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Machines X and Y work at their respective constant rates. Ho [#permalink]

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27 Jul 2009, 13:16

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

70% (01:43) correct
30% (01:06) wrong based on 76 sessions

HideShow timer Statistics

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

Re: Machines X and Y work at their respective constant rates. Ho [#permalink]

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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.

Re: Machines X and Y work at their respective constant rates. Ho [#permalink]

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26 Jul 2014, 01:23

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