Difference between revisions of "2021 AMC 10A Problems/Problem 5"
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− | ==Problem | + | ==Problem== |
− | The quiz scores of a class with <math>k > 12</math> students have a mean of <math>8</math>. The mean of a collection of <math>12</math> of these quiz scores is <math>14</math>. What is the mean of the remaining quiz scores | + | The quiz scores of a class with <math>k > 12</math> students have a mean of <math>8</math>. The mean of a collection of <math>12</math> of these quiz scores is <math>14</math>. What is the mean of the remaining quiz scores in terms of <math>k</math>? |
<math>\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}</math> | <math>\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}</math> | ||
− | ==Solution== | + | ==Solution 1 (Generalized Value of k)== |
− | The total score | + | The total score of the class is <math>8k,</math> and the total score of the <math>12</math> quizzes is <math>12\cdot14=168.</math> |
− | |||
Therefore, for the remaining quizzes (<math>k-12</math> of them), the total score is <math>8k-168.</math> Their mean score is <math>\boxed{\textbf{(B)} ~\frac{8k-168}{k-12}}.</math> | Therefore, for the remaining quizzes (<math>k-12</math> of them), the total score is <math>8k-168.</math> Their mean score is <math>\boxed{\textbf{(B)} ~\frac{8k-168}{k-12}}.</math> | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | ==Solution 2 ( | + | ==Solution 2 (Specified Values of k)== |
− | Set <math>k=13.</math> The answer is the same as the last student's quiz score, which is <math>8\cdot13-14\cdot12<0.</math> From the answer | + | Set <math>k=13.</math> The answer is the same as the last student's quiz score, which is <math>8\cdot13-14\cdot12<0.</math> From the answer choices, only <math>\boxed{\textbf{(B)} ~\frac{8k-168}{k-12}}</math> is negative at <math>k=13.</math> |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | == Video Solution (Using | + | ==Solution 3== |
+ | You know that the mean of the first <math>12</math> students is <math>14,</math> so that means all of them combined had a score of <math>12\cdot14 = 168.</math> Set the mean of the remaining students (in other words the value you are trying to solve for), to <math>a.</math> The total number of remaining students in a class of size <math>k</math> can be written as <math>k-12.</math> The total score <math>k-12</math> students got combined can be written as <math>a(k-12),</math> and the total score all of the students in the class got was <math>168 + a(k-12)</math> (the first twelve students, plus the remaining students). The mean of the whole class can be written as <math>\frac{168 + a(k-12)}{k}.</math> The mean of the class has already been given as <math>8,</math> so by just writing the equation <math>\frac{168 + a(k-12)}{k} = 8,</math> and solving for <math>a</math> (the mean of <math>k-12</math> students) will give you the answer in terms of <math>k,</math> which is <math>\boxed{\textbf{(B)} ~\frac{8k-168}{k-12}}.</math> | ||
+ | |||
+ | ~Ankitamc (Solution) | ||
+ | |||
+ | ~MRENTHUSIASM (<math>\LaTeX</math> Adjustments) | ||
+ | |||
+ | ==Video Solution (Simple and Quick)== | ||
+ | https://youtu.be/STPoBU6A3yU | ||
+ | |||
+ | ~ 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) == | ||
https://youtu.be/jocfZVNGU3o | https://youtu.be/jocfZVNGU3o | ||
~ pi_is_3.14 | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/wacb0roj20A | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/50CThrk3RcM/t=399 | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==Video Solution by The Learning Royal== | ||
+ | https://youtu.be/slVBYmcDMOI | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021|ab=A|num-b=4|num-a=6}} | {{AMC10 box|year=2021|ab=A|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:54, 14 September 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 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.