Difference between revisions of "2016 AMC 8 Problems/Problem 6"
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<math>\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7</math> | <math>\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7</math> | ||
| − | + | == Solution 1 == | |
| − | |||
| − | |||
We first notice that the median name will be the <math>(19+1)/2=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>(19+1)/2=10^{\mbox{th}}</math> name. The <math>10^{\mbox{th}}</math> name is <math>\boxed{\textbf{(B)}\ 4}</math>. | ||
| − | + | == Solution 2 == | |
To find the median length of a name from a bar graph, we must add up the number of names. Doing so gives us <math>7 + 3 + 1 + 4 + 4 = 19</math>. Thus the index of the median length would be the 10th name. Since there are <math>7</math> names with length <math>3</math>, and <math>3</math> names with length <math>4</math>, the <math>10</math>th name would have <math>4</math> letters. Thus our answer is <math>\boxed{\textbf{(B)}\ 4}</math>. | To find the median length of a name from a bar graph, we must add up the number of names. Doing so gives us <math>7 + 3 + 1 + 4 + 4 = 19</math>. Thus the index of the median length would be the 10th name. Since there are <math>7</math> names with length <math>3</math>, and <math>3</math> names with length <math>4</math>, the <math>10</math>th name would have <math>4</math> letters. Thus our answer is <math>\boxed{\textbf{(B)}\ 4}</math>. | ||
| − | ==Video Solution== | + | == Video Solution == |
https://youtu.be/M9Hooi5UwDg?si=4CPixqDwQ_9BCh6m | https://youtu.be/M9Hooi5UwDg?si=4CPixqDwQ_9BCh6m | ||
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~Elijahman~ | ~Elijahman~ | ||
| − | ==Video Solution (CREATIVE THINKING!!!)== | + | == Video Solution (CREATIVE THINKING!!!) == |
https://youtu.be/Xab3qcUUDRY | https://youtu.be/Xab3qcUUDRY | ||
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~ pi_is_3.14 | ~ pi_is_3.14 | ||
| − | ==Video Solution== | + | == Video Solution == |
https://youtu.be/800KF_3XSmM | https://youtu.be/800KF_3XSmM | ||
Latest revision as of 19:01, 21 August 2024
Contents
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 1
We first notice that the median name will be the 
name. The 
name is 
.
Solution 2
To find the median length of a name from a bar graph, we must add up the number of names. Doing so gives us 
. Thus the index of the median length would be the 10th name. Since there are 
names with length 
, and names with length 
, the 
th name would have letters. Thus our answer is .
Video Solution
https://youtu.be/M9Hooi5UwDg?si=4CPixqDwQ_9BCh6m
A solution so simple a 12-year-old made it!
~Elijahman~
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/TkZvMa30Juo?t=1830
~ pi_is_3.14
Video Solution
~savannahsolver
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
