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Difference between revisions of "2015 AMC 8 Problems/Problem 15"

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==Problem==
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At Euler Middle School, <math>198</math> students voted on two issues in a school referendum with the following results: <math>149</math> voted in favor of the first issue and <math>119</math> voted in favor of the second issue. If there were exactly <math>29</math> students who voted against both issues, how many students voted in favor of both issues?
 
At Euler Middle School, <math>198</math> students voted on two issues in a school referendum with the following results: <math>149</math> voted in favor of the first issue and <math>119</math> voted in favor of the second issue. If there were exactly <math>29</math> students who voted against both issues, how many students voted in favor of both issues?
  
 
<math>\textbf{(A) }49\qquad\textbf{(B) }70\qquad\textbf{(C) }79\qquad\textbf{(D) }99\qquad \textbf{(E) }149</math>
 
<math>\textbf{(A) }49\qquad\textbf{(B) }70\qquad\textbf{(C) }79\qquad\textbf{(D) }99\qquad \textbf{(E) }149</math>
 +
 +
==Solutions==
  
 
==Solution 1==
 
==Solution 1==
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We can see that this is a Venn Diagram Problem.[SOMEBODY DRAW IT PLEASE]
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We can see that this is a Venn Diagram Problem.
  
 
First, we analyze the information given. There are <math>198</math> students. Let's use A as the first issue and B as the second issue.  
 
First, we analyze the information given. There are <math>198</math> students. Let's use A as the first issue and B as the second issue.  
  
−
<math>149</math> students were for the A, and <math>119</math> students were for B. There were also <math>29</math> students against both A and B.  
+
<math>149</math> students were for A, and <math>119</math> students were for B. There were also <math>29</math> students against both A and B.  
  
 
Solving this without a Venn Diagram, we subtract <math>29</math> away from the total, <math>198</math>. Out of the remaining <math>169</math> , we have <math>149</math> people for A and  
 
Solving this without a Venn Diagram, we subtract <math>29</math> away from the total, <math>198</math>. Out of the remaining <math>169</math> , we have <math>149</math> people for A and  
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<math>119</math> people for B. We add this up to get <math>268</math> . Since that is more than what we need, we subtract <math>169</math> from <math>268</math> to get  
 
<math>119</math> people for B. We add this up to get <math>268</math> . Since that is more than what we need, we subtract <math>169</math> from <math>268</math> to get  
  
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<math>\boxed{\textbf{(D)}~99}</math>
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<math>\boxed{\textbf{(D)}~99}</math>.
  
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<asy>
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defaultpen(linewidth(0.7));
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draw(Circle(origin, 5));
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draw(Circle((5,0), 5));
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label("$A$", (0,5), N);
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label("$B$", (5,5), N);
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label("$99$", (2.5, -0.5), N);
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label("$50$", (-2.5,-0.5), N);
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label("$20$", (7.5, -0.5), N);
 +
</asy>
  
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Note: One could use the Principle of Inclusion-Exclusion in a similar way to achieve the same result.
  
 +
~ cxsmi (note)
  
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                                              Venn Diagram (I couldn't make circles)
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  <!--(to editors: this looks really weird)Venn Diagram (I couldn't make circles),
−
                                                
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                                               We need to know how many voted in favor for both
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                                          We need to know how many voted in favor for both
 
  
  
 
                                 Issue A              Against both issues        Issue B
 
                                 Issue A              Against both issues        Issue B
 
                                 149 students                29 students          119 students
 
                                 149 students                29 students          119 students
 +
  
 
                                                           149+29+119=297
 
                                                           149+29+119=297
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                                                 297-198=99 students in favor for both
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                                                 297-198=99 students in favor for both -->
 +
 
 +
<!--made into comment because there is a venn diagram available now-->
 +
 
 +
==Video Solution (HOW TO THINK CRITICALLY!!!)==
 +
https://youtu.be/skOXiXCZVK0
  
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==Solution 2==
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~Education, the Study of Everything
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There are <math>198</math> people. We know that <math>29</math> people voted against both the first issue and the second issue. That leaves us with <math>169</math> people that voted for at least one of them. If <math>119</math> people voted for both of them, then that would leave <math>20</math> people out of the vote, because <math>149</math> is less than <math>169</math> people. <math>169-149</math> is <math>20</math>, so to make it even, we have to take <math>20</math> away from the <math>119</math> people, which leaves us with <math>\boxed{\textbf{(D)}~99}</math>
 
  
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==Solution 3==
 
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Divide the students into four categories:
 
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* A. Students who voted in favor of both issues.
 
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* B. Students who voted against both issues.
 
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* C. Students who voted in favor of the first issue, and against the second issue.
 
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* D. Students who voted in favor of the second issue, and against the first issue.
 
  
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We are given that:
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===Video Solution===
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* A + B + C + D = 198.
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https://youtu.be/OOdK-nOzaII?t=827
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* B = 29.
 
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* A + C = 149 students voted in favor of the first issue.
 
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* A + D = 119 students voted in favor of the second issue.
 
  
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We can quickly find that:
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https://youtu.be/ATpixMaV-z4
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* 198 - 119 = 79 students voted against the second issue.
 
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* 198 - 149 = 49 students voted against the first issue.
 
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* B + C = 79, B + D = 49, so C = 50, D = 20, A = 99.
 
  
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The answer is <math>\boxed{\textbf{(D)}~99}</math>.
+
~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 00:41, 15 January 2024

Problem

At Euler Middle School, $198$ students voted on two issues in a school referendum with the following results: $149$ voted in favor of the first issue and $119$ voted in favor of the second issue. If there were exactly $29$ students who voted against both issues, how many students voted in favor of both issues?

$\textbf{(A) }49\qquad\textbf{(B) }70\qquad\textbf{(C) }79\qquad\textbf{(D) }99\qquad \textbf{(E) }149$

Solutions

Solution 1

We can see that this is a Venn Diagram Problem.

First, we analyze the information given. There are $198$ students. Let's use A as the first issue and B as the second issue.

$149$ students were for A, and $119$ students were for B. There were also $29$ students against both A and B.

Solving this without a Venn Diagram, we subtract $29$ away from the total, $198$. Out of the remaining $169$ , we have $149$ people for A and

$119$ people for B. We add this up to get $268$ . Since that is more than what we need, we subtract $169$ from $268$ to get

$\boxed{\textbf{(D)}~99}$.

[asy] defaultpen(linewidth(0.7)); draw(Circle(origin, 5)); draw(Circle((5,0), 5)); label("$A$", (0,5), N); label("$B$", (5,5), N); label("$99$", (2.5, -0.5), N); label("$50$", (-2.5,-0.5), N); label("$20$", (7.5, -0.5), N); [/asy]

Note: One could use the Principle of Inclusion-Exclusion in a similar way to achieve the same result.

~ cxsmi (note)


Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/skOXiXCZVK0

~Education, the Study of Everything


Video Solution

https://youtu.be/OOdK-nOzaII?t=827

https://youtu.be/ATpixMaV-z4

~savannahsolver

See Also

2015 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions

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