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Two water pumps, working simultaneously at their respective

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Two water pumps, working simultaneously at their respective  [#permalink]

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New post Updated on: 12 Jan 2014, 04:44
12
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

84% (01:23) correct 16% (01:25) wrong based on 237 sessions

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Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?

A. 5
B. 16/3
C. 11/2
D. 6
E. 20/3

OPEN DISCUSSION OF THIS QUESTION IS HERE: two-water-pumps-working-simultaneously-at-their-respective-155865.html

Originally posted by netcaesar on 14 Nov 2006, 12:56.
Last edited by Bunuel on 12 Jan 2014, 04:44, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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New post 14 Nov 2006, 13:06
1
4
E from me...

R(A) and T(A)= rate and time of 1st pump.
R(B) and T(B)= rate and time of 2nd pump.

R(A)= 1.5 R(B)

Hence T(B)=1.5 T(A)

1/T(A) + 1/T(B)= 1/4
Substituting

T(A)= 20/3
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New post 14 Nov 2006, 13:51
E for me as well


d=r*t

1 = 7/2r*4

1/10 = r

Faster rate = 1/10*3/2 = 3/20

1 = 3/20t

20/3 = t
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New post 17 Nov 2006, 11:31
You are right.

OA is E
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New post 17 Nov 2006, 13:02
3
1
Work = r * t

Work A = 1 * 4

Work B = 3/2 * 4

Total work = (1*4) + (3/2 * 4) = 10

10/rate B = 10/(3/2) or 10*2/3 = 20/3
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New post 19 Nov 2006, 18:49
2
Let the rate be:
Pump A - 1 pool in A hours
Pump B - 1 pool in B hours

Then Pump A - 1/A pool in 1 hour
And Pump B - 1/B pool in 1 hour

Together: (A+B)/AB pool in 1 hour -> 1 pool in AB/(A+B) hours = 4 hours
4A + 4B = AB -- (1)

Assuming pump B is faster and A = 1.5B, then 4A+4B = AB becomes
1.5B^2 = 10B
B(1.5B - 10) = 0
B = 0 or B = 20/3 hours

So 1 pool in 20/3 hours

Ans E
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Re: Two water pumps, working simultaneously at their respective  [#permalink]

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New post 11 Jan 2014, 15:09
I felt like I understood "work" problems until I saw this one on my GMATPrep Exam.

In my head the problem looks like this

4hrs = (3/2)X + X

or

1/4 = (3/2)*(1/X) + (1/X)

I simply dont know how to solve this problem....I'm confused!

I've seen how others have solved this problem, but it just isn't clicking.


Is someone willing to help?
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Re: Two water pumps, working simultaneously at their respective  [#permalink]

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New post 12 Jan 2014, 04:47
4
7
TroyfontaineMacon wrote:
I felt like I understood "work" problems until I saw this one on my GMATPrep Exam.

In my head the problem looks like this

4hrs = (3/2)X + X

or

1/4 = (3/2)*(1/X) + (1/X)

I simply dont know how to solve this problem....I'm confused!

I've seen how others have solved this problem, but it just isn't clicking.


Is someone willing to help?

Work rate problems are my weakest area. For whatever reason I get mixed up on these problem. Even knowing the answer to this question I can't figure out how it comes out to that answer. Any help would be great, thanks!


Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?

A. 5
B. 16/3
C. 11/2
D. 6
E. 20/3

Say the rate of the faster pump is x pool/hour, then the rate of the slower pump would be x/1.5=2x/3 pool/hour.

Since, the combined rate is 1/4 pool/hour, then we have that x+2x/3=1/4 --> x=3/20 pool hour.

The time is reciprocal of the rate, therefore it would take 20/3 hours the faster pump to fill the pool working alone.

Answer: E.

Theory on work/rate problems: work-word-problems-made-easy-87357.html

All DS work/rate problems to practice: search.php?search_id=tag&tag_id=46
All PS work/rate problems to practice: search.php?search_id=tag&tag_id=66


OPEN DISCUSSION OF THIS QUESTION IS HERE: two-water-pumps-working-simultaneously-at-their-respective-155865.html
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Re: Two water pumps, working simultaneously at their respective  [#permalink]

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