Difference between revisions of "2013 AMC 8 Problems/Problem 5"
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The average weight of the five kids is <math>\dfrac{5+5+6+8+106}{5} = \dfrac{130}{5} = 26</math>. | The average weight of the five kids is <math>\dfrac{5+5+6+8+106}{5} = \dfrac{130}{5} = 26</math>. | ||
| − | Therefore, the average weight is bigger, by <math>26-6 = 20</math> pounds, making the answer <math>\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>. |
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2013|num-b=4|num-a=6}} | {{AMC8 box|year=2013|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:11, 27 November 2013
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?
Solution
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 
.
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. 
