Difference between revisions of "2017 AMC 8 Problems/Problem 2"
Nukelauncher (talk | contribs) (→Solution) |
Nukelauncher (talk | contribs) m (→Solution) |
||
| Line 8: | Line 8: | ||
==Solution== | ==Solution== | ||
| − | Let x 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> | + | 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> |
~pegasuswa | ~pegasuswa | ||
Revision as of 14:21, 22 November 2017
Problem 2
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?
![[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
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 
~pegasuswa
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. 
