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

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03 Nov 2007, 08:32
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Running at their respective constant rates, machine X takes 2 days longer to produc ew widgets than machine Y. At these rates, if 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

I seem to have difficult time with work problem !
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03 Nov 2007, 08:42
hey i answered this one not too long ago!
http://www.gmatclub.com/forum/t52676
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Re: PS - Work problem [#permalink]

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03 Nov 2007, 09:22
alohagirl wrote:
Running at their respective constant rates, machine X takes 2 days longer to produc ew widgets than machine Y. At these rates, if 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

I seem to have difficult time with work problem !

Backsolving strategy worked out best for me; didn't have to deal w/ quadratics or anything.

let w = 36 (number divisible by 4); 2w = 72

Speed of A & B = 5/4 w in 3 days or 15 w/day.

Now backsolve starting w/ choice C.

Let time taken by A to produce 2w=72 widgets be 8 days
Therefore, speed of A = 9 widgets/day
Therefore, speed of B = 15-9 = 6 widgets/day
Therefore, time taken by B to produce 2w = 72 widgets = 12 days.

Time taken by A to produce 1 w = 4 days
Time taken by B to produce 1 w = 6 days
Since B is taking longer than A, C is wrong.

Backsolve using E.

Let time taken by A to produce 2w=72 widgets be 12 days
Therefore, speed of A = 6 widgets/day
Therefore, speed of B = 15-6 = 9 widgets/day
Therefore, time taken by B to produce 2w = 72 widgets = 8 days.

Time taken by A to produce 1 w = 6 days
Time taken by B to produce 1 w = 4 days

This is our answer.
Re: PS - Work problem   [#permalink] 03 Nov 2007, 09:22
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# Running at their respective constant rates, machine X takes

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