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Difference between revisions of "2016 AMC 12A Problems/Problem 4"

(Created page with "==Solution== <math>x</math> must be in the middle of the list. Order the known list first: <math>\{40,50,60,90,100,200\}</math>. <math>x</math> occurs the most times, and it i...")
 
(Redirected to matching 10A problem)
 
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==Solution==
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#REDIRECT [[2016 AMC 10A Problems/Problem 7]]
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<math>x</math> must be in the middle of the list. Order the known list first: <math>\{40,50,60,90,100,200\}</math>.
 
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<math>x</math> occurs the most times, and it is the average of the list.
 
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<math>\#1:</math>
 
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<math>x</math> is either <math>60</math> or <math>90</math> because it is the median and the mode.
 
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<math>\#2:</math>
 
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<math>x</math> is the average of the set values.
 
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<cmath>x=\frac{40+50+60+90+100+200+x}{7}</cmath>
 
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<cmath>x=\frac{540+x}{7}</cmath>
 
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<cmath>7x=540+x</cmath>
 
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<cmath>6x=540</cmath>
 
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<cmath>x=\boxed{\textbf{(D)}90}</cmath>
 

Latest revision as of 11:54, 4 February 2016

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