2000 AMC 10 Problems/Problem 8

Revision as of 15:01, 7 January 2009 by BOGTRO (talk | contribs) (New page: Let <math>f</math> be the number of freshman and s be the number of sophomores. <math>\frac{2}{5}f=\frac{4}{5}s</math>. <math>f=2s</math>. There are twice as many freshman as sophomores.)
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Let $f$ be the number of freshman and s be the number of sophomores.

$\frac{2}{5}f=\frac{4}{5}s$. $f=2s$.

There are twice as many freshman as sophomores.

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