Question 325954
<font face="Times New Roman" size="+2">


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 *[tex \Large x\ =\ 7] and *[tex \Large y\ =\ 10]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{1}{\frac{7 + 10}{70}}\ =\ \frac{70}{7\ +\ 10}\ \approx\ 4.11] hrs.


Or 4 hrs 7 minutes give or take a few seconds.  That means that if they start at 7AM, they will be done by 11:07 and Moe takes his lunch break AFTER they are done.



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