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# Running at their respective constant rates, machine X takes

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Senior Manager
Joined: 11 Feb 2007
Posts: 350

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Running at their respective constant rates, machine X takes [#permalink]

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11 Mar 2007, 17:07
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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

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Manager
Joined: 09 Jan 2007
Posts: 240

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11 Mar 2007, 17:35
Answer is (E), i.e., 12 days.

Solution:

Lets assume Machine Y takes y days to produce w widgets.
Therefore Machine X takes (y+2) days to produce w widgets.

From above we can write:

Machine X produces w/(y+2) and Machine Y produces w/y widgets in 1 day.
So, if both Machines run together, both will produce w/(y+2) + w/y widgets.
In three days both machines will produce 3*(w/(y+2) + w/y) = 3w*(1/y + 1/(y+2)) widgets.

It's given that in 3 days both together also produce 5w/4
Therefore
3w*(1/y + 1/(y+2)) = 5w/4
=> 1/y + 1/(y+2) = 5/12, by cancelling out w from both sides.
=> 5y^2 - 14y - 24 = 0
=> (y-4)*(5y+6) = 0

As y can not be -ve, so y = 4.

Now we can write

Machine X produces w widgets in (y+2)=6 days
so Machine X produces ww widgets in 2*6 = 12 days

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Senior Manager
Joined: 11 Feb 2007
Posts: 350

Kudos [?]: 191 [0], given: 0

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11 Mar 2007, 20:49

Thanks rdg for your thorough explanation!

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11 Mar 2007, 20:49
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