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

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== Problem ==
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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>
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== Solution ==
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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>
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==See Also==
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{{AMC10 box|year=2016|ab=A|num-b=6|num-a=8}}
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{{MAA Notice}}

Revision as of 19:37, 3 February 2016

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

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

See Also

2016 AMC 10A (ProblemsAnswer KeyResources)
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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