Difference between revisions of "2021 AMC 10A Problems/Problem 5"
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<math>\textbf{(A)} ~\frac{14-8}{k-12} \qquad\textbf{(B)} ~\frac{8k-168}{k-12} \qquad\textbf{(C)} ~\frac{14}{12} - \frac{8}{k} \qquad\textbf{(D)} ~\frac{14(k-12)}{k^2} \qquad\textbf{(E)} ~\frac{14(k-12)}{8k}</math> | <math>\textbf{(A)} ~\frac{14-8}{k-12} \qquad\textbf{(B)} ~\frac{8k-168}{k-12} \qquad\textbf{(C)} ~\frac{14}{12} - \frac{8}{k} \qquad\textbf{(D)} ~\frac{14(k-12)}{k^2} \qquad\textbf{(E)} ~\frac{14(k-12)}{8k}</math> | ||
| − | ==Solution 1== | + | ==Solution 1 (Generalized)== |
The total score in the class is <math>8k.</math> | The total score in the class is <math>8k.</math> | ||
The total score on the <math>12</math> quizzes is <math>12\cdot14=168.</math> | The total score on the <math>12</math> quizzes is <math>12\cdot14=168.</math> | ||
Revision as of 06:22, 16 February 2021
Contents
Problem
The quiz scores of a class with 
students have a mean of 
. The mean of a collection of 
of these quiz scores is 
. What is the mean of the remaining quiz scores of terms of 
?
Solution 1 (Generalized)
The total score in the class is 
The total score on the quizzes is 
Therefore, for the remaining quizzes (
of them), the total score is 
Their mean score is 
~MRENTHUSIASM
Solution 2 (Convenient Values and Observations)
Set 
The answer is the same as the last student's quiz score, which is 
From the answer choices, only 
yields a negative value for
~MRENTHUSIASM
Video Solution
https://www.youtube.com/watch?v=S4q1ji013JQ&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=5
~ North America Math Contest Go Go Go
Video Solution (Using average formula)
~ pi_is_3.14
Video Solution (Simple and Quick)
~ Education, the Study of Everything
Video Solution 4
~savannahsolver
See Also
| 2021 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
