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

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==Problem==
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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?
 
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>
 
<math>\textbf{(A) }49\qquad\textbf{(B) }70\qquad\textbf{(C) }79\qquad\textbf{(D) }99\qquad \textbf{(E) }149</math>
  
==Solution==
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==Solutions==
We can see that this is a Venn Diagram Problem.  
+
 
 +
===Solution 1===
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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.  
 
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 the 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 <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{\text{(D)}~99}</math>
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 +
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  
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 +
<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  
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<math>\boxed{\textbf{(D)}~99}</math>.
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<asy>
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defaultpen(linewidth(0.7));
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draw(Circle(origin, 5));
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draw(Circle((5,0), 5));
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label("$A$", (0,5), N);
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label("$B$", (5,5), N);
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label("$99$", (2.5, -0.5), N);
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label("$50$", (-2.5,-0.5), N);
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label("$20$", (7.5, -0.5), N);
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</asy>
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 +
 
 +
 
 +
  (to editors: this looks really weird)Venn Diagram (I couldn't make circles),
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                                              We need to know how many voted in favor for both
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 +
 
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                                Issue A              Against both issues        Issue B
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                                149 students                29 students          119 students
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 +
 
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                                                          149+29+119=297
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                                                297-198=99 students in favor for both
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===Solution 2===
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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>
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===Solution 3===
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Divide the students into four categories:
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* A. Students who voted in favor of both issues.
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* B. Students who voted against both issues.
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* C. Students who voted in favor of the first issue, and against the second issue.
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* D. Students who voted in favor of the second issue, and against the first issue.
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We are given that:
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* <math>A + B + C + D = 198.</math>
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* <math>B = 29.</math>
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* <math>A + C = 149</math> students voted in favor of the first issue.
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* <math>A + D = 119</math> students voted in favor of the second issue.
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We can quickly find that:
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* <math>198 - 119 = 79</math> students voted against the second issue.
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* <math>198 - 149 = 49</math> students voted against the first issue.
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* <math>B + C = 79, B + D = 49, \text{so} C = 50, D = 20, A = 99.</math>
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The answer is <math>\boxed{\textbf{(D)}~99}</math>.
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 +
===Video Solution===
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https://youtu.be/OOdK-nOzaII?t=827
  
 
==See Also==
 
==See Also==

Latest revision as of 11:05, 10 July 2021

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 the 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]


 (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
                               149 students                29 students          119 students


                                                         149+29+119=297
                                               297-198=99 students in favor for both

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.
  • $B + C = 79, B + D = 49, \text{so} C = 50, D = 20, A = 99.$

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

Video Solution

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

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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