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> | ||
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<asy> | <asy> | ||
unitsize(0.9cm); | unitsize(0.9cm); | ||
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</asy> | </asy> | ||
− | ==Solution== | + | ==Solutions== |
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+ | === Solution 1 === | ||
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>. | ||
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+ | === 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>. | ||
== See Also == | == See Also == |
Revision as of 16:28, 14 January 2021
Contents
[hide]Problem
The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?
Solutions
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 .
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.