2003 paper 4. 1.(a) use your calculator to work out 2 (2.3 + 1.8) x 1.07 write down all the figures...

2003 Paper 4

1.(a) Use your calculator to work out

2 (2.3 + 1.8) x 1.07

Write down all the figures on your calculator display.

Put brackets in the expression below so that its value is 45.024

1.6 + 3.8 x 2.4 x 4.2

Use your calculator to work out

2 (2.3 + 1.8) x 1.07

2(2.3+1.8)X x1.07=

ANSWER 1798.67

Put brackets in the expression below so that its value is 45.024

1.6 + 3.8 x 2.4 x 4.2

(1.6 + 3.8) x 2.4 x 4.2= = 54.432

1.6 + (3.8 x 2.4) x 4.2 = 39.904

1.6 + 3.8 x (2.4 x 4.2) = 39.904

(1.6 + 3.8 x 2.4) x 4.2 = 45.024

ANSWER (1.6 + 3.8 x 2.4) x 4.2

2. Simon repairs computers. He charges

£56.80 for the first hour he works on a computer and £42.50 for each extra hour’s work.

(a) Yesterday Simon repaired a computer and charged a total of £269.30 Work out how many hours Simon worked yesterday on this computer.

Simon reduces his charge by 5% when he is paid promptly.He was paid promptly for yesterday’s work on the computer.

(b) Work out how much he was paid.

£56.80 for the first hour he works on a computer and £42.50 for each extra hour’s work.

(a) Yesterday Simon repaired a computer and charged a total of £269.30 Work out how many hours Simon worked yesterday on this computer.

1 hour at £5.60

£269.30 - £56.80 = £212.50

Number of hour worked at £42.50 per hour = £212.50÷£42.50

= 5hours

Total hours worked = 5hours + 1 hour = 6 hours

ANSWER 6 hours

Simon reduces his charge by 5% when he is paid promptly.He was paid promptly for yesterday’s work on the computer.

(b) Work out how much he was paid.

5% of £269.30 = 5 x £269.30 =100

£13.465

£13.47

Amount he was paid = £269.30 - £13.47

ANSWER £255.83

o

X o

y

This is part of a design of a pattern found at the theatre of Diana at AlexandriaIt is made up of a regular hexagon, squares and equilateral triangles.

o

(a) Write down the size of the angle marked x

o(b) Write down the size of the angle marked y

2The area of each triangle is 2cm .(c) Work out the area of the regular hexagon.

(d) In the space below use a ruler and compass to construct an equilateral triangle with sides of length 4 centimetres. You must show all construction lines.

3.

o

X o

y

This is part of a design of a pattern found at the theatre of Diana at AlexandriaIt is made up of a regular hexagon, squares and equilateral triangles. oWrite down the size of the angle marked x The triangle is an equilateral triangle.

All the sides and angles of the equilateral triangle are equal triangle.

o Angles in a triangle add up to 180 .

oAngle x = 180 ÷ 3 .

O

ANSWER X = 60

1 23

4

o

(b) Write down the size of the angle marked y

o

y

0

Sum of angles in the hexagon = 4x180

0Sum of angles in the hexagon = 720

0Angle Y = 720 ÷ 6

O

ANSWER Y = 120

2The area of each triangle is 2cm .

(c) Work out the area of the regular hexagon.

Area Hexagon = 6 x Area Triangle

2

Area Hexagon = 6 x 2 cm O

ANSWER Area hexagon = 12 cm

(d) In the space below use a ruler and compass to construct an equilateral triangle with sides of length 4 centimetres. You must show all construction lines.

4 cm

4. In 2002, Shorebridge Chess Club’s total income came from a council grant and members’ fees.

Council grant £50 Members fees 240 x £5 each

(a) (i) Work out the total income of the club for the year 2002.

(ii) Find the council grant as a fraction of the club’s total income for the year 2002

Give your answer in the simplest form. In 2001, the club’s total income was £1000. The club spent 60% of its total income on a hall. It spent a further £250 on prizes. (b) Work out the ratio

The amount spent on the hall : the amount spent on pizzas.

Give your answer in its simplest form.

4. In 2002, Shorebridge Chess Club’s total income came from a council grant and members’ fees.

Council grant £50 Members fees 240 x £5 each

(a) (i) Work out the total income of the club for the year 2002.

Total income = Council grant + Member’s fees

Total income = £50 + 240x£5

Total income = £50 + £1200

ANSWER Total income = £1250

(ii) Find the council grant as a fraction of the club’s total income for the year 2002

Give your answer in the simplest form.

council grant =club’s total income

£50 £1250

ANSWER 1 25

In 2001, the club’s total income was £1000. The club spent 60% of its total income on a hall. It spent a further £250 on prizes. (b) Work out the ratio

The amount spent on the hall : the amount spent on pizzas.

Give your answer in its simplest form.

amount spent on the hall = 60% of £1250

amount spent on the hall = 60 x £1250 100

amount spent on the hall = 6 x £125 amount spent on the hall = £750

The amount spent on the hall : the amount spent on prizes.

£750:£250 = 3:1

ANSWER 3:1

5.

Flowerbed

The diagram represents a garden in the shape of a rectangle. All measurements are given in metres. The garden has a flowerbed in one corner. The flowerbed is a square of side x.

(a) Write down an expression, in terms of x, for the shortest side of the garden.

(b) Find an expression, in terms of x, for the perimeter of the garden. Give your answer in its simplest form.

The perimeter of the garden is 20 metres.(c) Find the value of x.

2

x

x 5

5.

Flowerbed

(a) Write down an expression, in terms of x, for the shortest side of the garden.

2

x

x 5

ANSWER Shortest side = x + 2(b) Find an expression, in terms of x, for the perimeter of the garden. Give your answer in its simplest form.

Perimeter = (x + 2) + (x + 5) + (x + 2) + (x + 5)

ANSWER Perimeter = 4X + 14

5.

Flowerbed

The perimeter of the garden is 20 metres.

(c) Find the value of x.

2

x

x 5

4X + 14 = 20

4X = 20 - 14

4X = 6

X = 6 4 ANSWER X = 1.5

6. (a) Simplify 5p + 2q – 3p – 3q

y = 5x – 3

(b) Find the value of y when x = 4

6. (a) Simplify 5p + 2q – 3p – 3q

5p + 2q – 3p – 3q

5p – 3p = 2p

+ 2q – 3q = - 1q

= 2p – 1q = 2p – q

ANSWER 2p – q

5p + 2q – 3p – 3q

y = 5x – 3

(b) Find the value of y when x = 4

y = 5x4 – 3

y = 20 – 3

y = 17

ANSWER y = 17

7. The table shows some rows of a number pattern.

Row 1 1 = 1 x 2 2

Row 2 1 + 2 = 2 x 3 2

Row 3 1 + 2 + 3 = 3 x 4 2

Row 4 1 + 2 + 3 + 4

Row 8

(a) In the table, complete row 4 of the number pattern.(b) In the table, complete row 8 of the number pattern.(c) Work out the sum of the first 100 whole numbers. (d) Write down an expression, in terms of n, for the sum of the n whole numbers.


Row 1 1 = 1 x 2 2

Row 2 1 + 2 = 2 x 3 2

Row 3 1 + 2 + 3 = 3 x 4 2

Row 4 1 + 2 + 3 + 4

Row 8

(a) In the table, complete row 4 of the number pattern.

= 4x5 2

(b) In the table, complete row 8 of the number pattern.

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 8x9 2


Row 1 1 = 1 x 2 2

Row 2 1 + 2 = 2 x 3 2

Row 3 1 + 2 + 3 = 3 x 4 2

Row 4 1 + 2 + 3 + 4

Row 8

= 4x5 2

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 8x9 2

(c) Work out the sum of the first 100 whole numbers.

= 1 + 2 + 3 + 4 + ………………………. + 98 + 99 + 100

ANSWER 100x101 2


Row 1 1 = 1 x 2 2

Row 2 1 + 2 = 2 x 3 2

Row 3 1 + 2 + 3 = 3 x 4 2

Row 4 1 + 2 + 3 + 4

Row 8

= 4x5 2

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 8x9 2

= 1 + 2 + 3 + 4 + ………………………. + n

ANSWER n(n+1) 2

(d) Write down an expression, in terms of n, for the sum of the n whole numbers.

8.

The diagram represents a large tank in the shape of a cuboid.The tank has a base.It does not have a top.The width of the tank is 2.8 metres.The length of the tank is 3.2 metres.The height of the tank is 4.5 metres.

The outside of the tank is going to be painted. 21 litre of paint will cover 2.5 mThe cost of the paint is £2.99 per litre.

Calculate the cost of the paint needed to paint the outside of the tank.

4.5 m

2.8 m3.2 m

3.2 m

2.8 m

4.5 m

2 Area = 3.2x4.5 m

2 Area = 14.4 m

2 Area = 14.4 m

2 Area = 2.8x4.5 m

2 Area = 12.6 m

2 Area = 12.6 m

2 Area = 3.2x2.8 m 2 Area = 8.96 m

2 2 2 2 2

8.96 m + 12.6 m + 12.6 m 14.4 m + 14.4 mTotal Area =

2

Total Area = 52.96 m

Number of litres of paint = 52.96 ÷ 2.5 litres

Number of litres of paint = 21.184 litres

Cost of paint = 21.184 x £2.99

ANSWER Cost of paint = £63.34

2 2

9. Change 2.5 m to cm

1 m = 100 cm

1 m x 1m = 100 cm x 100 cm

2 2

1 m = 10 0000 cm

2 2

2.5 m = 2.5 x 10 0000 cm

2 2

ANSWER 2.5 m = 10 000 cm

10.Bhavana asked some people which region their favourite football team came from.

The table shows her results.

Region Frequency

Midlands 22

London 36

Southern England 8

Northern England 24

(a) Complete the accurate pie chart to show these results. Use the circle given below.

Midlands

Region Frequency

Midlands 22

London 36

Southern England 8

Northern England 24

Angle

22 x 360 = 22 x 4 = 8890

Total = 90

36 x 360 = 36 x 4 = 14490 8 x 360 = 8 x 4 = 329024 x 360 = 24 x 4 = 9690

London

MidlandsNorthern England

South England

Four teams, City, Rovers, Town and United play a competition to win a cup.Only one team can win the cup.The table below shows the probabilities of City or Rovers or Town winning the cup.

City Rovers Town United

0.38 0.27 0.15 X

0.38 + 0.27 + 0.15 = 0.8

x = 1- 0.8

ANSWER x = 0.2

Here are the times, in minutes, taken to change some tyres.

5 10 15 12 8 7 20 35 24 15 20 33 15 25 10 8 10 20 16 10 In the space below, draw a stem and leaf diagram to show these times.

0

1

2

3

5 7 8

0 0 0 0 2 5 5 5 6

0 0 0 4 5

3 5

14.

10 cm

4 cm

The diagram shows a cylinder with a height of 10 cm and radius 4 cm.Calculate the volume of the cylinder.

Give your answer correct to 3 significant figures.

The length of a pencil is 13 cm.The pencil cannot be broken

(b) Show that this pencil cannot fit inside the cylinder.

2 Volume = πr h =

2 3

Πx4x10 cm

3

Volume = 160π cm

3

ANSWER Volume = 502 cm

14.

10 cm

4 cm

The length of a pencil is 13 cm.The pencil cannot be broken

(b) Show that this pencil cannot fit inside the cylinder.

14.

10 cm

8 cm 8 cm

10 cm

Pythagoras Theorem 2 2 2

c = a + b 2 2 2

c = 8 + 10 2 c = 64 + 100 2 c = 164

c = 164

c = 233

Pencil will not fit into box

c = 12.8 cm

12.8 cm

13. (a) Express the following numbers as products of their prime factors.

(i) 60

(ii) 96

(b) Find the Highest Common Factor of 60 and 96.

(c) Find the Lowest Common Multiple of 60 and 96.

Prime numbers = 2, 3, 5, 7, 11, 13, 17, 19


(i) 60

60 ÷ 2 = 30 60 = 2 x 30

30 ÷ 2 = 15 60 = 2 x 2 x 15

15 ÷ 3 = 5 60 = 2 x 2 x 3 x 5

2

ANSWER 60 = 2x3x5

Prime numbers = 2, 3, 5, 7, 11, 13, 17, 19


(ii) 96

96 ÷ 2 = 48 96 = 2 x 48

48 ÷ 2 = 24 96 = 2 x 2 x 24

24 ÷ 2 = 12 96 = 2 x 2 x 2 x 12

5

ANSWER 96 = 2x3

12 ÷ 2 = 6 96 = 2 x 2 x 2 x 2 x 6

6 ÷ 2 = 3 96 = 2 x 2 x 2 x 2 x 2 x 3

(b) Find the Highest Common Factor of 60 and 96.

96 = 2 x 2 x 2 x 2 x 2 x 3

60 = 2 x 2 x 3 x 5

60 = 2 x 2 x 3 x 5

96 = 2 x 2 x 2 x 2 x 2 x 3

Highest Common Factor = 2 x 2 x 3

ANSWER Highest Common Factor = 12

(c) Find the Lowest Common Multiple of 60 and 96.

1x60 = 60

2x60 = 120

3x60 = 180

4x60 = 240

5x60 = 300

6x60 = 360

7x60 = 420

1x96 = 96

2x96 = 192

3x96 = 288

4x96 = 384

5x96 = 480

8x60 = 480

5x96 = 480

8x60 = 480

ANSWER Lowest Common multiple = 480

14. A garage keeps records of the costs of repairs to its customers’ cars. The table gives information about the costs of all repairs which were less than £250 in one week.

Cost, (£C) Frequency

0<C≤50 4

50<C≤100 8

100<C≤150 7

150<C≤200 10

200<C≤250 11

(a) Find the class interval in which the median lies.

There was only one further repair that week, not included in the table. That repair cost £1000.

Dave says “The class interval in which the median lies will change.”

(b) Is Dave correct? Explain your answer.


Cost, (£C) Frequency 0<C≤50 4

50<C≤100 8

100<C≤150 7

150<C≤200 10

200<C≤250 11

Cumulative Frequency

4

4 + 8 =12

4 + 8 + 7 =19

4 + 8 + 7 + 10 =29

4 + 8 + 7 + 10 + 11 = 40

40 + 1 = 20.5 2

ANSWER Median class interval

150<C≤200


Cost, (£C) Frequency 0<C≤50 4

50<C≤100 8

100<C≤150 7

150<C≤200 10

200<C≤250 11

Cumulative Frequency

4

4 + 8 =12

4 + 8 + 7 =19

4 + 8 + 7 + 10 =29

4 + 8 + 7 + 10 + 11 = 40

41 + 1 = 21 2

ANSWER Median class interval same Dave is wrong

There was only one further repair that week, not included in the table. That repair cost £1000.

Dave says “The class interval in which the median lies will change.”

(b) Is Dave correct? Explain your answer.

250<C 1 4 + 8 + 7 + 10 + 11 + 1 = 41

The garage also sells cars.It offers discounts at 20% of the normal price for cash.Dave pays £5200 cash for a car.

(c) Calculate the normal price of the car.

Normal Price x Normal Percent = 100%

Discount Price £5200 Discount Percent = 100% - 20% = 80%

X = 100 £5200 80

X = 100 x £5200 80

ANSWER Normal Price £6500

15.

x

x

A cuboid has a square of side x cm. The height of the cuboid is 1 cm more than the length cm. 3The volume of the cuboid is 230 cm . 3 2 (a) Show that x + x = 230

x

x

x + 1

Volume = length x width x height

Volume = X x X x (X+1) = 230

2

Volume = X x (X+1)

3 2

Volume = X + X

3 2 The equation x + x = 230

has a solution between x = 5 and x = 6

(b) Use a trial and improvement method to find this solution. Give your answer correct to 1 decimal place. You must show all your working.

X= 5 3 2 X + X =

3 2 5 + 5 =

125 + 25 =

150 Less than 230

X= 6 3 2 X + X =

3 2 6 + 6 =

216 + 36 =

216 More than 230

3 2 The equation x + x = 230

has a solution between x = 5 and x = 6

X= 5.7 3 2 X + X =

3 2 5.7 + 5.7 =

217.683 Less than 230

X= 5.8 3 2 X + X =

3 2 5.8 + 5.8 =

228.752

More than 230

Less than 230

X= 5.9 3 2 X + X =

3 2 5.9 + 5.9 =

240.189

X= 5.85 3 2 X + X =

3 2 5.85 + 5.85 =

231.134 More than 230

X= 5.8 3 2 X + X =

228.752

More than 230

Less than 230

X= 5.9 3 2 X + X =

240.189

X= 5.85 3 2 X + X =

231.134 More than 230

Actual value between 5.8 and 5.85

ANSWER 5.8

16. 15 cm The diagrams shows a semi-circle.The diameter of the semi-circle is 15 cm.Calculate the area of the semi-circle.Give your answer correct to 3 significant figures.

2

Area circle = πr

2

Area circle = πx7.5

Diameter = 15 cm

radius = 15 cm ÷ 2

radius = 7.5 cm 2

Area circle = 3.14x7.5

2

Answer Area circle = 177 cm

17. A straight line has equation y = 1 x + 1 2 The point P lies on the straight line. P has a y-coordinate of 5.

(a) Find the x-coordinate of P.

y = 5

y = 1 x + 1 25 = 1 x + 1 2

5 - 1 = 1 x 2

4 = 1 x 2

1x = 2 x 4

ANSWER X = 8

(b) Write down the equation of a different straight line that is parallel to

y = 1 x + 1 2

y = 1 x + 3 2

y = 1 x - 2 2

y = 1 x + 5 2

(c) Rearrange y = 1 x + 1 to make x the subject. 2 y = 1 x + 1

2 y - 1 = 1 x 2 2 (y – 1) = 1 x 2y – 2x1 = 1 x 2y – 2 = 1 x 2y – 2 = x ANSWER x = 2y – 2

Solve 2x – 3y = 11 5x + 2y = 18

18.

2x – 3y = 11 x 2 5x + 2y = 18 x 3

4x – 6y = 2215x + 6y = 54

19x = 76 x = 76 19

x = 4

5x + 2y = 18

5x4 + 2y = 18 x = 4 20 + 2y = 18

2y = 18 - 20 2y = – 2 y = – 1

ANSWER X = 4 Y = - 1

BE is parallel to CD

AE = 6cm, ED = 4 cm, AB = 4.5 cm, BE = 4.8 cm

Calculate the length of CD.

(b) Calculate the perimeter of the trapezium EBCD.

A

B E

C D

4.5 cm 6 cm

4 cm 4.8 cm

CD = ADBE AE

A

B E

C D

4.5 cm 6 cm

4 cm 4.8 cm

CD = 104.8 6

CD = 10 X 4.8 cm 6

ANSWER CD = 8 cm

(b) Calculate the perimeter of the trapezium EBCD.

A

B E

C D

4.5 cm 6 cm

4 cm 4.8 cm

AC = ADAB AE

perimeter EBCD = BE + ED + CD + CB

perimeter EBCD = 4.8 cm + 4 cm + 8cm + CB

AC = 104.5 6

AC = 10 x 4.5 6

AC = 7.5 cm

CB = AC - AB = 7.5 cm – 4.5 cm

CB = 3 cm perimeter EBCD = 4.8 cm + 4 cm + 8cm + 3 cm

ANSWER perimeter EBCD = 19.8 cm

20. 2 y = ab a+b

6 a = 3 x 10 7 b = 2 x 10 Find y. Give your answer in standard form correct to 3 significant figures

2 y = ab a+b

2 6 7 y = 3x10x 2x10 6 7 3x10+2x10 2 13

y = 6x10 3000000 + 20000000

2 13 y = 6x10 3000000 + 20000000

2 13 y = 6x10 23000000

2 13 y = 6x10 7 2.3x10 2 6

y = 2.6x10

6 y = √2.6x10

½ x 6 y = 1.62x10

3

ANSWER y = 1.62x10

21.

The diagram shows triangle ABC. BC = 9.5 cm O Angle ABC = 90 O

Angle ABC = 38 Work out length AB Give your answer to 3 significant figures.

A

B C38O

8.5 cm

21.

Work out length AB

A

B C38O

8.5 cm

opp

adj

opp = adj tanθ

o

AB = 6.5x tan38

ANSWER AB = 5.8 cm

Julie does a statistical experiment. She throws a dice 600 times.She scores six 200 times.

Is the dice fare? Explain your answer.

Julie then throws a fair red dice once and a fair blue dice.Complete the probability tree diagram to show the outcomes.Label clearly the branches of the probability tree diagram.

22.

The probability tree diagram has been started in the space below.

Red Blue Dice Dice

six

not six

Julie does a statistical experiment. She throws a dice 600 times.She scores six 200 times.

Is the dice fare? Explain your answer.

22.

Fare dice p(six) =1/6

Numbers on dice 1, 2, 3, 4, 5, 6

Expected number of sixes =1/6 x 600

Expected number of sixes =100

Julie scores six 200 times

ANSWER Dice is not fare

Julie then throws a fair red dice once and a fair blue diceComplete the probability tree diagram to show the outcomes.Label clearly the branches of the probability tree diagram.

22.

The probability tree diagram has been started in the space below.

Red Blue Dice Dice

six

not six

six

not six

six

not six

1/6

5/6

1/6

5/6

1/6

5/6

six and six

not six and six

six and not six

not six and not six

1/6x1/6 = 1/36

5/6x1/6 = 5/36

1/6x5/6 = 5/36

5/6x1/6 = 25/36

2003 paper 4. 1.(a) use your calculator to work out 2 (2.3 + 1.8) x 1.07 write down all the figures...

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