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Difference between revisions of "2023 AMC 12A Problems/Problem 8"

(→‎Solution 2 (similar method to solution 1))
(→‎Solution 2 (similar method to solution 1))
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Let <math>s</math> represent the sum of Maureen's test scores previously and <math>t</math> be the number of scores taken previously.
 
Let <math>s</math> represent the sum of Maureen's test scores previously and <math>t</math> be the number of scores taken previously.
  
−
So, <math>\frac{x+11}{n+1} = \frac{x}{n}+1</math>
+
So, <math>\frac{x+11}{n+1} = \frac{x}{n}+1</math> and <math>\frac{x+33}{n+3} = \frac{x}{n}+2</math>
 +
 
 +
We can use the first equation to write <math>s</math> in terms of <math>n</math>.
  
 
==See Also==
 
==See Also==

Revision as of 22:52, 9 November 2023

Problem

Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$. If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$. What is the mean of her quiz scores currently? $\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8$

Solution 1

Let $a$ represent the amount of tests taken previously and $x$ the mean of the scores taken previously.

We can write the equations $\frac{ax+11}{a+1} = x+1$ and $\frac{ax+33}{a+3} = x+2$.

Expanding, $ax+11 = ax+a+x+1$ and $ax+33 = ax+2a+3x+6$.

This gives us $a+x = 10$ and $2a+3x = 27$. Solving for each variable, $x=7$ and $a=3$. The answer is $\boxed{\textbf{(D) }7}$

~walmartbrian ~Shontai ~andyluo

Solution 2 (similar method to solution 1)

Let $s$ represent the sum of Maureen's test scores previously and $t$ be the number of scores taken previously.

So, $\frac{x+11}{n+1} = \frac{x}{n}+1$ and $\frac{x+33}{n+3} = \frac{x}{n}+2$

We can use the first equation to write $s$ in terms of $n$.

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

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