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> | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/Oe-QLPIuTeY | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Revision as of 14:09, 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
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.