2023 AMC 10A Problem 10

P1-10~3 minPrint

Ratios, Percents & Averages Systems of Equations Set up the equation/formula and compute; no special trick needed

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?

Show solution without answering

Key insight

Turn each mean into a total: n quizzes with mean m add up to n times m. Each story then gives a simple equation.

Solution

A mean is hard to work with directly, so switch to totals: if Maureen has taken $n$ quizzes with mean $m$ , her scores add up to $n \times m$ .

1. First story. One more quiz with $11$ points. Now she has $n + 1$ quizzes, and the mean is $m + 1$ . So the new total is $(n+1)(m+1)$ :
$nm + 11 = (n+1)(m+1) = nm + n + m + 1.$
The $nm$ on both sides cancels, leaving $n + m = 10$ .
2. Second story. Three more quizzes with $11$ each. Now she has $n + 3$ quizzes, and the mean is $m + 2$ :
$nm + 33 = (n+3)(m+2) = nm + 2n + 3m + 6.$
Again $nm$ cancels, leaving $2n + 3m = 27$ .
3. Solve. From the first equation, $n = 10 - m$ . Put that into the second:
$2(10 - m) + 3m = 27 \quad\Longrightarrow\quad 20 + m = 27 \quad\Longrightarrow\quad m = 7.$
Then $n = 3$ .
4. Check. Three quizzes with mean $7$ total $21$ .
- Add one $11$ : total $32$ over $4$ quizzes, mean $8$ . That is $1$ more. ✓
- Add three $11$ s: total $54$ over $6$ quizzes, mean $9$ . That is $2$ more. ✓

The answer is $\boxed{\textbf{(D)}\ 7}$ .

Why this works

Averages become easy once you change them into totals: mean times count equals total. Each "if she scores..." sentence becomes one equation, the $nm$ terms cancel, and you are left with two ordinary equations in two unknowns. Checking the answer with real numbers at the end catches any slip.

Alternative approach

Test the answer choices. The first story alone gives $n + m = 10$ , so each choice for the mean also tells you how many quizzes there are.

(C) mean $6$ , so $4$ quizzes, total $24$ . Three $11$ s make $57$ over $7$ quizzes, and $57 \div 7$ is not a whole number, let alone $8$ . ✗
(D) mean $7$ , so $3$ quizzes, total $21$ . Three $11$ s make $54$ over $6$ quizzes, mean $9 = 7 + 2$ . ✓

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Let Maureen have taken $n$ quizzes with mean $m$ , so her scores total $nm$ .

One more quiz scoring $11$ gives $n + 1$ quizzes with mean $m + 1$ :
$nm + 11 = (n+1)(m+1) = nm + n + m + 1 \quad\Longrightarrow\quad n + m = 10.$

Three more quizzes scoring $11$ each give $n + 3$ quizzes with mean $m + 2$ :
$nm + 33 = (n+3)(m+2) = nm + 2n + 3m + 6 \quad\Longrightarrow\quad 2n + 3m = 27.$

Substituting $n = 10 - m$ into the second equation: $20 - 2m + 3m = 27$ , so $m = 7$ (and $n = 3$ ).

Check: scores total $21$ ; one $11$ gives $32/4 = 8$ ; three $11$ s give $54/6 = 9$ .

The answer is $\boxed{\textbf{(D)}\ 7}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 6: Ratios and Percents
Introduction to Algebra, Chapter 5: Multi-Variable Linear Equations (§5.2, §5.3, §5.6)
Introduction to Algebra, Chapter 7: Proportion (§7.1, §7.2, §7.3)
Intermediate Algebra, Chapter 1: Basic Techniques for Solving Equations (§1.4)