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2021 AMC 12B Problems/Problem 4: Difference between revisions

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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 1==
==Solution 1 (One Variable)==
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>
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>.


==Solution 2==
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>.
<br><br>
~ {TSun} ~
~ {TSun} ~


==Solution 3 (Two Variables)==
==Solution 2 (Two Variables)==
Suppose the morning class has <math>m</math> students and the afternoon class has <math>a</math> students. We have the following chart:
Suppose the morning class has <math>m</math> students and the afternoon class has <math>a</math> students. We have the following table:
<cmath>\begin{array}{c|c|c|c}
<cmath>\begin{array}{c|c|c|c}
& & & \\ [-2.5ex]
& & & \\ [-2.5ex]
Line 26: Line 23:
\textbf{Afternoon} & a & 70 & 70a
\textbf{Afternoon} & a & 70 & 70a
\end{array}</cmath>
\end{array}</cmath>
We are also given that <math>\frac ma=\frac34,</math> which rearranges as <math>m=\frac34a.</math>
We are also given that <math>\frac ma=\frac34,</math> which rearranges as <math>m=\frac34a.</math>


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>
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>
~MRENTHUSIASM
~MRENTHUSIASM


==Solution 4 (Ratio)==
==Solution 3 (Ratio)==
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>
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>


~Kinglogic
~Kinglogic
==Solution 4 (Convenient Values)==
WLOG, assume there are <imath>3</imath> students in the morning class and <imath>4</imath> in the afternoon class. Then the average is <imath>\frac{3\cdot 84 + 4\cdot 70}{7}=\boxed{\textbf{(C)} ~76}.</imath>
== Solution 5 (Basic manipulation) ==
Let <imath>S_1</imath> be the sum of the morning class's scores, and <imath>S_2</imath> be the sum of the afternoon class's scores. Let <imath>3x</imath> be the number of students in the morning class, and <imath>4x</imath> be the number of students in the afternoon class. We can write <imath>\frac{S_1}{3x}=84</imath> and <imath>\frac{S_2}{4x}=70</imath>, so <imath>S_2 = 70 \cdot 4x</imath> and <imath>S_1 = 84 \cdot 3x</imath>. Adding these, we get <imath>S_1 + S_2 = x(70 \cdot 4 + 84 \cdot 3)</imath>. We want to find the mean, which is <imath>\frac{S_1+S_2}{7x}</imath>, so dividing by <imath>7x</imath> on both sides of the equation, we get the mean to be <imath>10 \cdot 4 + 12 \cdot 3 = \boxed{\textbf{(C)} ~76}.</imath>
~vaishnav


==Video Solution by Punxsutawney Phil==
==Video Solution by Punxsutawney Phil==
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https://www.youtube.com/watch?v=VzwxbsuSQ80
https://www.youtube.com/watch?v=VzwxbsuSQ80


== Video Solution by OmegaLearn (Clever application of Average Formula) ==
== Video Solution by OmegaLearn (Clever Application of Average Formula) ==
https://youtu.be/lE8v7lXT8Go
https://youtu.be/lE8v7lXT8Go


Line 59: Line 61:


~Interstigation
~Interstigation
==Video Solution (Under 2 min!)==
https://youtu.be/EgBKBCOn9Mo
~Education, the Study of Everything


==See Also==
==See Also==

Latest revision as of 11:42, 9 November 2025

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 $\frac{3\cdot 84 + 4\cdot 70}{7}=\boxed{\textbf{(C)} ~76}.$

Solution 5 (Basic manipulation)

Let $S_1$ be the sum of the morning class's scores, and $S_2$ be the sum of the afternoon class's scores. Let $3x$ be the number of students in the morning class, and $4x$ be the number of students in the afternoon class. We can write $\frac{S_1}{3x}=84$ and $\frac{S_2}{4x}=70$, so $S_2 = 70 \cdot 4x$ and $S_1 = 84 \cdot 3x$. Adding these, we get $S_1 + S_2 = x(70 \cdot 4 + 84 \cdot 3)$. We want to find the mean, which is $\frac{S_1+S_2}{7x}$, so dividing by $7x$ on both sides of the equation, we get the mean to be $10 \cdot 4 + 12 \cdot 3 = \boxed{\textbf{(C)} ~76}.$

~vaishnav

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

These problems are copyrighted © by the Mathematical Association of America.

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