Question 374925
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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.


However, since John has been working for an hour before Sam starts to help him, John has completed *[tex \Large \frac{1}{20}] of the job already.  That means that there are *[tex \Large \frac{19}{20}] of the job remaining when Sam starts.  If it takes Sam 15 hours to do the entire job by himself, then it would take him *[tex \Large \frac{19}{20}\ \times\ 15\ =\ \frac{57}{4}] hours to do the remaining part of the job by himself.  And quite obviously it would take John 19 hours to finish by himself.


So just do the process outlined using *[tex \Large 19] hours and *[tex \Large \frac{57}{4}] hours as the times for the whole job.  And don't forget to add back the hour that John has already been working.


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