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

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https://youtu.be/STPoBU6A3yU
 
https://youtu.be/STPoBU6A3yU
  
~Education, the Study of Everything
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~ Education, the Study of Everything
  
 
==Video Solution 4==
 
==Video Solution 4==

Revision as of 22:30, 15 February 2021

Problem

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 $\boxed{\textbf{(B)} ~\frac{8k-168}{k-12}}.$

~MRENTHUSIASM

Solution 2 (Convenient Values and Observations)

Set $k=13.$ The answer is the same as the last student's quiz score, which is $8\cdot13-14\cdot12<0.$ From the answer choices, only $\boxed{\textbf{(B)} ~\frac{8k-168}{k-12}}$ yields a negative value for $k=13.$

~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)

https://youtu.be/jocfZVNGU3o

~ pi_is_3.14

Video Solution (Simple and Quick)

https://youtu.be/STPoBU6A3yU

~ Education, the Study of Everything

Video Solution 4

https://youtu.be/wacb0roj20A

~savannahsolver

See Also

2021 AMC 10A (ProblemsAnswer KeyResources)
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

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