2010 AMC 10B Problems/Problem 13: Difference between revisions

Revision as of 20:13, 24 January 2011

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

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

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 $20+60+12=92$

@@ Line 16: / Line 16: @@
 == Solution ==
 '''Case 1''':
 <math>
 x-|60-2x|=x
 x=|60-2x|
 </math>
-''Case 1a'':
+''Case 1a'':
 <math>
 x=60-2x
@@ Line 26: / Line 29: @@
 x=20
 </math>
 ''Case 1b'':
 <math>
 -x=60-2x
 x=60
 </math>
 '''Case 2''':
 <math>
 x-|60-2x|=-x
 x=|60-2x|
 </math>
 ''Case 2a'':
+<math>
 x=60-2x
 x=60
 x=12
+</math>
+''Case 2b'':
 <math>
-''Case 2b'':
-</math>
 -3x=60-2x
 -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=92$
+Since an absolute value cannot be negative, we exclude <math>x=-60</math>. The answer is <math>20+60+12=92</math>