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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>.
 +
 +
== Solution 2 ==
 +
We can plug in each answer choice and find our what is the greatest one that works. We start of with Answer Choice <math>E</math> since it is the largest one.
 +
 +
E is right, Jesse got <math>33</math> questions right.
 +
 +
If Jesse got <math>33</math> questions right, then he gains <math>132</math> points, he then needs to get another <math>33</math> wrong to achieve a score of <math>99</math>. However, this is impossible as the test only contains <math>50</math> questions, and he needs <math>66</math> questions in order to achieve this.
 +
 +
D is right, Jesse got <math>31</math> questions right.
 +
 +
If Jesse got <math>31</math> questions right, then he gains <math>124</math> points, he then needs to get another <math>25</math> wrong in order to achieve a score of <math>99</math>. However, this is impossible as the test only contains <math>50</math> questions, and he needs <math>56</math> questions in order to achieve this.
 +
 +
C is right, Jesse got <math>29</math> questions right.
 +
 +
If Jesse got <math>29</math> questions right, then he gains <math>116</math> points, he then needs to get another <math>17</math> wrong in order to achieve a score of <math>99</math>. This is possible since this only requires <math>46</math> questions. The other <math>4</math> questions remain blank and earn him <math>0</math> points.
 +
 +
- SAMANTAP
 +
 +
==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}}
 
{{MAA Notice}}

Latest revision as of 22:07, 10 November 2024

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 $\boxed{\textbf{(C)}\ 29}$.

Solution 2

We can plug in each answer choice and find our what is the greatest one that works. We start of with Answer Choice $E$ since it is the largest one.

E is right, Jesse got $33$ questions right.

If Jesse got $33$ questions right, then he gains $132$ points, he then needs to get another $33$ wrong to achieve a score of $99$. However, this is impossible as the test only contains $50$ questions, and he needs $66$ questions in order to achieve this.

D is right, Jesse got $31$ questions right.

If Jesse got $31$ questions right, then he gains $124$ points, he then needs to get another $25$ wrong in order to achieve a score of $99$. However, this is impossible as the test only contains $50$ questions, and he needs $56$ questions in order to achieve this.

C is right, Jesse got $29$ questions right.

If Jesse got $29$ questions right, then he gains $116$ points, he then needs to get another $17$ wrong in order to achieve a score of $99$. This is possible since this only requires $46$ questions. The other $4$ questions remain blank and earn him $0$ points.

- SAMANTAP

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

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