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

Revision as of 22:38, 26 November 2011

The following problem is from both the 2000 AMC 12 #5 and 2000 AMC 10 #9, so both problems redirect to this page.

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

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

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

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

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

2000 AMC 10 (Problems • Answer Key • Resources)
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

@@ Line 1: / Line 1: @@
+{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #5]] and [[2000 AMC 10 Problems|2000 AMC 10 #9]]}}
 == Problem ==
@@ Line 12: / Line 13: @@
 == See also ==
 {{AMC12 box|year=2000|num-b=4|num-a=6}}
+{{AMC10 box|year=2000|num-b=8|num-a=10}}
 [[Category:Introductory Algebra Problems]]