Question 181560
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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.


For your problem, <i>A</i> and <i>B</i> are the new machine and the old machine, <i>x</i> and <i>y</i> are 10 hours and 18 hours, your time period is 1 hour and 1 entire job is paving 1 km of roadway.


So take the results of the calculations outlined above which will be the number of hours for both machines to pave 1 km (leave it expressed as an improper fraction) and multiply by 21 to get the total hours required for the entire 21 km stretch of highway.


<i><b>Super Double Plus Extra Credit:</b></i> How would the formula derived above change if you considered three entities working together at three different rates?


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