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Gary’s Car Sales - Mathematics Coursework

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Introduction

Josh Anderson 11SPage 1

Gary’s Car Sales

Mathematics Coursework

For this experiment I was given a table of second hand cars for sale showing how the factors eg. age and mileage, effect the cost of a second hand car, compared with the same cars price when new. The table consists of 36 cars separated into 5 different manufacturers. For my investigation I will only use 3 of these manufacturers, Ford, Vauxhall, and Rover. I chose these 3 because they were the makes with the largest amount of information. Using the table I have been given I will try to find out which car is the best value for money. To do this I will have to draw two different types of graph, age against second hand price,

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Middle

8

Ford

9,140

108,000

1.6

19

2,950

8

Ford

17,750

96,000

2.9

20

3,250

7

Vauxhall

9,990

86,000

1.6

21

5,650

3

Vauxhall

11,150

34,000

1.6

22

4,600

2

Rover

7,300

17,000

1.1

23

5,400

1

Rover

7,300

11,000

1.1

24

4,800

1

Rover

7,300

26,000

1.1

25

2,700

5

Fiat

13,000

51,000

2.0

26

11,000

1

Peugeot

13,800

9,000

1.8

27

2,800

5

Fiat

6,500

43,000

1.0

28

8,000

4

Rover

21,000

142,000

2.3

29

6,495

2

Ford

8,800

23,000

1.3

30

4,050

4

Ford

8,400

48,000

1.3

31

6,300

2

Ford

10,300

26,000

1.3

32

4,100

4

Vauxhall

8,900

37,000

1.3

33

6,600

1

Vauxhall

8,500

9,000

1.3

34

7,800

1

Peugeot

10,500

13,000

1.4

35

8,700

3

Vauxhall

16,000

42,000

2.0

36

2,000

7

Peugeot

8,300

65,000

1.4

Page 2

Using the table I have drawn 4 graphs for age against second hand price. I have done 1 graph for each of my chosen manufacturers, Ford, Vauxhall, and Rover, and 1 with all three makes on it. I have used scatter graphs with lines of best fit, and have used the lines of best fit to work out the gradient using the formula x / y.

My first graph is for Ford. The line of best fit is a strong, negative correlation.

The price of a new Ford would be £7,900 and would be worth nothing when it is 10.9 years old

Using the line of best fit I have worked out the gradient:

1450 / 2 = 725

Therefore £725 is lost each year on a Ford car.

My second graph is for Vauxhall. The line of best fit is a strong, negative correlation. There is one odd point because that car is worth a lot more than most of the other Vauxhall’s when it is new.

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Conclusion

Because the Rover starts at £5,800 and the gradient is 400 the formula would be shown as:

Y = 5800 - (400 x 8)

Y = 5800 - 4000

Y = 2600

Page 5

Therefore the value of the car after 80,000 miles would be £2,600.

The table below shows the value of each make of car after various mileages using the formula.

Car Manufacturer

Value after:

Ford

Vauxhall

Rover

0 Miles (Cost when new)

£7,900

£8,200

£5,800

20,000 Miles

£6,600

£6,600

£5,000

40,000 Miles

£5,300

£5,000

£4,200

60,000 Miles

£4,000

£3,400

£3,400

80,000 Miles

£2,700

£1,800

£2,600

100,000 Miles

£1,400

£200

£1,800

Looking at this table Rover is the best value car as it loses the least amount of money after every 10,000 miles. After starting off the costliest the Vauxhall is only worth £200 after 100,000 miles, which makes it the car that loses the highest value by far. Whereas the Rover starts off the cheapest and finishes the highest priced of the three manufacturers after 100,000 miles.

In this experiment I discovered that Rover is the best value car out of the three that I chose to investigate. Vauxhall was the worst value car because it lost the most money on age and mileage. Ford was almost as bad value for money as the Vauxhall. My investigation could be improved by including more manufacturers and their cars. Also because the graphs are not 100% accurate. In real life as soon as you drive a new car off the forecourt it drops in value instantly. This means the graphs should have looked more like

...read more.

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