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# Work/Rate Problem

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Manager
Joined: 05 Dec 2005
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Location: Manhattan Beach, CA
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Work/Rate Problem [#permalink]  24 Apr 2006, 21:00
This is from the GMATPrep Practice Test (they don't provide answer exps, so I'm trying to figure out how to do this problem)

*************
Two water pumps, working simultaneously at their respective rates, took exactly 4 hrs to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other pump, how many hours would it have taken the faster pump to fill the pool if it had worked alone at its constant rate

a) 5
b) 16/3
c) 11/2
d) 6
e) 20/3

The answer is E. Just not sure how to set this up.

I figured it was

1.5x + x = 4 hrs, where x equaled the constant rate. But going down this path doesnâ€™t seem to workâ€¦
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GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5078
Location: Singapore
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I'm getting 10 as well...

Let the two rates be A and B.
Assuming the faster pump is A, then A = 1.5B

So 1/A + 1/B = 1/4
1/A + 1.5/A = 1/4
VP
Joined: 29 Apr 2003
Posts: 1405
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Let the pumps be p1 (faster) and p2 (slower)

p1 takes x hrs and p2 takes y hrs

thus x = y/1.5

in 1 hr:
p1: 1/x = 1.5/y
p2: =1/y
p1+p2 = 1.5/y + 1/y = 2.5/y

Now 2.5/y = 1/4

Thus y = 10

They want the time for the faster = 10/1.5 (see above) = 20/3

Hence E
Intern
Joined: 16 Apr 2006
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1/a + 1/b = 1/4

substituting, a=3/2b

1/a + 1/(3/2a) = 1/4

solving for a = 20/3
SVP
Joined: 14 Dec 2004
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x - faster pump (x - hours to fill the tank)
1.5x - slower pump

1/x + 1/(1.5x) = 1/4

x = 20/3.
Director
Joined: 13 Nov 2003
Posts: 793
Location: BULGARIA
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Hallo,
Think that it is 20/3.
(1/R1)+(1/R2)=1/4
R1=1,5*R2
substitute and get
[10r2-(1,5r2)^2]/(6r2)^2=0
now denomionator can not be 0, cause division by 0 is not defined.
Equate nominator with 0 and get r2*(10-1,5r2)=0 so r2=20/3
regards
Director
Joined: 05 Feb 2006
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Yes E...

1/x + 1/3/2x=1/4

(3/2x+x)/(3/2)x^2=1/4

5/2x/(3/2)x^2=1/4

10x=(3/2)x^2

10=(3/2)x

20=3x

x=20/3
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# Work/Rate Problem

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