Difference between revisions of "2023 AMC 12A Problems/Problem 8"
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| − | + | ==Problem== | |
| + | 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> | ||
| − | - | + | ==Solution 1== |
| + | |||
| + | Let <math>a</math> represent the amount of tests taken previously and <math>x</math> the mean of the scores taken previously. | ||
| + | |||
| + | We can write the equations <math>\frac{ax+11}{a+1} = x+1</math> and <math>\frac{ax+33}{a+3} = x+2</math>. | ||
| + | |||
| + | Expanding, <math>ax+11 = ax+a+x+1</math> and <math>ax+33 = ax+2a+3x+6</math>. | ||
| + | |||
| + | This gives us <math>a+x = 10</math> and <math>2a+3x = 27</math>. Solving for each variable, <math>x=7</math> and <math>a=3</math>. The answer is <math>\boxed{\textbf{(D) }7}</math> | ||
| + | |||
| + | ~walmartbrian ~Shontai ~andyluo | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC10 box|year=2023|ab=A|num-b=9|num-a=11}} | ||
| + | {{AMC10 box|year=2023|ab=A|num-b=7|num-a=9}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Revision as of 23:26, 9 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
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 10A (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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
