2017 AMC 8 Problems/Problem 2

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?

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

Solution 2

We're being asked for the total number of votes cast -- that represents $100\%$ of the total number of votes. 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 $\boxed{\textbf{(E)}\ 120}.$

Solution 3

If $36$ votes is $\frac{3}{10}$ of all the votes, we can divide that by $3$ to get $12$ as 10%, and then we can multiply the $12$ by $10$ to get to $120$. So, the answer is $\boxed{\textbf{(E)}\ 120}.$

~AllezW

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

2017 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png