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

Revision as of 08:47, 28 March 2023

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]$

Solution 2

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 who 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 $169$ people. $169-149$ is $20$ , so to make it even; we have to take $20$ away from the $119$ people, which leaves us with $\boxed{\textbf{(D)}~99}$ .

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:

$A + B + C + D = 198$ .
$B = 29$ .
$A + C = 149$ students voted in favor of the first issue.
$A + D = 119$ students voted in favor of the second issue.

We can quickly find that:

$198 - 119 = 79$ students voted against the second issue.
$198 - 149 = 49$ students voted against the first issue.
$B + C = 79, B + D = 49,$ so $C = 50, D = 20, A = 99.$

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

Solution 4 (PIE)

Using PIE (Principle of Inclusion-Exclusion), we find that the students who voted in favor of both issues are $149+119+29-198=\boxed{\textbf{(D)}~99}$ .

~MrThinker

Video Solution

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

https://youtu.be/ATpixMaV-z4

~savannahsolver

@@ Line 47: / Line 47: @@
 ===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 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>
+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===

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