Difference between revisions of "2021 AMC 12B Problems/Problem 4"
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==Solution 3 (Two Variables)== | ==Solution 3 (Two Variables)== | ||
| − | Suppose the morning class has <math>m</math> students and the afternoon class has <math>a</math> students. We have the following | + | 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 26: | ||
\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 | ||
Revision as of 22:58, 21 June 2021
- The following problem is from both the 2021 AMC 10B #6 and 2021 AMC 12B #4, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (Two Variables)
- 5 Solution 4 (Ratio)
- 6 Video Solution by Punxsutawney Phil
- 7 Video Solution by Hawk Math
- 8 Video Solution by OmegaLearn (Clever application of Average Formula)
- 9 Video Solution by TheBeautyofMath
- 10 Video Solution by Interstigation
- 11 See Also
Problem
Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is 
, and the afternoon class's mean score is 
. The ratio of the number of students in the morning class to the number of students in the afternoon class is 
. What is the mean of the scores of all the students?
Solution 1
WLOG, assume there are 
students in the morning class and 
in the afternoon class. Then the average is 
Solution 2
Let there be 
students in the morning class and 
students in the afternoon class. The total number of students is 
. The average is 
. Therefore, the answer is 
.
~ {TSun} ~
Solution 3 (Two Variables)
Suppose the morning class has 
students and the afternoon class has 
students. We have the following table:
![\[\begin{array}{c|c|c|c} & & & \\ [-2.5ex] & \textbf{\# of Students} & \textbf{Mean} & \textbf{Total} \\ \hline & & & \\ [-2.5ex] \textbf{Morning} & m & 84 & 84m \\ \hline & & & \\ [-2.5ex] \textbf{Afternoon} & a & 70 & 70a \end{array}\]](http://latex.artofproblemsolving.com/d/8/c/d8cfccc9e2bb5552c530ce9b1368e0f7f21acee0.png)
We are also given that 
which rearranges as 
The mean of the scores of all the students is ![\[\frac{84m+70a}{m+a}=\frac{84\left(\frac34a\right)+70a}{\frac34a+a}=\frac{133a}{\frac74a}=133\cdot\frac47=\boxed{\textbf{(C)} ~76}.\]](http://latex.artofproblemsolving.com/b/1/c/b1c4a3b5c094c6a9ed4c9ff7e9b5e886bba6b6de.png)
~MRENTHUSIASM
Solution 4 (Ratio)
Of the average, 
of the score came from the morning class and 
came from the afternoon class. The average is 
~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)
~ 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 (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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
