# Vegtown Elections

What percentage of people voted for the Broccoli party in the Vegtown election?

## Problem

Elections in Vegtown were held recently.

46% of voters ate broccoli.

Everyone who voted for the Broccoli Party ate broccoli.

Of those who voted for other parties, 90% never ate broccoli.

What percentage of voters voted for the Broccoli Party?

This problem is taken from the UKMT Mathematical Challenges.

## Student Solutions

**Using a two-way table and ratios**

Ate broccoli | Never ate broccoli | |

Voted Broccoli party | a | b |

Voted for other parties | c | d |

b = 0

Ratio of c:d = 1:9

d = 54% of total

so c = 6% of total

so a must be the remaining 40% of total

So 40% voted for the Broccoli Party.

**Using a two-way table and equations**

The information can be shown in a two-way table, where $x\%$ of voters voted for the Broccoli party and $y\%$ of voters voted for other parties:

Image

From the 'Never eaten broccoli' column, $0+0.9y=54$, so $y=54\div0.9=60$.

$x+y=100$ so $x=40$. So $40\%$ of voters voted for the Broccoli party.

**Using a tree diagram**

Some of the information from the question is shown on the tree diagram below:

Image

$46\%$ of voters had eaten broccoli before, so $100\%$ of $x$ added to $10\%$ of $y$ is $46,$ so $x+0.1y=46$.

All voters either voted for the Broccoli party or for another party, so $x+y=100$.

__Solving by elimination__

Subtracting $x+0.1y=46$ from $x+y=100$ gives $$\begin{align}x+y-(x+0.1y)=&100-46\\

\Rightarrow x+y-x-0.1y=&54\\

\Rightarrow 0.9y=&54\\

\Rightarrow y=&54\div0.9=60\end{align}$$ So 60% of voters voted for other parties, so 40% of voters voted for the Broccoli party.

__Solving by substitution__

$x+y=100\Rightarrow y=100-x$. Substituting this into $x+0.1y=46$ gives $$\begin{align}x+0.1(100-x)&=46\\

\Rightarrow x+ 10 - 0.1x&=46\\

\Rightarrow 0.9x&=36\\

\Rightarrow x&=36\div0.9=40\end{align}$$