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Work Rate problem

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Source: — Problem Solving |
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by parallel_chase » Wed Aug 06, 2008 12:17 pm
A can complete the job in x hours, therefore A's rate = 1/x
B can complete the job in 1/x hours, therefore B's rate = x

A and B working simultaneously = x+(1/x) = (x^2+1)/x
W/(x^2+1)/x = T

T(x^2+1)/x = W

In y hours A can complete = y/x job.
Now at this point B joins A and they work together for till 100 decks are manufactured.

y/x + W = 100

y/x + [(x^2+1)/x]T =100

[(x^2+1)/x]T =100 - (y/x)
[(x^2+1)/x]T = (100x-y)/x
t = (100x-y)/(x^2+1)

Hence B is the answer.

I know it looks complicated but its not. I was able to do this in less than 1 minute. I find plugging in options very time consuming specially when one can do it easily with variables.

Let me know if you still have any doubts.
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by pepeprepa » Wed Aug 06, 2008 1:07 pm
Explanations of Chase are perfect, here is how I did:

Number of objects which have to be made by A and B together: (100-y/x)
Number of hours for A and B together to make 1 object: x+(1/x)
Therefore,
(100-y/x) / (x+(1/x)) = 100x - y / x^2 + 1
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by [email protected] » Thu Aug 07, 2008 2:18 am
Thanks very much for the explanations guys!
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