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

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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]]}}
==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?
Line 4: Line 6:
<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 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 table:
<cmath>\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}</cmath>
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>
~MRENTHUSIASM
==Solution 3 (Ratio)==
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
==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==
Line 18: Line 46:
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


~ pi_is_3.14
~ 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==
==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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