# Percentages

• July 4th 2012, 05:35 AM
lm320
Percentages
Hello all

I have another percentages question that I think that you could help me with as it looks fairly basic but I am having trouble understanding the logic behind it.

Q. What percentage of total eligible adults in the 18–24 age category turned out to vote in Thundersley?

a) 3.8% b) 6% c) 9.5% d) 12.8% e) 15%

All the information is below in the table.

Attachment 24220
Any help would be great.

Cheers!
• July 4th 2012, 06:03 AM
HallsofIvy
Re: Percentages
What exactly was your difficulty with this? The table says that 6% of the 18-24 voters turned out to vote in Thundersley.
• July 4th 2012, 06:10 AM
emakarov
Re: Percentages
Quote:

Originally Posted by HallsofIvy
The table says that 6% of the 18-24 voters turned out to vote in Thundersley.

No, I think the table says that 6% of those who turned out to vote in Thundersley were in the 18-24 age category. Since we don't know from the table which share of the total population constitutes this age category, I don't see how we can answer the question.
• July 4th 2012, 07:19 AM
lm320
Re: Percentages
Hi, I have just found the answer booklet and it suggests that the answer is A 3.8%. Does anybody know how they got this number? All the information is above.

Many thanks
• July 4th 2012, 07:34 AM
daigo
Re: Percentages
Well what I did was: in Thundersley, subtract the percentage of voters who did not vote from the total eligible, so 22,397 minus 36% of 22,397 is 14,334.08 voters who actually voted.

Then 18-24 year olds were 6% of the total voters who did vote so the total amount of 18-24 year olds who voted is 6% of the 14,334.08 voters who did vote, which is 860.0448 18-24 year old voters.

So what percent of all eligible voters (including ones who didn't vote) in Thundersley (22,397) were 18-24 year olds who did vote (860.0448 as we found earlier)?

(x% of total possible voters 22,397 = 860.0448 18-24 year old voters)
• July 4th 2012, 07:36 AM
ReallyStupid
Re: Percentages
Out of the Adults 36% did not vote at all

100% of all elidgable voter munus that figure = 64% voted

6 percent of 64% = 3.84
• July 4th 2012, 08:45 AM
emakarov
Re: Percentages
To avoid confusion, let's introduce some notation. We are talking about Thundersley only in this question. Let $v_{\text{ac}}$ be the number of people in the age category ac who voted, and let $e_{\text{ac}}$ be the number of eligible voters in the age category ac. For example, $v_{\text{18-24}}$ is the number of people between 18 and 24 who voted, and $e_{\text{18-24}}$ is the number of people between 18 and 24 who could have voted (maybe they did, maybe they didn't). Obviously, $v_{\text{ac}}\le e_{\text{ac}}$ for each age category ac.

Also, let

$e = e_{\text{18-24}} + e_{\text{25-34}} + e_{\text{35-44}}+e_{\text{45-54}}+e_{\text{55-74}}+e_{\text{75+}}$

be the total number of eligible voters and let

$v=v_{\text{18-24}} + e_{\text{25-34}} + v_{\text{35-44}}+v_{\text{45-54}}+v_{\text{55-74}}+v_{\text{75+}}$

be the total number of people who voted. Then the table provides the following facts:

$e=22,397$

$v_{\text{18-24}} / e = 6\%$

$v_{\text{25-34}} / e = 12\%$

...

$v_{\text{75+}} / e = 4\%$

$(e - v) / e = 36\%$

Since we know $e$, we can find $v_{\text{ac}}$ for each category ac. However, we can't find $e_{\text{ac}}$ for each ac because we don't know how people who didn't vote are distributed over the age categories. One possible scenario is that all eligible voters who did not vote are between 18 and 24. Then $e_{\text{18-24}} = v_{\text{18-24}} + (e - v) = (6 + 36)\%$ of $e$, i.e., $e_{\text{18-24}}=0.42\cdot22,397=9,407$ people. Another scenario is that all eligible voters who did not vote are between 25 and 34. Then everybody who is eligible between 18 and 24 voted, i.e., $e_{\text{18-24}} = v_{\text{18-24}} = 6\%$ of $e$, i.e., $e_{\text{18-24}}=0.06\cdot 22,397=1,344$ people.

As far as I understand the question:
Quote:

Originally Posted by lm320
Q. What percentage of total eligible adults in the 18–24 age category turned out to vote in Thundersley?

it asks to find $v_{\text{18-24}}/e_{\text{18-24}}$. I don't think this question is answerable. Indeed, in the first scenario above, $v_{\text{18-24}}/e_{\text{18-24}}=6/42=14\%$, while in the second scenario $v_{\text{18-24}}/e_{\text{18-24}}=100\%$.

I found the book "How to Pass Numerical Reasoning Tests" by Heidi Smith in Google Books. The answer 3.6% to this question appears on p. 190. It says,
Quote:

The percentage of all voting adults in Thundersley = 64%. 6% of 64% = 0.06 x 64 = 3.8% of eligible adults who voted were aged 18-24.
However, this number is $v_{\text{18-24}}/v$ and not $v_{\text{18-24}}/e_{\text{18-24}}$, for which the original question is asking.

Quote:

Originally Posted by daigo
Well what I did was: in Thundersley, subtract the percentage of voters who did not vote from the total eligible, so 22,397 minus 36% of 22,397 is 14,334.08 voters who actually voted.

Then 18-24 year olds were 6% of the total voters who did vote so the total amount of 18-24 year olds who voted is 6% of the 14,334.08 voters who did vote, which is 860.0448 18-24 year old voters.

No, 18-24 year olds were 6% of the total eligible voters.

Quote:

Originally Posted by daigo
So what percent of all eligible voters (including ones who didn't vote) in Thundersley (22,397) were 18-24 year olds who did vote (860.0448 as we found earlier)?

The answer to this question is given in the table: 6%.
• July 4th 2012, 08:56 AM
daigo
Re: Percentages
Quote:

Originally Posted by emakarov
No, 18-24 year olds were 6% of the total eligible voters.

I didn't think the graph clearly stated that, so I guess I read the graph incorrectly. Since all the voters in Thundersley add up to 64% (6% + 12% + 13% + 13% + 16% + 4%), and the people who didn't vote make up the remaining 36%, it would seem like the 6% of the total 100% which include those who didn't vote (6% + 12% + 13% + 13% + 16% + 4% + 36%) were 18-24 year olds.

Quote:

Originally Posted by emakarov
The answer to this question is given in the table: 6%.

See my reasoning above
• July 4th 2012, 09:02 AM
emakarov
Re: Percentages
Quote:

Originally Posted by daigo
Since all the voters in Thundersley add up to 64% (6% + 12% + 13% + 13% + 16% + 4%), and the people who didn't vote make up the remaining 36%, it would seem like the 6% of the total 100% which include those who didn't vote (6% + 12% + 13% + 13% + 16% + 4% + 36%) were 18-24 year olds.

Yes, the fact that the percentages in the Thundersley row of the table add up to 100% made me also conclude that the first number is $v_{\text{18-24}}/e$, not $v_{\text{18-24}}/v$.