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

Revision as of 21:18, 17 January 2025

Problem

Hammie is in $6^\text{th}$ 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

Fat Baby

@@ Line 18: / Line 18: @@
 https://youtu.be/ATZp9dYIM_0 ~savannahsolver
-==See Also==
+Fat Baby
-{{AMC8 box|year=2013|num-b=4|num-a=6}}
-{{MAA Notice}}

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