Difference between revisions of "2014 AMC 10A Problems/Problem 5"
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Thus, the solution is <cmath>90-87=3\implies\boxed{\textbf{(C)} \ 3}</cmath> | Thus, the solution is <cmath>90-87=3\implies\boxed{\textbf{(C)} \ 3}</cmath> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/Oe-QLPIuTeY | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
Revision as of 13:09, 25 June 2020
- The following problem is from both the 2014 AMC 12A #5 and 2014 AMC 10A #5, so both problems redirect to this page.
Contents
Problem
On an algebra quiz, 
of the students scored 
points, 
scored 
points, 
scored 
points, and the rest scored 
points. What is the difference between the mean and median score of the students' scores on this quiz?
Solution
Without loss of generality, let there be 
students (the least whole number possible) who took the test. We have 
students score points, 
students score points, 
students score points and 
students score points. Therefore, the mean
is 
and the median is .
Thus, the solution is ![\[90-87=3\implies\boxed{\textbf{(C)} \ 3}\]](http://latex.artofproblemsolving.com/9/c/c/9cc0a8bdca61ca93db41147a74ef754c1822910c.png)
Video Solution
~savannahsolver
See Also
| 2014 AMC 10A (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 AMC 10 Problems and Solutions | ||
| 2014 AMC 12A (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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
