2017 AMC 8 Problems/Problem 2

Revision as of 09:32, 23 August 2019 by Odinthunder (talk | contribs) (Solution 1)

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 1

Let $DUBS$ 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 $x=\boxed{\textbf{(E)}\ 120}.$

Solution 2

We're being asked for the total number of votes cast -- that represents $100\%$ of the total number of votes (obviously). Brenda received $36$ votes, which is $\frac{30}{100} = \frac{3}{10}$ of the total number of votes. Multiplying $36$ by $\frac{10}{3},$ we get the total number of votes, which is $\boxed{\textbf{(E)}\ 120}.$

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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