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Difference between revisions of "2016 AMC 8 Problems/Problem 6"

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== Problem ==
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The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?
 
The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?
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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>
 
 
<asy>
 
<asy>
 
unitsize(0.9cm);
 
unitsize(0.9cm);
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</asy>
 
</asy>
  
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==Solution==
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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>
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We first notice that the median name will be the <math>10^{\mbox{th}}</math> name. We take all the <math>3</math> letter names away from the list to see that the <math>3^{\mbox{rd}}</math> name in the new table is the desired length. Since there are <math>3</math> names that are <math>4</math> letters long, the median name length is <math>\text{(B)}</math> <math>4</math>.  
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== Solution 1 ==
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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>.
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== Solution 2 ==
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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>.
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 +
== Video Solution ==
 +
 
 +
https://youtu.be/M9Hooi5UwDg?si=4CPixqDwQ_9BCh6m
 +
 
 +
A solution so simple a 12-year-old made it!
 +
 
 +
~Elijahman~
 +
 
 +
== Video Solution (CREATIVE THINKING!!!) ==
 +
https://youtu.be/Xab3qcUUDRY
 +
 
 +
~Education, the Study of Everything
 +
 
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/TkZvMa30Juo?t=1830
 +
 
 +
~ pi_is_3.14
 +
 
 +
== Video Solution ==
 +
https://youtu.be/800KF_3XSmM
 +
 
 +
~savannahsolver
 +
 
 +
== See Also ==
  
 
{{AMC8 box|year=2016|num-b=5|num-a=7}}
 
{{AMC8 box|year=2016|num-b=5|num-a=7}}
 +
{{MAA Notice}}

Latest revision as of 19:01, 21 August 2024

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]

$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7$

Solution 1

We first notice that the median name will be the $(19+1)/2=10^{\mbox{th}}$ name. The $10^{\mbox{th}}$ name is $\boxed{\textbf{(B)}\ 4}$.

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 $7 + 3 + 1 + 4 + 4 = 19$. Thus the index of the median length would be the 10th name. Since there are $7$ names with length $3$, and $3$ names with length $4$, the $10$th name would have $4$ letters. Thus our answer is $\boxed{\textbf{(B)}\ 4}$.

Video Solution

https://youtu.be/M9Hooi5UwDg?si=4CPixqDwQ_9BCh6m

A solution so simple a 12-year-old made it!

~Elijahman~

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/Xab3qcUUDRY

~Education, the Study of Everything

Video Solution by OmegaLearn

https://youtu.be/TkZvMa30Juo?t=1830

~ pi_is_3.14

Video Solution

https://youtu.be/800KF_3XSmM

~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

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