2010 AMC 10B Problems/Problem 13

Problem

What is the sum of all the solutions of $x = \left|2x-|60-2x|\right|$ ?

$\textbf{(A)}\ 32 \qquad \textbf{(B)}\ 60 \qquad \textbf{(C)}\ 92 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 124$

Solution

We evaluate this in cases:

Case 1 $x<30$

When $x<30$ we are going to have $60-2x>0$ . When $x>0$ we are going to have $|x|>0\implies x>0$ and when $-x>0$ we are going to have $|x|>0\implies -x>0$ . Therefore we have $x=|2x-(60-2x)|$ . $x=|2x-60+2x|\implies x=|4x-60|$

Subcase 1 $30>x>15$

When $30>x>15$ we are going to have $4x-60>0$ . When this happens, we can express $|4x-60|$ as $4x-60$ . Therefore we get $x=4x-60\implies -3x=-60\implies x=20$ . We check if $x=20$ is in the domain of the numbers that we put into this subcase, and it is, since $30>20>15$ . Therefore $20$ is one possible solution.

Subcase 2 $x<15$

When $x<15$ we are going to have $4x-60<0$ , therefore $|4x-60|$ can be expressed in the form $60-4x$ . We have the equation $x=60-4x\implies 5x=60\implies x=12$ . Since $12$ is less than $15$ , $12$ is another possible solution. $x=|2x-|60-2x||$

Case 2 : $x>30$

When $x>30$ , $60-2x<0$ . When $x<0$ we can express this in the form $-x$ . Therefore we have $-(60-2x)=2x-60$ . This makes sure that this is positive, since we just took the negative of a negative to get a positive. Therefore we have $x=|2x-(2x-60)|$

$x=|2x-2x+60|$

$x=|60|$

$x=60$

We have now evaluated all the cases, and found the solution to be $\{60,12,20\}$ which have a sum of $\boxed{\textbf{(C)}\ 92}$

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2010 AMC 10B Problems/Problem 13

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