Question 1117368
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Let *[tex \Large p\ =\ \frac{x}{100}]


Lee paid 300


Wong paid  *[tex \Large 300(1\ +\ p)]


Fei paid *[tex \Large 300(1\ +\ p)^2]


Fei paid *[tex \Large 300\ +\ 600p +\ 0.75]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 300(1\ +\ p)^2\ =\ 300\ +\ 600p\ +\ 0.75]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 300\ +\ 600p\ +\ 300p^2\ =\ 300\ +\ 600p\ +\ 0.75]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p^2\ =\ \frac{0.75}{300}\ =\ \frac{0.25}{100}\ =\ 0.0025]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p\ =\ 0.05]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 5]


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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