2016 AMC 12A Problems/Problem 4

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. $x=\frac{40+50+60+90+100+200+x}{7}$ $x=\frac{540+x}{7}$ $7x=540+x$ $6x=540$ $x=\boxed{\textbf{(D)}\text{ 90}}$

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2016 AMC 12A Problems/Problem 4

Solution