Thinking Skills: Critical Thinking and Problem Solving

Answers to assignments 329

As a further exercise you might consider

how the problem could be tackled if

the distance between pit stops was not

constant (for example, it might be worth

filling the car right up at the start to save

on refuelling time, although this would

make it slower).

4    It is more straightforward to work in

proportions than percentages. Suppose

the proportions are as follows: x Brazil

nuts, y walnuts and (1 − x − y) hazelnuts.

The cost to the shopkeeper for this mix

is 40x + 35y + 20(1 − x − y). She wishes to

make 50% profit selling it at 60¢, so 60¢

represents twice this value.

We now have a model:

40 x + 35y + 20(1 − x − y) = 30

Simplifying:

20 x + 15y = 10

This cannot be solved explicitly for x and y,

so we must investigate different values. We

can note that y = ()^24

3

− x, so this gives a

relationship between the two (and implies

the proportion of the third ingredient).

Putting some values into this:

x (Brazil nuts) y (walnuts) z (hazelnuts)

0.0 0.67 0.33

0.1 0.53 0.37

0.2 0.40 0.40

0.3 0.27 0.43

0.4 0.13 0.47

0.5 0.00 0.50

0.6 −0.13 0.53

Thus there is a range of mixes that fulfil

the conditions, from 0 to 50% Brazil nuts.

We can test one of these answers: 1kg

can be made up of 20% Brazil nuts,

costing 8¢, 40% walnuts costing 14¢ and

The quickest way of transporting

between 61 and 70 m^3 of stone is to move

two 24 m^3 blocks and one 20 m^3 block:

Blocks sized 24 + 24 + 20 will take

(132 mins 10 secs) + 5 mins

+ (132 mins 10 secs) + 5 mins

+ (139 mins 24 secs)

= 413 mins 44 secs.

Award yourself full marks if you correctly

identified the correct combination of

blocks: 24 + 24 + 20.

Award yourself one mark for 24 + 20 + 20,

or 24 + 24 + 16.

3    We need to calculate the total race time

for the various numbers of pit stops. For 1

pit stop, 150 litres of fuel are required for

each half of the race. The average lap time

(0.12 seconds slower than 75 seconds for

each 5 litres of fuel) is, therefore:

75 + 0.12^75

5









= 76.8 seconds

So 60 laps takes 60 × 76.8 = 4608 seconds.

The time for the pit stop is

10 +^150

15

= 20 seconds

so the total race time is 4628 seconds

(77 minutes 8 seconds).

For two stops, the calculation is based on

an average fuel load of 50 litres, so the

average lap time is 76.2 seconds and the

pit stop time is 16.7 seconds.

The total time is

60 × 76.2 + 2 × 16.7 = 4605.4 seconds

or 76 minutes 45.4 seconds.

For three stops, the average fuel load is 37.5

litres, the average lap time is 75.9 seconds

and the pit stop time is 15 seconds.

The total time is

60 × 75.9 + 3 × 15 = 4599 seconds

or 76 minutes 39 seconds. Therefore three

pit stops is optimum. Should you consider

four?