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

Revision as of 20:17, 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 $20+60+12=$ \boxed{\mathrm {(C)} 92}$

Revision as of 20:16, 24 January 2011 view source Jexmudder (talk \| contribs) 64 edits →Solution ← Older edit		Revision as of 20:17, 24 January 2011 view source Jexmudder (talk \| contribs) 64 edits →Solution Newer edit →
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	Since an absolute value cannot be negative, we exclude <math>x=-60</math>. The answer is <math>20+60+12= </math>\boxed{\mathrm {(C)}\ 92}$		Since an absolute value cannot be negative, we exclude <math>x=-60</math>. The answer is <math>20+60+12= </math>\boxed{\mathrm {(C)} 92}$