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A pool which was 2/3 full to begin with, was filled at a

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macjas
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A pool which was 2/3 full to begin with, was filled at a [#permalink] New post   02 Jun 2012, 23:51
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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 .
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Re: A pool which was 2/3 full to begin with, was filled at a [#permalink] New post   14 Nov 2012, 05:07
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Calculate the rate of the pipe:
\(Output = \frac{6}{7}-\frac{2}{3}=\frac{4}{21}\)
\(t = \frac{5}{3} hours\)
\(R = \frac{Output}{t} = 4/35\)

time it takes to fill the pool = reciprocal of rate
\(\frac{35}{4}hours = 8 \frac{3}{4}hours = 8 hours and 45 minutes\)
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Re: A pool which was 2/3 full to begin with, was filled at a con [#permalink] New post   03 Jun 2012, 02:56
A.

The pool was 2/3 full initially.

In 5/3 hrs, it was filled to 6/7 of its capacity.

Therefore, in 5/3 hrs, 6/7 - 2/3 was the amount of volume filled.

So, 5/3 hrs = 6/7 - 2/3 = 4/21

Therefore, for filling the vessel from empty to full, you would require 5/3 * 21/4 = 35/4 = 8 Hrs 45 Mins.

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

Answer: A.
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Re: A pool which was 2/3 full to begin with, was filled at a [#permalink] New post   24 Jan 2014, 11:45
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.
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Re: A pool which was 2/3 full to begin with, was filled at a [#permalink] New post   24 Jan 2014, 22:14
Amount currently filled + time*rate = final amount filled
2/3+5x/3=6/7

Solve for x to get the rate at which the pool was being filled.
x = 4/35.

Invert x to get how long it will take for one job to get finished.
35/4 = 8 3/4 = 8 hours and 45 minutes.
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Re: A pool which was 2/3 full to begin with, was filled at a [#permalink] New post   28 Jan 2014, 08:04
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)

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Re: A pool which was 2/3 full to begin with, was filled at a [#permalink] New post   04 May 2014, 07:02
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 ?
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Re: A pool which was 2/3 full to begin with, was filled at a [#permalink] New post   04 May 2014, 08:10
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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.
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Re: A pool which was 2/3 full to begin with, was filled at a [#permalink] New post   14 Feb 2015, 13:18
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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A pool which was 2/3 full to begin with, was filled at a [#permalink] New post   16 Jun 2016, 12:32
t=time to fill entire pool
6/7-2/3=4/21
t=(5/3)/(4/21)
t=8 hrs 45 min
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Re: A pool which was 2/3 full to begin with, was filled at a [#permalink] New post   20 Jun 2018, 09:34
I solved this question by picking easy number to work with, which is 21 because its multiple of two denominators 2/3 adn 6/7.
We know that 2/3 is 14
and it was filled with pipe to 6/7 of capacilty or 18, so pipe filled 4 liters(gallons, whatever) for 5/3h which is 100 min (5/3*60)
We have proportion, 4 l = 100 min
21l=x
4x=2100
x=525, which is 8h 45 min (A)
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Re: A pool which was 2/3 full to begin with, was filled at a [#permalink] New post   04 Apr 2023, 12:23
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