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Work/Rate Problem

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Manager
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Work/Rate Problem [#permalink] New post 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 doesn’t seem to work…
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 [#permalink] New post 24 Apr 2006, 21:16
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
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 [#permalink] New post 24 Apr 2006, 22:26
Answer is E

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
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 [#permalink] New post 24 Apr 2006, 22:31
1/a + 1/b = 1/4

substituting, a=3/2b

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

solving for a = 20/3
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 [#permalink] New post 24 Apr 2006, 22:33
x - faster pump (x - hours to fill the tank)
1.5x - slower pump

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

x = 20/3.
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 [#permalink] New post 24 Apr 2006, 22:36
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
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 [#permalink] New post 24 Apr 2006, 23:39
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
  [#permalink] 24 Apr 2006, 23:39
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