Question 183405
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This method works for all sorts of shared work problems.  Painting rooms, pumping water in and out of a tank, mowing lawns, etc.


If <i>A</i> can do an entire job in <i>x</i> time periods, then <i>A</i> can do *[tex \LARGE \frac {1}{x}] of the job in 1 time period.


Likewise, if <i>B</i> can do an entire job in <i>y</i> time periods, then <i>B</i> can do *[tex \LARGE \frac {1}{y} ] of the job in 1 time period.


Together, <i>A</i> and <i>B</i> can do *[tex \LARGE \frac {1}{x} + \frac {1}{y}] of the job in 1 time period.


Therefore, <i>A</i> and <i>B</i> can do the entire job in


*[tex \LARGE \frac {1}{\frac {1}{x} + \frac {1}{y}} = \frac {1}{\frac{x + y}{xy}} = \frac {xy}{x + y}]


time periods.


If you have three entities working together at three different rates, then add a factor to the numerator and a term to the denominator:


*[tex \LARGE \frac {1}{\frac {1}{x} + \frac {1}{y} + \frac {1}{z}} = \frac {1}{\frac{x + y + z}{xyz}} = \frac {xyz}{x + y + z}]


Said pattern can be extended indefinately.


Hint:  If you have times given in different units, convert them all to the same unit before proceeding.


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