Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2004 AMC 10A Problems/Problem 4

Revision as of 13:18, 12 November 2006 by Lotrgreengrapes7926 (talk | contribs) (→‎Solution)

Problem

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

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

Solution

$|x-1|$ is equivalent to the distance between $x$ and $1$; $|x-2|$ is equivalent to the distance between $x$ and $2$.

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

See Also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10A_Problems/Problem_4&oldid=11674"