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Work rates - fraction of a job

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Work rates - fraction of a job

by AleksandrM » Mon Jun 30, 2008 5:03 pm
The following problem appeared on my MGMAT CAT. MGMAT only uses the plugging in method for this problem. However, I would like to see an algebraic solution.

Machine A can complete a certain job in x hours. Machine B can complete the same job in y hours. If A and B work together at their respective rates to complete the job, which of the following represents the fraction of the job that B will not have to complete because of A's help?

A) x - y/x + y
B) x/y - x
C) x + y/xy
D) y/x - y
E) y/x + y
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by jay2007 » Mon Jun 30, 2008 8:07 pm
If both A and B work together they can complete the job in xy/(x+y) hours.

In "y" hours B completes "1" job
Therefore in xy/(x+y) hours B completes x/(x+y) job.

So the remaining portion of the job is: 1-x/(x+y) which is y/(x+y).

So, the answer is E.

Could you please post the OA?
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by AleksandrM » Tue Jul 01, 2008 9:30 am
Answer is E.

Can you please elaborate. I would like to see the setup and the actual steps. Thanks.
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by szapiszapo » Tue Jul 01, 2008 11:56 am
A does the job in x hours, meaning that his rate is rA = 1 / x
B does the job in y hours, meaning that his rate is rB = 1 / y

In z hours (the time needed to complete the job), A will have performed z / x of the job and B will have performed z / y of the job

therefore z/x + z/y = 1 job, i.e. z = xy / (x+y)

Given that B needs y hours to complete the job, the number of hours that B will not have to work (thanks to A) is (y - z) which, multiplied by his rate, give us a fraction F = (y - z) / y = 1 - (z/y)

F = 1 - (x/(x+y))
thus
F = y / (x + y)

Answer E


hope it helps
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