2013 AMC 8 Problems/Problem 5: Difference between revisions
Megaboy6679 (talk | contribs) |
Megaboy6679 (talk | contribs) |
||
| Line 5: | Line 5: | ||
==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 == | == Video Solution by OmegaLearn == | ||
Revision as of 23:02, 1 February 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?
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,
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 | ||
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