Printable

2012 AMC 10A Problem 3

2012 AMC 10A Problem 3AMC 10 Step by Step

Name: Date: September 13, 2026

Absolute Value & Inequalities · difficulty 1 of 5 · about 1 min · position P1-10

A bug crawls along a number line, starting at $-2$ . It crawls to $-6$ , then turns around and crawls to $5$ . How many units does the bug crawl altogether?

A)
$9$
B)
$11$
C)
$13$
D)
$14$
E)
$15$

Answer

Key insight

Distance traveled is the sum of the absolute differences of consecutive positions: 4 units left, then 11 units right, total 15.

Solution

Distance traveled along a number line is the absolute value of the change in position, added up leg by leg.

First leg, from $-2$ to $-6$ : $|-6 - (-2)| = |-4| = 4$ units.

Second leg, from $-6$ to $5$ : $|5 - (-6)| = |11| = 11$ units.

Total: $4 + 11 = 15$ units.

The answer is $\boxed{\textbf{(E)}\ 15}$ .

Why this works

Total distance is a sum of unsigned lengths, unlike net displacement, which would be $5 - (-2) = 7$ . Whenever a path reverses direction, break it at the turning point and add the absolute values of the pieces.

The trap

Computing the net displacement from -2 to 5 (7 units) or mishandling the negative signs when subtracting.

Common mistakes

Computing the net displacement from -2 to 5 (7 units) or mishandling the negative signs when subtracting.
Getting $-6$ to $5$ as $9$ or $10$ by miscounting across zero; $6 + 5 = 11$ .

Techniques

Set up the equation/formula and compute; no special trick needed