# 1987 AJHSME Problems/Problem 24

## Problem

A multiple choice examination consists of $20$ questions. The scoring is $+5$ for each correct answer, $-2$ for each incorrect answer, and $0$ for each unanswered question. John's score on the examination is $48$. What is the maximum number of questions he could have answered correctly? $\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 11 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 16$

## Solution

### Solution 1

Let $c$ be the number of questions correct, $w$ be the number of questions wrong, and $b$ be the number of questions left blank. We are given that \begin{align} c+w+b &= 20 \\ 5c-2w &= 48 \end{align}

Adding equation $(2)$ to double equation $(1)$, we get $$7c+2b=88$$

Since we want to maximize the value of $c$, we try to find the largest multiple of $7$ less than $88$. This is $84=7\times 12$, so let $c=12$. Then we have $$7(12)+2b=88\Rightarrow b=2$$

Finally, we have $w=20-12-2=6$. We want $c$, so the answer is $12$, or $\boxed{\text{D}}$.

### Solution 2

If John answered 16 questions correctly, then he answered at most 4 questions incorrectly, giving him at least $16 \cdot 5 - 4 \cdot 2 = 72$ points. Therefore, John did not answer 16 questions correctly. If he answered 12 questions correctly and 6 questions incorrectly (leaving 2 questions unanswered), then he scored $12 \cdot 5 - 6 \cdot 2 = 48$ points. As all other options are less than 12, we conclude that 12 is the most questions John could have answered correctly, and the answer is $\boxed{\text{D}}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 