2010 AMC 10B Problem 15

P11-15~3 minPrint

Linear Equations & Word Problems Diophantine Equations Bound the quantity above/below or estimate to pin it down Consider the largest/smallest element or boundary case

On a $50$ -question multiple choice math contest, students receive $4$ points for a correct answer, $0$ points for an answer left blank, and $-1$ point for an incorrect answer. Jesse’s total score on the contest was $99$ . What is the maximum number of questions that Jesse could have answered correctly?

Show solution without answering

Key insight

With c correct answers, Jesse needs 4c - 99 wrong ones to get 99 points. Correct plus wrong is at most 50, so c is at most 29.

Solution

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.

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

Correct	Wrong needed	Questions used
$25$	$1$	$26$
$27$	$9$	$36$
$29$	$17$	$46$
$30$	$21$	$51$
$31$	$25$	$56$

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 $\boxed{\textbf{(C)}\ 29}$ .

Why this works

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.

Alternative approach

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.

Concise solution

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

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 $\boxed{\textbf{(C)}\ 29}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Number Theory, Chapter 7: Algebra With Integers
Introduction to Algebra, Chapter 7: Proportion (§7.4)
Introduction to Algebra, Chapter 11: Special Factorizations (§11.5)