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

(Solution)
(18 intermediate revisions by 8 users not shown)
Line 1: Line 1:
 
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?
  
 +
<math>\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7</math>
 +
<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>
  
 
==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>.
 +
 +
{{AMC8 box|year=2016|num-b=5|num-a=7}}
 +
{{MAA Notice}}

Revision as of 21:45, 25 March 2019

The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?

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

Solution

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

2016 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png