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1987 AJHSME Problems/Problem 24

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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

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}}.

See Also

1987 AJHSME (ProblemsResources)
Preceded by
Problem 23
Followed by
Problem 25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
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