2012 AMC 10A Problem 3

P1-10~1 minPrint

Absolute Value & Inequalities Set up the equation/formula and compute; no special trick needed

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?

Show solution without answering

Key insight

Split the trip at the turning point. Find the length of each part (4 units and 11 units) and add them.

Solution

The bug changes direction at $-6$ , so measure the trip in two parts and add them.

From $-2$ to $-6$ : the bug moves left. The distance between the two points is $6 - 2 = 4$ units.
From $-6$ to $5$ : the bug moves right. It covers $6$ units to get from $-6$ to $0$ , then $5$ more units to get from $0$ to $5$ . That is $6 + 5 = 11$ units.
Altogether: $4 + 11 = 15$ units.

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

Why this works

"How far did it crawl" counts every step, no matter which direction. So split the path wherever it turns around and add up the lengths of the pieces. Going straight from start to finish, $5 - (-2) = 7$ , answers a different question: how far the bug ended up from where it began.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(E)}\ 15}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 20: Special Functions (§20.2, §20.5)
Introduction to Algebra, Chapter 15: More Inequalities (§15.1, §15.3, §15.4)
Intermediate Algebra, Chapter 12: Inequalities (§12.3, §12.4)