Difference between revisions of "2021 AMC 10A Problems/Problem 5"
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==Solution 2 (Convenient Values and Observations)== | ==Solution 2 (Convenient Values and Observations)== | ||
| − | Set <math>k=13.</math> The answer is the same as the last student's quiz score, which is <math>8\cdot13-14\cdot12<0.</math> From the answer | + | Set <math>k=13.</math> The answer is the same as the last student's quiz score, which is <math>8\cdot13-14\cdot12<0.</math> From the answer choices, only <math>\boxed{\textbf{(B)} ~\frac{8k-168}{k-12}}</math> yields a negative value for <math>k=13.</math> |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
Revision as of 18:49, 12 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
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 (Using average formula)
~ pi_is_3.14
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. 
