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Working at their respective rates machine A takes 2 days

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damit
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Working at their respective rates machine A takes 2 days [#permalink] New post   Updated on: 02 Sep 2004, 14:06
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Working at their respective rates machine A takes 2 days longer than machine B to produce X objects. At these rates if two machines together produce 5/4 X objects in 3 days. how many days would it take machine a alone to produce 2w widgets

a. 4
b. 6
c. 8
d. 10
e. 12

my answer differs from oa on this and i think oa maybe wrong.. so just checking



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Originally posted by damit on 02 Sep 2004, 13:01.
Last edited by damit on 02 Sep 2004, 14:06, edited 1 time in total.
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Re: Working at their respective rates machine A takes 2 days [#permalink] New post   02 Sep 2004, 13:27
Maybe a stupid question, but I cannot define machine x in this question.

Is it A or B


Regards,

Alex
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Re: Working at their respective rates machine A takes 2 days [#permalink] New post   02 Sep 2004, 14:05
sorry alex it should be machine A. how long does it take machine A
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Re: Working at their respective rates machine A takes 2 days [#permalink] New post   02 Sep 2004, 14:12
Hmmm, I tried to figure this out algebraically but it's not worth it. It took me a crazy time to do so but plugging in numbers took me only 25 seconds.
Answer should be A
3*(1/x + 1/(x+2)) = 5x/4
3*(1/4 + 1/6) = 5x/4
Hence, x=4
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Re: Working at their respective rates machine A takes 2 days [#permalink] New post   02 Sep 2004, 14:32
paul OA is 12 not sure how

also i am not sure if i follow your logic. if you subs x = 4 in LHS must you not do that in RHS as well. if you do then rhs not equall to lhs
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Re: Working at their respective rates machine A takes 2 days [#permalink] New post   02 Sep 2004, 14:36
Shouldn't this be E.

If it takes machine A 12 days to produce 2X, it will take 6 days to produce X and in 3 days it will produce 3X/6 or 2X/4. Similarly machine B will take 4 days to produce X since 6-2=4. So in 3 days machine B produces 3X/4.
And so both collectively produce 3X/4 + 2X/4 = 5X/4 in 3 days.

This is kind of backsolving that I did and E works for it.
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Re: Working at their respective rates machine A takes 2 days [#permalink] New post   02 Sep 2004, 18:06
Finally had some time to look at this one over. Maybe I should have taken a bit more time to think :beat

Let's start over
A's speed: 1/(x+2)
B's speed: 1/x

3*[1/(x+2) + 1/x] = 5/4

By backsolving, we actually find that x=4
3*[1/(4+2) + 1/4] = 5/4
3*[1/6 + 1/4] = 5/4
3*[(2+3)/12] = 5/4
3 * 5/12 = 5/4
15/12 = 5/4
5/4 = 5/4
Bingo! We know that x=4

However, we are looking for A's speed. A's speed is 1/(x+2) = 1/6
This means that A takes 6 days to do 1 widget
For 2 widgets, we have 6*2 = 12 days!

:oops:
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Re: Working at their respective rates machine A takes 2 days [#permalink] New post   05 Sep 2004, 19:38
To produce X objects, machine A takes A days
machine B takes B days or A-2 days

So rate of machine A = x/A
Rate of machine B = x/A-2

In 3 days,

(x/A + x/A-2)*3 = 5/4x

solving we get
5A^2x - 34Ax + 24A = 0
(5A-4)(A-6) = 0

A is either 4/5 or 6

If A takes 4/5 days to produce x widgets, so 2x will take 8/5 days

If A takes 6 days to produce x widgets, so 2x will take 12 days

I'm assuming it's not 2w widgets as stated in the question, but 2x widgets.



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Re: Working at their respective rates machine A takes 2 days [#permalink]
05 Sep 2004, 19:38
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