Copyright © University of Cambridge. All rights reserved.
'Couples' printed from https://nrich.maths.org/
Good solutions were received from Luke from
St George's College, Ruth from Manchester High School for Girls,
Daniel from Junction City High School, Max from Bishop Ramsey C of
E Secondary School and Rohan from Long Bay Primary.
Luke wrote:
Since two thirds of the adult men are married, the total number of
men must be a multiple of 3.
Possible totals are:
Total Men |
Total Married Men |
3 |
2 |
6 |
4 |
9 |
6 |
12 |
8 |
Since three quarters of the adult women are married, the total
number of women must be a multiple of 4.
Possible totals are:
Total Women |
Total Married Women |
4 |
3 |
8 |
6 |
12 |
9 |
16 |
12 |
The smallest number of couples is 6, when there are 9 men and
8 women.
There would be 17 adults in the smallest community of this
type.
Ruth wrote:
There are $m$ men and $w$ women in the community. $\frac{2}{3}$ of
the men and $\frac{3}{4}$ of the women are married and there must
be the same number of married men as married women, so
\begin{eqnarray} \frac{2}{3}m &=& \frac{3}{4}w \\ 8m
&=& 9w \end{eqnarray} Now $m$ and $w$ are both integers so
the smallest solution to this equation is $$m = 9 , w = 8
$$so there are 17 people in the community, and 6
married couples.
There are other solutions gained by multiplying all the numbers by
a constant for example you can double them all to get 18 men, 16
women and 12 married couples but 9 men and 8 women is the smallest
solution.
Max wrote:
First of all, I needed the amount of men and women to be
married to be equal. Therefore, I needed to find the lowest common
numerator.
Taking $\frac{2}{3}$ and $\frac{3}{4}$ and converting them
into $\frac{6}{9}$ and $\frac{6}{8}$ shows us that the smallest
number of married men and women must be 6.
By adding both of the denominators together, I found that the
lowest number of people needed in the community must be 17.