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# Running at their respective and constant rates, machine x

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Joined: 03 Dec 2007
Posts: 82
Schools: NYU STERN
Running at their respective and constant rates, machine x [#permalink]

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29 Dec 2008, 17:14
Running at their respective and constant rates, machine x takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce 5/4w widgets in 3 days, hoe many days would it take machine X aline to produce 2w widgets.

So far I have
X's rate = w/(d+2)
Y's rate = w/d

[w/d + w/(d+2)] * 3 = 5w/4

Not sure what to do with those fractions - how do you get a common denominator??

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Joined: 29 Aug 2007
Posts: 2426
Re: Rate Question! FUN [#permalink]

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30 Dec 2008, 00:05
hardaway7 wrote:
Running at their respective and constant rates, machine x takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce 5/4w widgets in 3 days, hoe many days would it take machine X aline to produce 2w widgets.

So far I have
X's rate = w/(d+2)
Y's rate = w/d

[w/d + w/(d+2)] * 3 = 5w/4

Not sure what to do with those fractions - how do you get a common denominator??

[w/d + w/(d+2)] * 3 = 5w/4
w[1/d + 1/(d+2)] * 3 = 5w/4
[1/d + 1/(d+2)] = 5/12
[d+2+d]/d(d+2) = 5/12
[2d+2]/d(d+2) = 5/12
24[d+1] = 5d(d+2)
24d + 24 = 5d^2 + 10d
5d^2 -14d -24 = 0
5d^2 -20d + 6d - 24 = 0
5d (d - 4) + 6 (d - 4) = 0
d = 4 or -6/5 but -ve value, -6/5 is not possible. so d = 4

x's daily rate = w/(d+2) = w/6
or, x can produce w/(d+2) or w/6 in a day
x can produce 2w in 12 day [(w/6) *12)].

so days that x takes to produce 2w is 12 days
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Posts: 1345
Re: Rate Question! FUN [#permalink]

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30 Dec 2008, 10:34
hardaway7 wrote:
Running at their respective and constant rates, machine x takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce 5/4w widgets in 3 days, hoe many days would it take machine X aline to produce 2w widgets.

So far I have
X's rate = w/(d+2)
Y's rate = w/d

[w/d + w/(d+2)] * 3 = 5w/4

Not sure what to do with those fractions - how do you get a common denominator??

Strictly speaking, one doesn't need a formula for rates problems. The well-known formula works by expressing the amount of work each machine does in the *same* amount of time- one hour. If you know this, you can do any rates problem without a formula; just get the same time (rather than the same number of widgets) for each machine. Algebraically, this produces the same equation as the formula (though without the fractions), but is more flexible in case you encounter some kind of twist. Applied to this problem, it's still quite abstract, but it works:

Y produces w widgets in d days
X produces w widgets in d+2 days
X+Y produce w widgets in 12/5 days

Getting the same time for each:

Y produces dw + 2w widgets in d(d+2) days
X produces dw widgets in d(d+2) days
X+Y produce (5/12)*d*(d+2)*w widgets in d(d+2) days

So:
dw + 2w + dw = (5/12)*d*(d+2)*w
(24/5)w(d+1) = dw(d+2)
24d + 24 = 5d^2 + 10d
5d^2 - 14d - 24 = 0
(5d + 6)(d - 4) = 0

And since d > 0, we find d = 4.

The question asks us to find 2d+4, which is 12.

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Re: Rate Question! FUN   [#permalink] 30 Dec 2008, 10:34
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# Running at their respective and constant rates, machine x

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