2010 AMC 10B Problem 13

Since the right side is an absolute value, $x \ge 0$. Handle the inner absolute value by splitting at $x = 30$.

Case $x \le 30$. Then $|60 - 2x| = 60 - 2x$, and the equation becomes
$$x = |2x - (60 - 2x)| = |4x - 60|.$$
Either $4x - 60 = x$, giving $x = 20$, or $4x - 60 = -x$, giving $x = 12$. Both satisfy $x \le 30$, and both check: $|80 - 60| = 20$ and $|48 - 60| = 12$.

Case $x > 30$. Then $|60 - 2x| = 2x - 60$, so
$$x = |2x - (2x - 60)| = |60| = 60.$$
So $x = 60$, which is indeed greater than $30$.

The solutions are $12, 20, 60$, with sum $92$.

The answer is $\boxed{\textbf{(C)}\ 92}$.

Nested absolute values are handled from the inside out: the inner one determines a breakpoint ($x = 30$), and on each side the equation becomes $|{\rm linear}| = x$, which has at most two solutions. Verifying that each solution lies in its case interval is what keeps extraneous roots out.

Graph $y = |2x - |60 - 2x||$: it is the V-shape $|4x - 60|$ for $x \le 30$ and the horizontal line $y = 60$ beyond. The line $y = x$ crosses the V twice ($12$ and $20$) and the horizontal piece once ($60$).

• Not checking candidates against the case assumption, or forgetting the x > 30 case where the expression collapses to the constant 60.
• Summing only $12 + 20 = 32$ (choice A) or including a spurious root like $x = -60$.