2005 AMC 12B Problems/Problem 1

Revision as of 21:26, 17 April 2009 by M1sterzer0 (talk | contribs) (Solution)

Problem

Two is $10\%$ of $x$ and $20\%$ of $y$. What is $x-y$?

$\mathrm{(A)}\ 1      \qquad \mathrm{(B)}\ 2      \qquad \mathrm{(C)}\ 5      \qquad \mathrm{(D)}\ 10      \qquad \mathrm{(E)}\ 20$

Solution

$2=0.1x \Rightarrow x=20$.

$2=0.2y \Rightarrow y=10$.

$\therefore x-y=20-10=\boxed{10}$.

See also

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