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

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

@@ Line 1: / Line 1: @@
-==Problem==
+== 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?
@@ Line 5: / Line 5: @@
 <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 ==
+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>
+* 29 students voted against both issues, so the number of students who voted for at least one issue is:
+<cmath>
+- 29 = 169
+</cmath>
+By the principle of inclusion and exclusion:
+<cmath>
+A \cup B = A + B - A \cap B
+</cmath>
+Substitute known values:
+<cmath>
+= 149 + 119 - A \cap B
+</cmath>
+<cmath>
+= 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.
 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
@@ Line 31: / Line 61: @@
 </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),
                                                 We need to know how many voted in favor for both
@@ Line 42: / Line 74: @@
 +29+119=297
--198=99 students in favor for both
+-198=99 students in favor for both -->
+<!--made into comment because there is a venn diagram available now-->
+== Solution 3 ==
+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.
+== Video Solutions ==
+=== Video Solution by OmegaLearn ===
-===Solution 2===
+[//www.youtube.com/watch?v=OOdK-nOzaII&t=829s www.youtube.com/watch?v=OOdK-nOzaII&t=829s]
-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>
-===Solution 3===
+~ pi_is_3.14
-Divide the students into four categories:
-* A. Students who voted in favor of both issues.
-* B. Students who voted against both issues.
-* C. Students who voted in favor of the first issue, and against the second issue.
-* D. Students who voted in favor of the second issue, and against the first issue.
-We are given that:
+=== Video Solution (HOW TO THINK CRITICALLY!!!) ===
-* <math>A + B + C + D = 198.</math>
-* <math>B = 29.</math>
-* <math>A + C = 149</math> students voted in favor of the first issue.
-* <math>A + D = 119</math> students voted in favor of the second issue.
-We can quickly find that:
+https://youtu.be/skOXiXCZVK0
-* <math>198 - 119 = 79</math> students voted against the second issue.
-* <math>198 - 149 = 49</math> students voted against the first issue.
-* <math>B + C = 79, B + D = 49,</math> so <math>C = 50, D = 20, A = 99.</math>
-The answer is <math>\boxed{\textbf{(D)}~99}</math>.
+~ Education, the Study of Everything
-===Video Solution===
+=== Video Solution 4 ===
-https://youtu.be/OOdK-nOzaII?t=827
 https://youtu.be/ATpixMaV-z4
-~savannahsolver
+~ savannahsolver
-==See Also==
+== See Also ==
 {{AMC8 box|year=2015|num-b=14|num-a=16}}
 {{MAA Notice}}
+[[Category:Introductory Algebra Problems]]

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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