Difference between revisions of "2010 AMC 10B Problems/Problem 15"

Revision as of 12:01, 4 July 2013

Problem

On a $50$ -question multiple choice math contest, students receive $4$ points for a correct answer, $0$ points for an answer left blank, and $-1$ point for an incorrect answer. Jesse’s total score on the contest was $99$ . What is the maximum number of questions that Jesse could have answered correctly?

$\textbf{(A)}\ 25 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 29 \qquad \textbf{(D)}\ 31 \qquad \textbf{(E)}\ 33$

Solution

Let $a$ be the amount of questions Jesse answered correctly, $b$ be the amount of questions Jesse left blank, and $c$ be the amount of questions Jesse answered incorrectly. Since there were $50$ questions on the contest, $a+b+c=50$ . Since his total score was $99$ , $4a-c=99$ . Also, $a+c\leq50 \Rightarrow c\leq50-a$ . We can substitute this inequality into the previous equation to obtain another inequality: $4a-(50-a)\leq99 \Rightarrow 5a\leq149 \Rightarrow a\leq \frac{149}5=29.8$ . Since $a$ is an integer, the maximum value for $a$ is $\boxed{\textbf{(C)}\ 29}$ .

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AMC 10 Problems and Solutions

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

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 ==See Also==
 {{AMC10 box|year=2010|ab=B|num-b=14|num-a=16}}
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Difference between revisions of "2010 AMC 10B Problems/Problem 15"

Revision as of 12:01, 4 July 2013

Problem

Solution

See Also