Difference between revisions of "1987 AJHSME Problems/Problem 24"

Revision as of 22:54, 4 July 2013

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 $\boxed{\text{D}}$ .

1987 AJHSME (Problems • Answer Key • Resources)
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
All AJHSME/AMC 8 Problems and Solutions

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

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 {{AJHSME box|year=1987|num-b=23|num-a=25}}
 [[Category:Introductory Algebra Problems]]
+{{MAA Notice}}

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Difference between revisions of "1987 AJHSME Problems/Problem 24"

Revision as of 22:54, 4 July 2013

Problem

Solution

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