### Chocolate

There are three tables in a room with blocks of chocolate on each. Where would be the best place for each child in the class to sit if they came in one at a time?

### F'arc'tion

At the corner of the cube circular arcs are drawn and the area enclosed shaded. What fraction of the surface area of the cube is shaded? Try working out the answer without recourse to pencil and paper.

### Do Unto Caesar

At the beginning of the night three poker players; Alan, Bernie and Craig had money in the ratios 7 : 6 : 5. At the end of the night the ratio was 6 : 5 : 4. One of them won $1 200. What were the assets of the players at the beginning of the evening? # Couples ##### Stage: 3 Challenge Level: Good solutions were received from Luke from St George's College, Ruth from Manchester High School for Girls, Daniel from Junction City High School, Max from Bishop Ramsey C of E Secondary School and Rohan from Long Bay Primary. Luke wrote: Since two thirds of the adult men are married, the total number of men must be a multiple of 3. Possible totals are:  Total Men Total Married Men 3 2 6 4 9 6 12 8 Since three quarters of the adult women are married, the total number of women must be a multiple of 4. Possible totals are:  Total Women Total Married Women 4 3 8 6 12 9 16 12 The smallest number of couples is 6, when there are 9 men and 8 women. There would be 17 adults in the smallest community of this type. Ruth wrote: There are$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.