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

(Solution 2)
m (Solution 1)
Line 15: Line 15:
  
 
<math>\boxed{\textbf{(D)}~99}</math>
 
<math>\boxed{\textbf{(D)}~99}</math>
 +
 +
 +
 +
 +
                                              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==
 
==Solution 2==

Revision as of 18:15, 12 October 2019

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.[SOMEBODY DRAW IT PLEASE]

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

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



                                             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, so C = 50, D = 20, A = 99.

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS