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Difference between revisions of "2021 AMC 12B Problems/Problem 4"

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The 2021 AMC 12B will be held on February 10th, 2021. The problems will not be made public until 24 hours after that.
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{{duplicate|[[2021 AMC 10B Problems#Problem 6|2021 AMC 10B #6]] and [[2021 AMC 12B Problems#Problem 4|2021 AMC 12B #4]]}}
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==Problem==
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Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is <math>84</math>, and the afternoon class's mean score is <math>70</math>. The ratio of the number of students in the morning class to the number of students in the afternoon class is <math>\frac{3}{4}</math>. What is the mean of the scores of all the students?
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<math>\textbf{(A)} ~74 \qquad\textbf{(B)} ~75 \qquad\textbf{(C)} ~76 \qquad\textbf{(D)} ~77 \qquad\textbf{(E)} ~78</math>
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==Solution 1 (One Variable)==
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Let there be <math>3x</math> students in the morning class and <math>4x</math> students in the afternoon class. The total number of students is <math>3x + 4x = 7x</math>. The average is <math>\frac{3x\cdot84 + 4x\cdot70}{7x}=76</math>. Therefore, the answer is <math>\boxed{\textbf{(C)} ~76}</math>.
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~ {TSun} ~
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==Solution 2 (Two Variables)==
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Suppose the morning class has <math>m</math> students and the afternoon class has <math>a</math> students. We have the following table:
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<cmath>\begin{array}{c|c|c|c}
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& & & \\ [-2.5ex]
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& \textbf{\# of Students} & \textbf{Mean} & \textbf{Total} \\
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\hline
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& & & \\ [-2.5ex]
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\textbf{Morning} & m & 84 & 84m \\
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\hline
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& & & \\ [-2.5ex]
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\textbf{Afternoon} & a & 70 & 70a
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\end{array}</cmath>
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We are also given that <math>\frac ma=\frac34,</math> which rearranges as <math>m=\frac34a.</math>
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The mean of the scores of all the students is <cmath>\frac{84m+70a}{m+a}=\frac{84\left(\frac34a\right)+70a}{\frac34a+a}=\frac{133a}{\frac74a}=133\cdot\frac47=\boxed{\textbf{(C)} ~76}.</cmath>
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~MRENTHUSIASM
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==Solution 3 (Ratio)==
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Of the average, <math>\frac{3}{3+4}=\frac{3}{7}</math> of the scores came from the morning class and <math>\frac{4}{7}</math> came from the afternoon class. The average is <math>\frac{3}{7}\cdot 84+\frac{4}{7}\cdot 70=\boxed{\textbf{(C)} ~76}.</math>
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~Kinglogic
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==Solution 4 (Convenient Values)==
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WLOG, assume there are <math>3</math> students in the morning class and <math>4</math> in the afternoon class. Then the average is <math>\frac{3\cdot 84 + 4\cdot 70}{7}=\boxed{\textbf{(C)} ~76}.</math>
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==Video Solution by Punxsutawney Phil==
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https://youtube.com/watch?v=qpvS2PVkI8A&t=249s
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==Video Solution by Hawk Math==
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https://www.youtube.com/watch?v=VzwxbsuSQ80
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== Video Solution by OmegaLearn (Clever Application of Average Formula) ==
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https://youtu.be/lE8v7lXT8Go
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~ pi_is_3.14
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==Video Solution by TheBeautyofMath==
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https://youtu.be/GYpAm8v1h-U (for AMC 10B)
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https://youtu.be/EMzdnr1nZcE?t=608 (for AMC 12B)
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 +
~IceMatrix
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==Video Solution by Interstigation==
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https://youtu.be/DvpN56Ob6Zw?t=426
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~Interstigation
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==Video Solution (Under 2 min!)==
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https://youtu.be/EgBKBCOn9Mo
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 +
~Education, the Study of Everything
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==See Also==
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{{AMC12 box|year=2021|ab=B|num-b=3|num-a=5}}
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{{AMC10 box|year=2021|ab=B|num-b=5|num-a=7}}
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{{MAA Notice}}

Latest revision as of 22:50, 18 July 2023

The following problem is from both the 2021 AMC 10B #6 and 2021 AMC 12B #4, so both problems redirect to this page.

Problem

Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84$, and the afternoon class's mean score is $70$. The ratio of the number of students in the morning class to the number of students in the afternoon class is $\frac{3}{4}$. What is the mean of the scores of all the students?

$\textbf{(A)} ~74 \qquad\textbf{(B)} ~75 \qquad\textbf{(C)} ~76 \qquad\textbf{(D)} ~77 \qquad\textbf{(E)} ~78$

Solution 1 (One Variable)

Let there be $3x$ students in the morning class and $4x$ students in the afternoon class. The total number of students is $3x + 4x = 7x$. The average is $\frac{3x\cdot84 + 4x\cdot70}{7x}=76$. Therefore, the answer is $\boxed{\textbf{(C)} ~76}$.

~ {TSun} ~

Solution 2 (Two Variables)

Suppose the morning class has $m$ students and the afternoon class has $a$ students. We have the following table: \[\begin{array}{c|c|c|c} & & & \\ [-2.5ex] & \textbf{\# of Students} & \textbf{Mean} & \textbf{Total} \\ \hline & & & \\ [-2.5ex] \textbf{Morning} & m & 84 & 84m \\ \hline & & & \\ [-2.5ex] \textbf{Afternoon} & a & 70 & 70a \end{array}\] We are also given that $\frac ma=\frac34,$ which rearranges as $m=\frac34a.$

The mean of the scores of all the students is \[\frac{84m+70a}{m+a}=\frac{84\left(\frac34a\right)+70a}{\frac34a+a}=\frac{133a}{\frac74a}=133\cdot\frac47=\boxed{\textbf{(C)} ~76}.\] ~MRENTHUSIASM

Solution 3 (Ratio)

Of the average, $\frac{3}{3+4}=\frac{3}{7}$ of the scores came from the morning class and $\frac{4}{7}$ came from the afternoon class. The average is $\frac{3}{7}\cdot 84+\frac{4}{7}\cdot 70=\boxed{\textbf{(C)} ~76}.$

~Kinglogic

Solution 4 (Convenient Values)

WLOG, assume there are $3$ students in the morning class and $4$ in the afternoon class. Then the average is $\frac{3\cdot 84 + 4\cdot 70}{7}=\boxed{\textbf{(C)} ~76}.$

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=qpvS2PVkI8A&t=249s

Video Solution by Hawk Math

https://www.youtube.com/watch?v=VzwxbsuSQ80

Video Solution by OmegaLearn (Clever Application of Average Formula)

https://youtu.be/lE8v7lXT8Go

~ pi_is_3.14

Video Solution by TheBeautyofMath

https://youtu.be/GYpAm8v1h-U (for AMC 10B)

https://youtu.be/EMzdnr1nZcE?t=608 (for AMC 12B)

~IceMatrix

Video Solution by Interstigation

https://youtu.be/DvpN56Ob6Zw?t=426

~Interstigation

Video Solution (Under 2 min!)

https://youtu.be/EgBKBCOn9Mo

~Education, the Study of Everything

See Also

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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 12 Problems and Solutions
2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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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