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
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
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/cMgngeSmFCY?si=MHL95YihFdxKROrU&t=2280 ~Math-X
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.