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

m (Added in the duplicates.)
m (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:
 
Suppose the morning class has <math>m</math> students and the afternoon class has <math>a</math> students. We have the following chart:
 
<cmath>\begin{array}{c|c|c|c}
 
<cmath>\begin{array}{c|c|c|c}
 +
& & & \ [-2.5ex]
 
& \textbf{\# of Students} & \textbf{Mean} & \textbf{Total} \
 
& \textbf{\# of Students} & \textbf{Mean} & \textbf{Total} \
 
\hline
 
\hline
 +
& & & \ [-2.5ex]
 
\textbf{Morning} & m & 84 & 84m \
 
\textbf{Morning} & m & 84 & 84m \
 
\hline
 
\hline
 +
& & & \ [-2.5ex]
 
\textbf{Afternoon} & a & 70 & 70a
 
\textbf{Afternoon} & a & 70 & 70a
 
\end{array}</cmath>
 
\end{array}</cmath>

Revision as of 02:34, 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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