Difference between revisions of "2015 AMC 8 Problems/Problem 15"
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| + | ==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 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> | ||
| + | |||
| + | Note: One could use the Principle of Inclusion-Exclusion in a similar way to achieve the same result. | ||
| + | |||
| + | ~ cxsmi (note) | ||
| + | |||
| + | <!--(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 --> | ||
| + | |||
| + | <!--made into comment because there is a venn diagram available now--> | ||
| + | |||
| + | ==Video Solution (HOW TO THINK CRITICALLY!!!)== | ||
| + | https://youtu.be/skOXiXCZVK0 | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | |||
| + | ===Video Solution=== | ||
| + | https://youtu.be/OOdK-nOzaII?t=827 | ||
| + | |||
| + | https://youtu.be/ATpixMaV-z4 | ||
| + | |||
| + | ~savannahsolver | ||
==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 00:41, 15 January 2024
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 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
.
![[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]](http://latex.artofproblemsolving.com/c/f/5/cf5b45df5f86a8c2b3ba4e2f566d64e6ff019761.png)
Note: One could use the Principle of Inclusion-Exclusion in a similar way to achieve the same result.
~ cxsmi (note)
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
Video Solution
https://youtu.be/OOdK-nOzaII?t=827
~savannahsolver
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. 
