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

Revision as of 20:23, 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 $20+60+12= \boxed{\mathrm {(C)}\ 92}$

@@ Line 18: / Line 18: @@
 <math>
-x-|60-2x|=x,
+x-|60-2x|=x
+</math>
+<math>
 x=|60-2x|
 </math>
@@ Line 25: / Line 28: @@
 <math>
-x=60-2x,
+x=60-2x
-x=60,
+</math>
+<math>
+x=60
+</math>
+<math>
 x=20
 </math>
@@ Line 33: / Line 42: @@
 <math>
--x=60-2x,
+-x=60-2x
+</math>
+<math>
 x=60
 </math>
@@ Line 40: / Line 52: @@
 <math>
-x-|60-2x|=-x,
+x-|60-2x|=-x
+</math>
+<math>
 x=|60-2x|
 </math>
@@ Line 47: / Line 62: @@
 <math>
-x=60-2x,
+x=60-2x
-x=60,
+</math>
+<math>
+x=60
+</math>
+<math>
 x=12
 </math>
@@ Line 55: / Line 76: @@
 <math>
--3x=60-2x,
+-3x=60-2x
--x=60,
+</math>
+<math>
+-x=60
+</math>
+<math>
 x=-60
 </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>