Couples
In a certain community two thirds of the adult men are married to
three quarters of the adult women. How many adults would there be
in the smallest community of this type?
Age
11 to 14
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In a certain community two thirds of the adult men are married to three quarters of the adult women.
How many adults would there be in the smallest community of this type?
The number of married men is equal to the number of married
women.
We know that two thirds of the adult men are married.
List how many men there could be, and how many of them would be married.
We know that three quarters of the adult women are married.
List how many women there could be, and how many of them would be married.
We know that two thirds of the adult men are married.
List how many men there could be, and how many of them would be married.
We know that three quarters of the adult women are married.
List how many women there could be, and how many of them would be married.
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.