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