Difference between revisions of "2010 AMC 10B Problems/Problem 13"

(Solution)
Line 50: Line 50:
 
3x=60-2x,  
 
3x=60-2x,  
 
5x=60,  
 
5x=60,  
x=12,
+
x=12
 
</math>
 
</math>
  
Line 58: Line 58:
 
-3x=60-2x,  
 
-3x=60-2x,  
 
-x=60,  
 
-x=60,  
x=-60,
+
x=-60
 
</math>
 
</math>
  
 
Since an absolute value cannot be negative, we exclude <math>x=-60</math>. The answer is <math>20+60+12= \boxed{\mathrm {(C)}\ 92}</math>
 
Since an absolute value cannot be negative, we exclude <math>x=-60</math>. The answer is <math>20+60+12= \boxed{\mathrm {(C)}\ 92}</math>

Revision as of 21:18, 24 January 2011

Problem

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

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

Solution

Case 1:

$2x-|60-2x|=x,  x=|60-2x|$

Case 1a:

$x=60-2x,  3x=60,  x=20$

Case 1b:

$-x=60-2x,  x=60$

Case 2:

$2x-|60-2x|=-x,  3x=|60-2x|$

Case 2a:

$3x=60-2x,  5x=60,  x=12$

Case 2b:

$-3x=60-2x,  -x=60,  x=-60$

Since an absolute value cannot be negative, we exclude $x=-60$. The answer is $20+60+12= \boxed{\mathrm {(C)}\ 92}$