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 ==
 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>
-==Solution 1==
-We can see that this is a Venn Diagram Problem.
+== 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 14: / Line 48: @@
 <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>\boxed{\textbf{(D)}~99}</math>
+<math>\boxed{\textbf{(D)}~99}</math>.
+<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)
+  <!--(to editors: this looks really weird)Venn Diagram (I couldn't make circles),
+                                               We need to know how many voted in favor for both
+                                 Issue A               Against both issues        Issue B
+students                29 students          119 students
++29+119=297
+-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 ===
+[//www.youtube.com/watch?v=OOdK-nOzaII&t=829s 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
-==Solution 2==
+~ savannahsolver
-There are 198 people. We know that 29 people voted against both the first issue and the second issue. That leaves us with 169 people that voted for at least one of them. If 119 people voted for both of them, then that would leave 20 people out of the vote, because 149 is less than 198 people. 198-149 is 20, so to make it even, we have to take 20 away from the 119 people, which leaves us with <math>\boxed{\textbf{(D)}~99}</math>
-==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
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

Page

Toolbox

Search