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Working simultaneously at their respective constant rates, M

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Working simultaneously at their respective constant rates, M [#permalink] New post 06 Dec 2012, 09:58
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Working simultaneously at their respective constant rates, Machines A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 nails in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

(A) x/(x+y)
(B) y/(x+y)
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)
[Reveal] Spoiler: OA
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Re: Working simultaneously at their respective constant rates, M [#permalink] New post 06 Dec 2012, 10:06
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Walkabout wrote:
Working simultaneously at their respective constant rates, Machines A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 nails in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

(A) x/(x+y)
(B) y/(x+y)
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)


Pick some smart numbers for x and y.

Say x=1 hour and y=2 hours (notice that y must be greater than x, since the time for machine A to do the job, which is y hours, must be more than the time for machines A and B working together to do the same job, which is x hours).

In this case, the time needed for machine B to do the job must also be 2 hours: 1/2+1/2=1.

Now, plug x=1 and y=2 in the options to see which one yields 2. Only option E fits.

Answer: E.
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Re: Working simultaneously at their respective constant rates, M [#permalink] New post 06 Dec 2012, 12:28
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Walkabout wrote:
Working simultaneously at their respective constant rates, Machines A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 nails in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

(A) x/(x+y)
(B) y/(x+y)
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)


The above sol is awesome.... but i did it the longer way, algebraically...

rate of A be a and B be b

a + b = 800/x ..... 1
a = 800/y ..... 2


Use 2 in 1... we get

b = 800 (y-x) / xy

Finally

Rate of B * time = Work done by B (we want time)

800 (y-x) / xy * t = 800

t = xy / (y-x)
Re: Working simultaneously at their respective constant rates, M   [#permalink] 06 Dec 2012, 12:28
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