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Running at their respective constant rates, machine X takes

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

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07 Dec 2005, 15:43
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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?

Please tell how much time it took to reach ans, honestly
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07 Dec 2005, 19:38
SVP
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07 Dec 2005, 19:42
12 days is correct

1/x+1/x+2 = 5/12
x = 4
so the slowest will take 6*2 (widgets) = 12
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07 Dec 2005, 20:13
5 minutes, and I gave up.

* The maximum time you may assign to one math question is said to be 5 minutes.
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VP
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08 Dec 2005, 10:07
Good job guys!!! OA is 12.

I would think its hard bin question. Moreever, it might have been a little faster for some if ACs were given (those who are good at backsolving)
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Whether you think you can or think you can't. You're right! - Henry Ford (1863 - 1947)

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09 Dec 2005, 12:42
Does anyone has any link/resource for Divisibility rules for different intergers?
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hey ya......

VP
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09 Dec 2005, 14:32
Took 2 minutes..

x's output per day = 1/(y+2)
y's output per day = 1/y

Working together, they produce (5/4)W in 3 days ==> 5/12W in 1 day

1/(y+2) + 1/y = 5/12 ==> Solve for y, y = 4 or -(6/5). Since y cannot be negative...y = 4

X's output per day is therefore 1/6 ==> it takes x 6 days to produce W ==> 12 days to produce 2W
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