Machine A can complete a certain job in x hours. Machine B

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Machine A can complete a certain job in x hours. Machine B [#permalink]

23 Sep 2007, 23:16

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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?

(x - y)/(x + y)

x/(y – x)

(x + y)/xy

y/(x – y)

y/(x + y)

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23 Sep 2007, 23:54

Machine A -> 1 job in x hours --> 1/x job per hour
Machine B -> 1 job in y hours --> 1/y job per hour

Toegther:
1/x + 1/y = (x+y)/xy job per hour --> 1 job in xy/(x+y) hours

In this amount of time, Machien B does x/(x+y) of the job. So it does not have to do 1 - x/(x+y) = y/(x+y) work.

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Re: PS: Job hours [#permalink]

24 Sep 2007, 01:05

carpeD wrote:

Just plug in some values. X=2hrs Y=4hrs.

So rates are 1/2 and 1/4. Combined they are 3/4. So it takes 4/3 hrs to complete the job working together. Or 1hr 20min.

Subtract 4hrs from 1hr 20 min. Difference is 2 2/3 or 2 hrs 40min.

So calculate how much work Y does in 2 2/3 hrs (8/3hrs). 1/4* 8/3 = 2/3 of the job.

So look for the choice that gives you this value.

A. 2-4/4+2 = -2/3 no.
try em all.

E: 4/2+4 = 2/3

This method is a bit slower, and its not completely full proof. Ex/ if you chose x=4 and y=2. Then A and E work. In this situation ud have to go back and calculate the difference of the time, then calculate how much work Y did. But in any case if u didn't know wut to do, u at least only have 2 choices left.

Re: PS: Job hours [#permalink] 24 Sep 2007, 01:05

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Machine A can complete a certain job in x hours. Machine B

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Machine A can complete a certain job in x hours. Machine B

Machine A can complete a certain job in x hours. Machine B

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