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

(Solution 3)
m (Solution 1)
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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.  
 
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  
 
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  

Revision as of 09:46, 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 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 $169$ 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

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
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

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