# 2016 AMC 10A Problems/Problem 7

## 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 1

Since $x$ is the mean, \begin{align*} x&=\frac{60+100+x+40+50+200+90}{7}\\ &=\frac{540+x}{7}. \end{align*}

Therefore, $7x=540+x$, so $x=\boxed{\textbf{(D) }90}.$

## Solution 2

Note that $x$ must be the median so it must equal either $60$ or $90$. You can see that the mean is also $x$, and by intuition $x$ should be the greater one. $x=\boxed{\textbf{(D) }90}.$ ~bjc

## Check

Order the list: $\{40,50,60,90,100,200\}$. $x$ must be either $60$ or $90$ because it is both the median and the mode of the set. Thus $90$ is correct.

~IceMatrix

~savannahsolver

## See Also

 2016 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
 2016 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

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