Difference between revisions of "2010 AMC 10B Problems/Problem 15"
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== Solution == | == Solution == | ||
Let <math>a</math> be the amount of questions Jesse answered correctly, <math>b</math> be the amount of questions Jesse left blank, and <math>c</math> be the amount of questions Jesse answered incorrectly. Since there were <math>50</math> questions on the contest, <math>a+b+c=50</math>. Since his total score was <math>99</math>, <math>4a-c=99</math>. Also, <math>a+c\leq50 \Rightarrow c\leq50-a</math>. We can substitute this inequality into the previous equation to obtain another inequality: <math>4a-(50-a)\leq99 \Rightarrow 5a\leq149 \Rightarrow a\leq \frac{149}5=29.8</math>. Since <math>a</math> is an integer, the maximum value for <math>a</math> is <math>\boxed{\textbf{(C)}\ 29}</math>. | Let <math>a</math> be the amount of questions Jesse answered correctly, <math>b</math> be the amount of questions Jesse left blank, and <math>c</math> be the amount of questions Jesse answered incorrectly. Since there were <math>50</math> questions on the contest, <math>a+b+c=50</math>. Since his total score was <math>99</math>, <math>4a-c=99</math>. Also, <math>a+c\leq50 \Rightarrow c\leq50-a</math>. We can substitute this inequality into the previous equation to obtain another inequality: <math>4a-(50-a)\leq99 \Rightarrow 5a\leq149 \Rightarrow a\leq \frac{149}5=29.8</math>. Since <math>a</math> is an integer, the maximum value for <math>a</math> is <math>\boxed{\textbf{(C)}\ 29}</math>. | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/vYXz4wStBUU?t=549 | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2010|ab=B|num-b=14|num-a=16}} | {{AMC10 box|year=2010|ab=B|num-b=14|num-a=16}} | ||
+ | {{MAA Notice}} |
Latest revision as of 02:44, 26 September 2020
Contents
Problem
On a -question multiple choice math contest, students receive
points for a correct answer,
points for an answer left blank, and
point for an incorrect answer. Jesse’s total score on the contest was
. What is the maximum number of questions that Jesse could have answered correctly?
Solution
Let be the amount of questions Jesse answered correctly,
be the amount of questions Jesse left blank, and
be the amount of questions Jesse answered incorrectly. Since there were
questions on the contest,
. Since his total score was
,
. Also,
. We can substitute this inequality into the previous equation to obtain another inequality:
. Since
is an integer, the maximum value for
is
.
Video Solution
https://youtu.be/vYXz4wStBUU?t=549
~IceMatrix
See Also
2010 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.