Vegtown elections

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 $$\begin{array}{rl}x+y-(x+0.1y)=& 100-46\\ \Rightarrow x+y-x-0.1y=& 54\\ \Rightarrow 0.9y=& 54\\ \Rightarrow y=& 54÷0.9=60\end{array}$$ 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 $$\begin{array}{rl}x+0.1(100-x)& =46\\ \Rightarrow x+10-0.1x& =46\\ \Rightarrow 0.9x& =36\\ \Rightarrow x& =36÷0.9=40\end{array}$$