2004 AMC 10A Problem 4 — Solution | LIVE by Po-Shen Loh

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2004 AMC 10A Problem 4

Attempt Problem 4 of the 2004 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

4.

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

$-\dfrac{1}{2}$

$\dfrac{1}{2}$

$1$

$\dfrac{3}{2}$

$2$

Answer: D

Concepts:absolute value midpoint

Difficulty rating: 1030

Solution:

Since $|x - 1|$ and $|x - 2|$ are the distances from $x$ to $1$ and $2,$ the point $x$ is equidistant from $1$ and $2.$

That midpoint is $x = \dfrac{1 + 2}{2} = \dfrac{3}{2}.$

Thus, the correct answer is D.

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