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Question 1a part (i)
[41
3− 1
2
5] ÷
4
15= [
13
3−
7
5] ÷
4
15
= [(5×13)−(7×3)
15] ÷
4
15
= [(65)−(21)
15] ÷
4
15
= [44
15] ×
15
4
= 11
Question 1a part (ii)
(3.1−1.15)2
0.005 =
(1.95)2
0.005 = 760.5
Question 1b part (i)
Total cost of phone under Plan A =Deposit +(Monthly Installments × # of months) + Tax
= 400 + (65 × 12) + 0
= $1180
Question 1b
Total cost of phone under Plan B =Deposit +(Monthly Installments × # of months) + Tax
= 600 + (80×6) + 0.05(600+(80×6))
= 600+480+ 54
= $1134
Plan B is the better deal as is has an overall cost (1180-1134 = $46) cheaper than that of
Plan A.
Question 1c part (i)
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No. of kWh used = 0 3 3 0 3 0 0 7 1 1 -
0 0 2 9 6
= 296 kWh
1 kWh = $5.10
296 kWh = 5.10 × 296
= $1509.60
Question 1c part (ii)
At $5.10 per kWh, $2351.10 = 2351.10
5.10
= 461 kWh
At the end of April, the meter should read: = 0 3 3 0 0 4 0 7 6 1 +
0 3 7 6 8
Question 2a part (i)
6y2 – 18xy = 6y×y – 6y × 3x
= 6y (y-3x)
Question 2a part (ii)
4m2 – 1 = (2m-1) (2m+1) (Difference of 2 squares)
Question 2a part (iii)
2t2 -3t – 2 = (2t + 1) (t – 2)
Question 2b
5𝑝+2
3−
3𝑝−1
4=
4(5𝑝+2)−3(3𝑝−1)
12
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= 20𝑝+8−9𝑝+3
12
=11𝑝+11
12
= 11(𝑝+1)
12
Question 2c part (i)
𝑑 = √4ℎ
5 h=29
𝑑 = √4(29)
5= √23.2= 4.82
Question 2c part (ii)
𝑑 = √4ℎ
5
d2 = 4ℎ
5
5d2 = 4h
h = 5𝑑2
4
Question 3a part (i)
U = {3, 4, 5, 6, 7, 8, 9, 10, 11}
M = {3, 5, 7, 9, 11}
Question 3a part (ii)
R = {4, 9}
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Question 3a part (iii)
Question 3b part (i)
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Question 3b part (ii)
Length of BQ = 6.9cm
Question 4a part (i)
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f(x) = 1
3𝑥 − 2
f(3) + f(-3) = [1
33 − 2] + [
1
3(−3) − 2 ]
= (1 – 2) + (-1-2)
= -1 + -3
= -4
Question 4a part (ii)
f(x) = 5
1
3𝑥 − 2 = 5
1
3𝑥 = 7
x = 21
Question 4a part (iii)
f(x) = 1
3𝑥 − 2
Let y = f(x)
y = 1
3𝑥 − 2
Making x the subject of the formula
x = 3y + 6
f-1(x) = 3x + 6 = 3( x+2)
Question 4b part (i)
Line l1:
Points on l1: (0,1) and (2,5)
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Gradient = 5−1
2−0=
4
2= 2
Line l2:
Points on l2: (12,0) and (0,6)
Gradient = 6−0
0−12=
6
−12= −
1
2
Question 4b part (ii)
y intercept of l1 = 1 (c)
Gradient of l1 = 2 (m)
y = mx + c
Equation: y = 2x + 1
Question 4b part (iii)
Gradient of l1 = 2, Gradient of l2 = −1
2
The lines are the negative inverses of each other. This means they are perpendicular.
Question 5a part (i)
<RQT
Angles QRT and QTR are equal, due to it being an isosceles triangle.
So, Angle RQT = 180 – (76×2)
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Angle RQT = 28o
Question 5a part (ii)
<PRT
First, we must find angle PQR.
Angle PQR = 180 – 28
= 152o
Now, we have another isosceles triangle.
Angle PRQ = (180 – 152) ÷2
= 14o
Angle PRT = Angle PRQ +Angle QRT
= 14 + 76
= 90o
Question 5a part (iii)
Given: Angle SRT = 145o, Angle PSR = 100o
Angle PRS = SRT - PRT
= 145 – 90
= 55o
Angle SPR = 180 – (100 + 55)
= 25o
Now, Angle SPT = 25 + 14
= 39o
Question 5b part (i)
A’B’C’ is a 90o clockwise rotation of ABC about the origin, O.
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Question 5b part (ii)
Translating A’B’C through the vector (4
−5)
Question 6a part (i)
Radius = 28
2 = 14 𝑚
Area = 90
360×
22
7× 142
Area = 154 m2
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Question 6a part (ii)
Perimeter = (2×radius) + ( 90
360𝜋𝑑)
= 28 + (90
360×
22
7× 28)
= 50 m
Question 6b part (i)
Using Pythagoras’ Theorem:
BC2 = AC2 + AB2
AC2 = 102 – 62
AC2 = 100 – 36 = 64
AC = √64 = 8𝑐𝑚
Area = 𝑏ℎ
2=
6×8
2= 24𝑐𝑚2
Question 6b part (ii)
Volume of Prism = Cross Sectional Area × Length
540 = 24 x Length
Length = 540
24= 22.5𝑐𝑚
Question 6b part (iii)
Surface Area = (2×Cross Sectional Area) + Areas of Rectangles (ABED + ADFE + BEFC)
2 × Cross Sectional Area = 48cm2
Area of ABED = 6 × 22.5 = 135 cm2
Area of ADFE = 8 × 22.5 = 180 cm2
Area of BEFC = 10 × 22.5 = 225 cm2
Surface Area = 48 + 135 + 180 + 225
Surface Area = 588cm2
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Question 7a part (i)
The upper-class limit is 39
Question 7a part (ii)
The class width is (39.5-19.5) = 20
Question 7a part (iii)
Sixteen vehicles passed a checkpoint at no more than 39.5 kmh-1
Question 7b
Speed (in km-1) Frequency Cumulative Frequency
0-19 5 5
20-39 11 16
40-59 26 42
60-79 37 79
80-99 9 88
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100-119 2 90
Question 7c
Question 7d part (i)
50% of vehicles = 90/2 = 45 vehicles
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Question 7d part (ii)
The estimated speed is 62 km/h
Question 8a
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Question 8b
1 + 2 + 3 + 4 + 5 + 6 = 21
Question 8c
Figure, n Number of Dots, d, in terms of n Number of dots used,
d
1 1
2× 1 × (1 + 1)
1
2 1
2× 2 × (2 + 1)
3
3 1
2× 3 × (3 + 1)
6
11 1
2× 11 × (11 + 1)
66
n
Question 8d
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Figure, n Number of Dots, d, in terms of n Number of dots used,
d
1 1
2× 1 × (1 + 1)
1
2 1
2× 2 × (2 + 1)
3
3 1
2× 3 × (3 + 1)
6
11 1
2× 11 × (11 + 1)
66
n 1
2× 𝑛 × (𝑛 + 1)
1
2𝑛(𝑛 + 1)
Question 8e
Let 1
2𝑛(𝑛 + 1) = 1000
n(n+1) = 2000
This solution will not return an integer. Hence, 1000 is not a valid number of dots
in any diagram of this sequence.
Question 9a part (i) part a
O (0,0) A (25,10)
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Gradient of OA = 10−0
25−10=
2
5
Question 9a part (i) part b
A (25,10) B (40,10)
Gradient of AB = 10−10
40−25= 0
Question 9a part (ii)
The cyclist started from rest, where his velocity was 0 m/s, and steadily increased his velocity
by 2
5 m/s each second during the first 25 seconds. During the next 15 seconds, his velocity
remained constant, that is his acceleration was 0 ms-2
Question 9a part (iii)
Average speed =𝑇𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒
Total distance = Area under graph
= 1
2[(40 − 25) + (40 − 0)] × 10
= 1
2[15 + 40] × 10
= 275 m
Average speed = 275
40=
6.875𝑚
𝑠
Question 9b part (i)
x2 + 2xy = 5 --- 1
x + y = 3 --------2
Substituting (1,2) in Eqn 1:
(1)2 + 2(1)(2) = 5
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1 + 4 = 5
True for equation 1
Substituting (1,2) in Eqn 2:
1 + 2 = 3
True for both equations, hence proven.
Question 9b part (ii)
x2 + 2xy = 5 --- 1
x + y = 3 --------2
From Eqn 2:
y = 3 – x ---------3
Substituting 3 in 1:
x2 + 2x(3-x) = 5
x2 + 6x - 2x2 = 5
- x2 + 6x = 5
x2 - 6x + 5 = 0
(x-1) (x-5) = 0
Question 10a part (i)
<SPQ = 180 – 58 = 122o (Opposite angles of a cyclic quadrilateral are supplementary)
Question 10a part (ii)
x = 1:
y = 3-1
y = 2
x = 5:
y = 3-5
y = -2
When x = 1, y = 2
When x = 5, y = -2
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<SOQ = 2 (SRQ) = 2(58) = 116o
<OQS = 180−116
2= 32o
Question 10b part (i)
Question 10b part (ii)
<ABS = 44o (Alternate angles are equal)
<CBS = 180 – 105 = 75o (2 angles making a straight line are supplementary.)
<ABC = 11+75 = 119o
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Question 10b part (iii)
AC2 = 522 + 722 – 2(52)(72) cos119
= 2704 + 5184 + 3630.25
= 11518.25
AC = √11518.25 = 107.3 𝑘𝑚
Question 10b part (iv)
<BCN = 75o (Co-interior angles are supplementary)
Using the sine rule:
52
𝑠𝑖𝑛𝐴𝐶𝐵=
107.3
𝑠𝑖𝑛119
sinACB = 52𝑠𝑖𝑛119
107.3 = 0.4238
<ACB = sin-1(0.4238) = 25.07o
Bearing of A from C = 360 – (75+25.07) = 259.9o = 260o (To the nearest degree) – Hence
Proven
Question 11a part (i)
A×B = (3 2 5 4 )(4 0 3 − 1 )
= (𝑒11 𝑒12 𝑒21 𝑒22 )
e11 = (3×4) +(2×3) = 18
B×A = (4 0 3 − 1 )(3 2 5 4 )
= (𝑒11 𝑒12 𝑒21 𝑒22 )
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e11 = (4×3) +(0×5) = 12
The first number in both matrix products do not match, hence A×B ≠ B×A
Question 11a part (ii)
det A = (3×4) – (2×5) = 12-10 = 2
A-1 = 1
2(4 − 2 − 5 3 ) = (2 − 1
−5
2 3
2 )
Question 11a part (iii)
A×A-1 = I
I = (1 0 0 1 )
Question 11b part (i)
(3 2 5 4 ) (𝑥
𝑦) = (
1
5)
Question 11b part (ii)
(3 2 5 4 ) (𝑥
𝑦) = (
1
5)
A-1× 𝐴 (𝑥
𝑦) = (
1
5) × A-1
(𝑥
𝑦) = (
1
5) × A-1
(𝑥
𝑦) = (
1
5) (2 − 1
−5
2 3
2 )
Question 11c part (i)
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OS:
OQ = (5
0)
QS = 3 (5
0) = (
15
0)
OS = OQ +QS
= (5
0) + (
15
0)
= (20
0) x = 20, y = 0
PQ:
PQ = PO + OQ
=- (4
3) + (
5
0)
PQ = (1
−3) x = 1, y = -3
RS:
PR = 3OP
= 3 (4
3)
= (12
9)
OR = OP + PR
= (4
3) + (
12
9)
= (16
12)
RS = RO + OS
= -(16
12)+(
20
0)
= (4
−12) x= 4, y = -12
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Question 11c part (ii)
PQ = (1
−3)
RS = (4
−12)
= 4 (1
−3) = 4PQ
RS is a scalar multiple of PQ, hence PQ and RS are parallel.
RS = 4(1
−3)
|RS| = 4|PQ| - This means that RS is quadruple the length of PQ.
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