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2013 AMC 8 Problems/Problem 5

Revision as of 23:02, 1 February 2023 by Megaboy6679 (talk | contribs) (→‎Solution)

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

Listing the elements from least to greatest, we have $(5, 5, 6, 8, 106)$, we see that the median weight is 6 pounds. The average weight of the five kids is $\frac{5+5+6+8+106}{5} = \frac{130}{5} = 26$.

Hence,\[26-6=\boxed{\textbf{(E)}\ \text{average, by 20}}.\]

Video Solution by OmegaLearn

https://youtu.be/TkZvMa30Juo?t=1709

~ pi_is_3.14

Video Solution

https://youtu.be/ATZp9dYIM_0 ~savannahsolver

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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