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2000 AMC 12 Problems/Problem 5

Revision as of 13:22, 16 November 2006 by JBL (talk | contribs)

Problem

If $\displaystyle |x - 2| = p,$ where $\displaystyle x < 2,$ then $\displaystyle x - p =$

$\mathrm{(A) \ -2 } \qquad \mathrm{(B) \ 2 } \qquad \mathrm{(C) \ 2-2p } \qquad \mathrm{(D) \ 2p-2 } \qquad \mathrm{(E) \ |2p-2| }$

Solution

When $\displaystyle x < 2,$, $x-2$ is negative so $|x - 2| = 2-x$ and $\displaystyle x - 2 = -p$.

Therefore:

$x=2-p$

$\displaystyle x-p = (2-p)-p = 2-2p \Longrightarrow \mathrm{(C)}$

See also

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