Difference between revisions of "2004 AMC 10A Problems/Problem 4"

Revision as of 11:08, 5 November 2006

What is the value of $x$ if $|x-1|=|x-2|$ ?

$\mathrm{(A) \ } -\frac12 \qquad \mathrm{(B) \ } \frac12 \qquad \mathrm{(C) \ } 1 \qquad \mathrm{(D) \ } \frac32 \qquad \mathrm{(E) \ } 2$

$|x-1|$ is the same thing as saying the distance between $x$ and $1$ ; $|x-2|$ is the same thing as saying the distance between $x$ and $2$ .

Therefore, $x$ is the same distance from $1$ and $2$ , so $x=\frac{1+2}2=\frac32\Rightarrow\mathrm{(D)}$ .

@@ Line 2: / Line 2: @@
 What is the value of <math>x</math> if <math>|x-1|=|x-2|</math>?
-<math> \mathrm{(A) \ } -\frac12 \qquad \mathrm{(B) \ } \frac12 \qquad \mathrm{(C) \ } 1 \qquad \mathrm{(D) \ } \frac32 {2}\qquad \mathrm{(E) \ } 2 </math>
+<math> \mathrm{(A) \ } -\frac12 \qquad \mathrm{(B) \ } \frac12 \qquad \mathrm{(C) \ } 1 \qquad \mathrm{(D) \ } \frac32 \qquad \mathrm{(E) \ } 2 </math>
 ==Solution==