Difference between revisions of "2017 AMC 8 Problems/Problem 2"
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− | ==Problem | + | ==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 36 votes, then how many votes were cast all together? | + | 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? |
<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> | <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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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> | 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> | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ~AllezW | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING!!!)== | ||
+ | https://youtu.be/WgCRI4xaSTI | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/cY4NYSAD0vQ | https://youtu.be/cY4NYSAD0vQ | ||
+ | |||
+ | https://youtu.be/-YMInDAHjcg | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
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?
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.