Difference between revisions of "2021 AMC 12B Problems/Problem 4"
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
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+ | ==Solution 4 (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> | ||
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+ | ~Kinglogic | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== |
Revision as of 18:09, 22 February 2021
Contents
[hide]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 chart:
We are also given that which rearranges as
The mean of the scores of all the students is
~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
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.