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

m (Solution 1)
(Solution 3)
(18 intermediate revisions by 16 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>
  
==Solution 1==
+
==Solutions==
We can see that this is a Venn Diagram Problem.[SOMEBODY DRAW IT PLEASE]
+
 
 +
===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.  
 
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.  
Line 14: Line 18:
 
<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>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>
+
<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 -->
 +
 
 +
<!--made into comment because there is a venn diagram available now-->
 +
 
 +
===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,</math> so <math>C = 50, D = 20, A = 99.</math>
 +
 
 +
The answer is <math>\boxed{\textbf{(D)}~99}</math>.
 +
 
 +
===Solution 4 (PIE)===
 +
 
 +
Using [[PIE]] (Principle of Inclusion-Exclusion), we find that the students who voted in favor of both issues are <math>149+119+29-198=\boxed{\textbf{(D)}~99}</math>.
 +
 
 +
~MrThinker
 +
 
 +
===Video Solution===
 +
https://youtu.be/OOdK-nOzaII?t=827
 +
 
 +
https://youtu.be/ATpixMaV-z4
  
==Solution 2==
+
~savannahsolver
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>
 
  
 
==See Also==
 
==See Also==

Revision as of 14:37, 2 November 2022

Problem

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?

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

Solutions

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

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

[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]



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

Solution 4 (PIE)

Using PIE (Principle of Inclusion-Exclusion), we find that the students who voted in favor of both issues are $149+119+29-198=\boxed{\textbf{(D)}~99}$.

~MrThinker

Video Solution

https://youtu.be/OOdK-nOzaII?t=827

https://youtu.be/ATpixMaV-z4

~savannahsolver

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