Difference between revisions of "2021 AMC 12B Problems/Problem 4"
MRENTHUSIASM (talk | contribs) m (→Video Solution by OmegaLearn (Clever application of Average Formula)) |
MRENTHUSIASM (talk | contribs) m |
||
Line 7: | Line 7: | ||
==Solution 1== | ==Solution 1== | ||
− | 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> | + | 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> |
==Solution 2== | ==Solution 2== | ||
Line 32: | Line 32: | ||
==Solution 4 (Ratio)== | ==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> | + | 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> |
~Kinglogic | ~Kinglogic |
Revision as of 23:03, 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
[hide]- 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: 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
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.