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Machines X and Y work at their respective constant rates. Ho

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Machines X and Y work at their respective constant rates. Ho [#permalink] New post 27 Jul 2009, 13:16
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Question Stats:

62% (01:36) correct 38% (00:58) wrong based on 37 sessions
Machines X and Y work at their respective constant rates. How many more hours does it take machine Y, working alone, to fill a production order of a certain size than it takes machine X, working alone?

Let x and y be the times needed for machines X and Y respectively working alone to fill a production order of this size.

Question: y-x=?

(1) Machines X and Y, working together, fill a production order of this size in 2/3 the time that machine X, working alone, does --> general relationship: \frac{1}{x}+\frac{1}{y}=\frac{1}{total \ time} --> Total time needed for machines X and Y working together is total \ time=\frac{xy}{x+y} (general formula) --> given \frac{xy}{x+y}=x*\frac{2}{3} --> 2x=y. Not sufficient

(2) Machine Y, working alone, fills a production order of this size in twice the time that machine X, working alone, does --> 2x=y, the same info. Not sufficient

(1)+(2) Nothing new. Not Sufficient.

Answer: E.

[color=#ff0000]OPEN DISCUSSION OF THIS QUESTION IS HERE: machines-x-and-y-work-at-their-respective-constant-rates-128011.html
[Reveal] Spoiler: OA

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Last edited by Bunuel on 26 Jul 2014, 01:39, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Re: Machines X and Y work at their respective constant rates. Ho [#permalink] New post 27 Jul 2009, 15:44
D.

assuming X takes x amount of time to finish a size of order, and Y takes y amount of time to finish the same size, we need the relationship of x and y

so in statement 1, we have 1/(1/x + 1/y) = 2x/3 =simplify=> xy/(x+y) = 2x/3 => 3y = 2x+2y => y = 2x

and in statement 2, we have y = 2x.

so D. each is sufficient

-edit
after reading whatdahell's reasoning, I neglected the word "hours" in the original question, if the question is looking for an actual number, then we cant calculate the number by any means...the only result we can get is how much faster is x than y, so E probably is a better choice if that's the case

Last edited by sprtng on 27 Jul 2009, 15:55, edited 1 time in total.
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Re: Machines X and Y work at their respective constant rates. Ho [#permalink] New post 27 Jul 2009, 15:47
IMO : E - both not suffis

I see that you've chosen C, that means you know why I alone and II alone is not sufficient.

From I, you get the equation: (2/3)x = (xy)/(x+y) ----- (assuming X takes x hours and Y takes y hours to complete the job)

From II, you get x = 2y,

We need to get-> y-x,

from 1 and 2, replacing x= 2y in (i),

(2/3)(2y) = (2yy)/(3y)
-> (4y/3) = (2y/3)
y cancels out so no way of knowing the variable's value. There (E)
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Re: Machines X and Y work at their respective constant rates. Ho [#permalink] New post 17 Aug 2009, 07:51
I agree with whatthehell

From the original statement and statement 1 I got the following equation:

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

Which is still not sufficent.

Statement 2 alone is not sufficient either...

and when you substitute 2*X for Y :

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

So the answer is E. Both statements together are not sufficient.
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Re: Machines X and Y work at their respective constant rates. Ho [#permalink] New post 17 Aug 2009, 21:12
it is E, by both we arrive at Y =2X, X can take any values.. not sufficient.
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Re: Machines X and Y work at their respective constant rates. Ho [#permalink] New post 26 Jul 2014, 01:23
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Re: Machines X and Y work at their respective constant rates. Ho   [#permalink] 26 Jul 2014, 01:23
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