Question 763951
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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.


So you could just do it in the formulatic way:  *[tex \LARGE \frac{2\,\cdot\,3}{2\,+\,3}\ =\ \frac{6}{5}\ =\ 1.2]


Or use the logic laid out above:


Whole job in 2 hours, then *[tex \LARGE \frac{1}{2}] of the job in one hour.  Likewise, for the other guy, *[tex \LARGE \frac{1}{3}] of the job in one hour.


Then *[tex \LARGE \frac{1}{2}\ +\ \frac{1}{3}\ =\ \frac{3}{6}\ +\ \frac{2}{6}\ =\ \frac{5}{6}] of the job in one hour, or the whole job in *[tex \LARGE \frac{6}{5}\ =\ 1.2] hours.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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