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

Revision as of 13:22, 17 October 2007

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

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

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

Therefore:

$x=2-p$

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

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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 12 Problems and Solutions

@@ Line 15: / Line 15: @@
 == See also ==
-* [[2000 AMC 12 Problems]]
+{{AMC12 box|year=2000|num-b=4|num-a=6}}
-*[[2000 AMC 12 Problems/Problem 4|Previous Problem]]
-*[[2000 AMC 12 Problems/Problem 6|Next problem]]
 * [[Absolute value]]
 [[Category:Introductory Algebra Problems]]