Difference between revisions of "2023 AMC 10A Problems/Problem 10"

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==Problem==
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#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}}
 

Latest revision as of 22:26, 9 November 2023