Thinking Skills: Critical Thinking and Problem Solving

330 Answers to assignments

40% hazels costing 8¢, a total of 30¢.

The most even mix is around 30% Brazil

nuts – can you define it more closely?

5.3 Carrying out investigations

1    a This must be carried out by looking

at amounts successively. There is more

than one way of doing some of the

amounts; only one is shown:

1¢ = 1¢; 2¢ = 2¢; 3¢ = 1¢ + 2¢;

4¢ = 2¢ + 2¢; 5¢ = 5¢; 6¢ = 5¢ + 1¢;

7¢ = 5¢ + 2¢; 8¢ = 10¢ − 2¢;

9¢ = 10¢ − 1¢; 10¢ = 10¢;

11¢ = 10¢ + 1¢; 12¢ = 10¢ + 2¢;

13¢ cannot be done in two coins.

b    This part of the investigation is open-

ended. A systematic approach should

be taken, possibly starting with the

1, 3, 5... example (in order to make

7, a 7¢, 8¢, 10¢ or 12¢ coin would be

needed and others follow from this). If

a set does not include a 1¢ coin, then

two denominations must differ by 1¢.

2    A 2 × 2 box contains 4 + 1 = 5 oranges;

a 3 × 3 box has 9 + 4 + 1 = 14. Each

successive size can be worked out as

the square number plus the sum of the

square numbers below it, so a 5 × 5 box

has 25 + 16 + 9 + 4 + 1 = 55 oranges.

As advanced level mathematics is not

expected for this paper, this answer

would be sufficient. The general

formula is

nn(2 + 1)( + 1n )

6

where n represents the number of

oranges on one side of the square.

For a rectangular box, students should

tabulate a series of values for n × m boxes.

They would be expected to recognise

that the pattern depends on the smallest

square that would be fitted into this box

(i.e. an n × n square if n<m). For rectangles

based on a given value of n, each extra

row adds a further number which is the

nth triangular number. So, for a 4 × 4

square, there are 30 oranges, and we then

add 10 (the 4th triangular number) for

each extra row, so a 4 × 5 box contains 40,

a 4 × 6 box contains 50, and so on.

3    There are 36 combinations of two dice.

Winning combinations are:

1, 4 1, 5 1, 6 2, 5 2, 6 3, 6

and the reverse of these (4, 1; 5, 1 etc.),

so 12 of the 36 combinations win, or^13.

If 200 people play, Milly takes $200 and

will expect to pay out 2 ×^2003 or $133,

so she should raise $67.

Investigating alternatives is again quite

open-ended. We can look at the two

options suggested. Multiples give:

First die

1 2 3 4 56

Second

die

1 1 2 3 4 5 6

2 2 4 6 8 10 12

3 3 6 9 12 15 18

4 4 8 12 16 20 24

5 5 10 15 20 25 30

6 6 12 18 24 30 36

She needs players to win less than half

the time to make a profit. There are

17 values in the table of 12 or over, so 12

is the minimum winning score which

would guarantee her a profit. If players

have to score over 12 to win, this would

give odds of^1336 – similar to those in the

original game.

One can similarly investigate the two

values written as a two-digit number (it

may be necessary to colour the dice to

define which is the first digit).