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

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== Solution ==
== Solution ==
'''Case 1''': <math>2x-|60-2x|=x</math>
'''Case 1''':  
<math>
2x-|60-2x|=x
x=|60-2x|
</math>
''Case 1a'':
<math>
x=60-2x
3x=60
x=20
</math>
''Case 1b'':
<math>
-x=60-2x
x=60
</math>
'''Case 2''':
<math>
2x-|60-2x|=-x
3x=|60-2x|
</math>
''Case 2a'':
3x=60-2x
5x=60
x=12
<math>
''Case 2b'':
</math>
-3x=60-2x
-x=60
x=-60
<math>
 
Since an absolute value cannot be negative, we exclude </math>x=-60<math>. The answer is </math>20+60+12=92$

Revision as of 20:12, 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$20+60+12=92$

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