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

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Since we want to maximize the value of <math>c</math>, we try to find the largest multiple of <math>7</math> less than <math>88</math>.  This is <math>84=7\times 12</math>, so let <math>c=12</math>.  Then we have <cmath>7(12)+2b=88\Rightarrow b=2</cmath>  
 
Since we want to maximize the value of <math>c</math>, we try to find the largest multiple of <math>7</math> less than <math>88</math>.  This is <math>84=7\times 12</math>, so let <math>c=12</math>.  Then we have <cmath>7(12)+2b=88\Rightarrow b=2</cmath>  
  
Finally, we have <math>b=20-12-2=6</math>.  We want <math>c</math>, so the answer is <math>12</math>, or <math>\boxed{\text{D}}</math>.
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Finally, we have <math>w=20-12-2=6</math>.  We want <math>c</math>, so the answer is <math>12</math>, or <math>\boxed{\text{D}}</math>.
 
 
(As a side note, the extra work done at the end was to make sure the given situation is possible for c=12)
 
  
 
==See Also==
 
==See Also==
  
[[1987 AJHSME Problems]]
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{{AJHSME box|year=1987|num-b=23|num-a=25}}
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[[Category:Introductory Algebra Problems]]

Revision as of 08:02, 31 May 2009

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 (ProblemsAnswer KeyResources)
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