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

Revision as of 16:08, 6 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]$

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 45: / Line 45: @@
 <!--made into comment because there is a venn diagram available now-->
-===Solution 2===
-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>.
-===Solution 3===
-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:
-* <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:
-* <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>.
-===Solution 4 (PIE)===
-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>.
-~MrThinker
 ==Video Solution (HOW TO THINK CRITICALLY!!!)==

Page

Toolbox

Search