2008 May/June 43005 1H

1

non- calculator paper1 compare fractions; equivalent fractions; factorise an expression2 draw plan and elevations of 3D shape made from cuboids3 circumference of circle; area of circle; area of compound shape4 algebraic expressions; make and solve an equation5 multiply out an expression; evaluate an expression6 square numbers7 multiply out and simplify expressions involving brackets8 solve a factorised quadratic equation9 calculate the length of a side in a right-angled triangle; similar triangles10 change the subject of a formula; evaluate an expression; negative numbers11 inequality on a number line; solve an inequality; find a non-integer solution12 use circle theorems13 sector of a circle; perimeter; answer “in terms of π”14 volume of compound shape; density and mass15 add algebraic fractionsgrade boundaries for this paper

2008 May/June 43005 1H

(a) Emma says that is greater than Is she correct?

Explain your answer.

€

37

48

€

3

4

Three quarters of 48 is 36.

€

34=3648

37 is greater than 36

(b) Complete the following

€

(i)3

4=88

66

€

(ii)4

3=11687

(c) Factorise 33x +44

11 ( 3x + 4 )

2 marks

2

1. Here is a multiplication table

X 11 12 13 14 15

3 33 36 39 42 45

4 44 48 52 56 60

Yes!

1 mark 1 mark 1 mark

2008 May/June 43005 1H

Plan view

Side elevation

Front elevation 3 marks

3

2. The diagram shows a solid made from two cuboids

The large cuboid is 5cm by 4cm by 3cmThe small cuboid is 3cm by 1cm by 1cm

On the centimetre grids draw the plan view, side elevation and front elevation

2008 May/June 43005 1H

Answer = 45 cm

4

3 (a) Use π = 3 to work out an estimate for the circumference of a circle with diameter 15 cm

circumference = π × diameter

= 3 × 15

3 (b) (i) Use π = 3.14 to work out the area of a circle with radius 10 cm

area of circle = π × radius2

= 3.14 × 102

Answer = 314 cm2

2 marks

2 marks

2008 May/June 43005 1H

area rectangle = 20 × 30 = 600

area shape = 600 + 314 = 914 cm2

5

3 (b) (ii) The diagram shows a shape made of two semicircles and a rectangle

Use your answer to part (b) (i) to work out the area of the shape

3 marks

2008 May/June 43005 1H

Kaz’s age = x − 3

x + x − 3 = 91

2x = 94 Matias is 47 years old

6

4 Matias is x years old

Kaz is three years younger than Matias

4 (a) Write down an expression, in terms of x, for Kaz’s age

2 marks

Matias’ age = x

(b) The sum of the ages of Matias and Kaz is 91

Use this information to write down an equation in terms of x

2x − 3 = 91

1 mark

(c) Solve your equation formed in part (b) to work out the age of Matias

2 marks

2008 May/June 43005 1H

5 (a) Multiply out a(b + c)

= ab + ac

= 27 ( 3 + 7 )

7

(b) Work out the value of xy +xz when x = 27, y = 3 and z = 7

xy + xz = x(y + z)

= 27 × 10

= 270

1 mark

3 marks

2008 May/June 43005 1H

Answer = 169

42 ends in a 6 not a 2

Or 10 × 14 = 140 4 × 14 = 56Therefore 14 × 14 = 196 which is not 192

8

1 mark

1 mark

6 (a) write down the value of 132

(b) explain how you know that 142 is not equal to 192

2008 May/June 43005 1H

= −6a + 2b − 10

9

7(a) Multiply out −2(3a − b + 5)

2 marks

(b) Multiply out and simplify 4(8e − 9) + 2e

= 32e − 36 + 2e

= 34e − 36 2 marks

2008 May/June 43005 1H

either x − 13 = 0 or x + 1 = 0

so x = 13 or x = -1 2 marks

10

8 Solve (x − 13)(x + 1) = 0

when two expressions are multiplied together and the answer is zero, then one of the expressions must be zero

2008 May/June 43005 1H

9 (a) The diagram shows a right-angled triangle ABC

AC = 10cm and BC = 3cm

Calculate the length of AB

Leave your answer as a square root

Think – PYTHAGORAS h2 = x2 + y2

102 = x2 + 32

x2 = 100 – 9

AB2 = 91

AB =√91 3 marks

11

h

x

y

2008 May/June 43005 1H

DF = 1.5 × AC x = 1.5 × 3 EF = 4.5 cm

3 marks

12

9 (b) Triangles ABC and DEF are similar

Work out the length of EF marked x on the diagramDEF is an enlargement of ABC with scale factor 15÷10 = 1.5

2008 May/June 43005 1H10 You are given the formula

(a) Make x the subject of the formula

5y = x2 - 49

x2 = 5y + 49

x = ± √( 5y + 49 )

Must have ± sign to score

full marks

x = ± √( 5 × −9 + 49 )

x = ± √( 4 )x = 2 and −2

3 marks

13

€

y =x2 −49

5

x2 - 49 = 5y

(b) Work out the values of x when y = −9

x = ± √( −45 + 49 )

3 marks

2008 May/June 43005 1H

11 (a) Write down the integers that satisfy this inequality diagram

integers -2, -1, 0, 1

14 – 12 ≤ 3x - x

2 ≤ 2x x ≥ 1

1 < answer < 2

2 marks

14

-2 ≤ x < 2

(b) Solve the inequality 14 + x ≤ 12 + 3x

2 marks

(c) Write down a non-integer value that satisfies both the inequality diagram and part (b)

1 markx = 1.5 for example

2008 May/June 43005 1H

x = 180 – 86 = 94⁰

0pposite angles of a cyclic quadrilateral add up to 180

180 – 86 = 94 angle BAC = 94 ÷ 2 = 47⁰

** Alternate Segment Theorem

15

12 ABCD is a cyclic quadrilateralPCQ is a tangent at CO is the centre of the circleTriangle ABC is isosceles

(a) work out the value of x

(b) (i) Work out the value of y

angle BAC is equal to y **

triangle BAC is isosceles

y = 47⁰

(b) (ii) Write down the name of the circle theorem used in part (b)(i)

2 marks

3 marks

1 mark

2008 May/June 43005 1H

240⁰

€

arclength=240

360×π ×36

Perimeter = 24π + 18 + 18 = 24π + 36 cm 4 marks

16

13 The diagram shows a major sector of a circle

The radius of the sector is 18 cm

The angle of the minor sector is 120°

Work out the perimeter of the major sector

Leave your answer in terms of π

Simplify your answer as fully as possible

€

lengthof arcof sector =angle

360×Circumference

€

arclength=24 π

2008 May/June 43005 1H

Area of square base = 25cm2 means length of each edge is 5cm

17

14 The diagram shows a solid metal object made from two cubes and a square-based pyramid

The area of the base of each cube is 25cm2

The height of the pyramid is equal to the height of each cube

The density of the metal is 9g/cm3

You are given the formula

Volume of pyramid = ⅓ × area of base × height

Work out the mass of the solid metal object

volume of each cube = 125cm3

volume of pyramid = ⅓ × 25 × 5 = 41⅔ cm3 volume of solid metal object = 125 + 125 + 41⅔ = 291⅔ cm3

mass = density × volume

= 9 × 291⅔ = 2625g 7 marks

2008 May/June 43005 1H

4 marks

18

€

15 Provethatx −3

x−

x −2

x +2=

x −6

x(x +2)

€

x −3

x−

x −2

x +2=(x −3)(x +2) −x(x −2)

x(x +2)

€

=(x2 +2x −3x −6) −(x2 −2x)

x(x +2)

€

=x2 −x −6 −x2 +2x)

x(x +2)

€

=x −6

x(x +2)

2008 May/June 43005 1H

1919

Total: out of 70

grade D C B A A*

score 19 25 37 49 61