2016 AMC 8 Problems/Problem 6: Difference between revisions
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==Solution== | ==Solution== | ||
We first notice that the median name will be the <math>10^{\mbox{th}}</math> name. The <math>10^{\mbox{th}}</math> name is <math>\boxed{\textbf{(B)}\ 4}</math>. | We first notice that the median name will be the <math>10^{\mbox{th}}</math> name. The <math>10^{\mbox{th}}</math> name is <math>\boxed{\textbf{(B)}\ 4}</math>. | ||
== See Also == | |||
{{AMC8 box|year=2016|num-b=5|num-a=7}} | {{AMC8 box|year=2016|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:23, 14 January 2021
Problem
The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?
![[asy] unitsize(0.9cm); draw((-0.5,0)--(10,0), linewidth(1.5)); draw((-0.5,1)--(10,1)); draw((-0.5,2)--(10,2)); draw((-0.5,3)--(10,3)); draw((-0.5,4)--(10,4)); draw((-0.5,5)--(10,5)); draw((-0.5,6)--(10,6)); draw((-0.5,7)--(10,7)); label("frequency",(-0.5,8)); label("0", (-1, 0)); label("1", (-1, 1)); label("2", (-1, 2)); label("3", (-1, 3)); label("4", (-1, 4)); label("5", (-1, 5)); label("6", (-1, 6)); label("7", (-1, 7)); filldraw((0,0)--(0,7)--(1,7)--(1,0)--cycle, black); filldraw((2,0)--(2,3)--(3,3)--(3,0)--cycle, black); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, black); filldraw((6,0)--(6,4)--(7,4)--(7,0)--cycle, black); filldraw((8,0)--(8,4)--(9,4)--(9,0)--cycle, black); label("3", (0.5, -0.5)); label("4", (2.5, -0.5)); label("5", (4.5, -0.5)); label("6", (6.5, -0.5)); label("7", (8.5, -0.5)); label("name length", (4.5, -1)); [/asy]](http://latex.artofproblemsolving.com/7/3/9/7396694d92e8f9794065daafaf571852434d0b08.png)
Solution
We first notice that the median name will be the 
name. The name is 
.
See Also
| 2016 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
