Work Rate Problem - 4

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Work Rate Problem - 4

by saadishah » Tue Sep 27, 2016 10:48 am
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)

This question seems like Greek to me. Not even able to start. Please help.

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by GMATGuruNY » Tue Sep 27, 2016 12:07 pm
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?

(x - y)/(x + y)

x/(y - x)

(x + y)/xy

y/(x - y)

y/(x + y)
Let the job = 10 units.

Let x = 5 hours.
Rate for A alone = w/t = 10/5 = 2 units per hour.

Let y = 2 hours.
Rate for B alone = w/t = 10/2 = 5 units per hour.

The combined rate for A+B = 2+5 = 7 units per hour.
Of the 7 units produced each hour, 2 will produced by A and 5 will be produced by B.
Thus, the fraction of the job produced by A = 2/7. This is our target.
Now we plug x=5 and y=2 into the answers to see which yields our target of 2/7.

A quick scan of the answers reveals that only E works:
y/(x+y) = 2/(5+2) = 2/7.

The correct answer is E.
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by Matt@VeritasPrep » Thu Sep 29, 2016 7:00 pm
Work = Rate * Time

We're only doing one job, so Work = 1. If Machine A did the job alone, we know

Work = Rate * Time
1 = Rate * x
1/x = Rate

If Machine B did the job alone, we know

Work = Rate * Time
1 = Rate * y
1/y = Rate

So we have each Machine's rate per hour: A's is 1/x, B's is 1/y.

If they work together, we have

Work = Rate * Time
1 = (1/x + 1/y) * Time
xy/(x + y) = Time

Each machine works for the same amount of time. The fraction that A does is

A's Rate * A's Time =>

(1/x) * xy/(x+y) =>

y/(x + y)

so our answer is E.