Parallelogram 16 • Level 3 • 21 Dec 2023Chrismaths - Part 3

To find Åle’s age we need to work out: 2014 − 1859 = 155.

So Åle’s age when it died was 155.

Alternatively, we can see that from 1859 to 1900 is 41 years, from 1900 to 2000 is 100 years, and from 2000 to 2014 is 14 years. So Åle’s age in years is 41 + 100 + 14, that is, 155.

We see that:

• 20 + 16 = 36.
• 201 − 6 = 195.
• 201 + 6 = 207.
• 20 × 16 = 320.

So we can obtain all four of the given numbers.

Since AB=AC, the triangle ABC is isosceles. Therefore ∠ACB=∠CBA. Since the angles in a triangle add up to 180°, we have ∠BAC+∠ACB+∠CBA = 180°. Hence 40° + 2∠ACB = 180°.

It follows that 2∠ACB = 180° − 40° = 140°. Therefore ∠ACB = 70°.

Since BD=BC, the triangle BCD is isosceles. Therefore ∠BDC=∠DCB. Since ∠DCB is the same as ∠ACB, it follows that ∠BDC=∠ACB = 70°.

By the External Angle Theorem, ∠BDC=∠BAD+∠ABD. That is, 70° = 40° + ∠ABD. It follows that ∠ABD = 30°.

The first box contains one-ninth of the pears. Therefore, the other two boxes contain, between them, eight-ninths of the pears. So, since they contain the same number of pieces of fruit, but no apples, they each contain four-ninths of the pears.

So the difference between the number of pears in each of the second and third boxes and the number of pears in the first box is equal to three-ninths of the number of pears, that is, to one-third of the number of pears.

Therefore, as the first box holds the same number of pieces of fruit as the other two boxes, the 12 apples in the first box match the extra one-third of the number of pears in the other two boxes. Since one-third of the pears amount to 12 pears, it follows that altogether there are 36 pears.

Hence the total number of pieces of fruit is 48. Since these are equally divided between the boxes, each of the three boxes contains 16 pieces of fruit.

Alternatively, Since the first box contains one-ninth of the pears, we avoid fractions by supposing that there are 9p pears. It follows that there are p pears in the first box and altogether 8p pears in the other boxes. As they contain equal numbers of pieces of fruit, there are 4p pears in each of these boxes.

Hence there are 12+p pieces of fruit in the first box, and 4p in each of the other two boxes.

Since each box contains the same number of pieces of fruit: 12+p=4p, and hence 12=3p, from which it follows that p=4.

We have seen that there are 12+p pieces of fruit in the first box. Since p=4, we deduce that the number of pieces of fruit in each box is 16.

We will suppose that the colours of the tiles that are available are red, blue and green.

We need to count the number of different ways in which we can choose to places the tiles, so that we use at least one tile of each of the three colours. To do this we will need two tiles of one colour and one tile of each of the other two colours.

The method we use is to split our choice of how to arrange the tiles into a sequence of independent choices, and to count the number of different ways in which we can make these choices.

The first choice is to decide the colour of the two tiles with the same colour. There are 3 ways to choose this colour. Once we have made this choice, we have no further choice of colours as we must have one tile of each of the two other colours.

We now need to count the number of different ways to place the four tiles in a row after we have chosen their colours.

Suppose, for the sake of argument, that we have decided to have two red tiles and hence one blue tile and one green tile.

We can decide how to place these tiles by first choosing the place for the blue tile, and then the place for the green tile. Then there will be no further choice, as we will need to put the two red tiles in the two remaining places.

There are 4 choices for the position of the blue tile.

Then, whichever position we choose for the blue tile, there will remain 3 choices of position for the green tile.

We thus see that placing the tiles involves three independent choices and that the number of possibilities for these choices are 3, 4 and 3 respectively.

Each of the 3 choices of colour shared by two of the tiles may be combined with any of the 4 choices of position for the tile of the second colour. So there are 3 × 4 ways to combine these choices.

Each of these 3 × 4 ways of making the first two choices may be combined with any of the 3 choices of position for the tile of the third colour.

Therefore the total number of different ways of combining these choices is 3 × 4 × 3 = 36. Hence there are 36 different ways in which the row of four tiles can be chosen.