Difference between revisions of "2010 AMC 10B Problems/Problem 13"
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| − | + | We evaluate this in cases: | |
| − | <math> | + | ''Case 1'' |
| − | + | <math>x<30</math> | |
| − | </math> | ||
| − | <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> |
| − | x=|60-2x| | + | <math>x=|2x-60+2x|\implies x=|4x-60|</math> |
| − | </math> | ||
| − | '' | + | ''Subcase 1 ''<math>30>x>15</math> |
| − | |||
| − | <math> | ||
| − | x | ||
| − | </math> | ||
| − | <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> |
| − | 3x=60 | + | 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. |
| − | </math> | ||
| − | <math> | + | '' Subcase 2 '' <math>x<15</math> |
| − | x | ||
| − | </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> | ||
| − | <math> | + | ''Case 2 '': <math>x>30</math> |
| − | |||
| − | </math> | ||
| − | <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 |
| − | x=60 | + | <math>(x=|2x-(2x-60)|</math> |
| − | </math> | ||
| − | + | <math>x=|2x-2x+60|</math> | |
| − | <math> | + | <math>x=|60|</math> |
| − | |||
| − | </math> | ||
| − | <math> | + | <math>x=60</math> |
| − | |||
| − | </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> | |
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Revision as of 14:12, 7 June 2011
Problem
What is the sum of all the solutions of 
?
Solution
We evaluate this in cases:
Case 1
When we are going to have 
. When 
we are going to have 
and when 
we are going to have 
. Therefore we have 
Subcase 1 
When we are going to have 
when this happens, we can express 
as 
Therefore we get 
We check if 
is in the domain of the numbers that we put into this subcase, and it is, since 
Therefore 
is one possible solutions.
Subcase 2 
When we are going to have 
, therefore can be expressed in the form 
We have the equation 
Since 
is less than 
, is another possible solution.
Case 2 : 
When , 
When 
we can express this in the form 
Therefore we have 
This makes sure that this is positive, since we just took the negative of a negative to get a positive. Therefore we have
We have now evaluated all the cases, and found the solution to be 
which have a sum of 
