Difference between revisions of "2016 AMC 10A Problems/Problem 7"

Revision as of 18:37, 3 February 2016

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 $x$ ?

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

As $x$ is the mean, $x=\frac{60+100+x+40+50+200+90}{7}\implies x=\frac{540+x}{7}\implies 7x=540+x\implies x=\boxed{\textbf{(D) }90.}$

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

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+== Problem ==
 The mean, median, and mode of the <math>7</math> data values <math>60, 100, x, 40, 50, 200, 90</math> are all equal to <math>x</math>. What is the value of <math>x</math>?
 <math>\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100</math>
+== Solution ==
+As <math>x</math> is the mean, <cmath>x=\frac{60+100+x+40+50+200+90}{7}\implies x=\frac{540+x}{7}\implies 7x=540+x\implies x=\boxed{\textbf{(D) }90.}</cmath>
+==See Also==
+{{AMC10 box|year=2016|ab=A|num-b=6|num-a=8}}
+{{MAA Notice}}