Difference between revisions of "2016 AMC 12A Problems/Problem 4"
(→Solution) |
(→Solution) |
||
| Line 21: | Line 21: | ||
<cmath>6x=540</cmath> | <cmath>6x=540</cmath> | ||
<cmath>x=\boxed{\textbf{(D)}\text{ 90}}</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}} | ||
Revision as of 22:46, 3 February 2016
Problem
The mean, median, and mode of the 7 data values 
are all equal to 
.
What is the value of ?
Solution
must be in the middle of the list. Order the known list first: 
.
occurs the most times, and it is the average of the list.
is either 
or 
because it is the median and the mode.
is the average of the set values.
![\[x=\frac{40+50+60+90+100+200+x}{7}\]](http://latex.artofproblemsolving.com/e/3/a/e3aaaa3c3d3bbc3b6c0e069af4175e70364be48a.png)
![\[x=\frac{540+x}{7}\]](http://latex.artofproblemsolving.com/6/f/e/6fea05ca3b0890dad230d1efe0e7fd712c6337fd.png)
![\[7x=540+x\]](http://latex.artofproblemsolving.com/2/5/0/25079bdfd04955e060ee8d0acc51fb4e891038b0.png)
![\[6x=540\]](http://latex.artofproblemsolving.com/6/b/7/6b7e58a412c03874549049365a1e237aeda0de52.png)
![\[x=\boxed{\textbf{(D)}\text{ 90}}\]](http://latex.artofproblemsolving.com/1/3/a/13a5a44206b6f69d0a76c2e47b9fa8fd78ce7398.png)
See Also
| 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. 
