Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 1)
m (→‎Solution 1)
Line 7: Line 7:
 
Let <math>a</math> represent the amount of tests taken previously and <math>x</math> the mean of the scores taken previously.  
 
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>.  
+
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>.  
 
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}
+
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
 
~walmartbrian ~Shontai ~andyluo

Revision as of 21:19, 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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_10A_Problems/Problem_10&oldid=201261"