2021 AMC 10A Problems/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 in 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 1 (Generalized Value of k)

The total score of the class is $8k,$ and the total score of 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 $\boxed{\textbf{(B)} ~\frac{8k-168}{k-12}}.$


Solution 2 (Specified Values of k)

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 $\boxed{\textbf{(B)} ~\frac{8k-168}{k-12}}$ is negative at $k=13.$


Solution 3

You know that the mean of the first $12$ students is $14,$ so that means all of them combined had a score of $12\cdot14 = 168.$ Set the mean of the remaining students (in other words the value you are trying to solve for), to $a.$ The total number of remaining students in a class of size $k$ can be written as $k-12.$ The total score $k-12$ students got combined can be written as $a(k-12),$ and the total score all of the students in the class got was $168 + a(k-12)$ (the first twelve students, plus the remaining students). The mean of the whole class can be written as $\frac{168 + a(k-12)}{k}.$ The mean of the class has already been given as $8,$ so by just writing the equation $\frac{168 + a(k-12)}{k} = 8,$ and solving for $a$ (the mean of $k-12$ students) will give you the answer in terms of $k,$ which is $\boxed{\textbf{(B)} ~\frac{8k-168}{k-12}}.$

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See Also

2021 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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All AMC 10 Problems and Solutions

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