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Difference between revisions of "2021 AMC 10A Problems/Problem 5"

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==Problem==
 
==Problem==
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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 of terms of <math>k</math>?
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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>
  
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==Solution 1 (Generalized)==
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==Solution 1 (Generalized Value of k)==
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The total score of the class is <math>8k.</math>
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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>
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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
  
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==Solution 2 (Convenient Values and Observations)==
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==Solution 2 (Specified Values of k)==
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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> yields a negative value for <math>k=13.</math>
+
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
  
 
==Solution 3==
 
==Solution 3==
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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>
+
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)
 
~Ankitamc (Solution)
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~ North America Math Contest Go Go Go
 
~ North America Math Contest Go Go Go
  
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== Video Solution (Using average formula) ==
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== Video Solution (Using Average Formula) ==
 
https://youtu.be/jocfZVNGU3o
 
https://youtu.be/jocfZVNGU3o
  
 
~ pi_is_3.14
 
~ pi_is_3.14
  
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==Video Solution 4==
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==Video Solution==
 
https://youtu.be/wacb0roj20A
 
https://youtu.be/wacb0roj20A
  
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~IceMatrix
 
~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 21:54, 14 September 2021

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 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 $\boxed{\textbf{(B)} ~\frac{8k-168}{k-12}}.$

~MRENTHUSIASM

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 $k=13.$

~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 $12\cdot14 = 168.$ 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 $\boxed{\textbf{(B)} ~\frac{8k-168}{k-12}}.$

~Ankitamc (Solution)

~MRENTHUSIASM ($\LaTeX$ 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

~ 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

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

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