2016 AMC 12A Problems/Problem 4

Revision as of 23:46, 3 February 2016 by FractalMathHistory (talk | contribs) (Solution)

Problem

The mean, median, and mode of the 7 data values $60,100,x,40,50,200,90$ are all equal to $x$. What is the value of $x$?

$\textbf{(A)}\ 50\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 100$


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

See Also

2016 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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All AMC 12 Problems and Solutions

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