Difference between revisions of "2015 AMC 8 Problems/Problem 15"
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==Solution 2== | ==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 <math>\boxed{\textbf{(D)}~99}</math> | + | 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> |
==Solution 3== | ==Solution 3== | ||
Revision as of 22:11, 28 July 2019
At Euler Middle School, 
students voted on two issues in a school referendum with the following results: 
voted in favor of the first issue and 
voted in favor of the second issue. If there were exactly 
students who voted against both issues, how many students voted in favor of both issues?
Contents
Solution 1
We can see that this is a Venn Diagram Problem.[SOMEBODY DRAW IT PLEASE]
First, we analyze the information given. There are students. Let's use A as the first issue and B as the second issue.
students were for the A, and students were for B. There were also students against both A and B.
Solving this without a Venn Diagram, we subtract away from the total, . Out of the remaining 
, we have people for A and
people for B. We add this up to get 
. Since that is more than what we need, we subtract from to get
Solution 2
There are people. We know that people voted against both the first issue and the second issue. That leaves us with people that voted for at least one of them. If people voted for both of them, then that would leave 
people out of the vote, because is less than people. 
is , so to make it even, we have to take away from the people, which leaves us with
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 .
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
