2023 AMC 12A Problems/Problem 8

The following problem is from both the 2023 AMC 10A #10 and 2023 AMC 12A #8, so both problems redirect to this page.

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 following 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 1 with slight variation

n tests with an average of m

When she takes another test her new average, m+1, is (nm + 11)/(n+1)

Cross-multiplying, nm + 11 = nm + n + m + 1; n+m = 10 -- I'll use this in a moment

When she takes 3 more tests, the situation is (nm + 33)/(n+3) = m+2

Cross-multiplying, nm + 33 = nm + 2n + 3m + 6; 2n + 3m = 27

But 2n + 3m, which = 27, is also 2(n+m) + m = 20 + m, so m = $\boxed{\textbf{(D) 7}}$

~Dilip

Solution 2

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

So, $\frac{s+11}{t+1} = \frac{s}{t}+1$ and $\frac{s+33}{t+3} = \frac{s}{t}+2$

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

We then substitute this into the second equation: $\frac{-t^2+10t+33}{t+3} = \frac{-t^2+10}{t}+2$

From here, we solve for t, getting $t=3$ .

We substitute this to get $s=21$ .

Therefore, the solution to the problem is $\frac{21}{3}=$ $\boxed{\textbf{(D) }7}$

~milquetoast

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

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

Page

Toolbox

Search

2023 AMC 12A Problems/Problem 8

Contents

Problem

Solution 1

Solution 1 with slight variation

Solution 2

See Also