2000 AMC 10 Problems/Problem 9

Problem

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

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

$|x-2|=p$

$x<2$ , so $2-x=p$ .

$x+p=2$ .

$x-p=2-2p$ .

$\boxed{\text{C}}$

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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