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

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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==
 
==Problem==
 
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?
 
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>
 
<math>\textbf{(A)} ~74 \qquad\textbf{(B)} ~75 \qquad\textbf{(C)} ~76 \qquad\textbf{(D)} ~77 \qquad\textbf{(E)} ~78</math>
  
==Solution==
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==Solution 1==
WLOG assume there <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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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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==Solution 2==
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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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<br><br>
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~ {TSun} ~
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==Solution 3 (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 chart:
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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 4 (Ratio)==
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Of the average, <math>\frac{3}{3+4}=\frac{3}{7}</math> of the score 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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==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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==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 03:34, 25 March 2021

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

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

Solution 2

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 3 (Two Variables)

Suppose the morning class has $m$ students and the afternoon class has $a$ students. We have the following chart: \[\begin{array}{c|c|c|c} & & & \\ [-2.5ex] & \textbf{\# of Students} & \textbf{Mean} & \textbf{Total} \\ \hline & & & \\ [-2.5ex] \textbf{Morning} & m & 84 & 84m \\ \hline & & & \\ [-2.5ex] \textbf{Afternoon} & a & 70 & 70a \end{array}\]

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 4 (Ratio)

Of the average, $\frac{3}{3+4}=\frac{3}{7}$ of the score 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

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

See Also

2021 AMC 12B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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