Difference between revisions of "2016 AMC 10A Problems/Problem 7"
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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 1 == | ||
+ | |||
+ | Since <math>x</math> is the mean, | ||
+ | <cmath>\begin{align*} | ||
+ | x&=\frac{60+100+x+40+50+200+90}{7}\\ | ||
+ | &=\frac{540+x}{7}. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Therefore, <math>7x=540+x</math>, so <math>x=\boxed{\textbf{(D) }90}.</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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> | ||
+ | ~bjc | ||
+ | |||
+ | ==Check== | ||
+ | |||
+ | 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. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/XXX4_oBHuGk?t=163 | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | https://youtu.be/joLWmbpvrCw | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2016|ab=A|num-b=6|num-a=8}} | ||
+ | {{AMC12 box|year=2016|ab=A|num-b=3|num-a=5}} | ||
+ | {{MAA Notice}} |
Latest revision as of 14:54, 29 November 2020
Problem
The mean, median, and mode of the data values are all equal to . What is the value of ?
Solution 1
Since is the mean,
Therefore, , so
Solution 2
Note that must be the median so it must equal either or . You can see that the mean is also , and by intuition should be the greater one. ~bjc
Check
Order the list: . must be either or because it is both the median and the mode of the set. Thus is correct.
Video Solution
https://youtu.be/XXX4_oBHuGk?t=163
~IceMatrix
~savannahsolver
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 |
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.