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 $$\begin{align}x+0.1(100-x)&=46\\
\Rightarrow x+ 10 - 0.1x&=46\\
\Rightarrow 0.9x&=36\\
\Rightarrow x&=36\div0.9=40\end{align}$$