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

Revision as of 18:46, 4 January 2008

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

@@ Line 1: / Line 1: @@
 == Problem ==
-If <math>\displaystyle |x - 2| = p,</math> where <math>\displaystyle x < 2,</math> then <math>\displaystyle x - p =</math>
+If <math>|x - 2| = p,</math> where <math>x < 2,</math> then <math>x - p =</math>
 <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 ==
-When <math>\displaystyle x < 2,</math>, <math>x-2</math> is negative so <math>|x - 2| = 2-x</math> and <math>\displaystyle x - 2 = -p</math>.
+When <math>x < 2,</math> <math>x-2</math> is negative so <math>|x - 2| = 2-x = p</math> and <math>x = 2-p</math>.
-Therefore:
+Thus <math>x-p = (2-p)-p = 2-2p \Longrightarrow \mathrm{(C)} </math>.
-<math>x=2-p</math>
-<math>\displaystyle x-p = (2-p)-p = 2-2p \Longrightarrow \mathrm{(C)} </math>
 == See also ==
 {{AMC12 box|year=2000|num-b=4|num-a=6}}
-* [[Absolute value]]
 [[Category:Introductory Algebra Problems]]