Question 425168
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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.


For the problem you gave, you have to back into the answer.  You are given that *[tex \Large x\ =\ 4] (your cousin's time) and the final answer is *[tex \Large \frac{xy}{x\ +\ y}\ =\ 3].  So substitute 4 in place of  *[tex \Large x] and then solve the resulting equation for *[tex \Large y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4y}{4\ +\ y}\ =\ 3]


And man are you slow!  You had better pick up the pace or you will get replaced by a Roomba with a paint brush attachment.


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