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2013 AMC 8 Problems/Problem 5: Difference between revisions

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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>.
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>.
 
==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 11:11, 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 (Problems • Answer Key • Resources)
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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