Difference between revisions of "2016 AMC 10A Problems/Problem 7"
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== Solution == | == Solution == | ||
| − | + | Since <math>x</math> is the mean, | |
| − | x=\frac{60+100+x+40+50+200+90}{7} | + | <cmath>\begin{align*} |
| − | & | + | x&=\frac{60+100+x+40+50+200+90}{7}\\ |
| − | + | &=\frac{540+x}{7}. | |
| − | |||
| − | |||
\end{align*}</cmath> | \end{align*}</cmath> | ||
| + | |||
| + | Therefore, <math>7x=540+x</math>, so <math>x=\boxed{\textbf{(D) }90}.</math> | ||
==Check== | ==Check== | ||
| − | Order the list: <math>\{40,50,60,90,100,120\}</math>. <math>x</math> must be <math>60</math> or <math>90</math> because it is the median and mode of the set. Thus <math>90</math> is correct. | + | |
| + | Order the list: <math>\{40,50,60,90,100,120\}</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. | ||
==See Also== | ==See Also== | ||
Revision as of 16:47, 17 February 2016
Contents
Problem
The mean, median, and mode of the 
data values 
are all equal to 
. What is the value of ?
Solution
Since is the mean,
Therefore, 
, so 
Check
Order the list: 
. must be either 
or 
because it is both the median and the mode of the set. Thus is correct.
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. 
