2023 AMC 10A Problem 10

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

Averages become tractable the moment you convert them to totals: "mean $m$ over $n$ items" is a total of $nm$. Each scenario is then one linear equation, and the $nm$ terms cancel, leaving an ordinary two-variable system. The check at the end is cheap and catches sign errors.

Think of each new $11$ as spreading its excess over all quizzes. After one quiz the new mean is $m + 1$, so the $11$ exceeds it by $10 - m$, which must lift the other $n$ scores by $1$ each: $n = 10 - m$. After three quizzes the new mean is $m + 2$; each $11$ exceeds it by $9 - m$, and $3(9 - m) = 2n$. Same system, no expansion needed.

• Treating the mean increase as if the number of quizzes stays the same, or letting the two scenarios have different current means.
• Expanding $(n+3)(m+2)$ incorrectly, e.g. writing $3m + 2n$ as $2m + 3n$.