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Module 5 Paper1 Higher Non Calculator Specimen Paper (non-coursework) Mock exam 2009 1hr 15mins

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Module 5 Paper1 Higher Non Calculator Specimen Paper (non-coursework) Mock exam 2009 1hr 15mins. 1. In the diagram AB is parallel to CD. (a) State the value of x. Give a reason for your answer. Answer:- 130 degrees Reason:- Corresponding angle. 2 Marks.

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slide1

Module 5

Paper1

Higher

Non Calculator

Specimen Paper (non-coursework)

Mock exam 2009

1hr 15mins

slide2

1. In the diagram AB is parallel to CD.

(a) State the value of x. Give a reason for your answer.

Answer:- 130 degrees

Reason:- Corresponding angle

2 Marks

(b) State the value of y. Give a reason for your answer.

Answer:- 130 degrees

Reason:- Alternate angle or vertically opposite

2 Marks

( c) Find the value of z

Answer:- 50 degrees

1 Mark

slide3

2. Javed says that the triangle and parallelogram shown have the same area.

Is he correct?

You MUST show your working.

Triangle area = ½ x 4 x 12

= 24cm2

Parallelogram area = 8 x 3

= 24cm2

He is correct

4 Marks

slide4

3 (a) The nth term of a sequence is 5n-2

Write down the first three terms of the sequence

5 (1) – 2 = 3

5 (2) – 2 = 8

5 (3) – 2 = 13

Answer:- 3, 8, 13

2 Marks

(b) Matchsticks are used to make this pattern of triangles

Find an expression for the number of matchsticks needed to make Pattern n

Answer:- 2n + 1

2 Marks

+2

How many complete triangles can be made with 40 matchsticks?

2n + 1 = 40

2n = 39

n = 19.5

19 complete triangles

n 2n + 1

2 Marks

slide5

4. (a) Complete the table of values for y = x2 + x

Needs to be a smooth curve

2

2 Marks

6

(b) Draw the graph of y = x2 + x for values of x from -4 to +3

x

x

(c ) Write down the values of x where the line y = 5 crosses the graph

Y = 5

x

Answer : 1.8 and -2.8

x

2 Marks

(d) What happens top the value of y between x=-1 and x=0?

x

x

2 Marks

x

x

Answer: It is negative

1 Mark

slide6

Simplify

(a) 2(q + 3) + 3(q – 4)

2q + 6 + 3q - 12

2 Marks

= 5q - 6

(b) x2 X x5

Answer: x7

1 Mark

(c)

1 Mark

Answer: y3

slide7

6.

(a) Describe fully the single transformation that takes shape A onto shape B

Reflection in the line y = 2

2 Marks

(b) Triangle C is rotated onto triangle D.

(i) Write down the angle of rotation.

Answer 180º

(ii) Write down the coordinates of the centre of rotation

Answer ( 3, 0 )

2 Marks

slide8

6 (c)

Draw the new position of shape L

2 Marks

slide9

7. (a) In triangle ABC, angle B = 90º, AB = 9cm and AC = 15cm

Calculate the length of BC

152 = 92 + BC2

225 – 81 = BC2

BC2 = 144

BC =√ (144)

3 Marks

BC = 12 cm

slide10

7(b) In triangle DEF , angle E = 90º, DF = 10cm and angle F = 70º

hypotenuse

opposite

adjacent

Use the table of data to work out the length of EF

Given Hypotenuse

Need to find Adjacent

Use COSINE

Cos 70º =

Cos 70º=

10 x 0.342 = EF

3 Marks

EF = 3.42 cm

slide11

8(a) Expand and simplify (x + 3)(x – 4)

x2 + 3x - 4x - 12

x2 - x - 12

2 Marks

(b) Factorise fully 6a2 – 9ab

3(2)aa - 3(3)ab

3a(2a – 3b)

2 Marks

(c) Factorise x2 – 7x + 10

Hence solve x2 - 7x + 10 = 0

( x - 2 ) ( x - 5 )

Either (x – 2) = 0

Or (x - 5) = 0

3 Marks

x = 2 and x = 5

slide12

The line AB has equation 2x – 5y = 10

Show that the gradient of AB is

Need to write it in the form y=mx+c

Rearrange: 2x – 10 = 5y

÷ each side by 5

Gradient m is

2 Marks

9(b) A second line PQ has equation 2y = 6 - 5x.

Jade says that the line PQ is parallel to the line AB. Is she correct? You MUST show your working.

Divide both sides by 2

Jade is wrong

Gradient is

y = 3 -

2 Marks

slide13

10. The diagram shows two containers. One container is a hemisphere of radius 6cm. The other container is a cone of radius 5cm and height 18cm.

(a) Show that the volume of the hemisphere is 144π

Volume of a sphere is π r3

Which container has the larger volume?

Volume of a hemisphere is π r3

Volume of a cone is πr2h

V = π x 5 x 5 x 18

V = π x 6 x 6 x 6

V = 144π

V = 150π

The cone has the larger volume

5 Marks

slide14

11. Prove that (n+3)2 – (n-2)2 = 5(2n+1)

Multiply out

(n + 3)(n + 3) - (n – 2)(n – 2)

n2 + 3n + 3n + 9 - (n2 - 2n - 2n + 4)

n2 + 6n + 9 – n2 + 4n - 4

10n + 5

Factorise

5( 2n + 1) Hence shown

3 Marks

slide15

12. OABC is a parallelogram.

OA = p and OC = q

T is a point outside the parallelogram such that AT = 2p + 3q

(a) Find in terms of p and q, expressions for the following vectors. Give your answer in its simplest form.

OB = OA + AB

2 Marks

= p + q

Explain what your answers to part (a) tell you

BT = BA + AT

= -q + 2p + 3q

2 Marks

They are in a straight line and the length of BT is twice the length of OB

= 2q + 2p

2 Marks

slide16

13. The diagram shows the graph of y = cos x for 0º ≤ x ≤ 360º

One solution of the equation cos x = 0.809 is x = 36º

Work out the other solution in this range.

Other solution is 360 – 36 = 324º

Write down the number of solutions in the same range for each equation.

(i) 2cos x = -0.6

(ii)cos 2x = -0.6

cos x = -0.3

4 solutions

3 Marks

2 solutions

slide17

Find the values of a and b such that

x2 – 8x + 21 = (x – a) 2 + b

x2 – 8x + 21

= ( x - 4 )2 + 5

a = 4 and b = 5

2 Marks

Hence or otherwise write down

(i) The minimum value of x2 - 8x + 21

Minimum value is the coordinate (4,5)

(ii) The equation of the line of symmetry of x2 - 8x + 21

Answer : x = 4

2 Marks

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