|
|
(One intermediate revision by one other user not shown) |
Line 1: |
Line 1: |
− | ==Problem==
| + | #redirect[[2023 AMC 12A Problems/Problem 8]] |
− | 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 equation <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 \boxed{\textbf{(B) }7}
| |
− | | |
− | ~walmartbrian ~Shontai ~andyluo
| |
− | | |
− | ==See Also==
| |
− | {{AMC10 box|year=2023|ab=A|num-b=9|num-a=11}}
| |
− | {{MAA Notice}}
| |