Difference between revisions of "2016 AMC 10A Problems/Problem 7"
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== Solution == | == Solution == | ||
| − | 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> | + | As <math>x</math> is the mean, <cmath>\begin{align*} |
| + | x=\frac{60+100+x+40+50+200+90}{7} | ||
| + | &\implies x=\frac{540+x}{7} \\ | ||
| + | &\implies 7x=540+x \\ | ||
| + | &\implies 6x=540 \\ | ||
| + | &\implies x=\boxed{\textbf{(D) }90.} | ||
| + | \end{align*}</cmath> | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=A|num-b=6|num-a=8}} | {{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}} | {{MAA Notice}} | ||
Revision as of 11:55, 4 February 2016
Problem
The mean, median, and mode of the 
data values 
are all equal to 
. What is the value of ?
Solution
As is the mean, 
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. 
