Difference between revisions of "2017 AMC 8 Problems/Problem 2"

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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>
 
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
 
  
 
==See Also==
 
==See Also==

Revision as of 22:49, 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]

$\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }100\qquad\textbf{(D) }106\qquad\textbf{(E) }120$

Solution

Let $x$ be the total amount of votes casted. From the chart, Brenda received $30\%$ of the votes and had $36$ votes. We can express this relationship as $\frac{30}{100}x=36$. Solving for $x$, we get $x=\boxed{\textbf{(E)}\ 120}.$

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AJHSME/AMC 8 Problems and Solutions

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