# 2021 AMC 10A Problems/Problem 5

## 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 1 (Generalized)

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

## 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*14 = 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 \frac{8k-168}{k-12}.

## Video Solution (Simple and Quick)

~ Education, the Study of Everything

## Video Solution

~ North America Math Contest Go Go Go

~ pi_is_3.14

~savannahsolver

## Video Solution by TheBeautyofMath

~IceMatrix

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 