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 1 ==
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Since <math>x</math> is the mean,
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<cmath>\begin{align*}
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x&=\frac{60+100+x+40+50+200+90}{7}\\
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&=\frac{540+x}{7}.
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\end{align*}</cmath>
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Therefore, <math>7x=540+x</math>, so <math>x=\boxed{\textbf{(D) }90}.</math>
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==Solution 2==
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Note that <math>x</math> must be the median so it must equal either <math>60</math> or <math>90</math>. You can see that the mean is also <math>x</math>, and by intuition <math>x</math> should be the greater one. <math>x=\boxed{\textbf{(D) }90}.</math>
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~bjc
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==Check==
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Order the list: <math>\{40,50,60,90,100,200\}</math>. <math>x</math> must be either <math>60</math> or <math>90</math> because it is both the median and the mode of the set. Thus <math>90</math> is correct.
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==Video Solution==
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https://youtu.be/XXX4_oBHuGk?t=163
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~IceMatrix
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https://youtu.be/joLWmbpvrCw
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~savannahsolver
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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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{{AMC12 box|year=2016|ab=A|num-b=3|num-a=5}}
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{{MAA Notice}}

Latest revision as of 14:54, 29 November 2020

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

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
2016 AMC 12A (ProblemsAnswer KeyResources)
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

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