Difference between revisions of "2010 AMC 10B Problems/Problem 15"
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On a <math>50</math>-question multiple choice math contest, students receive <math>4</math> points for a correct answer, <math>0</math> points for an answer left blank, and <math>-1</math> point for an incorrect answer. Jesse’s total score on the contest was <math>99</math>. What is the maximum number of questions that Jesse could have answered correctly? | On a <math>50</math>-question multiple choice math contest, students receive <math>4</math> points for a correct answer, <math>0</math> points for an answer left blank, and <math>-1</math> point for an incorrect answer. Jesse’s total score on the contest was <math>99</math>. What is the maximum number of questions that Jesse could have answered correctly? | ||
| − | <math> | + | <math>\textbf{(A)}\ 25 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 29 \qquad \textbf{(D)}\ 31 \qquad \textbf{(E)}\ 33</math> |
| − | \ | ||
| − | \qquad | ||
| − | \ | ||
| − | \qquad | ||
| − | \ | ||
| − | \qquad | ||
| − | \ | ||
| − | \qquad | ||
| − | \ | ||
| − | </math> | ||
== 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{\ | + | 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>. |
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/vYXz4wStBUU?t=549 | ||
| + | |||
| + | ~IceMatrix | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC10 box|year=2010|ab=B|num-b=14|num-a=16}} | ||
| + | {{MAA Notice}} | ||
Revision as of 01:44, 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. 
