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by resilient » Sun May 11, 2008 11:57 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


qa is e

plugging in number and calculating away however, facing a bit of difficulty in translating what is actually being asked and finalizing the lest step.
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by VP_Tatiana » Sun May 11, 2008 1:22 pm
Ok, so we need to remember that work = rate*time. We can find out how much work a certain machine completed if we know its rate (given in the problem) and the time it worked. So, we need to figure out the time a machine will work when it works in tandem with another machine.

We calculate the time needed to do the job when both machines work together with the formula:

1/total time = (1/x) + (1/y).

How did I get that? Well suppose x takes 3 hours to do the job. It's rate would then be 1/3 of a job per hour. We can get the rate of two machines working together by summing their individual rates.

We solve for total time, and get xy/(y+x).

Now, we want to know the work that B didn't have to do because of A's help. To figure out the work A did, we take A's rate (1/x) and A's time working (xy/(y+x)), and multiply them together.

This gives us y/(y+x), which is E.

Let me know if you have questions,

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by California4jx » Sun Aug 17, 2008 3:22 pm
another explanation could be:

Rate of A = 1/x
Rate od B = 1/y

Work Together: 1/x + 1/y
= x+y / xy (completed in 1 hour)
= xy / x + y (total time it took to complete entire job)

job that B doesnt has to do because of A's help:
=> y - (work together)
=> y - xy / x+y
=> y (x+y ) - xy / x + y
=> y (square) / x + y ---------- (I)

original time B took = y -----------(II)

Fraction = job that B doesnt has to do / original time B took ----- (III)

=> plug values in (III) from (I) and (II)
=> y(square) / x + y * (1/y)

=> y / x + y -----> E
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by pepeprepa » Mon Aug 18, 2008 1:25 am
To answer you question concerning plugging in. Personally, I don't think it is a good method for work problem, often confusing and hard to apply.
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by California4jx » Mon Aug 18, 2008 5:05 am
the above solution is not plugging any values - read carefully.
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by pepeprepa » Mon Aug 18, 2008 6:10 am
Yep I know what you did is not plugging in, I was talking about what resilient tried to do.
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by pepeprepa » Mon Aug 18, 2008 6:28 am
Yep I know what you did is not plugging in, I was talking about what resilient tried to do.
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