Difference between revisions of "2000 AMC 12 Problems/Problem 5"

(Added problem and solution)
 
(fixed typo)
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<math>x=2-p</math>
 
<math>x=2-p</math>
  
<math>\displaystyle x-p = (2-p)-p = 2-2p \Rightarrow D </math>
+
<math>\displaystyle x-p = (2-p)-p = 2-2p \Rightarrow C </math>
  
 
== See Also ==
 
== See Also ==
 
[[2000 AMC 12 Problems]]
 
[[2000 AMC 12 Problems]]

Revision as of 19:12, 16 July 2006

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 $\displaystyle x - 2 = -p,$.

Therefore:

$x-2=-p$

$x=2-p$

$\displaystyle x-p = (2-p)-p = 2-2p \Rightarrow C$

See Also

2000 AMC 12 Problems