Difference between revisions of "2013 AMC 8 Problems/Problem 5"
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==Problem== | ==Problem== | ||
| + | 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? | ||
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| + | <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== | ==Solution== | ||
| + | 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>. | ||
| + | |||
| + | Hence,<cmath>26-6=\boxed{\textbf{(E)}\ \text{average, by 20}}.</cmath> | ||
| + | |||
| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/TkZvMa30Juo?t=1709 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/ATZp9dYIM_0 ~savannahsolver | ||
==See Also== | ==See Also== | ||
| − | {{AMC8 box|year=2013| | + | {{AMC8 box|year=2013|num-b=4|num-a=6}} |
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 19:22, 5 January 2024
Problem
Hammie is in 
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?
Solution
Listing the elements from least to greatest, we have 
, we see that the median weight is 6 pounds.
The average weight of the five kids is 
.
Hence,![\[26-6=\boxed{\textbf{(E)}\ \text{average, by 20}}.\]](http://latex.artofproblemsolving.com/c/5/c/c5c7346ee982f5b48ab4a6a7867063ac17a2704a.png)
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. 
