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

Latest revision as of 03:39, 14 December 2024

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

Negative marking will only decrease his score in even amounts. This means that immediately, we do not need to test options $A$ or $C$ . The lowest score that can be obtained from option $E$ is 72, which does not work. The lowest score that can be obtained from option $D$ is 44, which clearly means we can obtain a score of 48 this way, so the answer chosen is $\boxed{\text{D}}$ .

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 22: / Line 22: @@
 ===Solution 2===
-If John answered 16 questions correctly, then he answered at most 4 questions incorrectly, giving him at least <math>16 \cdot 5 - 4 \cdot 2 = 72</math> 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 <math>12 \cdot 5 - 6 \cdot 2 = 48</math> 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 <math>\boxed{\text{D}}</math>.
+Negative marking will only decrease his score in even amounts. This means that immediately, we do not need to test options <math>A</math> or <math>C</math>. The lowest score that can be obtained from option <math>E</math> is 72, which does not work. The lowest score that can be obtained from option <math>D</math> is 44, which clearly means we can obtain a score of 48 this way, so the answer chosen is <math>\boxed{\text{D}}</math>.
 ==See Also==
+[[2010 AMC 10B Problems/Problem 15|2010 AMC10B Problem 15]]
 {{AJHSME box|year=1987|num-b=23|num-a=25}}
 [[Category:Introductory Algebra Problems]]
 {{MAA Notice}}

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