2010 AMC 10B Problem 13

Below is the professionally curated solution for Problem 13 of the 2010 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 10B solutions, or check the answer key.

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Concepts:absolute valuecasework

Difficulty rating: 1660

13.

What is the sum of all the solutions of x=2x602x?x = |2x-|60-2x||?

3232

6060

9292

120120

124124

Solution:

We first take care of the outer absolute value. We have either x=2x602x x = 2x - |60 - 2x| or x=2x+602x. x = -2x + |60 - 2x|.

These simplify to x=602x and 3x=602x. x = |60 - 2x| \text{ and } 3x = |60 - 2x|.

We have 22 cases for each equation. For the first one, we have x=602x and x=2x60. x = 60 - 2x \text{ and } x = 2x - 60.

Solving both gives us 3x=60 3x = 60 x=20 x = 20 and x=60 -x = -60 x=60. x = 60.

For the other equation, we have 3x=602x and 3x=2x60. 3x = 60 - 2x \text{ and } 3x = 2x - 60.

Again solving both, we have 5x=60 5x = 60 x=12 x = 12 and x=60.x = -60. Note that xx cannot be negative since the right side of the original equation is nonnegative.

Adding up all the solutions gives us 20+60+12=92. 20 + 60 + 12 = 92.

Thus, C is the correct answer.

Problem 13 in Other Years