2013 AMC 8 Problems/Problem 5: Difference between revisions

Latest revision as of 09:01, 18 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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

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 ==Problem==
-Hammie is in the <math>6^\text{th}</math> 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?
+Hammie is in <math>6^\text{th}</math> 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?
-<math>\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}</math>
+<math>\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}</math>
 ==Solution==
-The median here is OBVIOUSLY less than the mean, so option (A) and (B) are out.
+Listing the elements from least to greatest, we have <math>(5, 5, 6, 8, 106)</math>, we see that the median weight is 6 pounds.
+The average weight of the five kids is <math>\frac{5+5+6+8+106}{5} = \frac{130}{5} = 26</math>.
-Lining up the numbers (5, 5, 6, 8, 106), we see that the median weight is 6 pounds.
+Hence,<cmath>26-6=\boxed{\textbf{(E)}\ \text{average, by 20}}.</cmath>
-The average weight of the five kids is <math>\dfrac{5+5+6+8+106}{5} = \dfrac{130}{5} = 26</math>.
+== Video Solution by OmegaLearn ==
+https://youtu.be/TkZvMa30Juo?t=1709
-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>.
+~ pi_is_3.14
+==Video Solution==
+https://youtu.be/ATZp9dYIM_0 ~savannahsolver
 ==See Also==
 {{AMC8 box|year=2013|num-b=4|num-a=6}}
 {{MAA Notice}}

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