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Groby 2 tier Module 5 p1 June 2004Question 1 solve equations

Question 2 substitute numbers for letters

Question 3 reflection and enlargement

Question 4 nth term of a sequence

Question 5 construct perpendicular bisector; locus

Question 6 quadrilaterals – facts about sides, angles and diagonals

Question 7 table of values; draw graph; solve quadratic

Question 8 solve RATs with Pythagoras and SOHCAHTOA

Question 9 circle theorems

Question 10 factorise quadratic expressions

Question 11 calculate perimeter of a sector of a circle; volume of a prism

Question 12 transformations of graph of y =sinx

Question 13 volume of frustum of a pyramid

Question 14 area of a triangle

Question 15 complete the square; solve quadratic equation

Grade boundaries

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Groby 2 tier Module 5 p1 June 20041. Solve these equations

(a) 4x – 7 = 5

(Add 7) 4x = 12

x = 32 Marks

(b) 2( y + 5 ) = 28

2y + 10 = 28

( - 10 ) 2y = 18

y = 9 3 marks

( c ) 7z + 2 = 9 – 3z

(Add 3z) 10z + 2 = 9

( - 2 ) 10z = 7

x = 710 3 Marks

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Groby 2 tier Module 5 p1 June 20042 (a) Find the value of 5p + 2q when p = 4 and q = -7

5 x 4 + 2 x -7

20 +-14

6 2 Marks

(b) Find the value of u2 – v2 when u = 5 and v = 3

52 – 32

25 – 9 = 162 Marks

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Groby 2 tier Module 5 p1 June 20043. The diagram shows two shapes P and Q

Describe fully the single transformation which takes shape P onto shape Q

Reflection in x = 32 Marks

x= 3

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Groby 2 tier Module 5 p1 June 2004

3 (b) the vertices of triangle T are ( 1, 1 ), ( 1,2 ) and ( 4,1 )

Enlarge triangle T by scale factor 2, with ( 0, 0 ) as the centre of enlargement.

3 Marks

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Groby 2 tier Module 5 p1 June 20044. A sequence of numbers is shown.

2 5 8 11 14

a) Find an expression for the nth term of the sequence.

1 2

2 5

3 8

4 11

5 14

Goes up by 3 each time

n 3n - 1

Answer 3n - 1 2 Marks

b) Explain why 99 will not be a term in this sequence.

The answer must be one less than a multiple of 3. 99 is a multiple of 3. 98 would be in the sequence so would 102. 3n – 1 = 99 does not give a whole number.

2 marks

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Groby 2 tier Module 5 p1 June 20045 (a) The line LM is drawn below. Use ruler and compass to construct the perpendicular bisector of LM. You must show clearly all your construction arcs.

b) Complete the sentence.

The perpendicular bisector of LM is the locus of points which are ………

Equidistant from the two fixed points L and M

2 Marks

1 Mark

L ML ML M

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Groby 2 tier Module 5 p1 June 20046. Here is a list of quadrilaterals.

Kite rectangle rhombus square trapezium

For each of the following description, choose the correct name from the list.

a) One pair of sides are parallel. The other two sides are not parallel.

Trapezium 1 Mark

b) All the angles are the same size. Only opposite sides are equal.

Rectangle 1 Mark

c) All the sides are the same length. The diagonals are not equal in length.

Rhombus 1 Mark

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Groby 2 tier Module 5 p1 June 2004

7. a) Complete the table of values for y = 2x2 – 4 x – 1

x -2 -1 0 1 2 3

y 15 -1 -1 55 -3

2 ( -1 )2 – 4 ( -1 ) - 1

2 + 4 – 1 = 5

2 ( 1 )2 – 4 (1) -1

2 – 4 – 1 = - 3

b) On the grid opposite, draw the graph of y = 2x2 – 4x – 1 for values of x from -2 to 3.

2 Marks

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Groby 2 tier Module 5 p1 June 2004

b) On the grid, draw the graph of y = 2x2 – 4x – 1 for values of x from -2 to 3.

–3 –2 –1 1 2 3 4

–4

–3

–2

–1

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

x

y 2 Marks

An approximate solution of the equation 2x2 - 4x – 1 = 0 is x = 2.2

i) Explain how you can find this from the graph and ii) use your graph to find another solution to this equation

Intersection with the x axis

1 Mark

x = - 0.2 1 Mark

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Groby 2 tier Module 5 p1 June 20048 (a) The diagram shows a right angled triangle ABC.

AB = 10cm and AC = 15cm.

Calculate the length of BC. Leave your answer as a square root.

15cm

10cmA B

C

102 + BC2 = 152

100 + BC2 = 225

BC2 = 225 – 100

BC2 = 125

BC = √ 125 or 5 √5

3 Marks

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Groby 2 tier Module 5 p1 June 20048 (b) The diagram shows a right – angled triangle DEF. EF = 10cm. Angle F = 50º . Use the table to of data to work out the length of DE.

E F

D

10cm

Not drawn accurately

Angle Sine Cosine Tangent

40º 0.643 0.766 0.839

50º 0.766 0.643 1.192

50º

Given Adjacent side Find Opposite side Use Tan

Tan 50 = DE ÷10

10 X 1.192 = DE

DE = 11 .92 cm 3 Marks

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Groby 2 tier Module 5 p1 June 20049 (a) In the diagram, O is the centre of the circle and P, q and R are points on the circumference. Angle P = 25º.

Work out the size of angle R.

O25º

P

Q

R

Remember angle in a semi circle = 90º

Angle R = 180 – ( 25 + 90)

Angle R = 65º

2 Marks

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Groby 2 tier Module 5 p1 June 20049 (b) A, B, C and D are four points on the circumference of another circle. AC meets BD at X.

Angle ABD = 56º and angle CXD = 80º

Work out the value of d.

You must show all your working.

X

A

BC

D

56º

80º

d

Angle ACD = 56ºAngles in the same segment

d = 180 – ( 80 + 56)

d = 44º

3 Marks

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Groby 2 tier Module 5 p1 June 200410(a) Factorise x2 - 10x + 25

Answer ( x – 5 ) ( x – 5 ) 2 Marks

( b ) Factorise 2x2 + 3x – 5

Answer ( 2x + 5 ) ( x – 1 ) 2 Marks

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Groby 2 tier Module 5 p1 June 200411 (a) The diagram shows a sector of a circle of radius 9 cm

Find the perimeter of the sector. Give your answer in terms of π

Perimeter of sector:

4π + 9 + 9

4π + 183 Marks

Length of arc:

x 2 x π x 9

= 4π

80360

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Groby 2 tier Module 5 p1 June 200411(b). The cross section of a prism is a sector of a circle, of radius 9cm as shown in the diagram. The height of the prism is 10cm. Calculate the volume of the prism. Give your answer in terms of π.

Volume of prism:

= 18π x 10

= 180π cm3

1 mark for units

4 Marks

Area of sector:

x π x 92

= 18π

80360

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Groby 2 tier Module 5 p1 June 2004

90 180 270 360

–3

–2

–1

1

2

3

x

y

90 180 270 360

–3

–2

–1

1

2

3

x

y

12 The diagram shows the graph of

y = sin xº for 0 x 360

(a) Sketch the graph of

y = 2 sin xº for 0 x 360

1 mark

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Groby 2 tier Module 5 p1 June 2004

90 180 270 360

–3

–2

–1

1

2

3

x

y

90 180 270 360

–3

–2

–1

1

2

3

x

y

12(b) Sketch the graph of

y= sin 2xº for 0 x 360

12(c) Sketch the graph of

y = 2 + sin xº for 0 x 360

1 mark

1 mark

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Groby 2 tier Module 5 p1 June 200413. A square – based pyramid A is divided into 2 parts: a square-based pyramid B and a frustum C, as shown.

Pyramid A is similar to pyramid B.

The base of pyramid A is a square of side 10cm. The base of pyramid B is a square of side 5cm.

The vertical height of pyramid A is 12cm.

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Groby 2 tier Module 5 p1 June 2004

13 You are given the formula Volume of a pyramid = ⅓ x area of base x vertical height

Calculate the volume of the frustum C.

Volume of A

⅓ x 10 x 10 x 12 = 400

Volume of B

⅓ x 5 x 5 x 6 = 50

Dimensions of B are ½

the dimensions

of A

Volume of C

400 – 50

= 350cm3

4 Marks

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Groby 2 tier Module 5 p1 June 200414 The diagram shows a triangle ABC. AB = 6cm, BC = 5cm and angle B = 75º

Calculate the area of the triangle.

Give your answer to a suitable degree of accuracy.

5cm

6cm

A

B C75º

You are given that sine 75º = 0.966 (3sf)

Area = ½ x 6 x 5 x sine75

= 15 x 0.966

=14.49

=14.5 cm2 ( 3sf)

3 Marks

Area of triangle = ½ ac sine B (from page 2)

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Groby 2 tier Module 5 p1 June 200415 (a) Find the values of a and b such that

x2 + 6x – 3 = ( x + a )2 + b

Completing the square

x2 + 6x – 3

= ( x + 3 )2 – 12 a= 3, b = –12

2 marks( b) Hence or otherwise, solve the equation

x2 + 6x – 3 = 0 giving your answer in surd form

x2 + 6x – 3 = 0

( x+ 3 )2 – 12 = 0

( x + 3 )2 = 12

x + 3 = ±√ 12

x = –3 ±√12 3 Marks

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Groby 2 tier Module 5 p1 June 2004

Total: out of 70 - a rough guide

grade D C B A A*

score 16 26 36 46 56