2023 AMC 10A Problem 10

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 $\boxed{\textbf{(D)}\ 7}$.

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.

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 $9 = 7 + 2$. ✓

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 $\boxed{\textbf{(D)}\ 7}$.