2016 AMC 12A Problems/Problem 4

Revision as of 23:26, 3 February 2016 by FractalMathHistory (talk | contribs) (Solution)

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. \[x=\frac{40+50+60+90+100+200+x}{7}\] \[x=\frac{540+x}{7}\] \[7x=540+x\] \[6x=540\]

\[x=\boxed{\textbf{(D)}\text{ 90}\] (Error compiling LaTeX. Unknown error_msg)