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

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==Solution 2==
 
==Solution 2==
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There are 119 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 198 people. 198-149 is 20, so to make it even, we have to take 20 away from the 119 people, which leaves us with
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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 198 people. 198-149 is 20, so to make it even, we have to take 20 away from the 119 people, which leaves us with <math>\boxed{\textbf{(D)}~99}</math>
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<math>\boxed{\textbf{(D)}~99}</math>
 
  
 
==See Also==
 
==See Also==

Revision as of 22:57, 7 November 2016

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?

$\textbf{(A) }49\qquad\textbf{(B) }70\qquad\textbf{(C) }79\qquad\textbf{(D) }99\qquad \textbf{(E) }149$

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

$\boxed{\textbf{(D)}~99}$

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 198 people. 198-149 is 20, so to make it even, we have to take 20 away from the 119 people, which leaves us with $\boxed{\textbf{(D)}~99}$

See Also

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

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