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

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A pool which was 2/3 full to begin with, was filled at a [#permalink]

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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 .
[Reveal] Spoiler: OA

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

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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 [#permalink]

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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 con [#permalink]

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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]

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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]

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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]

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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]

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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]

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New post 04 May 2014, 08:10
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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Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
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Re: A pool which was 2/3 full to begin with, was filled at a [#permalink]

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

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A pool which was 2/3 full to begin with, was filled at a [#permalink]

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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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A pool which was 2/3 full to begin with, was filled at a   [#permalink] 16 Jun 2016, 12:32
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