2023 AMC 12A Problems/Problem 8
- The following problem is from both the 2023 AMC 10A #10 and 2023 AMC 12A #8, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Variation on Solution 1)
- 4 Solution 3
- 5 Solution 4 (Trial and Error)
- 6 Video Solution by Power Solve (easy to digest!)
- 7 Video Solution (🚀 Just 3 min 🚀)
- 8 Video Solution by Math-X (First understand the problem!!!)
- 9 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 10 Video Solution
- 11 See Also
Problem
Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an on the next quiz, her mean will increase by . If she scores an on each of the next three quizzes, her mean will increase by . What is the mean of her quiz scores currently?
Solution 1
Let represent the amount of tests taken previously and the mean of the scores taken previously.
We can write the following equations:
Multiplying by and solving, we get:
Multiplying by and solving, we get:
Solving the system of equations for and , we find that and .
~walmartbrian ~Shontai ~andyluo ~megaboy6679
Solution 2 (Variation on Solution 1)
Suppose Maureen took tests with an average of .
If she takes another test, her new average is
Cross-multiplying: , so .
If she takes more tests, her new average is
Cross-multiplying: , so .
But can also be written as . Therefore
~Dilip ~megaboy6679 (latex)
Solution 3
Let represent the sum of Maureen's test scores previously and be the number of scores taken previously.
So, and
We can use the first equation to write in terms of .
We then substitute this into the second equation:
From here, we solve for t, getting .
We substitute this to get .
Therefore, the solution to the problem is
~milquetoast
Solution 4 (Trial and Error)
Let's consider all the answer choices. If the average is , then, we can assume that all her test choices were . We can see that she must have gotten twice, in order for another score of to bring her average up by one. However, adding three 's will not bring her score up to 10. Continuing this process for the answer choices, we see that the answer is ~andliu766
Video Solution by Power Solve (easy to digest!)
https://www.youtube.com/watch?v=ZUgo3-BKt30
Video Solution (🚀 Just 3 min 🚀)
~Education, the Study of Everything
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/cMgngeSmFCY?si=MHL95YihFdxKROrU&t=2280 ~Math-X
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=1QJ_BQOfZtg
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 |
2023 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.