=\frac{4}{21}\) --> \(Rate=\frac{4}{21}*\frac{3}{5}=\frac{4}{35}\) --> time = reciprocal of rate = 35/4 hours. _________________

Re: A pool which was 2/3 full to begin with, was filled at a [#permalink]
14 Feb 2015, 12:18

macjas wrote:

A pool which was 2/3 full to begin with, was filled at a constant rate for 5/3 hours until it was until it was 6/7 full. At this rate, how much time would it take to completely fill this pool if it was empty to begin with?

A. 8 hrs 45 mins. B. 9 hrs. C. 9 hrs 30 mins. D. 11 hrs 40 mins. E. 15 hrs 30 mins .

It 2/3 full

After 5/3 hour (T) it was at 6/7 . which means it took 5/3 hour to full = 6/7-2/3= 4/21

so rate= 4/21 x 3/5 = 4/35

Now work = 1 Rate = 4/35 so Time = 35/4 = 8 hour 45 minutes _________________

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