Difference between revisions of "2000 AMC 10 Problems/Problem 9"

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==Problem==
 
==Problem==
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If <math>|x-2|=p</math>, where <math>x<2</math>, then <math>x-p=</math>
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<math>\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 2-2p \qquad\mathrm{(D)}\ 2p-2 \qquad\mathrm{(E)}\ |2p-2|</math>
  
 
==Solution==
 
==Solution==

Revision as of 22:52, 8 January 2009

Problem

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

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

Solution

$|x-2|=p$

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

$x+p=2$.

$x-p=2-2p$.

$\boxed{\text{C}}$

See Also

2000 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions