Difference between revisions of "2013 AMC 8 Problems/Problem 5"
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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>. | 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>. | ||
| + | |||
| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/TkZvMa30Juo?t=1709 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 02:44, 16 January 2023
Problem
Hammie is in the 
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?
![\[\qquad \textbf{(D)}\ \text{average, by 15} \qquad \textbf{(E)}\ \text{average, by 20}\]](http://latex.artofproblemsolving.com/4/0/c/40c8595ceba5982b6c941c882dded67ff4ad18d9.png)
Solution
The median here is obviously less than the mean, so options 
and 
are out.
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 
.
Therefore, the average weight is bigger, by 
pounds, making the answer 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
