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

Latest revision as of 19:00, 22 March 2024

Problem

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

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

Solution 1

We evaluate this in cases:

Case 1 $x<30$

When $x<30$ we are going to have $60-2x>0$ . When $x>0$ we are going to have $|x|>0\implies x>0$ and when $-x>0$ we are going to have $|x|>0\implies -x>0$ . Therefore we have $x=|2x-(60-2x)|$ . $x=|2x-60+2x|\implies x=|4x-60|$

Subcase 1 $30>x>15$

When $30>x>15$ we are going to have $4x-60>0$ . When this happens, we can express $|4x-60|$ as $4x-60$ . Therefore we get $x=4x-60\implies -3x=-60\implies x=20$ . We check if $x=20$ is in the domain of the numbers that we put into this subcase, and it is, since $30>20>15$ . Therefore $20$ is one possible solution.

Subcase 2 $x<15$

When $x<15$ we are going to have $4x-60<0$ , therefore $|4x-60|$ can be expressed in the form $60-4x$ . We have the equation $x=60-4x\implies 5x=60\implies x=12$ . Since $12$ is less than $15$ , $12$ is another possible solution. $x=|2x-|60-2x||$

Case 2 : $x>30$

When $x>30$ , $60-2x<0$ . When $x<0$ we can express this in the form $-x$ . Therefore we have $-(60-2x)=2x-60$ . This makes sure that this is positive, since we just took the negative of a negative to get a positive. Therefore we have $x=|2x-(2x-60)|$

$x=|2x-2x+60|$

$x=|60|$

$x=60$

We have now evaluated all the cases, and found the solution to be $\{60,12,20\}$ which have a sum of $\boxed{\textbf{(C)}\ 92}$

Solution 2

From the equation $x = \left|2x-|60-2x|\right|$ , we have $x = 2x-|60-2x|$ , or $-x = 2x-|60-2x|$ . Therefore, $x=|60-2x|$ , or $3x=|60-2x|$ . From here we have four possible cases:

1. $x=60-2x$ ; this simplifies to $3x=60$ , so $x=20$ .

2. $-x=60-2x$ ; this simplifies to $x=60$ .

3. $3x=60-2x$ ; this simplifies to $5x=60$ , so $x=12$ .

4. $-3x=60-2x$ ; this simplifies to $-x=60$ , so $x=-60$ . However, this solution is extraneous because the absolute value of $2x-|60-2x|$ cannot be negative.

The sum of all of the solutions of $x$ is $20+60+12=\boxed{\textbf{(C)}\ 92}$

Video Solution

https://youtu.be/vYXz4wStBUU?t=272

~IceMatrix

https://youtu.be/1DjO74VEr3E

~savannahsolver

@@ Line 2: / Line 2: @@
 What is the sum of all the solutions of <math>x = \left|2x-|60-2x|\right|</math>?
-<math>
+<math>\textbf{(A)}\ 32 \qquad \textbf{(B)}\ 60 \qquad \textbf{(C)}\ 92 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 124</math>
-\mathrm{(A)}\ 32
-\qquad
-\mathrm{(B)}\ 60
-\qquad
-\mathrm{(C)}\ 92
-\qquad
-\mathrm{(D)}\ 120
-\qquad
-\mathrm{(E)}\ 124
-</math>
-== Solution ==
+== Solution 1==
-'''Case 1''': <math>2x-|60-2x|=x</math>
+We evaluate this in cases:
+''Case 1''
+<math>x<30</math>
+When <math>x<30</math> we are going to have <math>60-2x>0</math>.  When <math>x>0</math> we are going to have <math>|x|>0\implies x>0</math> and when <math>-x>0</math> we are going to have <math>|x|>0\implies -x>0</math>.  Therefore we have <math>x=|2x-(60-2x)|</math>.
+<math>x=|2x-60+2x|\implies x=|4x-60|</math>
+''Subcase 1 ''<math>30>x>15</math>
+When <math>30>x>15</math> we are going to have <math>4x-60>0</math>. When this happens, we can express <math>|4x-60|</math> as <math>4x-60</math>.
+Therefore we get <math>x=4x-60\implies -3x=-60\implies x=20</math>.  We check if <math>x=20</math> is in the domain of the numbers that we put into this subcase, and it is, since <math>30>20>15</math>. Therefore <math>20</math> is one possible solution.
+'' Subcase 2 '' <math>x<15</math>
+When <math>x<15</math> we are going to have <math>4x-60<0</math>, therefore <math>|4x-60|</math> can be expressed in the form <math>60-4x</math>.
+We have the equation <math>x=60-4x\implies 5x=60\implies x=12</math>.  Since <math>12</math> is less than <math>15</math>, <math>12</math> is another possible solution.
+<math>x=|2x-|60-2x||</math>
+''Case 2 '':  <math>x>30</math>
+When <math>x>30</math>, <math>60-2x<0</math>. When <math>x<0</math> we can express this in the form <math>-x</math>.  Therefore we have <math>-(60-2x)=2x-60</math>.  This makes sure that this is positive, since we just took the negative of a negative to get a positive.  Therefore we have
+<math>x=|2x-(2x-60)|</math>
+<math>x=|2x-2x+60|</math>
+<math>x=|60|</math>
+<math>x=60</math>
+We have now evaluated all the cases, and found the solution to be <math>\{60,12,20\}</math> which have a sum of <math>\boxed{\textbf{(C)}\ 92}</math>
+==Solution 2==
+From the equation <math>x = \left|2x-|60-2x|\right|</math> , we have <math>x = 2x-|60-2x|</math> , or <math>-x = 2x-|60-2x|</math>. Therefore, <math>x=|60-2x|</math> , or <math>3x=|60-2x|</math>. From here we have four possible cases:
+. <math>x=60-2x</math>; this simplifies to <math>3x=60</math>, so <math>x=20</math>.
+. <math>-x=60-2x</math>; this simplifies to <math>x=60</math>.
+. <math>3x=60-2x</math>; this simplifies to <math>5x=60</math>, so <math>x=12</math>.
+. <math>-3x=60-2x</math>; this simplifies to <math>-x=60</math>, so <math>x=-60</math>. However, this solution is extraneous because the absolute value of <math>2x-|60-2x|</math> cannot be negative.
+The sum of all of the solutions of <math>x</math> is <math>20+60+12=\boxed{\textbf{(C)}\ 92}</math>
+==Video Solution==
+https://youtu.be/vYXz4wStBUU?t=272
+~IceMatrix
+https://youtu.be/1DjO74VEr3E
+~savannahsolver
+==See Also==
+{{AMC10 box|year=2010|ab=B|num-b=12|num-a=14}}
+{{MAA Notice}}

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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