2021 AMC 12B Problems/Problem 4

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

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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