2005 AMC 12A Problems/Problem 1

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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

$2 = \frac {1}{10}x \Longrightarrow x = 20,\quad 2 = \frac{1}{5}y \Longrightarrow y = 10,\quad x-y = 20 - 10=10  \mathrm{(D)}$.

See also

2005 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
First question
Followed by
Problem 2
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All AMC 12 Problems and Solutions
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