Difference between revisions of "2013 AMC 8 Problems/Problem 5"

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The average weight of the five kids is <math>\dfrac{5+5+6+8+106}{5} = \dfrac{130}{5} = 26</math>.
 
The average weight of the five kids is <math>\dfrac{5+5+6+8+106}{5} = \dfrac{130}{5} = 26</math>.
  
Therefore, the average weight is bigger, by <math>26-6 = 20</math> pounds, making the answer <math>\textbf{(E)}\ \text{average, by 20}</math>.
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Therefore, the average weight is bigger, by <math>26-6 = 20</math> pounds, making the answer <math>\boxed{\textbf{(E)}\ \text{average, by 20}}</math>.
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==See Also==
 
==See Also==
 
{{AMC8 box|year=2013|num-b=4|num-a=6}}
 
{{AMC8 box|year=2013|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:11, 27 November 2013

Problem

Hammie is in the $6^\text{th}$ grade and weighs 106 pounds. His quadruplet sisters are tiny babies and weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these five children or the median weight, and by how many pounds?

$\textbf{(A)}\ \text{median, by 60} \qquad \textbf{(B)}\ \text{median, by 20} \qquad \textbf{(C)}\ \text{average, by 5} \qquad \textbf{(D)}\ \text{average, by 15} \qquad \textbf{(E)}\ \text{average, by 20}$

Solution

Lining up the numbers (5, 5, 6, 8, 106), we see that the median weight is 6 pounds.

The average weight of the five kids is $\dfrac{5+5+6+8+106}{5} = \dfrac{130}{5} = 26$.

Therefore, the average weight is bigger, by $26-6 = 20$ pounds, making the answer $\boxed{\textbf{(E)}\ \text{average, by 20}}$.

See Also

2013 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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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