Difference between revisions of "2021 AMC 10A Problems/Problem 5"

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==Problem 5==
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The quiz scores of a class with <math>k > 12</math> students have a mean of <math>8</math>. The mean of a collection of <math>12</math> of these quiz scores is <math>14</math>. What is the mean of the remaining quiz scores of terms of <math>k</math>?
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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>
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==Solution==
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The total score in the class is <math>8k.</math>
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The total score on the <math>12</math> quizzes is <math>12\cdot14=168.</math>
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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>
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== Video Solution (Using average formula) ==
 
== Video Solution (Using average formula) ==
 
https://youtu.be/jocfZVNGU3o
 
https://youtu.be/jocfZVNGU3o
  
 
~ pi_is_3.14
 
~ pi_is_3.14

Revision as of 22:40, 11 February 2021

Problem 5

The quiz scores of a class with $k > 12$ students have a mean of $8$. The mean of a collection of $12$ of these quiz scores is $14$. What is the mean of the remaining quiz scores of terms of $k$?

$\textbf{(A)} ~\frac{14-8}{k-12} \qquad\textbf{(B)} ~\frac{8k-168}{k-12} \qquad\textbf{(C)} ~\frac{14}{12} - \frac{8}{k} \qquad\textbf{(D)} ~\frac{14(k-12)}{k^2} \qquad\textbf{(E)} ~\frac{14(k-12)}{8k}$

Solution

The total score in the class is $8k.$ The total score on the $12$ quizzes is $12\cdot14=168.$ Therefore, for the remaining quizzes ($k-12$ of them), the total score is $8k-168.$ Their mean score is $\boxed{\textbf{(B)} ~\frac{8k-168}{k-12}}.$

Video Solution (Using average formula)

https://youtu.be/jocfZVNGU3o

~ pi_is_3.14