Difference between revisions of "2017 AMC 8 Problems/Problem 2"
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| − | + | ==Problem== | |
| − | ~ | + | Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received <math>36</math> votes, then how many votes were cast all together? |
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| + | <asy> draw((-1,0)--(0,0)--(0,1)); draw((0,0)--(0.309, -0.951)); filldraw(arc((0,0), (0,1), (-1,0))--(0,0)--cycle, lightgray); filldraw(arc((0,0), (0.309, -0.951), (0,1))--(0,0)--cycle, gray); draw(arc((0,0), (-1,0), (0.309, -0.951))); label("Colby", (-0.5, 0.5)); label("25\%", (-0.5, 0.3)); label("Alicia", (0.7, 0.2)); label("45\%", (0.7, 0)); label("Brenda", (-0.5, -0.4)); label("30\%", (-0.5, -0.6)); </asy> | ||
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| + | <math>\textbf{(A) }70 \qquad \textbf{(B) }84 \qquad \textbf{(C) }100 \qquad \textbf{(D) }106 \qquad \textbf{(E) }120</math> | ||
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| + | ==Solution 1== | ||
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| + | Let <math>x</math> be the total amount of votes casted. From the chart, Brenda received <math>30\%</math> of the votes and had <math>36</math> votes. We can express this relationship as <math>\frac{30}{100}x=36</math>. Solving for <math>x</math>, we get <math>x=\boxed{\textbf{(E)}\ 120}.</math> | ||
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| + | ==Solution 2== | ||
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| + | We're being asked for the total number of votes cast -- that represents <math>100\%</math> of the total number of votes. Brenda received <math>36</math> votes, which is <math>\frac{30}{100} = \frac{3}{10}</math> of the total number of votes. Multiplying <math>36</math> by <math>\frac{10}{3},</math> we get the total number of votes, which is <math>\boxed{\textbf{(E)}\ 120}.</math> | ||
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| + | ==Solution 3== | ||
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| + | If <math>36</math> votes is <math>\frac{3}{10}</math> of all the votes, we can divide that by <math>3</math> to get <math>12</math> as 10%, and then we can multiply the <math>12</math> by <math>10</math> to get to <math>120</math>. So, the answer is <math>\boxed{\textbf{(E)}\ 120}.</math> | ||
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| + | ~AllezW | ||
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| + | ==Video Solution (CREATIVE THINKING!!!)== | ||
| + | https://youtu.be/WgCRI4xaSTI | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/cY4NYSAD0vQ | ||
| + | |||
| + | https://youtu.be/-YMInDAHjcg | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2017|num-b=1|num-a=3}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 16:40, 25 May 2024
Contents
Problem
Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received 
votes, then how many votes were cast all together?
![[asy] draw((-1,0)--(0,0)--(0,1)); draw((0,0)--(0.309, -0.951)); filldraw(arc((0,0), (0,1), (-1,0))--(0,0)--cycle, lightgray); filldraw(arc((0,0), (0.309, -0.951), (0,1))--(0,0)--cycle, gray); draw(arc((0,0), (-1,0), (0.309, -0.951))); label("Colby", (-0.5, 0.5)); label("25\%", (-0.5, 0.3)); label("Alicia", (0.7, 0.2)); label("45\%", (0.7, 0)); label("Brenda", (-0.5, -0.4)); label("30\%", (-0.5, -0.6)); [/asy]](http://latex.artofproblemsolving.com/7/b/7/7b7b4b1940b8afd1669156f092481dd6fc6f5a83.png)
Solution 1
Let 
be the total amount of votes casted. From the chart, Brenda received 
of the votes and had votes. We can express this relationship as 
. Solving for , we get 
Solution 2
We're being asked for the total number of votes cast -- that represents 
of the total number of votes. Brenda received votes, which is 
of the total number of votes. Multiplying by 
we get the total number of votes, which is 
Solution 3
If votes is 
of all the votes, we can divide that by 
to get 
as 10%, and then we can multiply the by 
to get to 
. So, the answer is
~AllezW
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
| 2017 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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. 
