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Running at their respective constant rates, machine X takes

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

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New post 17 Oct 2008, 09:57
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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Re: widgets [#permalink]

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New post 17 Oct 2008, 13:59
E. 12.

If it takes machine X t days to produce w widgets then
1/t + 1/(t-2) = 5/12 or t = 6 and 2t = 12.

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Re: widgets [#permalink]

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New post 17 Oct 2008, 20:25
I think you missed 3 days...question says together produce 5/4 w widgets in 3 days, not in 1 days.. so 1/t + 1(t-2)=15/4...

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New post 19 Oct 2008, 22:12
vishalgc wrote:
I think you missed 3 days...question says together produce 5/4 w widgets in 3 days, not in 1 days.. so 1/t + 1(t-2)=15/4...


No. I have taken 5/12.

In 3 days, 5/4 hence in 1 day, 5/12.

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Re: widgets [#permalink]

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New post 20 Oct 2008, 00:45
scthakur wrote:
E. 12.

If it takes machine X t days to produce w widgets then
1/t + 1/(t-2) = 5/12 or t = 6 and 2t = 12.


I solved the problem but my method was tedious. Could you pls guide me in this step you used
1/t + 1/(t-2)

I followed the 5/12 part. However how did you narrow down to the equation ?

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Re: widgets [#permalink]

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New post 20 Oct 2008, 02:37
kandyhot27 wrote:
scthakur wrote:
E. 12.

If it takes machine X t days to produce w widgets then
1/t + 1/(t-2) = 5/12 or t = 6 and 2t = 12.


I solved the problem but my method was tedious. Could you pls guide me in this step you used
1/t + 1/(t-2)

I followed the 5/12 part. However how did you narrow down to the equation ?


If t days are taken to complete the work then in 1 day, 1/t work will be completed.
Thus, in the above case, 1/t + 1/(t-2) is the sum of work done individually and should be equal to the combined work done that is 5/12. Hope, this explains.

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Re: widgets   [#permalink] 20 Oct 2008, 02:37
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Running at their respective constant rates, machine X takes

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