Difference between revisions of "2021 AMC 10A Problems/Problem 5"
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==Solution 3== | ==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}. | 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}. | ||
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+ | ==Video Solution (Simple and Quick)== | ||
+ | https://youtu.be/STPoBU6A3yU | ||
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+ | ~ Education, the Study of Everything | ||
== Video Solution == | == Video Solution == | ||
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~ pi_is_3.14 | ~ pi_is_3.14 | ||
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==Video Solution 4== | ==Video Solution 4== |
Latest revision as of 17:45, 7 March 2021
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 of terms of ?
Solution 1 (Generalized)
The total score in the class is The total score on the quizzes is Therefore, for the remaining quizzes ( of them), the total score is Their mean score is
~MRENTHUSIASM
Solution 2 (Convenient Values and Observations)
Set The answer is the same as the last student's quiz score, which is From the answer choices, only yields a negative value for
~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
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 4
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/50CThrk3RcM/t=399
~IceMatrix
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.