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

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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>
\mathrm{(A)}\ 32
\qquad
\mathrm{(B)}\ 60
\qquad
\mathrm{(C)}\ 92
\qquad
\mathrm{(D)}\ 120
\qquad
\mathrm{(E)}\ 124
</math>


== Solution ==
== Solution ==
Line 20: Line 10:
<math>x<30</math>  
<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>
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>
<math>x=|2x-60+2x|\implies x=|4x-60|</math>


''Subcase 1 ''<math>30>x>15</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>
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 solutions.   
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>
'' 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>
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.
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>x=|2x-|60-2x||</math>


''Case 2 '':  <math>x>30</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  
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=|2x-2x+60|</math>
<math>x=|2x-2x+60|</math>
Line 45: Line 35:
<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>
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>


==See Also==
==See Also==
{{AMC10 box|year=2010|ab=B|num-b=12|num-a=14}}
{{AMC10 box|year=2010|ab=B|num-b=12|num-a=14}}

Revision as of 13:06, 26 November 2011

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

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 $\boxed{\textbf{(C)}\ 92}$

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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