2021 AMC 10A Problems/Problem 5: Difference between revisions
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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> | ||
Therefore, for the remaining quizzes (<math>k-12</math> of them), the total score is <math>8k-168.</math> Their mean score is <math>\boxed{\textbf{(B)} ~\frac{8k-168}{k-12}}.</math> | Therefore, for the remaining quizzes (<math>k-12</math> of them), the total score is <math>8k-168.</math> Their mean score is <math>\boxed{\textbf{(B)} ~\frac{8k-168}{k-12}}.</math> | ||
~MRENTHUSIASM | |||
== Video Solution (Using average formula) == | == Video Solution (Using average formula) == | ||
Revision as of 21:40, 11 February 2021
Problem 5
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
Video Solution (Using average formula)
~ pi_is_3.14
