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.
Specimen Paper (non-coursework)
Mock exam 2009
(a) State the value of x. Give a reason for your answer.
Answer:- 130 degrees
Reason:- Corresponding angle
(b) State the value of y. Give a reason for your answer.
Answer:- 130 degrees
Reason:- Alternate angle or vertically opposite
( c) Find the value of z
Answer:- 50 degrees
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
Parallelogram area = 8 x 3
He is correct
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
(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
How many complete triangles can be made with 40 matchsticks?
2n + 1 = 40
2n = 39
n = 19.5
19 complete triangles
n 2n + 1
Needs to be a smooth curve
(b) Draw the graph of y = x2 + x for values of x from -4 to +3
(c ) Write down the values of x where the line y = 5 crosses the graph
Y = 5
Answer : 1.8 and -2.8
(d) What happens top the value of y between x=-1 and x=0?
Answer: It is negative
(a) 2(q + 3) + 3(q – 4)
2q + 6 + 3q - 12
= 5q - 6
(b) x2 X x5
(a) Describe fully the single transformation that takes shape A onto shape B
Reflection in the line y = 2
(b) Triangle C is rotated onto triangle D.
(i) Write down the angle of rotation.
(ii) Write down the coordinates of the centre of rotation
Answer ( 3, 0 )
Draw the new position of shape L
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)
BC = 12 cm
7(b) In triangle DEF , angle E = 90º, DF = 10cm and angle F = 70º
Use the table of data to work out the length of EF
Need to find Adjacent
Cos 70º =
10 x 0.342 = EF
EF = 3.42 cm
x2 + 3x - 4x - 12
x2 - x - 12
(b) Factorise fully 6a2 – 9ab
3(2)aa - 3(3)ab
3a(2a – 3b)
(c) Factorise x2 – 7x + 10
Hence solve x2 - 7x + 10 = 0
( x - 2 ) ( x - 5 )
Either (x – 2) = 0
Or (x - 5) = 0
x = 2 and x = 5
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
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
y = 3 -
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
11. Prove that (n+3) hemisphere of radius 6cm. The other container is a cone of radius 5cm and height 18cm.2 – (n-2)2 = 5(2n+1)
(n + 3)(n + 3) - (n – 2)(n – 2)
n2 + 3n + 3n + 9 - (n2 - 2n - 2n + 4)
n2 + 6n + 9 – n2 + 4n - 4
10n + 5
5( 2n + 1) Hence shown
12. OABC is a parallelogram. hemisphere of radius 6cm. The other container is a cone of radius 5cm and height 18cm.
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
= p + q
Explain what your answers to part (a) tell you
BT = BA + AT
= -q + 2p + 3q
They are in a straight line and the length of BT is twice the length of OB
= 2q + 2p
13. The diagram shows the graph of y = hemisphere of radius 6cm. The other container is a cone of radius 5cm and height 18cm.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
Find the values of a and b such that hemisphere of radius 6cm. The other container is a cone of radius 5cm and height 18cm.
x2 – 8x + 21 = (x – a) 2 + b
x2 – 8x + 21
= ( x - 4 )2 + 5
a = 4 and b = 5
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