Difference between revisions of "2023 AMC 12A Problems/Problem 8"
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Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an <math>11</math> on the next quiz, her mean will increase by <math>1</math>. If she scores an <math>11</math> on each of the next three quizzes, her mean will increase by <math>2</math>. What is the mean of her quiz scores currently? | Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an <math>11</math> on the next quiz, her mean will increase by <math>1</math>. If she scores an <math>11</math> on each of the next three quizzes, her mean will increase by <math>2</math>. What is the mean of her quiz scores currently? | ||
<math>\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8</math> | <math>\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8</math> | ||
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==Solution 1== | ==Solution 1== | ||
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~walmartbrian ~Shontai ~andyluo | ~walmartbrian ~Shontai ~andyluo | ||
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==Solution 1 with slight variation== | ==Solution 1 with slight variation== | ||
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~Dilip | ~Dilip | ||
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==Solution 2== | ==Solution 2== | ||
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~milquetoast | ~milquetoast | ||
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| + | ==Video Solution by Math-X (First understand the problem!!!)== | ||
| + | https://youtu.be/cMgngeSmFCY?si=MHL95YihFdxKROrU&t=2280 ~Math-X | ||
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==See Also== | ==See Also== | ||
Revision as of 14:12, 10 November 2023
- 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:
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
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. 
