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

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==Solution==
 
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===Solution 1===
  
 
Let <math>c</math> be the number of questions correct, <math>w</math> be the number of questions wrong, and <math>b</math> be the number of questions left blank.  We are given that  
 
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==See Also==
 
==See Also==
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[[2010 AMC 10B Problems/Problem 15|2010 AMC10B Problem 15]]
  
 
{{AJHSME box|year=1987|num-b=23|num-a=25}}
 
{{AJHSME box|year=1987|num-b=23|num-a=25}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
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Latest revision as of 08:56, 11 June 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 $\boxed{\text{D}}$.

See Also

2010 AMC10B Problem 15

1987 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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