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

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==Problem==
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== Problem ==
  
 
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?
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<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>
  
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==Solutions==
 
  
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===Solution 1===
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== Solution 1 ==
 +
 
 +
Let:
 +
* <math> A </math> be the number of students who voted in favor of the first issue,
 +
* <math> B </math> be the number of students who voted in favor of the second issue,
 +
* <math> A \cap B </math> be the number of students who voted in favor of both issues.
 +
 
 +
We are given:
 +
* <math> A = 149 </math>
 +
* <math> B = 119 </math>
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* 29 students voted against both issues, so the number of students who voted for at least one issue is:
 +
<cmath>
 +
198 - 29 = 169
 +
</cmath>
 +
 
 +
By the principle of inclusion and exclusion:
 +
<cmath>
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A \cup B = A + B - A \cap B
 +
</cmath>
 +
Substitute known values:
 +
<cmath>
 +
169 = 149 + 119 - A \cap B
 +
</cmath>
 +
<cmath>
 +
169 = 268 - A \cap B
 +
</cmath>
 +
<cmath>
 +
A \cap B = 268 - 169 = \boxed{\textbf{(D) }99}
 +
</cmath>
 +
 
 +
== Solution 2 ==
 +
 
 
We can see that this is a Venn Diagram Problem.
 
We can see that this is a Venn Diagram Problem.
  
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</asy>
 
</asy>
  
 +
Note: One could use the Principle of Inclusion-Exclusion in a similar way to achieve the same result.
  
 +
~ cxsmi (note)
  
 
   <!--(to editors: this looks really weird)Venn Diagram (I couldn't make circles),
 
   <!--(to editors: this looks really weird)Venn Diagram (I couldn't make circles),
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<!--made into comment because there is a venn diagram available now-->
 
<!--made into comment because there is a venn diagram available now-->
  
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===Solution 2===
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== Solution 3 ==
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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 who 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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 +
There are <math>198-149=49</math> students who voted against the first issue. Out of these, <math>49-29=20</math> voted for the second issue. We also find that <math>50</math> students voted against the second issue but in favor of the first in a similar way. This means that <math>149-50=119-20=\boxed{\textbf{(D)}~99}</math> students voted in favor of both issues.
 +
 
  
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===Solution 3===
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== Video Solutions ==
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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 by OmegaLearn ===
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* <math>A + B + C + D = 198</math>.
 
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* <math>B = 29</math>.
 
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* <math>A + C = 149</math> students voted in favor of the first issue.
 
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* <math>A + D = 119</math> students voted in favor of the second issue.
 
  
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We can quickly find that:
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[//www.youtube.com/watch?v=OOdK-nOzaII&t=829s www.youtube.com/watch?v=OOdK-nOzaII&t=829s]
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* <math>198 - 119 = 79</math> students voted against the second issue.
 
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* <math>198 - 149 = 49</math> students voted against the first issue.  
 
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* <math>B + C = 79, B + D = 49,</math> so <math>C = 50, D = 20, A = 99.</math>
 
  
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The answer is <math>\boxed{\textbf{(D)}~99}</math>.
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~ pi_is_3.14
  
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===Solution 4 (PIE)===
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=== Video Solution (HOW TO THINK CRITICALLY!!!) ===
  
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Using [[PIE]] (Principle of Inclusion-Exclusion), we find that the students who voted in favor of both issues are <math>149+119+29-198=\boxed{\textbf{(D)}~99}</math>.
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https://youtu.be/skOXiXCZVK0
  
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~MrThinker
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~ Education, the Study of Everything
  
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===Video Solution===
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=== Video Solution 4 ===
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https://youtu.be/OOdK-nOzaII?t=827
 
  
 
https://youtu.be/ATpixMaV-z4
 
https://youtu.be/ATpixMaV-z4
  
−
~savannahsolver
+
~ savannahsolver
  
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==See Also==
 
  
 +
== See Also ==
 
{{AMC8 box|year=2015|num-b=14|num-a=16}}
 
{{AMC8 box|year=2015|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
[[Category:Introductory Algebra Problems]]

Latest revision as of 09:29, 10 August 2025

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?

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


Solution 1

Let:

  • $A$ be the number of students who voted in favor of the first issue,
  • $B$ be the number of students who voted in favor of the second issue,
  • $A \cap B$ be the number of students who voted in favor of both issues.

We are given:

  • $A = 149$
  • $B = 119$
  • 29 students voted against both issues, so the number of students who voted for at least one issue is:

\[198 - 29 = 169\]

By the principle of inclusion and exclusion: \[A \cup B = A + B - A \cap B\] Substitute known values: \[169 = 149 + 119 - A \cap B\] \[169 = 268 - A \cap B\] \[A \cap B = 268 - 169 = \boxed{\textbf{(D) }99}\]

Solution 2

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)


Solution 3

There are $198-149=49$ students who voted against the first issue. Out of these, $49-29=20$ voted for the second issue. We also find that $50$ students voted against the second issue but in favor of the first in a similar way. This means that $149-50=119-20=\boxed{\textbf{(D)}~99}$ students voted in favor of both issues.


Video Solutions

Video Solution by OmegaLearn

www.youtube.com/watch?v=OOdK-nOzaII&t=829s

~ pi_is_3.14

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/skOXiXCZVK0

~ Education, the Study of Everything

Video Solution 4

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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