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.

Video Solution

https://youtu.be/XXX4_oBHuGk?t=163

~IceMatrix

https://youtu.be/joLWmbpvrCw

~savannahsolver

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AMC 10 Problems and Solutions

2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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