Question 340393
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If A can do a job in <i>x</i> time periods, then A can do *[tex \Large \frac{1}{x}] of the job in 1 time period.  Likewise, if B can do the same job in <i>y</i> time periods, then B can do *[tex \Large \frac{1}{y}] of the job in 1 time period.


So, working together, they can do


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{1}{x}\ +\ \frac{1}{y}\ =\ \frac{x\ +\ y}{xy} ]


of the job in 1 time period.


Therefore, they can do the whole job in:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{1}{\frac{x + y}{xy}}\ =\ \frac{xy}{x\ +\ y}]


time periods.


This idea is easily expanded to three workers:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{x}\ +\ \frac{1}{y}\ =\ \frac{1}{z}\ =\ \frac{yz\ +\ xz\ +\ xy}{xyz}]


and so on...


Don't for get to change Tony's "do it by himself" time from 1 and 1/2 hours to 90 minutes so that you are working with the same time units for all three people.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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