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.
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[hide]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 n tests with an average of m
When she takes another test her new average, , is
Cross-multiplying, nm + 11 = nm + n + m + 1; n+m = 10 -- I'll use this in a moment.
When she takes 3 more tests, the situation is (nm + 33)/(n+3) = m+2
Cross-multiplying, nm + 33 = nm + 2n + 3m + 6; 2n + 3m = 27
But 2n + 3m, which = 27, is also 2(n+m) + m = 20 + m, so m =
~Dilip
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
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/cMgngeSmFCY?si=MHL95YihFdxKROrU&t=2280 ~Math-X
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.