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Difference between revisions of "2010 AMC 10B Problems/Problem 13"
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What is the sum of all the solutions of <math>x = \left|2x-|60-2x|\right|</math>?
 
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>
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\mathrm{(A)}\ 32
 
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\qquad
 
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\mathrm{(B)}\ 60
 
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\qquad
 
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\mathrm{(C)}\ 92
 
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\qquad
 
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\mathrm{(D)}\ 120
 
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\qquad
 
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\mathrm{(E)}\ 124
 
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</math>
 
  
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== Solution ==
+
== Solution 1==
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'''Case 1''':  
+
We evaluate this in cases:
  
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<math>
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''Case 1''
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2x-|60-2x|=x,
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<math>x<30</math>
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x=|60-2x|
 
  
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</math>
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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>
  
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''Case 1a'':
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''Subcase 1 ''<math>30>x>15</math>
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<math>
 
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x=60-2x,
 
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3x=60,
 
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x=20
 
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</math>
 
  
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''Case 1b'':
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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. 
  
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<math>
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'' Subcase 2 '' <math>x<15</math>
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-x=60-2x,
 
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x=60
 
−
</math>
 
  
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'''Case 2''':
+
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>
  
−
<math>
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''Case 2 '':  <math>x>30</math>
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2x-|60-2x|=-x,
 
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3x=|60-2x|
 
−
</math>
 
  
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''Case 2a'':
+
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>
  
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<math>
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<math>x=|2x-2x+60|</math>
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3x=60-2x,
 
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5x=60,
 
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x=12,
 
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</math>
 
  
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''Case 2b'':
+
<math>x=|60|</math>
  
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<math>
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<math>x=60</math>
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-3x=60-2x,
 
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-x=60,
 
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x=-60,
 
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</math>
 
  
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Since an absolute value cannot be negative, we exclude <math>x=-60</math>. The answer is <math>20+60+12=92</math>
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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:
 +
 
 +
1. <math>x=60-2x</math>; this simplifies to <math>3x=60</math>, so <math>x=20</math>.
 +
 
 +
2. <math>-x=60-2x</math>; this simplifies to <math>x=60</math>.
 +
 
 +
3. <math>3x=60-2x</math>; this simplifies to <math>5x=60</math>, so <math>x=12</math>.
 +
 
 +
4. <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}}

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

See Also

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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