2016 AMC 10A Problems/Problem 7: Difference between revisions
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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>? | 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> | <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}} | |||
Revision as of 18:37, 3 February 2016
Problem
The mean, median, and mode of the 
data values 
are all equal to 
. What is the value of ?
Solution
As is the mean, ![\[x=\frac{60+100+x+40+50+200+90}{7}\implies x=\frac{540+x}{7}\implies 7x=540+x\implies x=\boxed{\textbf{(D) }90.}\]](http://latex.artofproblemsolving.com/6/8/9/68912518f77eef951042ae85064154218345a4c4.png)
See Also
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
