Question 1163280
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C can do one paper in *[tex \Large x] minutes, and S can do one paper in *[tex \Large x\ +\ 1] minutes.  Therefore, C can do *[tex \Large \frac{1}{x}] of a paper in one minute, and S can do *[tex \Large \frac{1}{x\,+\,1}] of a paper in one minute.  Since they can do 27 papers in one hour, they can do *[tex \Large \frac{9}{20}] of one paper in one minute.  Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{x}\ +\ \frac{1}{x\,+\,1}\ =\ \frac{9}{20}]


Solve for *[tex \Large x] and then calculate *[tex \Large x\ +\ 1]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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