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

(Solution 1)
(Solution 1)
Line 14: Line 14:
 
<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>
 
<asy>
Line 22: Line 22:
 
label("$A$", (0,5), N);
 
label("$A$", (0,5), N);
 
label("$B$", (5,5), N);
 
label("$B$", (5,5), N);
label("$169$", (2.5, -0.5), N);
+
label("$99$", (2.5, -0.5), N);
label("$149$", (-2.5,-0.5), N);
+
label("$50$", (-2.5,-0.5), N);
label("$119$", (7.5, -0.5), N);
+
label("$20$", (7.5, -0.5), N);
 
</asy>
 
</asy>
  

Revision as of 14:13, 10 November 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

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

[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]

(Guys I tried)


                                             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.
  • $B + C = 79, B + D = 49, \text{so} C = 50, D = 20, A = 99.$

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

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