2021 AMC 10A Problems/Problem 5
Contents
Problem
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 in terms of 
?
Solution 1 (Generalized Value of k)
The total score of the class is 
and the total score of the quizzes is 
Therefore, for the remaining quizzes (
of them), the total score is 
Their mean score is 
~MRENTHUSIASM
Solution 2 (Specified Values of k)
Set 
The answer is the same as the last student's quiz score, which is 
From the answer choices, only 
is negative at
~MRENTHUSIASM
Solution 3
You know that the mean of the first students is 
so that means all of them combined had a score of 
Set the mean of the remaining students (in other words the value you are trying to solve for), to 
The total number of remaining students in a class of size can be written as 
The total score students got combined can be written as 
and the total score all of the students in the class got was 
(the first twelve students, plus the remaining students). The mean of the whole class can be written as 
The mean of the class has already been given as 
so by just writing the equation 
and solving for 
(the mean of students) will give you the answer in terms of 
which is
~Ankitamc (Solution)
~MRENTHUSIASM (
Adjustments)
Video Solution (Simple and Quick)
~ Education, the Study of Everything
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)
~ pi_is_3.14
Video Solution
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/50CThrk3RcM/t=399
~IceMatrix
Video Solution by The Learning Royal
See Also
| 2021 AMC 10A (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
