|
|
| Line 1: |
Line 1: |
| − | ==Problem==
| + | #REDIRECT [[2016 AMC 10A Problems/Problem 7]] |
| − | | |
| − | The mean, median, and mode of the 7 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==
| |
| − | <math>x</math> must be in the middle of the list. Order the known list first: <math>\{40,50,60,90,100,200\}</math>.
| |
| − | <math>x</math> occurs the most times, and it is the average of the list.
| |
| − | | |
| − | <math>\#1:</math>
| |
| − | <math>x</math> is either <math>60</math> or <math>90</math> because it is the median and the mode.
| |
| − | | |
| − | <math>\#2:</math>
| |
| − | <math>x</math> is the average of the set values.
| |
| − | <cmath>x=\frac{40+50+60+90+100+200+x}{7}</cmath>
| |
| − | <cmath>x=\frac{540+x}{7}</cmath>
| |
| − | <cmath>7x=540+x</cmath>
| |
| − | <cmath>6x=540</cmath>
| |
| − | <cmath>x=\boxed{\textbf{(D)}\text{ 90}}</cmath>
| |
| − | | |
| − | ==See Also==
| |
| − | {{AMC12 box|year=2016|ab=A|num-b=3|num-a=5}}
| |
| − | {{MAA Notice}}
| |
