# 2015 AMC 8 Problems/Problem 15

## 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 (HOW TO THINK CRITICALLY!!!)

~Education, the Study of Everything

### Video Solution

~savannahsolver

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