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...")
 
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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>.
 
<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 is the average of the list.
 
<math>x</math> occurs the most times, and it is the average of the list.
 +
 
<math>\#1:</math>
 
<math>\#1:</math>
 
<math>x</math> is either <math>60</math> or <math>90</math> because it is the median and the mode.
 
<math>x</math> is either <math>60</math> or <math>90</math> because it is the median and the mode.

Revision as of 22:22, 3 February 2016

Solution

$x$ must be in the middle of the list. Order the known list first: $\{40,50,60,90,100,200\}$. $x$ occurs the most times, and it is the average of the list.

$\#1:$ $x$ is either $60$ or $90$ because it is the median and the mode.

$\#2:$ $x$ is the average of the set values. \[x=\frac{40+50+60+90+100+200+x}{7}\] \[x=\frac{540+x}{7}\] \[7x=540+x\] \[6x=540\] \[x=\boxed{\textbf{(D)}90}\]