2010 AMC 10B Problem 15

More correct answers give more points than $99$, and each extra point must be cancelled by a wrong answer. Wrong answers use up questions, and there are only $50$. Try some numbers.

1. With $25$ correct Jesse has $100$ points, so he needs $1$ wrong answer to get to $99$.
2. Each extra correct answer adds $4$ points, so it needs $4$ more wrong answers:

3. In general, with $c$ correct the score is $4c$ minus the number wrong, so the number wrong is $4c - 99$. The questions used are
$$c + (4c - 99) = 5c - 99.$$

$29$ correct is possible. $29$ correct, $17$ wrong and $4$ blank: that is $50$ questions, and the score is $4 \cdot 29 - 17 = 116 - 17 = 99$.

$30$ or more is impossible. For $c = 30$ the questions used are $5 \cdot 30 - 99 = 51$, more than $50$. Each extra correct answer uses $5$ more questions, so larger $c$ is even worse.

The answer is $\boxed{\textbf{(C)}\ 29}$.

For a fixed score, correct and wrong answers must balance, and both use up questions. Writing the number wrong in terms of the number correct turns the $50$-question limit into a limit on the number correct. For a "maximum" question, show the best value really works and that the next one up does not.

Test the answer choices from the top. $33$ correct needs $4 \cdot 33 - 99 = 33$ wrong, $66$ questions. $31$ correct needs $25$ wrong, $56$ questions. $29$ correct needs $17$ wrong, $46$ questions, which fits.

Let $c$ be the number correct and $w$ the number wrong; the remaining $50 - c - w \ge 0$ questions are blank. The score condition is
$$4c - w = 99 \quad\Longrightarrow\quad w = 4c - 99.$$
Substituting into $c + w \le 50$:
$$c + 4c - 99 \le 50 \quad\Longrightarrow\quad 5c \le 149 \quad\Longrightarrow\quad c \le 29.$$
With $c = 29$ we get $w = 17$ and $4$ blanks, all nonnegative, and $4\cdot 29 - 17 = 99$.

The answer is $\boxed{\textbf{(C)}\ 29}$.