Difference between revisions of "2015 AMC 8 Problems/Problem 15"
(23 intermediate revisions by 18 users not shown) | |||
Line 1: | Line 1: | ||
+ | ==Problem== | ||
+ | |||
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> | ||
+ | |||
+ | ==Solutions== | ||
+ | |||
+ | ===Solution 1=== | ||
+ | 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. | ||
+ | |||
+ | <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{\textbf{(D)}~99}</math>. | ||
+ | |||
+ | <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 <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=== | ||
+ | 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: | ||
+ | * <math>A + B + C + D = 198.</math> | ||
+ | * <math>B = 29.</math> | ||
+ | * <math>A + C = 149</math> students voted in favor of the first issue. | ||
+ | * <math>A + D = 119</math> students voted in favor of the second issue. | ||
+ | |||
+ | We can quickly find that: | ||
+ | * <math>198 - 119 = 79</math> students voted against the second issue. | ||
+ | * <math>198 - 149 = 49</math> students voted against the first issue. | ||
+ | * <math>B + C = 79, B + D = 49, \text{so} C = 50, D = 20, A = 99.</math> | ||
+ | |||
+ | The answer is <math>\boxed{\textbf{(D)}~99}</math>. | ||
+ | |||
+ | ===Video Solution=== | ||
+ | https://youtu.be/OOdK-nOzaII?t=827 | ||
==See Also== | ==See Also== | ||
− | {{AMC8 box|year=2015|num-b=14| | + | {{AMC8 box|year=2015|num-b=14|num-a=16}} |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 16:41, 16 January 2021
Contents
Problem
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?
Solutions
Solution 1
We can see that this is a Venn Diagram Problem.
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
.
(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 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:
- students voted in favor of the first issue.
- students voted in favor of the second issue.
We can quickly find that:
- students voted against the second issue.
- students voted against the first issue.
The answer is .
Video Solution
https://youtu.be/OOdK-nOzaII?t=827
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.