Difference between revisions of "2023 AMC 12A Problems/Problem 8"
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But 2n + 3m, which = 27, is also 2(n+m) + m = 20 + m, so m = <math>\boxed{\textbf{(D) 7}}</math> | But 2n + 3m, which = 27, is also 2(n+m) + m = 20 + m, so m = <math>\boxed{\textbf{(D) 7}}</math> | ||
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| + | ~Dilip | ||
==Solution 2== | ==Solution 2== | ||
Revision as of 02:50, 10 November 2023
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 equations 
and 
.
Expanding, 
and 
.
This gives us 
and 
. Solving for each variable, 
and 
. The answer is 
~walmartbrian ~Shontai ~andyluo
Solution 1 with slight variation
n tests with an average of m
When she takes another test her new average, m+1, is (nm + 11)/(n+1)
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 2
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
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. 
