Running at their respective constant rates, machine X takes : PS Archive
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# Running at their respective constant rates, machine X takes

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VP
Joined: 26 Apr 2004
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Running at their respective constant rates, machine X takes [#permalink]

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22 Aug 2004, 02:18
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Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y.

At these rates, if the two machines together produce 5/4 w widgets in 3 days, how many days would it take machine X alone to produce 2w widgets?

A. 4
B. 6
C. 8
D. 10
E. 12

Senior Manager
Joined: 19 May 2004
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22 Aug 2004, 11:29
I got B.

If the two machines together produce 5/4 w widgets in 3 days,
They will complete w widgets together in 3*4/5 = 12/5 days.
Let t be the time it takes machine Y to complete the job alone.
t+2 is the time it takes machine X to complete the job alone.

[(12/5) / t] + [(12/5) / (t+2] = 1

t=4 -> t+2 =6.

It took me more than 2 minutes.
I hope there is a simpler solution.
VP
Joined: 26 Apr 2004
Posts: 1218
Location: Taiwan
Followers: 2

Kudos [?]: 609 [0], given: 0

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22 Aug 2004, 17:46
Dear buddy,

The OA is (E)

Two machines togerther produce 5/4w widgets in 3 days; together they produce w widgets in 12/5 days.

And, if we set the finished days of Y alone as N days, the finished days of Y alone are N+2 days.

So, 1/N + 1/(N+2) = 5/12 from this equation, we can acquire N = 4

Hence, machine X alone takes N+2 = 6 days to produce w widgets; that is, it takes 12 days to produce 2w widgets.
Senior Manager
Joined: 19 May 2004
Posts: 291
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22 Aug 2004, 21:09
Thanks, i didn't read it well enough.
Those traps get me all the time!
22 Aug 2004, 21:09
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