Question 187121
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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.


Your problem is stated backwards, so you can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{xy}{x + y}=40]


But you also know that *[tex \Large 2x = y] because one of them works twice as fast as the other, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x(2x)}{x + (2x)}=40]


Solve for <i>x</i> to get the faster person's time, and then multiply that by 2 to get the slower person's time.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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