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Two water pumps, working simlutaneosly at their respective

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Director
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Two water pumps, working simlutaneosly at their respective [#permalink] New post 12 Nov 2006, 04:19
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Two water pumps, working simlutaneosly at their respective rates, took exactly 4 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 its constant rate?

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

Finding x is easy, but i found difficulty in finding the total work done....
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 [#permalink] New post 12 Nov 2006, 05:40
Let Y be the rate of the faster pump in terms of the the slower pump:

3/2Y(Y)/3/2Y+Y=4 ---> 3/2Y^2=10Y ---> 3/2Y=10 ---> Y=20/3
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 [#permalink] New post 12 Nov 2006, 09:26
GMATT73 wrote:
Let Y be the rate of the faster pump in terms of the the slower pump:

3/2Y(Y)/3/2Y+Y=4 ---> 3/2Y^2=10Y ---> 3/2Y=10 ---> Y=20/3


A very unclear equation...

(3/2y)*y / (3/2y)+y=4?
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 [#permalink] New post 12 Nov 2006, 16:02
I set up two equations and used the substitution method for this problem:

equation 1

1/x + 1/y = 1/4

equation 2

x = 1.5y

Solve for Y and you get 20/3..
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 [#permalink] New post 12 Nov 2006, 17:24
Let x be the faster one, y is the slower machine.

Since the rate of the faster one is 3/2 of the slower machine and the rate is expressed as 1/x and 1/y for the 2 machine.
I have: 1/x = (3/2) (1/y) <=> y = (3x)/2 <1>
another equation is 1/x +1/y = 1/4 <2>

Substitute the value of y into the second equation and solve for x, that would be the number of hrs x will take to work alone (20/3)
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 [#permalink] New post 12 Nov 2006, 19:21
Pump A - A hours to fill 1 pool --> 1/A pool in 1 hour
Pump B - B hours to fill 1 pool --> 1/B pool in 1 hour

Together, they can fill (A+B)/AB pool in 1 hour. They can fill AB/A+B hours, which is 4 hours.

AB/A+B = 4

Assuming B is faster, then A = 1.5B
1.5B^2/2.5B = 4
1.5B = 10
B = 20/3 hours
  [#permalink] 12 Nov 2006, 19:21
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