workbook_full_solutions_2.pdf

WORKBOOK FULL SOLUTIONS
NEW SYLLABUS
MATHEMATICS
7th
EDITION
2
with
New Trend
Questions
1

1
1
ANSWERS
Chapter 1 Direct and Inverse Proportions
Basic
1. Cost of 15 l of petrol =
$14.70
7
× 15
= $31.50
2. (i) y = kx
When x = 200, y = 40,
40 = k(200)
k =
40
200
=
1
5
y =
1
5
x
(ii) When x = 15,
y =
1
5
(15)
= 3
(iii) When y = 8,
8=
1
5
x
x = 40
3. (i) s = kt2
When t = 4, s = 8,
8 = k(4)2
k =
8
16
=
1
2
s =
1
2
t2
(ii) When t = 3,
s =
1
2
(3)2
= 4
1
2
(iii) When s = 32,
32 =
1
2
t2
t2
= 64
t = ±8
4. (i) y = k(4x + 1)
When x = 2, y = 3,
3 = k(8 + 1)
k =
3
9
=
1
3
y =
1
3
(4x + 1)
(ii) When x = 5,
y =
1
3
(20 + 1)
= 7
(iii) When y = 11,
11 =
1
3
(4x + 1)
33 = 4x + 1
4x = 32
x = 8
5. (i) D3
= kL
When L = 6, D = 2,
23
= k(6)
k =
8
6
=
4
3
D3
=
4
3
L
(ii) When L = 48,
D3
=
4
3
(48)
= 64
D = 4
(iii) When D =
2
3
,
2
3






3
=
4
3
L
8
27
=
4
3
L
L =
8
27
÷
4
3
=
2
9

1 2
6. Time taken for 1 tap to fill the bath tub = 15 × 2
= 30 minutes
Time taken for 3 taps to fill the bath tub =
30
3
= 10 minutes
7. (i) When x = 5,
y = 100 × 2
= 200
(ii) y =
k
x
When x = 10, y = 100,
100 =
k
10
k = 1000
y =
x
1000
(iii) When y = 80,
80 =
x
1000
x =
1000
80
= 12.5
8. (i) y =
k
x
When x = 16, y = 5,
5 =
k
16
=
k
4
k = 20
y =
x
20
(ii) When x = 100,
y =
20
100
=
20
10
= 2
(iii) When y = 4,
4 =
x
20
x = 5
x = 52
= 25
Intermediate
9. (i) a = kb
When b = 15, a = 75,
75 = k(15)
k =
75
15
= 5
a = 5b
When b = 37.5,
a = 5(37.5)
= 187.5
(ii) When a = 195,
195 = 5b
b =
195
5
= 39
10. h = kl
When l = 36, h = 30,
30 = k(36)
k =
30
36
=
5
6
h =
5
6
l
When h = 15,
15 =
5
6
l
l =
6
5
× 15
= 18
When l = 72,
h =
5
6
(72)
= 60
When h = 75,
75 =
5
6
l
l =
6
5
× 75
= 90
h 15 30 60 75
l 18 36 72 90
11. (i) w = kt
When t = 0.3, w = 1.8,
1.8 = k(0.3)
k =
1.8
0.3
= 6
w = 6t

1
3
(ii) When t = 2.5,
w = 6(2.5)
= 15
15 g of silver will be deposited.
(iii) w
O
(0.3, 1.8)
w = 6t
t
12. (i) F = km
When m = 250, F = 60,
60 = k(250)
k =
60
250
=
6
25
F =
6
25
m
(ii) When m = 300,
F =
6
25
(300)
= 72
The net force required is 72 newtons.
(iii) When F = 102,
102 =
6
25
m
m =
25
6
× 102
= 425
The mass of the box is 425 kg.
(iv)
F
O
(250, 60)
F =
6
25
m
m
13. (i) C = an + b
When n = 200, C = 55 000,
55 000 = 200a + b —(1)
When n = 500, C = 62 500,
62 500 = 500a + b —(2)
(2) – (1): 300a = 7500
a =
7500
300
= 25
Substitute a = 25 into (1):
200(25) + b = 55 000
5000 + b = 55 000
b = 55 000 – 5000
= 50 000
a = 25, b = 50 000
(ii) C = 25n + 50 000
When n = 420,
C = 25(420) + 50 000
= 60 500
The total cost is $60 500.
(iii) When C = 70 000,
70 000 = 25n + 50 000
25n = 20 000
n =
20 000
25
= 800
(iv) C
50 000 (200, 55 000)
(500, 62 500)
C = 25n + 50 000
n
O
No, C is not directly proportional to n since the graph of
C against n does not pass through the origin.
14. (i) Annual premium payable = $25 +
$20 000
$1000
× $2
= $65
(ii) Face value = ($155 – $25) ×
$1000
$2
= $65 000
(iii) p = 25 +
n
1000
× 2
= 25 +
2n
1000
= 25 +
n
500

1 4
(iv) p
n
(20 000, 65)
p = 25 +
n
500
25
O
No, p is not directly proportional to n since the graph of
p against n does not pass through the origin.
15. (a) y and x5
(b) y3
and x
(c) (y – 2)2
and x
16. (i) n = km3
When m = 1
1
2
, n = 27,
27 = k
3
2






3
=
27
8
k
k = 8
n = 8m3
When m = 2,
n = 8(2)3
= 64
(ii) When n = 125,
125 = 8m3
m3
=
125
8
m =
5
2
= 2
1
2
17. y = k(x + 2)(x + 7)
When x = 1, y = 4,
4 = k(3)(8)
k =
4
24
=
1
6
y =
1
6
(x + 2)(x + 7)
When x = 5,
y =
1
6
(7)(12)
= 14
18. (i) h2
= kl
When l =
1
8
, h =
1
2
,
1
2






2
= k
1
8






1
4
=
1
8
k
k = 2
h2
= 2l
When l = 8,
h2
= 2(8)
= 16
h = ±4
(ii) When h = 6,
62
= 2l
36 = 2l
l =
36
2
= 18
19. y = k +
x 1
When x = 224, y = 5,
5 = k 225
= 15k
k =
5
15
=
1
3
y = +
x
1
3
1
When x = –1, y = p,
p =
1
3
0
= 0
When x = q, y = 3
1
3
,
3
1
3
= +
q
1
3
1
10 = +
q 1
q + 1 = 100
q = 99
p = 0, q = 99

1
5
20. (i) m = kr3
When r = 3, m = 54,
54 = k(3)3
= 27k
k =
54
27
= 2
m = 2r3
(ii) When r = 4,
m = 2(4)3
= 128
The mass of the sphere is 128 g.
21. (i) v = k r
When r = 121, v = 22,
22 = k 121
= 11k
k =
22
11
= 2
v = 2 r
When r = 81,
v = 2 81
= 18
The safe speed is 18 m/s.
(ii) When v = 11,
11 = 2 r
r =
11
2
r =
11
2






2
= 30.25
The radius is 30.25 m.
22. H = kd3
When d = 6, H = 120,
120 = k(6)3
= 216k
k =
120
216
=
5
9
H =
5
9
d3
When d = 9,
H =
5
9
(9)3
= 405
The shaft can transmit 405 horsepower.
23. Number of workers to complete in 1 day = 6 × 8
= 48
Number of workers to complete in 12 days =
48
12
= 4
24.
Number of
girls
Number of
paper cranes
Number of
minutes
8 5 6
8 120 144
1 120 1152
36 120 32
× 24 × 24
× 8
÷ 36
÷ 8
× 36
36 girls take 32 minutes to fold 120 paper cranes.
Assume that all the girls have the same rate of folding
paper cranes.
25. (i) f =
k
w
When w = 1.5 × 103
, f = 2.0 × 105
,
2.0 × 105
=
×
k
1.5 103
k = (2.0 × 105
) × (1.5 × 103
)
= 3.0 × 108
f =
×
w
3.0 108
When w = 480,
f =
×
3.0 10
480
8
= 625 000
The frequency is 625 000 Hz.
(ii) When f = 9.6 × 105
,
9.6 × 105
=
×
w
3.0 108
w =
×
×
3.0 10
9.6 10
8
5
= 312.5
The wavelength is 312.5 m.

1 6
26. P =
k
V
When V = 2, P = 500,
500 =
k
2
k = 500 × 2
= 1000
P =
V
1000
When V = 5,
P =
1000
5
= 200
The pressure of the gas is 200 pascals.
27. (a) y and x5
(b) y2
and x
(c) y – 1 and x
28. y =
k
x2
When x = 4, y = 5,
5 =
k
42
k = 5 × 16
= 80
y =
x
80
2
When x = 2,
y =
80
22
= 20
29. y =
+
k
r 1
2
When r = 1, y = 32,
32 =
+
k
1 1
=
k
2
k = 64
y =
+
r
64
1
2
When r = 7,
y =
+
64
7 1
2
= 1
7
25
30. u =
k
v
When v = 9, u = 10,
10 =
k
9
k = 10 × 3
= 30
u =
v
30
When v = 25,
u =
30
25
= 6
31. y =
k
x2
When x = 10, y = 2,
2 =
k
102
k = 2 × 100
= 200
y =
x
200
2
When x = 5,
y =
200
52
= 8
When y =
8
9
,
8
9
=
x
200
2
x2
= 225
x = ±15
x 5 10 15
y 8 2
8
9
32. (i) F =
k
R2
(ii) When R = 32, F = 50,
50 =
k
322
k = 50 × 322
= 51 200
(iii) F =
R
51 200
2
When F = 512,
512 =
R
51 200
2
R2
= 100
R = ±10
R = 10

1
7
Advanced
33. y = kx3
When x = a, y = p,
p = ka3
When x =
1
2
a,
y = k
1
2
a






3
=
1
8
ka3
=
1
8
p
34. (i) W =
k
d2
When d = 6500, W = 800,
800 =
k
65002
k = 800 × 65002
= 3.38 × 1010
W=
×
d
3.38 1010
2
When d = 2.5 × 104
+ 6500 = 31 500,
W=
×
3.38 10
31 500
10
2
= 34.1 (to 3 s.f.)
The weight of the astronaut is 34.1 N.
(ii) When W = 400,
400 =
×
d
3.38 1010
2
d2
= 8.45 × 107
d = ±9190 (to 3 s.f.)
The astronaut is 9190 km above the centre of the
earth.
New Trend
35. Number of pails Tap A can fill in 1 minute =
8
5
Number of pails Tap B can fill in 1 minute =
7
4
Number of pails Tap A and B can fill in 1 minute
=
8
5
+
7
4
= 3
7
20
Time taken to fill 16 pails =
16
3
7
20
= 4
52
67
= 4 min 47 s (to nearest
second)
36. (a) f = k T
3
When f = 320, T = 64,
320 = k 64
3
k =
320
4
= 80
f = 80 T
3
(b) When f = 450,
450 = 80 T
3
T
3
=
450
80
T =
91125
512
= 178 N (to 3 s.f.)
(c)
f1
f2
=
80 T1
3
80 T2
3
1
2
=
T1
3
T2
3
T1
T2
=
1
8
The ratio of the tensions in the string is 1 : 8.
37. (a) (i) P = kR2
When P = 200, R = 5,
200 = k(5)2
k =
200
25
= 8
P = 8R2
(ii) When R = 20,
P = 8(20)2
= 3200 kPa
(b)
PA
PB
=
1.8
( )
2
1
( )
2
= 1
4
5






2
=
9
5






2
=
81
25
The ratio of the pressure acting on disc A to the
pressure acting on disc B is 81 : 25.

1 8
38. (i) s = kt2
When t = 3, s = 45,
45 = k(3)2
= 9k
k =
45
9
= 5
s = 5t2
When t = 7,
s = 5(7)2
= 245
The distance is 245 m.
(ii) When s = 20,
20 = 5t2
t2
=
20
5
= 4
t = ±2
The time taken is 2 s.

1
Chapter 2 Linear Graphs and Simultaneous
Linear Equations
Basic
1. (a) Take two points (0, 2) and (7, 2).
Vertical change (or rise) = 2 – 2 = 0
Horizontal change (or run) = 7 – 0 = 7
Gradient =
rise
run
=
0
7
= 0
(b) Take two points (7, 0) and (7, 7).
Gradient =
rise
run
=
7
0
= undefined
(c) Take two points (0, 2) and (4, 6).
Since the line slopes upwards from the left to the right,
its gradient is positive.
Gradient =
rise
run
=
4
4
= 1
(d) Take two points (4, 6) and (7, 0).
Since the line slopes downwards from the left to the
right, its gradient is negative.
Gradient =
rise
run
= –
6
3
= –2
2. (a) Take two points (−3, 4) and (4, 4).
Horizontal change (or run) = 4 – (−3) = 7
Gradient =
rise
run
=
0
7
= 0
(b) Take two points (−3, −3) and (4, −3).
Vertical change (or rise) = −3 – (−3) = 0
Gradient =
rise
run
=
0
7
= 0
(c) Take two points (−3, 4) and (−3, −3).
Vertical change (or rise) = 4 – (−3) = 7
Horizontal change (or run) = −3 – (−3) = 0
Gradient =
rise
run
=
7
0
= undefined
(d) Take two points (−4, 4) and (0, −3).
Gradient =
rise
run
= –
7
4
(e) Take two points (0, −3) and (4, 4).
Gradient =
rise
run
=
7
4
3.
2
1
–1
–1 1
x
y
2
(a) y =
1
2
(c) y = –1
1
2
(b) y = –4
(d) y = 0
0
–2
–2
–3
–4
4.
x
y
4
3
2
1
–1
–2
–3 –2 –1 1 2
0
(b) x = –3 (c) x = –
1
2
(a) x = 2
(d) x = 0
9

1
5. (i) Line 1: x = 1
Line 2: x = –1.2
Line 3: y = 2
Line 4: y = –2.6
(ii) Area enclosed = (2.2)(4.6)
= 10.12 units2
6. (a) y = x + 2
x 0 1 2
y 2 3 4
y = –2x + 2
x 0 1 2
y 2 0 –2
y
x
0
–1
1 2
y = x + 2
y = –2x + 2
1
2
3
4
–2
From the graph,
x = 0 and y = 2
(b) 8x + 3y = 7
x 0 1 2
y 2.3 –0.3 –3
2x + y = 2
x 0 1 2
y 2 0 –2
x
–1
1 2
2x + y = 2
8x + 3y = 7
1
0
2
–2
–3
y
From the graph,
x =
1
2
and y = 1.
(c) 3x + y = 13
x 0 2 4
y 13 7 1
5x – y = 35
x 0 5 10
y –35 –10 15
y
15
10
5x – y = 35
3x + y = 13
5
–5
2 4 6 8 10
0
–10
–15
–20
–25
–30
–35
x
From the graph,
x = 6 and y = –5.
(d) 5x – 3y = 23
x 0 2 4
y –7.7 –4.3 –1
x – 7y = 11
x 0 2 4
y –1.6 –1.3 –1
y
–2
–1
1
2
x – 7y = 11
5x – 3y = 23
1 2 3 4
0
–3
–4
–5
–6
–7
–8
x
From the graph,
x = 4 and y = –1.
10

1
7. (a) x + y = 7 —(1)
x – y = 3 —(2)
(1) + (2): 2x = 10
x = 5
Substitute x = 5 into (1):
5 + y = 7
y = 2
x = 5, y = 2
(b) 5x – 4y = 18 —(1)
3x + 2y = 13 —(2)
(2) × 2: 6x + 4y = 26 —(3)
(1) + (3): 11x = 44
x = 4
5(4) – 4y = 18
20 – 4y = 18
4y = 2
y =
1
2
x = 4, y =
1
2
(c) x + 3y = 7 —(1)
x + y = 3 —(2)
(1) – (2): 2y = 4
y = 2
Substitute y = 2 into (2):
x + 2 = 3
x = 1
x = 1, y = 2
(d) 3x – 5y = 19 —(1)
5x + 2y = 11 —(2)
(1) × 2: 6x – 10y = 38 —(3)
(2) × 5: 25x + 10y = 55 —(4)
(3) + (4): 31x = 93
x = 3
5(3) + 2y = 11
15 + 2y = 11
2y = –4
y = –2
x = 3, y = –2
(e) 3x – 4y = 30 —(1)
2x – 7y = 33 —(2)
(1) × 2: 6x – 8y = 60 —(3)
(2) × 3: 6x – 21y = 99 —(4)
(3) – (4): 13y = –39
y = –3
Substitute y = –3 into (2):
2x – 7(–3) = 33
2x + 21 = 33
2x = 12
x = 6
x = 6, y = –3
8. (a) 3x + y = 17 —(1)
3x – y = 19 —(2)
From (1),
y = 17 – 3x —(3)
Substitute (3) into (2):
3x – (17 – 3x) = 19
3x – 17 + 3x = 19
6x = 36
x = 6
y = 17 – 3(6)
= –1
x = 6, y = –1
(b) 2x – y = 3 —(1)
x + y = 0 —(2)
From (1),
y = 2x – 3 —(3)
x + (2x – 3) = 0
x + 2x – 3 = 0
3x = 3
x = 1
y = 2(1) – 3
= 2 – 3
= –1
x = 1, y = –1
11

1
(c) 3x + 3 = 6y —(1)
x – y = 1 —(2)
From (2),
y = x – 1 —(3)
3x + 3 = 6(x – 1)
= 6x – 6
3x = 9
x = 3
y = 3 – 1
= 2
x = 3, y = 2
(d) 6x + 2y = –3 —(1)
4x – 7y = 23 —(2)
From (1),
y =
–3 – 6x
2





 —(3)
4x – 7
–3 – 6x
2





 = 23
8x + 21 + 42x = 46
50x = 25
x =
1
2
Substitute x =
1
2
into (3):
y =
–3 – 6
1
2
( )
2
= –3
x =
1
2
, y = –3
(e) 5x + y = 7 —(1)
3x – 5y = 13 —(2)
From (1),
y = 7 – 5x —(3)
3x – 5(7 – 5x) = 13
3x – 35 + 25x = 13
28x = 48
x = 1
5
7
Substitute x = 1
5
7
into (3):
y = 7 – 5 1
5
7






= –1
4
7
x = 1
5
7
, y = –1
4
7
9. (a) 3x – y = –1 —(1)
x + y = –3 —(2)
(1) + (2): 4x = –4
x = –1
Substitute x = –1 into (2):
–1 + y = –3
y = –2
x = –1, y = –2
(b) 2x – 3y = 13 —(1)
3x – 12y = 42 —(2)
From (2),
x – 4y = 14
x = 4y + 14 —(3)
2(4y + 14) – 3y = 13
8y + 28 – 3y = 13
5y = –15
y = –3
x = 4(–3) + 14
= –12 + 14
= 2
x = 2, y = –3
(c) 14x + 6y = 9 —(1)
6x – 15y = –2 —(2)
(1) × 5: 70x + 30y = 45 —(3)
(2) × 2: 12x – 30y = –4 —(4)
(3) + (4): 82x = 41
x =
1
2
Substitute x =
1
2
into (2):
6
1
2





 – 15y = –2
3 – 15y = –2
15y = 5
y =
1
3
x =
1
2
, y =
1
3
(d) 8x + y = 24 —(1)
4x – y = 6 —(2)
(1) + (2): 12x = 30
x = 2
1
2
Substitute x = 2
1
2
into (2):
4 2
1
2





 – y = 6
10 – y = 6
y = 4
x = 2
1
2
, y = 4
12

1
(e) 3x + 7y = 17 —(1)
3x – 6y = 4 —(2)
(1) – (2): 13y = 13
y = 1
3x + 7(1) = 17
3x + 7 = 17
3x = 10
x = 3
1
3
x = 3
1
3
, y = 1
(f) 7x – 3y = 6 —(1)
7x – 4y = 8 —(2)
(1) – (2): y = –2
7x – 3(–2) = 6
7x + 6 = 6
7x = 0
x = 0
x = 0, y = –2
Intermediate
10. For L1:
Since the line slopes upwards from the left to the right, its
gradient is positive.
m = gradient of line
=
4
4
= 1
c = y-intercept
= 2
For L2:
= 8
4
= 2
c = y-intercept
= −2
For L3:
Since the line slopes downwards from the left to the right,
its gradient is negative.
= –
4
4
= −1
c = y-intercept
= 4
11. (i)
x –4 0 2 4
y =
1
2
x + 1
y =
1
2
(–4) + 1
= –1
y =
1
2
(0) + 1
= 1
y =
1
2
(2) + 1
= 2
y =
1
2
(4) + 1
= 3
(ii)
–2
–3
–4 1
–1 4
2 3
0
1
2
3
y
–1
x
y =
1
2 x + 1
(iii) From the graph, the point (3, 2.5) lies on the line
but the point –1, –
1
2





 does not lie on the line.
(iv) From the graph, the line cuts the x-axis at
x = −2. The coordinates are (−2, 0).
(v) Vertical change (or rise) = 3 – (−1) = 4
=
4
8
=
1
2
12. The equation of a straight line is in the form of
y = mx + c, where m is the gradient. So, to find the gradient
of the lines, express the equation of the given lines to be
in the form of the equation of a straight line.
(a) y + x = 5
y = –x + 5
From the equation, the value of the gradient
m is –1.
13

1
(b) 3y + x = 6
3y = –x + 6
y
3
3
=
+
x
– 6
3
y =
x
–
3
+ 2
= –
1
3
x + 2
From the equation, the value of m is –
1
3
.
(c) 2y + 3x = 7
2y = –3x + 7
y
2
2
=
+
x
–3 7
2
y =
x
–3
2
+
7
2
From the equation, the value of m is –
3
2
.
(d) 2x − 5y = 9
2x = 9 + 5y
2x – 9 = 5y
5y = 2x – 9
y
5
5
=
x
2 – 9
5
y =
x
2
5
–
9
5
From the equation, the value of m is
2
5
.
(e) 4x – 6y + 1 = 0
4x + 1 = 6y
6y = 4x + 1
y
6
6
=
+
x
4 1
6
y =
x
4
6
–
1
6
y =
x
2
3
–
1
6
2
3
.
(f)
1
2
x –
2
3
y – 5= 0
2
3
y =
1
2
x – 5
y =
3
4
x – 7
1
2
3
4
.
13. (a)
–1 0 1 2
1
2
3
4
5
6
7
8
4y + 2x = 0
2y + 6x = 10
y
x
For 4y + 2x = 0,
Vertical change (or rise) =
1
2
− 0
=
1
2
Horizontal change (or run) = 0 − (–1)
= 1
Since the line slopes downwards from the left to
the right, its gradient is negative.
=
1
2
1
= −
1
2
For 2y + 6x = 10,
Vertical change (or rise) = 5 − 2
= 3
Horizontal change (or run) = 1 − 0
= 1
= –
3
1
= −3
14

1
(b)
–1
1
2
3
4
5
–2
–3
–4
–5
0 1 2 3 4
5x – 2y = 10
2y = x + 2
y
x
For 2y = x + 2,
Vertical change (or rise) = 2
1
2
− 1
1
2
= −1
= −2
Since the line slopes upwards from the left to the
right, its gradient is positive.
=
1
2
For 5x − 2y = 10,
Vertical change (or rise) = 2
1
2
− Q
−2
1
2 R
= 5
= 2
=
5
2
= 2
1
2
(c)
–1 1 2 3 4 5 6 7
–8
–6
–4
–2
2
4
6
8
10
12
0
7x + y = 12
5y + 6x = 2
y
x
For 7x + y = 12,
= 7
= 1
=
7
1
= −7
For 5y + 6x = 2,
Vertical change (or rise) = –2 − (−8)
= 6
= 5
= −
6
5
15

1
(d)
–6 –4 –2 2 4 6 8 10
–1
1
2
–2
–3
–4
0
y
x
1
2
x +
1
2
y = 1
1
5
x –
1
2
y = 1
1
10
For
1
2
x +
1
2
y = 1,
Vertical change (or rise) = 0 − (−4)
= 4
= 4
= –
4
4
= −1
For
1
5
x −
1
2
y = 1
1
10
,
Vertical change (or rise) = 1 − (−1)
= 2
= 5
=
2
5
14. (i) From the graph, the value of x can be obtained by
taking the value of the y-intercept, i.e. when the
number of units used is zero.
x = 14
The value of y can be obtained by find the gradient of
the line since the gradient, in this case, represents the
cost for every unit of electricity used.
y = m = gradient of line
=
40
400
=
1
10
(ii) From the graph, the cost of using 300 units of
electricity is $44.
(iii) From the graph, the number of units of electricity used
if the cost is $32 is 180.
15. (a)
–2
–3
–4
P(–4, 3)
S(4, 1)
R(1, 7)
Q(–2, 6)
1
–1 4
2 3
0
2
1
4
3
6
7
8
5
y
x
(b) (i) Vertical change (or rise) = 6 – 3 = 3
Horizontal change (or run) = –2 – (–4) = 2
Gradient of line =
3
2
(ii) Vertical change (or rise) = 7 – 6 = 1
Horizontal change (or run) = 1 – (–2) = 3
Gradient of line =
1
3
(iii) Vertical change (or rise) = 7 – 1 = 6
Gradient of line = –
6
3
= –2
(iv) Vertical change (or rise) = 3 – 1 = 2
2
8
= –
1
4
(c) From the graph, the coordinates of the point is
(0, 2).
16

1
16. (a)
–2
–3 1
–1
–1
2
0
1
2
y
–2
x
Y(2, 2)
X(–3, 1)
W(–3, –2)
Z(2, –1)
(b) (i) Vertical change (or rise) = 1 – (–2) = 3
Horizontal change (or run) = −3 – (−3) = 0
Gradient of line =
3
0
= undefined
Gradient of line =
1
5
(iii) Vertical change (or rise) = 2 – (−1) = 3
Gradient of line =
3
0
= undefined
(iv) Vertical change (or rise) = –1 – (–2)
Gradient of line =
1
5
(c) The quadrilateral WXYZ is a parallelogram.
17. (a)
x –2 0 2
y 1 5 9
(b)
x
y
–2 –1 0
2
y = 3
d(i)
y – 2x – 5 = 0
4
6
8
1 2
(c) From the graph, k = 8.
(d) (ii) x = –1
18. (a) y – 2x = 4
y = 2x + 4
When x = –6,
y = –8
a = –8
When x = 4,
y = 12
b = 12
(b)
–6 –4 –2
–2
2
4
6
8
10
12
–4
–6
–8
2
y – 2x = 4
x = 3
0 4
x
y
d(i)
(c) h = –4
(d) (ii) Area of triangle
=
1
2
× 5 × 10
= 25 units2
17

1
19. (i) y
x
0
–2
2
2
x = 4
y = –2
y = x – 2
4
(ii) Area of triangle =
1
2
× 4 × 4
= 8 units2
20. (i)
–4 –2
–2
2
0
4
6
y =
2
3
x + 3
y = –
1
2
(x + 1)
x = 3
y
2
x
(ii) Area of triangle =
1
2
× 7 × 6
= 21 units2
21. (i)
2 4
2
0
4
6
y
6 x + y = 6
y = 2
y = x
x
(ii) Area of trapezium =
1
2
× (2 + 6) × 2
= 8 units2
22. (a) y = –5x —(1)
y = 5 —(2)
5 = –5x
x = –1
Coordinates of vertices of triangle are (0, 0),
(–1, 5) and (5, 5).
Area of shaded region =
1
2
[5 – (–1)](5)
= 15 units2
(b) Coordinates of vertices of triangle are (0, 0), (4, 2)
and (4, 4).
1
2
(4 – 2)(4)
= 4 units2
(c) y = –x – 3
When y = 0,
–x – 3 = 0
x = –3
2y = x – 6
When y = 0,
x – 6 = 0
x = 6
Coordinates of vertices of triangle are (–3, 0),
(6, 0) and (0, –3).
1
2
[6 – (–3)](3)
= 13.5 units2
(d) 2y = x – 2
When x = 0,
2y = –2
y = –1
y = –
1
8
x + 4 —(1)
2y = x – 2 —(2)
2 –
1
8
x + 4





 = x – 2
–
1
4
x + 8 = x – 2
5
4
x = 10
x = 8
y = 3
Coordinates of vertices of triangle are (0, –1),
(0, 4) and (8, 3).
1
2
[4 – (–1)](8)
= 20 units2
18

1
23. (a) 4x – 6y = 12 —(1)
2x + 4y = –4.5 —(2)
(1) ÷ 2: 2x – 3y = 6 —(3)
(2) – (3): 7y = –10.5
y = –1.5
Substitute y = –1.5 into (3):
2x – 3(–1.5) = 6
2x + 4.5 = 6
2x = 1.5
x = 0.75
x = 0.75, y = –1.5
(b) 3x – 5y = 2 —(1)
x – 2y =
4
15
—(2)
(2) × 3: 3x – 6y =
4
5
—(3)
(1) – (3): y =
6
5
= 1
1
5
Substitute y = 1
1
5
into (2):
x – 2 1
1
5





 =
4
15
x =
8
3
= 2
2
3
x = 2
2
3
, y = 1
1
5
(c) 5x – 8y = 23
1
2
—(1)
4x + y = 22
1
2
—(2)
(2) × 8: 32x + 8y = 180 —(3)
(1) + (3): 37x = 203
1
2
x = 5
1
2
Substitute x = 5
1
2
into (2):
4 5
1
2





 + y = 22
1
2
22 + y = 22
1
2
y =
1
2
x = 5
1
2
, y =
1
2
(d) 5x – 3y = 1.4 —(1)
2x + 5y = 14.2 —(2)
(1) × 2: 10x – 6y = 2.8 —(3)
(2) × 5: 10x + 25y = 71 —(4)
(4) – (3): 31y = 68.2
y = 2.2
Substitute y = 2.2 into (2):
2x + 5(2.2) = 14.2
2x + 11 = 14.2
2x = 3.2
x = 1.6
x = 1.6, y = 2.2
24. (a) 15x – 7y = 14
1
4
—(1)
5x – y = 3
3
4
—(2)
From (2),
y = 5x – 3
3
4
—(3)
15x – 7 5x – 3
3
4





 = 14
1
4
15x – 35x +
105
4
=
57
4
20x = 12
x =
3
5
Substitute x =
3
5
into (3):
y = 5
3
5





 – 3
3
4
= 3 – 3
3
4
= –
3
4
x =
3
5
, y = –
3
4
(b) 3x + 1.4y = 0.1 —(1)
x – 3.6y = 10.2 —(2)
From (2),
x = 3.6y + 10.2 —(3)
3(3.6y + 10.2) + 1.4y = 0.1
10.8y + 30.6 + 1.4y = 0.1
12.2y = –30.5
y = –2.5
Substitute y = –2.5 into (3):
x = 3.6(–2.5) + 10.2
= 1.2
x = 1.2, y = –2.5
19

1
(c)
1
2
x –
1
3
y –1 = 0 —(1)
x + 6y + 8 = 0 —(2)
From (2),
x = –6y – 8 —(3)
1
2
(–6y – 8) –
1
3
y – 1 = 0
–3y – 4 –
1
3
y – 1 = 0
–
10
3
y = 5
y = –
3
2
= –1
1
2
Substitute y = –1
1
2
into (3):
x = –6 –1
1
2





 – 8
= 9 – 8
= 1
x = 1, y = –1
1
2
(d) 3x – 2y = 8 —(1)
1
8
x +
1
2
y = 1.25 —(2)
From (2),
1
2
y = 1.25 –
1
8
x
y = 2.5 –
1
4
x —(3)
3x – 2 2.5 –
1
4
x





 = 8
3x – 5 +
1
2
x = 8
7
2
x = 13
x =
26
7
= 3
5
7
Substitute x = 3
5
7
into (3):
y = 2
1
2
–
1
4
3
5
7






= 1
4
7
x = 3
5
7
, y = 1
4
7
25. (a) 3x + 2y + 7 = 0 —(1)
5x – 2y + 1 = 0 —(2)
(1) + (2): 8x + 8 = 0
8x = –8
x = –1
3(–1) + 2y + 7 = 0
–3 + 2y + 7 = 0
2y = –4
y = –2
x = –1, y = –2
(b) 2y – 7x + 69 = 0 —(1)
4x – 3y – 45 = 0 —(2)
(1) × 3: 6y – 21x + 207 = 0 —(3)
(2) × 2: 8x – 6y – 90 = 0 —(4)
(3) + (4): –13x + 117 = 0
13x = 117
x = 9
2y – 7(9) + 69 = 0
2y – 63 + 69 = 0
2y = –6
y = –3
x = 9, y = –3
(c) 0.5x – 0.2y = 2 —(1)
2.5x + 0.6y = 2 —(2)
(1) × 3: 1.5x – 0.6y = 6 —(3)
(2) + (3): 4x = 8
x = 2
0.5(2) – 0.2y = 2
1 – 0.2y = 2
0.2y = –1
y = –5
x = 2, y = –5
(d) x +
1
2
y = 9 —(1)
3x – 2y = 13 —(2)
(1) × 4: 4x + 2y = 36 —(3)
(2) + (3): 7x = 49
x = 7
7 +
1
2
y = 9
1
2
y = 2
y = 4
x = 7, y = 4
20

1
(e)
1
3
(x + 1) + y – 8 = 0 —(1)
x + 4 =
+
y 1
3
—(2)
From (1),
x + 1 + 3y – 24 = 0
x = 23 – 3y —(3)
23 – 3y + 4 =
+
y 1
3
27 – 3y =
+
y 1
3
81 – 9y = y + 1
10y = 80
y = 8
x = 23 – 3(8)
= 23 – 24
= –1
x = –1, y = 8
(f)
1
5
x +
3
4
y = –1
1
2
—(1)
5
6
x –
1
8
y = 13
1
4
—(2)
(1) × 20: 4x + 15y = –30 —(3)
(2) × 24: 20x – 3y = 318 —(4)
From (4),
3y = 20x – 318 —(5)
4x + 5(20x – 318) = –30
4x + 100x – 1590 = –30
104x = 1560
x = 15
3y = 20(15) – 318
= –18
y = –6
x = 15, y = – 6
(g)
1
3
x –
2
3
y + 5 = 0 —(1)
1
2
x +
1
3
y –
1
2
= 0 —(2)
(1) × 3: x – 2y + 15 = 0 —(3)
(2) × 6: 3x + 2y – 3 = 0 —(4)
(3) + (4): 4x + 12 = 0
4x = –12
x = –3
–3 – 2y + 15 = 0
2y = 12
y = 6
x = –3, y = 6
(h)
+
x y
y
13 – 7
=
1
3
—(1)
+
x y
y x
4 – 4 – 3
6 – 3 2
=
4
3
—(2)
From (1),
3x + 3y = 13 – 7y
3x + 10y = 13 —(3)
From (2),
12x – 12y – 9 = 24y – 12x + 8
24x – 36y = 17 —(4)
From (3),
3x = 13 – 10y —(5)
8(13 – 10y) – 36y = 17
104 – 80y – 36y = 17
116y = 87
y =
3
4
Substitute y =
3
4
into (5):
3x = 13 – 10
3
4






=
11
2
x =
11
6
= 1
5
6
x = 1
5
6
, y =
3
4
26. (a) 4x + 4 = 5x = 60y – 100
4x + 4 = 5x —(1)
5x = 60y – 100 —(2)
From (1),
x = 4
5(4) = 60y – 100
20 = 60y – 100
60y = 120
y = 2
x = 4, y = 2
21

1
(b) 2x – 2 + 12y = 9 = 4x – 2y
2x – 2 + 12y = 9 —(1)
4x – 2y = 9 —(2)
From (1),
2x + 12y = 11
x =
y
11 – 12
2
—(3)
4
11 – 12y
2





 – 2y = 9
22 – 24y – 2y = 9
26y = 13
y =
1
2
Substitute y =
1
2
into (3):
x =
11 – 12
1
2
( )
2
=
5
2
= 2
1
2
x = 2
1
2
, y =
1
2
(c) 5x + 3y = 2x + 7y = 29
5x + 3y = 29 —(1)
2x + 7y = 29 —(2)
(1) × 2: 10x + 6y = 58 —(3)
(2) × 5: 10x + 35y = 145 —(4)
(4) – (3): 29y = 87
y = 3
2x + 7(3) = 29
2x + 21 = 29
2x = 8
x = 4
x = 4, y = 3
(d) 10x – 15y = 12x – 8y = 150
10x – 15y = 150 —(1)
12x – 8y = 150 —(2)
(1) ÷ 5: 2x – 3y = 30 —(3)
(2) ÷ 2: 6x – 4y = 75 —(4)
From (3),
2x = 3y + 30 —(5)
3(3y + 30) – 4y = 75
9y + 90 – 4y = 75
5y = –15
y = –3
2x = 3(–3) + 30
= –9 + 30
= 21
x =
21
2
= 10
1
2
x = 10
1
2
, y = –3
(e) x + y + 3 = 3y – 2 = 2x + y
x + y + 3 = 3y – 2 —(1)
x + y + 3 = 2x + y —(2)
From (2),
x = 3
3 + y + 3 = 3y – 2
2y = 8
y = 4
x = 3, y = 4
(f) 5x – 8y = 3y – x + 8 = 2x – y + 1
5x – 8y = 3y – x + 8 —(1)
5x – 8y = 2x – y + 1 —(2)
From (1),
6x – 11y = 8 —(3)
From (2),
3x – 7y = 1
3x = 7y + 1 —(4)
2(7y + 1) – 11y = 8
14y + 2 – 11y = 8
3y = 6
y = 2
3x = 7(2) + 1
= 15
x = 5
x = 5, y = 2
(g) 4x + 2y = x – 3y + 1 = 2x + y + 3
4x + 2y = x – 3y + 1 —(1)
4x + 2y = 2x + y + 3 —(2)
From (1),
3x + 5y = 1 —(3)
From (2),
2x + y = 3
y = 3 – 2x —(4)
3x + 5(3 – 2x) = 1
3x + 15 – 10x = 1
7x = 14
x = 2
22

1
y = 3 – 2(2)
= 3 – 4
= –1
x = 2, y = –1
(h) 3x – 4y – 7 = y + 10x – 10 = 4x – 7y
3x – 4y – 7 = y + 10x – 10 —(1)
3x – 4y – 7 = 4x – 7y —(2)
From (1),
7x + 5y = 3 —(3)
From (2),
x – 3y = –7
x = 3y – 7 —(4)
7(3y – 7) + 5y = 3
21y – 49 + 5y = 3
26y = 52
y = 2
x = 3(2) – 7
= 6 – 7
= –1
x = –1, y = 2
27. 6x – 3y = 4 —(1)
y = 2x + 5 —(2)
6x – 3(2x + 5) = 4
6x – 6x – 15 = 4
–15 = 4 (N.A.)
From (1),
3y = 6x – 4
y = 2x –
4
3
Since the gradients of the lines are equal, the lines are
parallel and have no solution.
28. 6y + 3x = 15 —(1)
y = –
1
2
x +
5
2
—(2)
From (1),
6y = –3x + 15
y = –
1
2
x +
5
2
Since the lines are identical, they overlap each other and
have an infinite number of solutions.
29. (a) x + y + 2 = 3y + 1 = 2x
x + y + 2 = 3y + 1 —(1)
3y + 1 = 2x —(2)
From (1),
x = 2y – 1 —(3)
3y + 1 = 2(y – 1)
= 4y – 2
y = 3
Perimeter = 3[3(3) + 1]
= 30 cm
(b) x + 5y + 9 = 2x + 3y – 3 = x + y + 1
x + 5y + 9 = 2x + 3y – 3 —(1)
x + 5y + 9 = x + y + 1 —(2)
From (2),
4y = – 8
y = –2
x + 5(–2) + 9 = 2x + 3(–2) – 3
x – 1 = 2x – 9
x = 8
Perimeter = 3[8 + (–2) + 1]
= 21 cm
30. (a) 2x + y + 1 = 12 —(1)
4x + y + 2 = 3x + 3y —(2)
From (1),
y = 11 – 2x —(3)
4x + 11 – 2x + 2 = 3x + 3(11 – 2x)
2x + 13 = 3x + 33 – 6x
= 33 – 3x
5x = 20
x = 4
y = 11 – 2(4)
= 11 – 8
= 3
Perimeter = 2[3(4) + 3(3) + 12]
= 66 cm
Area = 12[3(4) + 3(3)]
= 252 cm2
23

1
(b) 3x + y + 6 = 4x – y —(1)
5x – 2y + 1 = 6x + y —(2)
From (1),
x = 2y + 6 —(3)
5(2y + 6) – 2y + 1 = 6(2y + 6) + y
10y + 30 – 2y + 1 = 12y + 36 + y
8y + 31 = 13y + 36
5y = –5
y = –1
x = 2(–1) + 6
= –2 + 6
= 4
Perimeter = 2[6(4) + (–1) + 4(4) – (–1)]
= 80 cm
Area = [6(4) + (–1)][4(4) – (–1)]
= 391 cm2
31. y – 1 = x + 5 —(1)
2x + y + 1 = 3x – y + 18 —(2)
From (1),
y = x + 6 —(3)
2x + x + 6 + 1 = 3x – (x + 6) + 18
3x + 7 = 3x – x – 6 + 18
= 2x + 12
x = 5
y = 5 + 6
= 11
Perimeter = 2[2(5) + 11 + 1 + 11 – 1]
= 64 cm
32. y + 2 = x – 1 —(1)
2x + y = 3x – y + 12 —(2)
From (1),
y = x – 3 —(3)
2x + x – 3 = 3x – (x – 3) + 12
3x – 3= 3x – x + 3 + 12
= 2x + 15
x = 18
y = 18 – 3
= 15
y + 2 = 15 + 2
= 17
2x + y = 2(18) + 15
= 51
Since the lengths of the sides are not equal, the quadrilateral
is not a rhombus.
33. 0.3x + 0.4y = 7 —(1)
1.1x – 0.3y = 8 —(2)
(1) × 30: 9x + 12y = 210 —(3)
(2) × 40: 44x – 12y = 320 —(4)
(3) + (4): 53x = 530
x = 10
0.3(10) + 0.4y = 7
3 + 0.4y = 7
0.4y = 4
y = 10
p = 10, q = 10
34. 3x – y = 7 —(1)
2x + 5y = –1 —(2)
From (1),
y = 3x – 7 —(3)
2x + 5(3x – 7) = –1
2x + 15x – 35 = –1
17x = 34
x = 2
y = 3(2) – 7
= 6 – 7
= –1
Coordinates of point of intersection are (2, –1).
35. x2
+ ax + b = 0 —(1)
32
+ a(3) + b = 0
3a + b = –9 —(2)
(–4)2
+ a(–4) + b = 0
4a – b = 16 —(3)
(2) + (3): 7a = 7
a = 1
3(1) + b = –9
b = –9 – 3
= –12
a = 1, b = –12
24

1
36. ax – by = 1 —(1)
ay + bx = –7 —(2)
Substitute x = –1, y = 2 into (1):
a(–1) – b(2) = 1
–a – 2b = 1 —(3)
Substitute x = –1, y = 2 into (2):
a(2) + b(–1) = –7
2a – b = –7
b = 2a + 7 —(4)
–a – 2(2a + 7) = 1
–a – 4a – 14 = 1
5a = –15
a = –3
Substitute a = –3 into (4):
b = 2(–3) + 7
= 1
a = –3, b = 1
37. Using the same method,
4x – 3y = 48x + 8y
44x = –11y
4x = –y
This method cannot be used as we have one equation
with two unknowns at the end.
38. Let Khairul’s age be x years and his aunt’s age be y years.
y = 4x —(1)
y + 8 =
5
2
(x + 8) —(2)
4x + 8 =
5
2
(x + 8)
8x + 16 = 5x + 40
3x = 24
x = 8
y = 4(8)
= 32
His aunt’s present age is 32 years.
39. (i) Let Jun Wei’s age be x years and his mother’s age be
y years.
x + y = 61 —(1)
y – x = 29 —(2)
(1) – (2): 2x = 32
x = 16
Jun Wei’s present age is 16 years.
(ii) Substitute x = 16 into (2):
y – 16 = 29
y = 45
y + 5 = 45 + 5
= 50
Jun Wei’s mother will be 50 years old.
40. Let the numbers be x and y.
y + 7 = 4x —(1)
x + 28 = 2y —(2)
From (1),
y = 4x – 7 —(3)
x + 28 = 2(4x – 7)
= 8x – 14
7x= 42
x = 6
y = 4(6) – 7
= 17
The numbers are 17 and 6.
41. Let the original fraction be
x
y
.
x
y
– 1
– 1
=
3
4
—(1)
+
+
x
y
1
1
=
4
5
—(2)
From (1),
4x – 4 = 3y – 3
4x – 3y = 1 —(3)
From (2),
5x + 5 = 4y + 4
4y = 5x + 1
y =
1
4
(5x + 1) —(4)
4x –
3
4
(5x + 1) = 1
16x – 15x – 3 = 4
x = 7
y =
1
4
(35 + 1)
= 9
The fraction is
7
9
.
25

1 26
42. Let the fractions be represented by x and y.
x + y = 3(y – x) —(1)
6x – y =
3
2
—(2)
From (2),
y = 6x –
3
2
—(3)
x + 6x –
3
2
= 3 6x –
3
2
– x






7x –
3
2
= 15x –
9
2
8x = 3
x =
3
8
Substitute x =
3
8
into (3):
y = 6
3
8





 –
3
2
=
3
4
The fractions are
3
4
and
3
8
.
43. Let the price of a chicken be $x and that of a duck be $y.
5x + 5y = 100 —(1)
10x + 17y = 287.5 —(2)
From (1),
x + y = 20
y = 20 – x —(3)
10x + 17(20 – x) = 287.5
10x + 340 – 17x = 287.5
7x = 52.5
x = 7.5
Substitute x = 7.5 into (3):
y = 20 – 7.5
= 12.5
3x + 2y = 3(7.5) + 2(12.5)
= 47.5
He will receive $47.50.
44. Let the number of chickens and goats be x and y
respectively.
x + y = 45 —(1)
2x + 4y = 150 —(2)
From (2),
x + 2y = 75 —(3)
(2) – (1): y = 30
x + 30 = 45
x = 15
y – x = 30 – 15
= 15
There are 15 more goats than chickens.
45. Let the cost of 1 can of condensed milk and 1 jar of instant
coffee be $x and $y respectively.
5x + 3y = 27 —(1)
12x + 5y = 49.4 —(2)
From (1),
3y = 27 – 5x
y = 9 –
5
3
x —(3)
12x + 5 9 –
5
3
x





 = 49.4
12x + 45 –
25
3
x = 49.4
11
3
x = 4.4
x = 1.2
Substitute x = 1.2 into (3):
y = 9 –
5
3
(1.2)
= 7
7x + 2y = 7(1.2) + 2(7)
= 22.4
The total cost is $22.40.
46. Let the cost of 1 kiwi fruit and 1 pear be $x and $y
respectively.
8x + 7y = 4.1 —(1)
4x + 9y = 3.7 —(2)
(2) × 2: 8x + 18y = 7.4 —(3)
(3) – (1): 11y = 3.3
y = 0.3
Substitute y = 0.3 into (1):
8x + 7(0.3) = 4.1
8x = 2.0
x = 0.25
2x + 2y = 2(0.25) + 2(0.3)
= 1.1
The cost is $1.10.

1
27
47. Let the number of research staff and laboratory assistants
be x and y respectively.
x + y = 540 —(1)
240x + 200y = 120 000 —(2)
From (2),
6x + 5y = 3000 —(3)
(1) × 5: 5x + 5y = 2700 —(4)
(3) – (4): x = 300
300 + y = 540
y = 240
The facility employs 300 research staff and
240 laboratory assistants.
48. Let the time taken to travel at 90 km/h and 80 km/h be
x hours and y hours respectively.
x + y = 8 —(1)
90x + 80y = 690 —(2)
From (2),
9x + 8y = 69 —(3)
(1) × 9: 9x + 9y = 72 —(4)
(4) – (3): y = 3
80y = 80(3)
= 240
The distance he covered was 240 km.
Advanced
49. For the line AC,
Vertical change (or rise) = 2 – n
Gradient of line =
n
2 –
5
=
4
5
2 – n = 4
n = 2 – 4
= –2
For the line AB,
Vertical change (or rise) = m – 2
Since the line slopes downwards from the left to the right,
its gradient is negative.
m – 2
5
= –
1
5
m – 2 = 1
m = 1 + 2
= 3
50. (a) From the graph, Jun Wei left home at 1100.
(b) He took 45 minutes to have lunch.
(c) The distance between Jun Wei’s house and his friend’s
house is 60 km.
(d) (i) Vertical change (or rise) = 40 – 0 = 40
Horizontal change (or run)
= 1145 – 1100
= 45 min
= 1
3
4
h
Gradient of line =
40
3
4
= 53
1
3
(to 3 s.f.)
The gradient represents the speed at which
Jun Wei travels to his lunch venue.
= 1230 – 1145
= 45 min
=
3
4
h
Gradient of line =
0
3
4
= 0
The gradient represents the speed. In this case,
Jun Wei is taking his lunch and so his speed is
zero.
(iii) Vertical change (or rise) = 60 – 40 = 20
= 1300 – 1230
= 30 min
=
1
2
h
Gradient of line =
20
1
2
= 40
The gradient represents the speed at which Jun Wei
travels from the lunch venue to Jurong West.

1 28
51. (a)
2
3
x –
3
5
y – 4 =
1
20
x – y +
17
30
= 2x – y – 18
14
15
2
3
x –
3
5
y – 4 =
1
20
x – y +
17
30
—(1)
1
20
x – y +
17
30
= 2x – y – 18
14
15
—(2)
From (1),
40x – 36y – 240 = 3x – 60y + 34
37x + 24y = 274 —(3)
From (2),
3x – 60y + 34 = 120x – 60y – 1136
117x = 1170
x = 10
37(10) + 24y = 274
24y = –96
y = –4
x = 10, y = –4
(b)
2
7
x +
3
4
y – 4 =
3
5
x –
2
7
y – 44 =
7
15
x + y – 3
1
3
2
7
x +
3
4
y – 4 =
3
5
x –
2
7
y – 44 —(1)
3
5
x –
2
7
y – 44 =
7
15
x + y – 3
1
3
—(2)
From (1),
40x + 105y – 560 = 84x – 40y – 6160
44x – 145y = 5600 —(3)
From (2),
63x – 30y – 4620 = 49x + 105y – 350
14x = 135y + 4270
x =
135
14
y + 305 —(4)
44
135
14
y + 305





 – 145y = 5600
2970
7
y + 13 420 – 145y = 5600
1955
7
y = –7820
y = –28
x =
135
14
(–28) + 305
= 35
x = 35, y = –28
52. Let the number be represented by 10x + y.
10x + y = 4(x + y) —(1)
(10y + x) – (10x + y) = 27 —(2)
From (1),
10x + y = 4x + 4y
6x = 3y
y = 2x —(3)
From (2),
9y – 9x = 27
y – x = 3 —(4)
2x – x = 3
x = 3
y = 2(3)
= 6
The original number is 36.
53. Let the digit in the tens place be x and the digit in the ones
place be y.
x =
1
2
y —(1)
(10y + x) – (10x + y) = 36 —(2)
From (2),
9y – 9x = 36
y – x = 4 —(3)
y –
1
2
y = 4
1
2
y = 4
y = 8
x =
1
2
(8)
= 4
The original number is 48.
54. Let the larger number be x and the smaller number be y.
x + y = 55 —(1)
x = 2y + 7 —(2)
2y + 7 + y = 55
3y = 48
y = 16
x = 2(16) + 7
= 39
Difference in the reciprocals =
1
16
–
1
39
=
23
624

1
29
55. Let the walking speed of Ethan and Michael be x m/s and
y m/s respectively.
8x + 8y = 64 —(1)
32x – 64 = 32y —(2)
From (1),
x + y = 8 —(3)
From (2),
32x – 32y = 64
x – y = 2 —(4)
(3) + (4): 2x = 10
x = 5
5 – y = 2
y = 3
Ethan’s walking speed is 5 m/s and Michael’s walking
speed is 3 m/s.
The assumption is that when they are walking in the same
direction, Ethan starts off 64 m behind Michael.
New Trend
56. 3x = y + 1 —(1)
y – x = 3 —(2)
From (1),
y = 3x – 1 —(3)
3x – 1 – x = 3
2x = 4
x = 2
y = 3(2) – 1
= 5
x = 2, y = 5
57. (a) Let the speed of the faster ship and slower ship be
x km/h and y km/h respectively.
x = y + 8 —(1)
60x + 60y = 4320 —(2)
From (2),
x + y = 72 —(3)
y + 8 + y = 72
2y = 64
y = 32
x = 32 + 8
= 40
The speeds of the faster ship and slower ship are
40 km/h and 32 km/h respectively.
(b)
1780
32
−
1780
40
= 55.625 − 44.5
= 11.125 h
= 11 h 8 min (nearest min)
58. 4x + 4(6) = 40
4x = 40 − 24
x = 16 ÷ 4
= 4
Since the rectangles are of equal area,
6z = 39x
z = 39(4) ÷ 6
= 26
y = 39 − z
= 39 − 26
= 13
x = 4 cm, y = 13 cm and z = 26 cm
59. At x-axis, y = 0
3x = 30
x = 10
At y-axis, x = 0
−5y = 30
y = −6
The coordinates of P are (10, 0) and of Q are (0, −6).
60. (i) 4x − 6 = 5y − 7 (isos. trapezium)
4x − 5y = −1 ––(1)
(4x − 6) + (5x + 6y + 33) = 180 (int. s)
9x + 6y = 153
3x + 2y = 51 ––(2)
(ii) (1) × 3: 12x − 15y = −3 ––(3)
(2) × 4: 12x + 8y = 204 ––(4)
(4) − (3): 23y = 207
y = 9
B = C
= [5(9) − 7]°
= 38°
A = 180° − B
= 180° − 38°
= 142°
A = 142° and B = 38°

1 30
61. (a) 4x − 2y − 5 = 0
2y = 4x − 5
y = 2x − 2
1
2
(i) Gradient of line l = 2
(ii) y-intercept of line l = −2
1
2
(b) 2x + 3y = –5 —(1)
4x – 2y = 5 —(2)
(1) × 2: 4x + 6y = –10 —(3)
(3) – (2): 8y = –15
y = –1
7
8
Substitute y = –1
7
8
into (1):
2x + 3 –1
7
8



 = –5
2x – 5
5
8
= –5
2x =
5
8
x =
5
16
The coordinates of C are
5
16
, −1
7
8





 .
62. (i) y = 7 − 2x —(1)
y = x + 10 —(2)
Substitute x = −9 into (1):
y = 7 − 2(−9)
= 7 + 18
= 25
Substitute x = −9 into (2):
y = −9 + 10
= 1
The coordinates of A are (−9, 25) and of B are (−9, 1).
(ii) y = 7 − 2x
From the equation, gradient of the line = −2.
(iii) (0, k) lies on the perpendicular bisector of AB.
k =
1+25
2
= 13

1
Chapter 3 Expansion and Factorisation of
Quadratic Expressions
Basic
1. (a) 5a2
+ 2a – 3a2
– a
= 2a2
+ a
(b) b2
– 3b + 4 – 2b2
+ 3b – 7
= –b2
– 3
(c) c2
+ 4c + 3 + (–2c2
) + (–c) – 2
= c2
+ 4c + 3 – 2c2
– c – 2
= –c2
+ 3c + 1
(d) 4d2
– d – 5 – (–2d2
) – (–d) + 6
= 4d2
– d – 5 + 2d2
+ d + 6
= 6d2
+ 1
(e) 8e2
+ 8e + 9 – (5e2
+ 2e – 3)
= 8e2
+ 8e + 9 – 5e2
– 2e + 3
= 3e2
+ 6e + 12
(f) 6f2
– 4f – 1 – (2f2
– 7f)
= 6f2
– 4f – 1 – 2f2
+ 7f
= 4f2
+ 3f – 1
(g) –(2 + g – g2
) + (6g – g2
)
= –2 – g + g2
+ 6g – g2
= 5g – 2
(h) –(1 + 5h – 3h2
) – (2h2
+ 4h – 7)
= –1 – 5h + 3h2
– 2h2
– 4h + 7
= h2
– 9h + 6
2. (a) 8 × 2h
= 16h
(b) 3h × 4h
= 12h2
(c) (–5h) × 6h
= –30h2
(d) (–10h) × (–7h)
= 70h2
3. (a) 5(2a + 3)
= 10a + 15
(b) –4(5b + 1)
= –20b – 4
(c) 8(c2
+ 2c – 3)
= 8c2
+ 16c – 24
(d) –2(4 – 6d2
)
= –8 + 12d2
= 12d2
– 8
(e) 3e(8e + 7)
= 24e2
+ 21e
(f) –f(9 – f)
= –9f + f2
= f2
– 9f
(g) –6g(5g – 1)
= –30g2
+ 6g
= 6g – 30g2
(h) –2h(–3h – 4)
= 6h2
+ 8h
4. (a) 3(a + 2) + 4(2a + 3)
= 3a + 6 + 8a + 12
= 11a + 18
(b) 11(5b – 7) + 9(2 – 3b)
= 55b – 77 + 18 – 27b
= 28b – 59
(c) 8(5c – 4) + 3(2 – 4c)
= 40c – 32 + 6 – 12c
= 28c – 26
(d) 2d(3d + 4) + d(5d – 2)
= 6d2
+ 8d + 5d2
– 2d
= 11d2
+ 6d
(e) e(6e – 1) + 2e(e – 2)
= 6e2
– e + 2e2
– 4e
= 8e2
– 5e
(f) 4f(1 – 2f) + f(3 – f)
= 4f – 8f2
+ 3f – f2
= 7f – 9f2
5. (a) (x + 5)(x + 7)
= x2
+ 7x + 5x + 35
= x2
+ 12x + 35
(b) (2x + 1)(x + 3)
= 2x2
+ 6x + x + 3
= 2x2
+ 7x + 3
(c) (x + 6)(3x + 4)
= 3x2
+ 4x + 18x + 24
= 3x2
+ 22x + 24
(d) (4x + 3)(5x + 6)
= 20x2
+ 24x + 15x + 18
= 20x2
+ 39x + 18
6. (a) a2
+ 20a + 75
× a 15
a a2
15a
5 5a 75
a2
+ 20a + 75 = (a + 15)(a + 5)
(b) b2
+ 19b + 18
× b 18
b b2
18b
1 b 18
b2
+ 19b + 18 = (b + 18)(b + 1)
31

1
(c) c2
– 11c + 28
× c –7
c c2
–7c
–4 –4c 28
c2
– 11c + 28 = (c – 7)(c – 4)
(d) d2
– 21d + 68
× d –17
d d2
–17d
–4 –4d 68
d2
– 21d + 68 = (d – 17)(d – 4)
(e) e2
+ 4e – 77
× e 11
e e2
11e
–7 –7e –77
e2
+ 4e – 77 = (e + 11)(e – 7)
(f) f2
+ 3f – 154
× f 14
f f2
14f
–11 –11f –154
f2
+ 3f – 154 = (f + 14)(f – 11)
(g) g2
– 2g – 35
× g –7
g g2
–7g
5 5g –35
g2
– 2g – 35 = (g – 7)(g + 5)
(h) h2
– 10h – 171
× h –19
h h2
–19h
9 9h –171
h2
– 10h – 171 = (h – 19)(h + 9)
7. (a) 6a2
+ 31a + 5
× 6a 1
a 6a2
a
5 30a 5
6a2
+ 31a + 5 = (6a + 1)(a + 5)
(b) 8b2
+ 30b + 27
× 4b 9
2b 8b2
18b
3 12b 27
8b2
+ 30b + 27 = (4b + 9)(2b + 3)
(c) 4c2
– 25c + 6
× 4c –1
c 4c2
–c
–6 –24c 6
4c2
– 25c + 6 = (4c – 1)(c – 6)
(d) 9d2
– 36d + 32
× 3d –8
3d 9d2
–24d
–4 –12d 32
9d2
– 36d + 32 = (3d – 8)(3d – 4)
(e) 15e2
+ 2e – 1
× 5e –1
3e 15e2
–3e
1 5e –1
15e2
+ 2e – 1 = (5e – 1)(3e + 1)
(f) 2g2
– 5g – 3
× 2g 1
g 2g2
g
–3 –6g –3
2g2
– 5g – 3 = (2g + 1)(g – 3)
(g) 12h2
– 31h – 15
× 12h 5
h 12h2
5h
–3 –36h –15
12h2
– 31h – 15 = (12h + 5)(h – 3)
32

1
Intermediate
8. (a) 6(a + 3) – 5(a – 4)
= 6a + 18 – 5a + 20
= a + 38
(b) 13(5b + 7) – 6(3b – 5)
= 65b + 91 – 18b + 30
= 47b + 121
(c) 9(3c – 2) – 5(2 + c)
= 27c – 18 – 10 – 5c
= 22c – 28
(d) 8(5 – 4d) – 7(7 – 5d)
= 40 – 32d – 49 + 35d
= 3d – 9
(e) 7(12 – 5e) – 3(9 – 7e)
= 84 – 35e – 27 + 21e
= 57 – 14e
(f) 5f(f + 3) – 4f(5 – f)
= 5f2
+ 15f – 20f + 4f2
= 9f2
– 5f
(g) –2g(4 – g) – 3g(2g + 1)
= –8g + 2g2
– 6g2
– 3g
= –4g2
– 11g
(h) –5h(3h + 7) – 4h(–h – 2)
= –15h2
– 35h + 4h2
+ 8h
= –11h2
– 27h
9. (a) (y + 7)(y – 11)
= y2
– 11y + 7y – 77
= y2
– 4y – 77
(b) (y – 6)(y + 8)
= y2
+ 8y – 6y – 48
= y2
+ 2y – 48
(c) (y – 9)(y – 4)
= y2
– 4y – 9y + 36
= y2
– 13y + 36
(d) (2y + 3)(4y – 5)
= 8y2
– 10y + 12y – 15
= 8y2
+ 2y – 15
(e) (5y – 9)(6y – 1)
= 30y2
– 5y – 54y + 9
= 30y2
– 59y + 9
(f) (4y – 1)(3 – 4y)
= 12y – 16y2
– 3 + 4y
= –16y2
+ 16y – 3
(g) (7 – 2y)(4 + y)
= 28 + 7y – 8y – 2y2
= 28 – y – 2y2
(h) (7 – 3y)(8 – 5y)
= 56 – 35y – 24y + 15y2
= 56 – 59y + 15y2
10. (a) 4 + (a + 2)(a + 5)
= 4 + a2
+ 5a + 2a + 10
= a2
+ 7a + 14
(b) 6b + (3b + 1)(b – 2)
= 6b + 3b2
– 6b + b – 2
= 3b2
+ b – 2
(c) (7c + 2)(3c – 8) + 9c(c + 1)
= 21c2
– 56c + 6c – 16 + 9c2
+ 9c
= 30c2
– 41c – 16
(d) (4d – 5)(8d – 7) + (2d + 3)(d – 3)
= 32d2
– 28d – 40d + 35 + 2d2
– 6d + 3d – 9
= 34d2
– 71d + 26
11. (a) –x2
– 4x + 21
× –x 3
x –x2
3x
7 –7x 21
–x2
– 4x + 21 = (–x + 3)(x + 7)
(b) –6x2
+ 2x + 20 = –2(3x2
– x – 10)
× 3x 5
x 3x2
5x
–2 –6x –10
–6x2
+ 2x + 20 = –2(3x + 5)(x – 2)
(c) 12hx2
– 25hx + 12h = h(12x2
– 25x + 12)
× 4x –3
3x 12x2
–9x
–4 –16x 12
12hx2
– 25hx + 12h = h(4x – 3)(3x – 4)
12. 3x2
+ 26x + 51
× 3x 17
x 3x2
17x
3 9x 51
3x2
+ 26x + 51 = (3x + 17)(x + 3)
32 651 = 3(100)2
+ 26(100) + 51
Let x = 100.
32 651 = 317 × 103
The factors are 317 and 103.
13. 4x2
+ 13x + 3
× 4x 1
x 4x2
x
3 12x 3
4x2
+ 13x + 3 = (4x + 1)(x + 3)
33
33

1
41 303 = 4(100)2
+ 13(100) + 3
Let x = 100.
41 303 = 401 × 103
The prime factors are 401 and 103.
Advanced
14. (a) 9a2
– (4a – 1)(a + 2)
= 9a2
– (4a2
+ 8a – a – 2)
= 9a2
– (4a2
+ 7a – 2)
= 9a2
– 4a2
– 7a + 2
= 5a2
– 7a + 2
(b) 3b(2 – b) – (1 + b)(1 – b)
= 6b – 3b2
– (1 – b + b – b2
)
= 6b – 3b2
– (1 – b2
)
= 6b – 3b2
– 1 + b2
= –2b2
+ 6b – 1
(c) (5c + 6)(6c – 5) – (3 – 2c)(1 – 15c)
= 30c2
– 25c + 36c – 30 – (3 – 45c – 2c + 30c2
)
= 30c2
+ 11c – 30 – (3 – 47c + 30c2
)
= 30c2
+ 11c – 30 – 3 + 47c – 30c2
= 58c – 33
(d) (2d – 8)
1
2
d – 4





 – (3d – 6)
1
3
d + 1






= d2
– 8d – 4d + 32 – (d2
+ 3d – 2d – 6)
= d2
– 12d + 32 – (d2
+ d – 6)
= d2
– 12d + 32 – d2
– d + 6
= 38 – 13d
15. (i) 3x2
+ 48x + 189 = 3(x2
+ 16x + 63)
× x 7
x x2
7x
9 9x 63
3x2
+ 48x + 189 = 3(x + 7)(x + 9)
(ii) 969 = 3(10)2
+ 48(10) + 189
Let x = 10.
969 = 3 × 17 × 19
Sum = 3 + 17 + 19
= 39
16. 2x2
– 2.9x – 3.6 = 0.1(20x2
– 29x – 36)
× 5x 4
4x 20x2
16x
–9 –45x –36
2x2
– 2.9x – 3.6 = 0.1(5x + 4)(4x – 9)
i.e. p = 5, q = 4, r = 4, s = –9
p + q + r + s = 5 + 4 + 4 – 9
= 4
New Trend
17. (a) 16a2
– 9b2
× 4a 3b
14a 16a2
12ab
–3b –12ab –9b2
16a2
− 9b2
= (4a + 3b)(4a − 3b)
(b) 3f2
+ 11f – 20
× 3f –4
f 3f2
–4f
5 15f –20
3f2
+ 11f – 20 = (3f – 4)(f + 5)
(c) 9x2
– 15x – 6 = 3(3x2
– 5x – 2)
× 3x 1
x 3x2
x
–2 –6x –2
9x2
– 15x – 6 = 3(3x + 1)(x – 2)
18. (i) Since x is a positive integer, 2x is a positive even
number and (2x − 1) is one less than an even number.
2x − 1 is not divisible by 2.
Hence, 2x − 1 is an odd number.
(ii) (2x − 1) + 2 = 2x + 1
(iii) (2x − 1)2
= 4x2
− 4x + 1
(2x + 1)2
= 4x2
+ 4x + 1
(iv) (4x2
+ 4x + 1) − (4x2
− 4x + 1)
= 4x + 4x
= 8x
Since 8x has a factor of 8, it is always divisible by 8.
34

1
Chapter 4 Further Expansion and
Factorisation of Algebraic
Expressions
Basic
1. (a) 7a × 3b
= 21ab
(b) 5c × (–4d)
= –20cd
(c) (–10e) × (–2f)
= 20ef
(d)
1
6
g × 24h
= 4gh
2. (a) 5a(2a + 3b)
= 10a2
+ 15ab
(b) 8c(5c – 2d)
= 40c2
– 16cd
(c) 9e(–4e + 7f)
= –36e2
+ 63ef
(d) 4h(–2g – 3h)
= –8gh – 12h2
(e) –6j(k – 4j)
= –6jk + 24j2
(f) –4m(2n + 5m)
= –8mn – 20m2
(g) –7p(–3p + 4q)
= 21p2
– 28pq
(h) –3r(–2r – s)
= 6r2
+ 3rs
(i) 2u(5u + v – w)
= 10u2
+ 2uv – 2uw
(j) –6x(3x – 2y + z)
= –18x2
+ 12xy – 6xz
3. (a) 4a(3a – b) + 2a(a – 5b)
= 12a2
– 4ab + 2a2
– 10ab
= 14a2
– 14ab
(b) 2c(4d – 3c) + 5c(5c – 2d)
= 8cd – 6c2
+ 25c2
– 10cd
= 19c2
– 2cd
(c) 3f(2e – 7f) + 2e(6f – 5e)
= 6ef – 21f2
+ 12ef – 10e2
= –10e2
+ 18ef – 21f2
(d) 5h(–2h – 3g) + 2h(–h + 3g)
= –10h2
– 15gh – 2h2
+ 6gh
= –12h2
– 9gh
4. (a) (x + y)(x + 4y)
= x2
+ 4xy + xy + 4y2
= x2
+ 5xy + 4y2
(b) (2x + y)(3x + y)
= 6x2
+ 2xy + 3xy + y2
= 6x2
+ 5xy + y2
(c) (x2
+ 1)(x + 1)
= x3
+ x2
+ x + 1
(d) (4x2
+ 3)(2x + 3)
= 8x3
+ 12x2
+ 6x + 9
5. (a) (a + 5)2
= a2
+ 10a + 25
(b) (2b + 3)2
= 4b2
+ 12b + 9
(c) (c + 6d)2
= c2
+ 12cd + 36d2
(d) (7e + 4f)2
= 49e2
+ 56ef + 16f2
6. (a) (a – 8)2
= a2
– 16a + 64
(b) (4b – 1)2
= 16b2
– 8b + 1
(c) (c – 3d)2
= c2
– 6cd + 9d2
(d) (9e – 2f)2
= 81e2
– 36ef + 4f2
7. (a) (a + 6)(a – 6)
= a2
– 36
(b) (4b + 3)(4b – 3)
= 16b2
– 9
(c) (9 + 4c)(9 – 4c)
= 81 – 16c2
(d) (5d + e)(5d – e)
= 25d2
– e2
8. (a) 9042
= (900 + 4)2
= 9002
+ 2(900)(4) + 42
= 810 000 + 7200 + 16
= 817 216
(b) 7912
= (800 – 9)2
= 8002
– 2(800)(9) + 92
= 640 000 – 14 400 + 81
= 625 681
(c) 603 × 597
= (600 + 3)(600 – 3)
= 6002
– 32
= 360 000 – 9
= 359 991
35

1
(d) 99 × 101
= (100 – 1)(100 + 1)
= 1002
– 12
= 10 000 – 1
= 9999
9. (a + b)2
= a2
+ 2ab + b2
73 = a2
+ b2
+ 2(65)
= a2
+ b2
+ 130
a2
+ b2
= 73 – 130
= –57
10. (a) a2
+ 12a + 36
= (a + 6)2
(b) 9b2
+ 12b + 4
= (3b + 2)2
(c) 4c2
+ 4cd + d2
= (2c + d)2
(d) 16e2
+ 40ef + 25f2
= (4e + 5f)2
11. (a) a2
– 18a + 81
= (a – 9)2
(b) 25b2
– 20b + 4
= (5b – 2)2
(c) 9c2
– 6cd + d2
= (3c – d)2
(d) 49e2
– 28ef + 4f2
= (7e – 2f)2
12. (a) a2
– 196
= a2
– 142
= (a + 14)(a – 14)
(b) 4b2
– 81
= (2b)2
– 92
= (2b + 9)(2b – 9)
(c) 289 – 36c2
= 172
– (6c)2
= (17 + 6c)(17 – 6c)
(d) 9d2
– e2
= (3d)2
– e2
= (3d + e)(3d – e)
13. (a) abc – a2
bc3
= abc(1 – ac2
)
(b) 2a2
b3
c – 8ab2
c3
= 2ab2
c(ab – 4c2
)
(c) 6k2
+ 8k3
– 10k5
= 2k2
(3 + 4k – 5k3
)
(d) m2
n – mn2
+ m2
n2
= mn(m – n + mn)
(e) p2
q – 2pq2
+ 4p2
q2
= pq(p – 2q + 4pq)
(f) 2s – 4s2
+ 8st2
= 2s(1 – 2s + 4t2
)
(g) 12x3
– 9x2
y2
+ 6xy3
= 3x(4x2
– 3xy2
+ 2y3
)
(h) 5y2
z – 3y3
z2
+ 6y2
z2
= y2
z(5 – 3yz + 6z)
14. (a) 4a(x + y) + 7(x + y)
= (4a + 7)(x + y)
(b) 5b(6x + y) – c(y + 6x)
= (5b – c)(6x + y)
(c) 8d(x – 3y) – e(3y – x)
= 8d(x – 3y) + e(x – 3y)
= (8d + e)(x – 3y)
(d) (x + 5)(x – 1) + a(x + 5)
= (x – 1 + a)(x + 5)
Intermediate
15. (a)
1
2
a ×
2
3
b
=
1
3
ab
(b)
2
5
c × –
3
8
d






= –
3
20
cd
(c) –
1
4
e





 ×
12
13
f
= –
3
13
ef
(d) –
6
7
g





 × –
7
12
h






=
1
2
gh
(e) 0.2p × 12q
= 2.4pq
(f) 3r × 0.9s
= 2.7rs
(g) 4w2
x × 5wx3
= 20w3
x4
(h) (–8xy2
z) × (–2xz3
)
= 16x2
y2
z4
16.
4
5
a2
bc3
×
15
16
ab2
=
3
4
a3
b3
c3
17. (a) 5ab(a – 4b)
= 5a2
b – 20ab2
(b) –3c(2c2
d + d2
)
= –6c3
d – 3cd2
(c) –8ef(6f – e2
)
= –48ef2
+ 8e3
f
(d) –10h2
(–7g2
h – 9h3
)
= 70g2
h3
+ 90h5
36

1
18. (a) 4a(3b + 5c) – 3b(8c – 9a)
= 12ab + 20ac – 24bc + 27ab
= 39ab + 20ac – 24bc
(b) 5d(2d + 5e) – 3e(2e – 7d)
= 10d2
+ 25de – 6e2
+ 21de
= 10d2
+ 46de – 6e2
(c) 7f(2f + 3g) – 3f(–4g + 3f)
= 14f2
+ 21fg + 12fg – 9f2
= 5f2
+ 33fg
(d) 4h(–3h + k) – 2h(–5k + h)
= –12h2
+ 4hk + 10hk – 2h2
= –14h2
+ 14hk
19. (a) (a + 6b)(a – 2b)
= a2
– 2ab + 6ab – 12b2
= a2
+ 4ab – 12b2
(b) (4c + 5d)(5c + 7d)
= 20c2
+ 28cd + 25cd + 35d2
= 20c2
+ 53cd + 35d2
(c) (4e – 3f)(2e + 7f)
= 8e2
+ 28ef – 6ef – 21f2
= 8e2
+ 22ef – 21f2
(d) (2g – 3h)(g – 2h)
= 2g2
– 4gh – 3gh + 6h2
= 2g2
– 7gh + 6h2
(e) (m2
– 4)(2m + 3)
= 2m3
+ 3m2
– 8m – 12
(f) (2n – 4)(n2
+ 3)
= 2n3
+ 6n – 4n2
– 12
= 2n3
– 4n2
+ 6n – 12
(g) (2p – 3q)(2p – 5r)
= 4p2
– 10pr – 6pq + 15qr
(h) (xy – 5)(xy + 8)
= x2
y2
+ 8xy – 5xy – 40
= x2
y2
+ 3xy – 40
20. (a) (a + 1)(a – 3) + (2a – 3)(5 – 7a)
= a2
– 3a + a – 3 + 10a – 14a2
– 15 + 21a
= –13a2
+ 29a – 18
(b) (7b + 1)(b – 5) – 3(4 – 2b – b2
)
= 7b2
– 35b + b – 5 – 12 + 6b + 3b2
= 10b2
– 28b – 17
(c) (3c – 8)(c + 1) – (2c – 1)(5 – c)
= 3c2
+ 3c – 8c – 8 – (10c – 2c2
– 5 + c)
= 3c2
– 5c – 8 – (–2c2
+ 11c – 5)
= 3c2
– 5c – 8 + 2c2
– 11c + 5
= 5c2
– 16c – 3
(d) (d + 3e)(d – 3e) – 2(d + 2e)(d – e)
= d2
– 9e2
– 2(d2
– de + 2de – 2e2
)
= d2
– 9e2
– 2(d2
+ de – 2e2
)
= d2
– 9e2
– 2d2
– 2de + 4e2
= –d2
– 2de – 5e2
21. (a) (a + 3)(a2
+ 3a + 9)
= a3
+ 3a2
+ 9a + 3a2
+ 9a + 27
= a3
+ 6a2
+ 18a + 27
(b) (b + c)(b2
+ bc + c2
)
= b3
+ b2
c + bc2
+ b2
c + bc2
+ c3
= b3
+ 2b2
c + 2bc2
+ c3
(c) (5 + 2d)(2 + 3d + d2
)
= 10 + 15d + 5d2
+ 4d + 6d2
+ 2d3
= 2d3
+ 11d2
+ 19d + 10
(d) (2e + f)(3e – 4f + g)
= 6e2
– 8ef + 2eg + 3ef – 4f2
+ fg
= 6e2
– 5ef – 4f2
+ 2eg + fg
22. (a) a2
+ 7ab + 6b2
× a b
a a2
ab
6b 6ab 6b2
a2
+ 7ab + 6b2
= (a + b)(a + 6b)
(b) c2
+ 11cd – 12d2
× c 12d
c c2
12cd
–d –cd –12d2
c2
+ 11cd – 12d2
= (c + 12d)(c – d)
(c) 2d2
– de – 15e2
× 2d 5e
d 2d2
5de
–3e –6de –15e2
2d2
– de – 15e2
= (2d + 5e)(d – 3e)
(d) 6f2
– 29fg + 28g2
× 3f –4g
2f 6f2
–8fg
–7g –21fg 28g2
6f2
– 29fg + 28g2
= (3f – 4g)(2f – 7g)
(e) 2m2
+ 2mn – 12n2
= 2(m2
+ mn – 6n2
)
× m –2n
m m2
–2mn
3n 3mn –6n2
2m2
+ 2mn – 12n2
= 2(m – 2n)(m + 3n)
37

1
(f) px2
– 11pxy + 24py2
= p(x2
– 11xy + 24y2
)
× x –3y
x x2
–3xy
–8y –8xy 24y2
px2
– 11pxy + 24py2
= p(x – 3y)(x – 8y)
23. 12x2
+ xy – 20y2
× 4x –5y
3x 12x2
–15xy
4y 16xy –20y2
12x2
+ xy – 20y2
= (4x – 5y)(3x + 4y)
Breadth of rectangle =
+
x y x y
x y
(4 – 5 )(3 4 )
4 – 5
= (3x + 4y) cm
24. (a) a +
b
3






2
= a2
+
ab
2
3
+
b
9
2
(b) (0.5c + d)2
= 0.25c2
+ cd + d2
(c) (ef + 2)2
= e2
f2
+ 4ef + 4
(d) g+
2
g






2
= g2
+ 4 +
g
4
2
(e) (h2
+ 3)2
= h4
+ 6h2
+ 9
(f) (k3
+ 4)2
= k6
+ 8k3
+ 16
(g)
2
p
+
3
q






2
=
p
4
2 +
pq
12
+
q
9
2
(h)
x
y
+ 3y






2
=
x
y
2
2
+ 6x + 9y2
25. (a) 3a –
1
4
b






2
= 9a2
–
3
2
ab +
1
16
b2
(b) (10c – 0.1d)2
= 100c2
– 2cd + 0.01d2
(c) (2ef – 1)2
= 4e2
f2
– 4ef + 1
(d) 2h –
1
h






2
= 4h2
– 4 +
h
1
2
(e) (p4
– 2)2
= p8
– 4p4
+ 4
(f)
x
y
–
y
x






2
=
x
y
2
2
– 2 +
y
x
2
2
26. (a)
1
2
a+b





 1
2
a – b






=
1
4
a2
– b2
(b) (0.2c + d)(d – 0.2c)
= (d + 0.2c)(d – 0.2c)
= d2
– 0.04c2
(c) (3ef + 4)(3ef – 4)
= 9e2
f2
– 16
(d)
g
2
–
h
4





 h
4
+
g
2






=
g
2
+
h
4





 g
2
–
h
4






=
g
4
2
–
h
16
2
27. x2
– y2
= 6
(x + y)(x – y) = 6
2(x + y) = 6
x + y = 3
(x + y)2
= 9
28. (i) (x + y)2
= x2
+ 2xy + y2
= 43 + 24
= 67
(ii) (2x – 2y)2
= 4x2
– 8xy + 4y2
= 4(43) – 2(48)
= 76
29. (i) x2
– 4y2
= (x + 2y)(x – 2y)
= (–2)(18)
= –36
(ii) x + 2y = –2 —(1)
x – 2y = 18 —(2)
(1) + (2): 2x = 16
x = 8
(1) – (2): 4y = –20
y = –5
x2
+ 4y2
= 82
+ 4(–5)2
= 164
38

1
30. (i) a2
– b2
= (a + b)(a – b)
(ii) 20302
– 20292
+ 20282
– 20272
= (2030 + 2029)(2030 – 2029)
+ (2028 + 2027)(2028 – 2027)
= 2030 + 2029 + 2028 + 2027
= 8114
31. (a) 4a2
+ 32a + 64
= 4(a2
+ 8a + 16)
= 4(a + 4)2
(b)
1
4
b2
+ 4bc + 16c2
=
1
2
b+ 4c






2
(c)
1
9
d2
+
4
15
de +
4
25
e2
=
1
3
d +
2
5
e






2
(d) f4
+ 8f2
+ 16
= (f2
+ 4)2
32. (a) 3a2
– 36a + 108
= 3(a2
– 12a + 36)
= 3(a – 6)2
(b) 64b2
– 4bc +
1
16
c2
= 8b –
1
4
c






2
(c) e2
f2
– 10ef + 25
= (ef – 5)2
(d)
1
4
g2
–
1
4
gh +
1
16
h2
=
1
2
g –
1
4
h






2
33. (a)
1
4
a2
– b2
=
1
2
a+b





 1
2
a – b






(b) 4c3
– 49c
= c(4c2
– 49)
= c(2c + 7)(2c – 7)
(c) 81ef2
– 4eg2
= e(81f2
– 4g2
)
= e(9f + 2g)(9f – 2g)
(d) 18h3
– 8hk2
= 2h(9h2
– 4k2
)
= 2h(3h + 2k)(3h – 2k)
(e) 81m5
n3
– 121m3
n5
= m3
n3
(81m2
– 121n2
)
= m3
n3
(9m + 11n)(9m – 11n)
(f) p4
– 81q4
= (p2
+ 9q2
)(p2
– 9q2
)
= (p2
+ 9q2
)(p + 3q)(p – 3q)
(g) (t2
– 1)2
– 9
= (t2
– 1 + 3)(t2
– 1 – 3)
= (t2
+ 2)(t2
– 4)
= (t2
+ 2)(t + 2)(t – 2)
(h) 9 – (a – b)2
= (3 + a – b)(3 – a + b)
(i) (d + 2c)2
– c2
= (d + 2c + c)(d + 2c – c)
= (d + 3c)(d + c)
(j) (e – 3)2
– 16f2
= (e – 3 + 4f)(e – 3 – 4f)
(k) (3g – h)2
– g2
= (3g – h + g)(3g – h – g)
= (4g – h)(2g – h)
(l) 4j2
– (k – 2)2
= (2j + k – 2)(2j – k + 2)
(m) 9m2
– (3m – 2n)2
= (3m + 3n – 2n)(3m – 3m + 2n)
= (6m – 2n)(2n)
= 4n(3m – n)
(n) 9p2
– 4(p – 2q)2
= (3p)2
– (2p – 4q)2
= (3p + 2p – 4q)(3p – 2p + 4q)
= (5p – 4q)(p + 4q)
(o) (3x – 2y)2
– (2x – 3y)2
= (3x – 2y + 2x – 3y)(3x – 2y – 2x + 3y)
= (5x – 5y)(x + y)
= 5(x + y)(x – y)
34. (a) 412
+ 738 + 81
= 412
+ 2(41)(9) + 92
= (41 + 9)2
= 502
= 2500
(b) 652
+ 650 + 25
= 652
+ 2(65)(5) + 52
= (65 + 5)2
= 702
= 4900
(c) 922
– 368 + 4
= 922
– 2(92)(2) + 22
= (92 – 2)2
= 902
= 8100
(d) 2012
– 402 + 1
= 2012
– 2(201)(1) + 12
= (201 – 1)2
= 2002
= 40 000
39

1
(e) 2012
– 992
= (201 + 99)(201 – 99)
= (300)(102)
= 30 600
(f) 1.0132
– 0.0132
= (1.013 + 0.013)(1.013 – 0.013)
= 1.026
35. (a) (2a + b)(x + y) + (a + b)(x + y)
= (2a + b + a + b)(x + y)
= (3a + 2b)(x + y)
(b) (4c + 3d)2
+ (4c + 3d)(c + d)
= (4c + 3d)(4c + 3d + c + d)
= (4c + 3d)(5c + 4d)
(c) 2p(5r – 7s) + 3q(7s – 5r)
= 2p(5r – 7s) – 3q(5r – 7s)
= (2p – 3q)(5r – 7s)
(d) 9w(y – x) – 8z(x – y)
= 9w(y – x) + 8z(y – x)
= (9w + 8z)(y – x)
36. (a) p2
+ pq + 3qr + 3pr
= p(p + q) + 3r(q + p)
= (p + 3r)(p + q)
(b) 3xy + 6y – 5x – 10
= 3y(x + 2) – 5(x + 2)
= (3y – 5)(x + 2)
(c) x2
z – 4y – x2
y + 4z
= x2
z – x2
y + 4z – 4y
= x2
(z – y) + 4(z – y)
= (x2
+ 4)(z – y)
(d) x3
+ xy – 3x2
y – 3y2
= x(x2
+ y) – 3y(x2
+ y)
= (x2
+ y)(x – 3y)
(e) x – 4x2
– 4 + x3
= x3
+ x – 4x2
– 4
= x(x2
+ 1) – 4(x2
+ 1)
= (x2
+ 1)(x – 4)
(f) h2
– 1 + hk + k
= (h + 1)(h – 1) + k(h + 1)
= (h – 1 + k)(h + 1)
(g) m – n – m2
+ n2
= (m – n) – (m2
– n2
)
= (m – n) – (m + n)(m – n)
= (1 – m – n)(m – n)
(h) a2
– 3bc – ab + 3ac
= a2
– ab + 3ac – 3bc
= a(a – b) + 3c(a – b)
= (a + 3c)(a – b)
(i) x2
y – 3y – 6 + 2x2
= y(x2
– 3) – 2(3 – x2
)
= y(x2
– 3) + 2(x2
– 3)
= (y + 2)(x2
– 3)
(j) a2
x – 12by – 3bx + 4a2
y
= a2
x + 4a2
y – 3bx – 12by
= a2
(x + 4y) – 3b(x + 4y)
= (x + 4y)(a2
– 3b)
Advanced
37. (a) (2h + 3)(h – 7) – (h + 4)(h2
– 1)
= 2h2
– 14h + 3h – 21 – (h3
– h + 4h2
– 4)
= 2h2
– 11h – 21 – h3
+ h – 4h2
+ 4
= –h3
– 2h2
– 10h – 17
(b) (3p2
+ q)(2p – q) – (2p + q)(3p2
– q)
= 6p3
– 3p2
q + 2pq – q2
– (6p3
– 2pq + 3p2
q – q2
)
= 6p3
– 3p2
q + 2pq – q2
– 6p3
+ 2pq – 3p2
q + q2
= 4pq – 6p2
q
38. (a) (2a + 1)(a2
– 3a – 4)
= 2a3
– 6a2
– 8a + a2
– 3a – 4
= 2a3
– 5a2
– 11a – 4
(b) (b + 2)(3b2
– 5b + 6)
= 3b3
– 5b2
+ 6b + 6b2
– 10b + 12
= 3b3
+ b2
– 4b + 12
(c) (7 – c)(5c2
– 2c + 1)
= 35c2
– 14c + 7 – 5c3
+ 2c2
– c
= –5c3
+ 37c2
– 15c + 7
(d) (d2
– 4)(d2
– 2d + 1)
= d4
– 2d3
+ d2
– 4d2
+ 8d – 4
= d4
– 2d3
– 3d2
+ 8d – 4
(e) (h – 2k)(2h + 3k – 1)
= 2h2
+ 3hk – h – 4hk – 6k2
+ 2k
= 2h2
– hk – 6k2
– h + 2k
(f) (m – n)(m2
+ mn + n2
)
= m3
+ m2
n + mn2
– m2
n – mn2
– n3
= m3
– n3
(g) (p + 1)(p3
– p2
+ p – 1)
= p4
– p3
+ p2
– p + p3
– p2
+ p – 1
= p4
– 1
(h) (q – 1)(q3
– 3q2
+ 3q – 1)
= q4
– 3q3
+ 3q2
– q – q3
+ 3q2
– 3q + 1
= q4
– 4q3
+ 6q2
– 4q + 1
39. (a) 2a2
b2
+ 4ab – 48 = 2(a2
b2
+ 2ab – 24)
× ab 6
ab a2
b2
6ab
–4 –4ab –24
2a2
b2
+ 4ab – 48 = 2(ab + 6)(ab – 4)
40

1
(b) 15c2
d2
e – 77cde + 10e = e(15c2
d2
– 77cd + 10)
× 15cd –2
cd 15c2
d2
–2cd
–5 –75cd 10
15c2
d2
e – 77cde + 10e = e(15cd – 2)(cd – 5)
(c) 12p2
q2
r – 34pqr – 28r = 2r(6p2
q2
– 17pq – 14)
× 3pq 2
2pq 6p2
q2
4pq
–7 –21pq –14
12p2
q2
r – 34pqr – 28r = 2r(3pq + 2)(2pq – 7)
(d) 3x2
+ 7xy +
15
4
y2
=
1
4
(12x2
+ 28xy + 15y2
)
× 6x 5y
2x 12x2
10xy
3y 18xy 15y2
3x2
+ 7xy +
15
4
y2
=
1
4
(6x + 5y)(2x + 3y)
40. (x2
– y)(x2
+ y)(x4
+ y2
)
= (x4
– y2
)(x4
+ y2
)
= x8
– y4
41. (a) 102
– 92
+ 82
– 72
+ 62
– 52
+ 42
– 32
+ 22
– 12
= (10 + 9)(10 – 9) + (8 + 7)(8 – 7) + (6 + 5)(6 – 5)
+ (4 + 3)(4 – 3) + (2 + 1)(2 – 1)
= 19 + 15 + 11 + 7 + 3
= 55
(b) 20082
– 20072
+ 20062
– 20052
+ 20042
– 20032
= (2008 + 2007)(2008 – 2007)
+ (2006 + 2005)(2006 – 2005)
+ (2004 + 2003)(2004 – 2003)
= 2008 + 2007 + 2006 + 2005 + 2004 + 2003
= 12 033
42. (a) a(b – c) + bc – a2
= ab – ac + bc – a2
= ab + bc – a2
– ac
= b(a + c) – a(a + c)
= (b – a)(a + c)
(b) 25x4
+
9
4
y2
z2
– x2
z2
–
225
4
x2
y2
=
1
4
[100x4
+ 9y2
z2
– 4x2
z2
– 225x2
y2
]
=
1
4
[100x4
– 4x2
z2
+ 9y2
z2
– 225x2
y2
]
=
1
4
[4x2
(25x2
– z2
) + 9y2
(z2
– 25x2
)]
=
1
4
[4x2
(25x2
– z2
) – 9y2
(25x2
– z2
)]
=
1
4
(4x2
– 9y2
)(25x2
– z2
)
=
1
4
(2x + 3y)(2x – 3y)(5x + z)(5x – z)
43. (i)
1
3
xy +
1
4
x2
y – y2
–
1
12
x3
=
1
12
[4xy + 3x2
y – 12y2
– x3
]
=
1
12
[4xy – 12y2
+ 3x2
y – x3
]
=
1
12
[4y(x – 3y) + x2
(3y – x)]
=
1
12
[4y(x – 3y) – x2
(x – 3y)]
=
1
12
(4y – x2
)(x – 3y)
(ii) Let x = 22 and y = 9:
1
3
× 22 × 9 +
1
4
× 484 × 9 – 81 –
1
12
× 10 648
=
1
12
[4(9) – 222
][22 – 3(9)]
= 186
2
3
New Trend
44. (a) 2ax – 4ay + 3bx – 6by
= 2a(x – 2y) + 3b(x – 2y)
= (2a + 3b)(x – 2y)
(b) 5ax − 10ay − 3bx + 6by
= 5a(x − 2y) − 3b(x − 2y)
= (5a − 3b)(x − 2y)
(c) 8ab – 6bc + 15cd – 20ad
= 2b(4a – 3c) + 5d(3c – 4a)
= 2b(4a – 3c) – 5d(4a – 3c)
= (2b – 5d)(4a – 3c)
45. (a) 27d3
– 48d
= 3d(9d2
– 16)
= 3d(3d + 4)(3d – 4)
(b) 3x2
− 75y2
= 3(x2
− 25y2
)
= 3(x + 5y)(x − 5y)
41

1
Revision Test A1
1. (a) 2x – y = 1 —(1)
8x – 3y = 9 —(2)
(1) × 3: 6x – 3y = 3 —(3)
(2) – (3): 2x = 6
x = 3
2(3) – y = 1
6 – y = 1
y = 5
x = 3, y = 5
(b) y =
1
2
x + 1 —(1)
x + y = 4 —(2)
x +
1
2
x + 1= 4
3
2
x = 3
x = 3 ×
2
3
= 2
y =
1
2
(2) + 1
= 2
x = 2, y = 2
2. (a) (2x – 3)2
– 3x(x + 7)
= 4x2
– 12x + 9 – 3x2
– 21x
= x2
– 33x + 9
(b) 3z(z + y – 4) – (y + 3)(z + 1)
= 3z2
+ 3yz – 12z – (yz + y + 3z + 3)
= 3z2
+ 3yz – 12z – yz – y – 3z – 3
= 3z2
+ 2yz – y – 15z – 3
3. (a) 8a2
– 12a + 12ab – 18b
= 4a(2a – 3) + 6b(2a – 3)
= (2a – 3)(4a + 6b)
= 2(2a – 3)(2a + 3b)
(b) 2(m – n)2
– 2m + 2n
= 2(m – n)2
– 2(m – n)
= 2(m – n)(m – n – 1)
(c) 343p4
– 7q2
= 7(49p4
– q2
)
= 7(7p2
+ q)(7p2
– q)
4. (i) 18x2
– 102x + 60 = 6(3x2
– 17x + 10)
× 3x –2
x 3x2
–2x
–5 –15x 10
18x2
– 102x + 60 = 6(3x – 2)(x – 5)
(ii) Breadth of rectangle =
x x
x
6(3 – 2)( – 5)
3 – 2
= (6x – 30) cm
5. (i) y = k x
When x = 16, y = 20,
20 = k 16
= 4k
k =
20
4
= 5
y = 5 x
(ii) When x = 25,
y = 5 25
= 25
(iii) When y = 8,
8 = 5 x
x =
8
5
x =
64
25
= 2
14
25
6. (a) P =
k
r2
When r = 0.1, P = 10,
10 =
k
0.12
k = 10 × 0.12
= 0.1
P =
r
0.1
2
(b) When r = 0.2,
P =
0.1
0.22
= 2.5
The pressure exerted is 2.5 pascals.
42

1
7. 2x + 3y + 5 = 0
x –2 0 2
y –0.33 –1.67 –3
3x + 2y = 0
x –2 0 2
y 3 0 –3
–3 –2 –1
–1
1
2
3
–2
–3
1
2x + 3y + 5 = 0
0 2
x
y
3x + 2y = 0
From the graph,
x = 2 and y = –3.
8. (a) R = a + bV2
When V = 27, R = 281,
281 = a + b(27)2
a + 729b = 281 —(1)
When V = 36, R = 344,
344 = a + b(36)2
a + 1296b = 344 —(2)
(b) (2) – (1): 567b = 63
b =
63
567
=
1
9
Substitute b =
1
9
into (1):
a + 729
1
9





 = 281
a + 81 = 281
a = 200
a = 200, b =
1
9
(c) (i) R = 200 +
1
9
V2
When V = 63,
R = 200 +
1
9
(63)2
= 641
The resistance is 641 newtons.
(ii) When R = 425,
425 = 200 +
1
9
V2
1
9
V2
= 225
V2
= 2025
V = 2025
= 45
The speed is 45 km/h.
43

1
Revision Test A2
1. (a) 7(a + 5) – 3(2 – 2a)
= 7a + 35 – 6 + 6a
= 13a + 29
(b) –4b(2b + 1) – 3b(5 – 3b)
= –8b2
– 4b – 15b + 9b2
= b2
– 19b
(c) (x + 3)(x2
+ x + 2)
= x3
+ x2
+ 2x + 3x2
+ 3x + 6
= x3
+ 4x2
+ 5x + 6
(d) (3y + 2z)(3y – 2z) – (y – z)2
= 9y2
– 4z2
– (y2
– 2yz + z2
)
= 9y2
– 4z2
– y2
+ 2yz – z2
= 8y2
+ 2yz – 5z2
2. (a) 2a4
– 32b2
c2
= 2(a4
– 16b2
c2
)
= 2(a2
+ 4bc)(a2
– 4bc)
(b) 64m2
n2
– 16mn + 1
= (8mn – 1)2
(c) p2
– 4q2
+ 3(p – 2q)
= (p + 2q)(p – 2q) + 3(p – 2q)
= (p + 2q + 3)(p – 2q)
3. 2x2
+ 25x + 63
× 2x 7
x 2x2
14x
9 18x 63
2x2
+ 25x + 63 = (2x + 7)(x + 9)
22 563 = 2(100)2
+ 25(100) + 63
Let x = 100:
22 563 = 207 × 109
The factors are 207 and 109.
4.
x
3
=
+
y
2 1
5
+ 2 —(1)
+
x y
x y
–
= 2
3
4
—(2)
From (1),
5x = 6y + 3 + 30
5x – 6y = 33 —(3)
From (2),
+
x y
x y
–
=
11
4
4x + 4y = 11x – 11y
15y = 7x
y =
7
15
x —(4)
5x – 6
7
15
x





 = 33
5x –
14
5
x = 33
11
5
x = 33
x =
5
11
× 33
= 15
y =
7
15
(15)
= 7
x = 15, y = 7
5. (a) y = k(x + 3)2
When x = 0, y = 36,
36 = 9k
k =
36
9
= 4
y = 4(x + 3)2
When x = 2,
y = 4(5)2
= 100
(b) H =
k
p
(2 – 3)3
When p = 1, H = –5,
–5 =
k
(–1)3
k = (–5)(–1)
= 5
H =
p
5
(2 – 3)3
(i) When p = 2.5,
H =
5
(5 – 3)3
=
5
8
(ii) When H =
5
27
,
5
27
=
p
5
(2 – 3)3
(2p – 3)3
= 27
2p – 3 = 3
2p = 6
p = 3
44

1
6. (a) Time taken to fry 1 pancake =
8
12
=
2
3
minutes
Time taken to fry 50 pancakes =
2
3
× 50
= 33
1
3
minutes
(b)
Number of
men
Number of
design projects
Number of
hours
8 12 9
1 12 72
6 12 12
6 32 32
× 8
÷ 6
×
8
3
÷ 8
× 6
×
8
3
6 men take 32 hours to complete 32 design projects.
7. x + y = 5
x 0 1 2
y 5 4 3
y = 2x – 1
x 0 1 2
y –1 1 3
–1
–1
1
2
3
4
5
1
0 2
x
x + y = 5
y = 2x – 1
y
From the graph,
x = 2 and y = 3.
8. x + y = 14
1
2
—(1)
40x + 50y = 660 —(2)
From (1),
y = 14
1
2
– x —(3)
From (2),
4x + 5y = 66 —(4)
4x + 5 14
1
2
– x





 = 66
4x + 72
1
2
– 5x = 66
x = 6
1
2
Substitute x = 6
1
2
into (3):
y = 14
1
2
– 6
1
2
= 8
Machine A was used for 6
1
2
hours and Machine B was
used for 8 hours.
45

1
Chapter 5 Quadratic Equations and Graphs
Basic
1. (a) a(a – 6) = 0
a = 0 or a = 6
(b) b(b + 4) = 0
b = 0 or b = –4
(c) 3c(c – 5) = 0
c = 0 or c = 5
(d) 5d(3d + 2) = 0
d = 0 or d = –
2
3
(e) –7e(9e – 4) = 0
e = 0 or e =
4
9
(f) –
8
3
f(7 – 5f) = 0
f = 0 or f =
7
5
= 1
2
5
2. (a) (a – 5)(2a – 7) = 0
a = 5 or a =
7
2
= 3
1
2
(b) (7c – 5)(2 – 9c) = 0
c =
5
7
or c =
2
9
(c) (6 – 5d)(15 + 11d) = 0
d =
6
5
or d = –
15
11
= 1
1
5
= –1
4
11
(d)
1
2
(e + 1)(2e – 5) = 0
e = –1 or e =
5
2
= 2
1
2
(e) –
3
4
(5f – 4)(1 + f)= 0
f =
4
5
or f = –1
3. (a) a2
+ 7a = 0
a(a + 7) = 0
a = 0 or a = –7
(b) b2
– 16b = 0
b(b – 16) = 0
b = 0 or b = 16
(c) 2c2
+ 5c = 0
c(2c + 5) = 0
c = 0 or c = –
5
2
= –2
1
2
(d) 3d2
– 12d = 0
3d(d – 4) = 0
d = 0 or d = 4
(e) 7e – 8e2
= 0
e(7 – 8e) = 0
e = 0 or e =
7
8
(f) –8f – 16f2
= 0
–8f(1 + 2f) = 0
f = 0 or f = –
1
2
4. (a) a2
+ 10a + 25 = 0
(a + 5)2
= 0
a = –5
(b) b2
– 20b + 100 = 0
(b – 10)2
= 0
b = 10
(c) c2
– 49 = 0
(c + 7)(c – 7) = 0
c = –7 or c = 7
(d) 9d2
+ 48d + 64 = 0
(3d + 8)2
= 0
d = –
8
3
= –2
2
3
(e) 36e2
– 132e + 121 = 0
(6e – 11)2
= 0
e =
11
6
= 1
5
6
(f) 2f2
– 288 = 0
f2
– 144 = 0
(f + 12)(f – 12) = 0
f = –12 or f = 12
46

1
5. (a) a2
+ 10a + 24 = 0
(a + 4)(a + 6) = 0
a = –4 or a = –6
(b) 5b2
– 17b + 6 = 0
(b – 3)(5b – 2) = 0
b = 3 or b =
2
5
(c) 2c2
+ 7c – 4 = 0
(2c – 1)(c + 4) = 0
c =
1
2
or c = –4
(d) 12d2
– d – 6 = 0
(4d – 3)(3d + 2) = 0
d =
3
4
or d = –
2
3
(e) 3 – 4e – 7e2
= 0
7e2
+ 4e – 3 = 0
(e + 1)(7e – 3) = 0
e = –1 or e =
3
7
(f) 8 – 5f2
– 18f = 0
5f2
+ 18f – 8 = 0
(f + 4)(5f – 2) = 0
f = –4 or f =
2
5
6. Let the number be x.
x + 2x2
= 36
2x2
+ x – 36 = 0
(2x + 9)(x – 4) = 0
x = –
9
2
or x = 4
= –4
1
2
(rejected)
The number is 4.
7. Let the numbers be x and x + 5.
x2
+ (x + 5)2
= 193
x2
+ x2
+ 10x + 25 = 193
2x2
+ 10x – 168 = 0
x2
+ 5x – 84 = 0
(x + 12)(x – 7) = 0
x = –12 or x = 7
(rejected) x + 5 = 12
x(x + 3) = 154
x2
+ 3x – 154 = 0
(x + 14)(x – 11) = 0
x = –14 or x = 11
x + 3 = –11 x + 3 = 14
The numbers are –14 and –11 or 11 and 14.
9. (i) (4x + 7)(5x – 4) = 209
20x2
– 16x + 35x – 28 = 209
20x2
+ 19x – 237 = 0
(20x + 79)(x – 3) = 0
x = –
79
20
or x = 3
= –3
19
20
(rejected)
x = 3
(ii) Perimeter of rectangle
= 2[4(3) + 7 + 5(3) – 4]
= 60 cm
10.
1
2
(x + 3 + x + 9)(3x – 4) = 80
1
2
(2x + 12)(3x – 4) = 80
(x + 6)(3x – 4) = 80
3x2
– 4x + 18x – 24 = 80
3x2
+ 14x – 104 = 0
(x – 4)(3x + 26) = 0
x = 4 or x = –
26
3
= –8
2
3
(rejected)
x = 4
47

1
11. (a) When x = 1, y = a,
a = 12
– 5(1) + 5
= 1
When x = 3, y = b,
b = 32
– 5(3) + 5,
= –1
When x = 4, y = c,
c = 42
– 5(4) + 5
= 1
a = 1, b = –1, c = 1
(b)
–1
1
2
3
4
5
1
0 2 3 4 5
x
y
y = x2
– 5x + 5
(c) (i) When x = 4.5, y = 2.75
(ii) When y = 2, x = 0.7 or x = 4.3
48

1
12. (a) When x = –2, y = a,
a = (–2 + 3)(–2 – 2)
= –4
When x = –1, y = b.
b = (–1 + 3(–1 – 2)
= –6
a = –4, b = –6
(b)
–3
–4 –2 –1
–2
2
4
6
–4
–6
1
y = (x + 3)(x – 2)
0 2 3
x
y
(c) (i) When x = 2.6, y = 3.4
(ii) When y = 1, x = –3.2 or x = 2.2
49

1
13. (a) When x = –1, y = a,
a = (–1)2
+ 1
= 2
When x = 3, y = b,
b = (3)2
+ 1
= 10
a = 2, b = 10
(b)
–3 –2 –1 0
6
8
10
12
14
16
4
2
1
y = x2
+ 1
2 3 4
y
x
(c) The minimum point is (0, 1).
(d) The equation of line of symmetry of the graph is x = 0.
50

1
14. (a) When x = –1, y = p,
p = (–1)2
– 4(–1)
= 5
When x = 3, y = q,
q = 32
– 4(3)
= –3
p = 5, q = –3
(b)
–2 –1 0
2
4
6
8
10
12
–2
–4
1
y = x2
– 4x
2 3 4
y
x
(c) The minimum point is (2, –4).
(d) The equation of line of symmetry of the graph is x = 2.
15. (a) When x = 2, y = a,
a = 6(2) – 22
= 8
When x = 5, y = b,
b = 6(5) – 52
= 5
a = 8, b = 5
(b)
1
2
3
4
5
6
7
8
9
1
0 2 3 4 5 6
x
y
y = 6x – x2
(c) (3, 9), maximum point
(d) The equation of the line of symmetry of the graph is
x = 3.
51

1
16. (a)
x –5 –4 –3 –2 –1 0 1 2 3
y 7 0 –5 –8 –9 –8 –5 0 7
(b)
–3
–4
–5 –2 –1
–2
2
4
6
–4
–6
–8
1
y = x2
+ 2x – 8
0 2 3
x
y
(c) (i) When y = 0, x = –4 or x = 2
(ii) When y = –2, x = –3.65 or x = 1.65
(iii) When y =
1
2
, x = –4.1 or x = 2.1
(d) The equation of the line of symmetry of the graph is
x = –1.
(e) Minimum value of y = –9,
Minimum value of y occurs when x = –1.
52

1
17. (a)
–2 –1 0
1
2
3
4
–1
–2
1
y = x2
– x – 2
2 3
y
x
(b) (i) When y = 1, x = –1.3 or x = 2.3.
(ii) Minimum value of y occurs when x = 0.5.
18. (i) x = –2 or x = –1
(ii) x =
+
–2 (–1)
2
= –1.5
Equation of line of symmetry is x = –1.5
19. (i) G1: y = –x2
G2: y = –x2
– 2
(ii) G3: y = –x2
– 4
Equation of line of symmetry is x = 0
Coordinates of maximum point are (0, –4)
Intermediate
20. (a) 2a2
= 3a + 14
2a2
– 3a – 14 = 0
(2a – 7)(a + 2) = 0
a =
7
2
or a = –2
= 3
1
2
(b) 12b2
– 12 = 7b
12b2
– 7b – 12 = 0
(3b – 4)(4b + 3) = 0
b =
4
3
or b = –
3
4
= 1
1
3
(c) c2
+ 4 = 8c – 8
c2
– 8c + 12 = 0
(c – 6)(c – 2) = 0
c = 6 or c = 2
(d) d2
=
+
d 15
6
6d2
= d + 15
6d2
– d – 15 = 0
(3d – 5)(2d + 3) = 0
d =
5
3
or d = –
3
2
= 1
2
3
= –1
1
2
(e) e(2e + 5) = 3
2e2
+ 5e – 3 = 0
(2e – 1)(e + 3) = 0
e =
1
2
or e = –3
(f) 3f(3f – 1) = 20
9f2
– 3f – 20 = 0
(3f – 5)(3f + 4) = 0
f =
5
3
or f = –
4
3
= 1
2
3
= –1
1
3
(g) 9g2
= 6(g + 20)
3g2
= 2(g + 20)
= 2g + 40
3g2
– 2g – 40 = 0
(g – 4)(3g + 10) = 0
g = 4 or g = –
10
3
= –3
1
3
(h) (6h + 5)(h – 1) = –3
6h2
– 6h + 5h – 5 = –3
6h2
– h – 2 = 0
(3h – 2)(2h + 1) = 0
h =
2
3
or h = –
1
2
21. Let the numbers be 2x, 2x + 2 and 2x + 4.
Sum = 2x + 2x + 2 + 2x + 4
= 6x + 6
= 6(x + 1), which is divisible by 6
22. Let the numbers be 2x + 1, 2x + 3
2x + 5 and 2x + 7.
Sum = 2x + 1 + 2x + 3 + 2x + 5 + 2x + 7
= 8x + 16
= 8(x + 2), which is divisible by 8
53

1
23. Let the integers be x and x + 2.
x2
+ (x + 2)2
= 340
x2
+ x2
+ 4x + 4 = 340
2x2
+ 4x – 336 = 0
x2
+ 2x – 168 = 0
(x – 12)(x + 14) = 0
x = 12 or x = –14
x + 2 = 14 x + 2 = –12
The integers are 12 and 14 or –14 and –12.
24. Let the integers be x – 1, x and x + 1.
(x – 1)2
+ x2
+ (x + 1)2
= 245
x2
– 2x + 1 + x2
+ x2
+ 2x + 1 = 245
3x2
= 243
x2
= 81
x = 9
x + 1 = 10
The largest number is 10.
(x + x + 2)2
– [x2
+ (x + 2)2
] = 126
(2x + 2)2
– (x2
+ x2
+ 4x + 4) = 126
4x2
+ 8x + 4 – 2x2
– 4x – 4 = 126
2x2
+ 4x – 126 = 0
x2
+ 2x – 63 = 0
(x + 9)(x – 7) = 0
x = –9 or x = 7
(rejected) x + 2 = 9
26. S =
1
2
n(n + 1)
When S = 325,
1
2
n(n + 1) = 325
n2
+ n = 650
n2
+ n – 650= 0
(n + 26)(n – 25) = 0
n = –26 or n = 25
(rejected)
25 integers must be taken.
27. Let Huixian’s age be x years.
x(x + 5) = 234
x2
+ 5x – 234 = 0
(x – 13)(x + 18) = 0
x = 13 or x = –18 (rejected)
Huixian’s current age is 13 years.
28.
+
p
p
8 5
5
=
+
p
p
3 4
2
16p2
+ 10p = 15p2
+ 20p
p2
– 10p = 0
p(p – 10) = 0
p = 0 or p = 10
p= 10
29. (i) (x + 2)2
+ (5x – 1)2
= (5x)2
x2
+ 4x + 4 + 25x2
– 10x + 1 = 25x2
x2
– 6x + 5 = 0 (shown)
(ii) x2
– 6x + 5 = 0
(x – 1)(x – 5) = 0
x = 1 or x = 5
(iii) Perimeter = x + 2 + 5x – 1 + 5x
= 11x + 1
Area =
1
2
(x + 2)(5x – 1)
When x = 1,
Perimeter = 12 cm
Area = 6 cm2
When x = 5,
Perimeter = 56 cm
Area = 84 cm2
30. (i) (3x + 1)(2x + 1) = 117
6x2
+ 3x + 2x + 1 = 117
6x2
+ 5x – 116 = 0
(x – 4)(6x + 29) = 0
x = 4 or x = –
29
6
= –4
5
6
x = 4
(ii) Perimeter = 2(3x + 1 + 2x + 1)
= 2(5x + 2)
When x = 4,
Perimeter = 44 cm
31. (i) 5x(4x + 2) = (6x + 3)(3x + 1)
20x2
+ 10x = 18x2
+ 6x + 9x + 3
2x2
– 5x – 3 = 0
(2x + 1)(x – 3) = 0
x = –
1
2
or x = 3
x = 3
(ii) Perimeter of A = 2(5x + 4x + 2)
= 2(9x + 2)
Perimeter of B = 2(6x + 3 + 3x + 1)
= 2(9x + 4)
B has a greater perimeter.
32. (i) Let the breadth of the original rectangle be x cm.
x(x – 8) –
x
2
(x – 8 + 6) = 36
2x(x – 8) – x(x – 2) = 72
2x2
– 16x – x2
+ 2x = 72
x2
– 14x – 72 = 0
(x – 18)(x + 4) = 0
x = 18 or x = –4
The length of the original rectangle is 18 cm.
54

1
(ii) Perimeter of original rectangle
= 2(18 + 18 – 8)
= 56 cm
33. (i) Let the length of the shorter side be x m.
x(x + 7) = 450
x2
+ 7x – 450 = 0
(x – 18)(x + 25) = 0
x = 18 or x = –25
The length of the shorter side is 18 m.
(ii) Perimeter of field = 2(18 + 18 + 7)
= 86 m
34. Let the length of the smaller field be 3x m.
(5x)2
– (3x)2
= 576
25x2
– 9x2
= 576
16x2
= 576
x2
= 36
x = 6
Area of smaller field = [3(6)]2
= 324 m2
35. (a)
x –2 –1 0 0.5 1 2 3 4
y 7 –2 –7 –8 –8 –5 2 13
(b)
–2 –1 0
5
10
15
–5
–10
1
y = 2x2
– 3x – 7
y = 2
2 3 4
y
x
(c) y = 8.1
(d) 2x2
– 3x – 7 = 2
Draw y = 2.
x = –1.5 or x = 3
55

1
36. (a)
x –2 –1 0 1 2 3 4 5
y –39 –14 3 12 13 6 –9 –32
(b)
–1
–2 1 2
–10
10
20
–20
–30
–40
4
3
y = 3 + 13x – 4x2
y = –10
0 5
x
y
(c) x = –0.2 or x = 3.5
(d) x = –0.8 or x = 4.05
37. (i) y = (x + 2)(x – 4)
When y = 0,
(x + 2)(x – 4) = 0
x = –2 or x = 4
A(–2, 0), C(4, 0)
When x = 0,
y = (0 + 2)(0 – 4)
= –8
B(0, –8)
A(–2, 0), B(0, –8), C(4, 0)
(ii) x =
+
–2 4
2
= 1
Equation of line of symmetry is x = 1.
(iii) When x = 1,
y = (1 + 2)(1 – 4)
= –9
Coordinates of minimum point are (1, –9).
56
38. (i) y = 12 + 4x – x2
= (2 + x)(6 – x)
When y = 0,
(2 + x)(6 – x) = 0
x = –2 or x = 6
A(–2, 0), D(6, 0)
When x = 0,
y = (2 + 0)(6 – 0)
= 12
B(0, 12)
x =
+
–2 6
2
= 2
When x = 2,
y = (2 + 2)(6 – 2)
= 16
C(2, 16)
A(–2, 0), B(0, 12), C(2, 16), D(6, 0)
(ii) Equation of line of symmetry is x = 2

1
39. (i) y = x(4 – x)
When y = 0,
x(4 – x) = 0
x = 0 or x = 4
R(4, 0)
(ii) Substitute x = –1, y = k into y = x(4 – x):
k = –1[4 – (–1)]
= –5
(iii) x =
+
0 4
2
= 2
Equation of line of symmetry is x = 2.
When x = 2,
y = 2(4 – 2)
= 4
M(2, 4)
(iv) Substitute x = 3, y = p into y = x(4 – x):
p = 3(4 – 3)
= 3
m =
3 – 0
3 – 0
= 1
p = 3, m = 1
Advanced
40. (i) Substitute x = 3 into 2x2
+ px = 15:
2(3)2
+ p(3) = 15
18 + 3p = 15
3p = –3
p = –1
(ii) 2x2
– x = 15
2x2
– x – 15 = 0
(2x + 5)(x – 3) = 0
x = –
5
2
or x = 3
= –2
1
2
The other solution is x = –2
1
2
.
41. x2
= 12(x – 3) + 1
= 12x – 36 + 1
x2
– 12x + 35 = 0
(x – 5)(x – 7) = 0
x = 5 or x = 7
42. Case I is true.
Case II is not true.
Case III is not true.
Case IV is not true.
43. (d) is true.
New Trend
44. (5b + 9)(8 – 3b) = 0
b = –
9
5
or b =
8
3
= –1
4
5
= 2
2
3
45. (i) T14 =
14 15
( )
6
= 35
(ii)
n n+1
( )
6
= 57
n2
+ n = 342
n2
+ n − 342 = 0
(n + 19)(n − 18) = 0
n = −19 (rejected) or n = 18
The 18th
term has the value 57.
46. (a) (i) Next line is the 6th
line: 62
– 6 = 30.
(ii) 8th
line: 82
– 8 = 56
(iii) From the number pattern, we observe that
12
– 1 = 1(1 – 1)
22
– 2 = 2(2 – 1)
32
– 3 = 3(3 – 1)
42
– 4 = 4(4 – 1)
52
– 5 = 5(5 – 1)
:
nth
line: n2
– n = n(n – 1)
(b) (i) 1392
– 139 = 139(139 – 1) = 19 182
(ii) x2
– x = 2450
x(x – 1) = 2450
We need to find the product of two numbers where
the difference of the two numbers is 1 that gives
2450. By trial and error, 2450 = 50 × 51 and
x = 50.
(c) Pn = 3n − 8
(d)
3n – 8
n2
– n
=
1
3
9n − 24 = n2
− n
n2
– 10n + 24 = 0
(n − 4)(n − 6) = 0
n = 4 or n = 6
57

1
Chapter 6 Algebraic Fractions and Formulae
Basic
1. (a)
a b
a
45
3
2
= 15ab
(b)
c d
cd
35
7
7 3
4
=
c
d
5 6
(c)
ef g
e fg
64
24
3 4
3 2
=
f g
e
8
3
2 2
2
(d)
h jk
hjk
8
(2 )
3 4
4
=
h jk
h j k
8
16
3 4
4 4 4
=
hj
1
2 3
(e)
8mn2
x3
(4mnx)2
=
8mn2
x3
16m2
n2
x2
=
x
2m
(f)
p q r
pq r
9
(3 )
3 4
2 3
=
p q r
p q r
9
27
3 4
3 6 3
=
q r
1
3 2 2
2. (a)
a b
ab
(5 )
25
3 4 3
3
=
a b
ab
125
25
9 12
3
= 5a8
b9
(b)
c d e
cde
(4 )
8
2 2 3
4
=
c d e
cde
16
8
4 3
4
=
c d
e
2 3 2
3
(c)
f g h
gh
(7 )
21
2 2 4
=
f g h
gh
49
21
4 2 4
=
f gh
7
3
4 3
(d)
jkl
j k
(2 )
8
2 4
2 3 =
j k l
j k
16
8
4 4 8
2 3
= 2j2
kl8
3. (a)
+
+
a b
a b
4 8
6 12
=
+
+
a b
a b
4( 2 )
6( 2 )
=
2
3
(b)
c cd
c d
8 – 16
5 – 10
2
=
c c d
c d
8 ( – 2 )
5( – 2 )
=
c
8
5
(c)
+
+
e ef
ef f
2
2
=
+
+
e e f
f e f
( )
( )
=
e
f
(d)
gh h
g h
–
( – )
2
2
=
h g h
g h
( – )
( – )2
=
h
g h
–
(e)
j jk
k jk
–
–
2
2
=
j j k
k k j
( – )
( – )
=
j j k
k j k
( – )
– ( – )
= –
j
k
(f)
mn m
m
4 – 8
6
2
2
=
m n m
m
4 ( – 2 )
6 2
=
n m
m
2( – 2 )
3
4. (a)
a b
a b
–
( – )
2 2
2
=
+
a b a b
a b
( )( – )
( – )2
=
+
a b
a b
–
(b)
c c
c
– 4
– 16
2
2
=
+
c c
c c
( – 4)
( 4)( – 4)
=
+
c
c 4
(c)
+ +
+
d d
d d
4 4
2
2
2
=
+
+
d
d d
( 2)
( 2)
2
=
+
d
d
2
(d)
+
e
e e
– 2
– 5 6
2
=
e
e e
– 2
( – 2)( – 3)
=
e
1
– 3
(e)
+
f
f f
5 – 15
3 – 13 12
2
=
f
f f
5( – 3)
(3 – 4)( – 3)
=
f
5
3 – 4
(f)
+
+ +
gh h
g g
7 6
2
=
+
+ +
h g
g g
( 1)
( 1)( 6)
=
+
h
g 6
5. (a)
ab
c
6
7
2
×
a
bc
56
48
3
=
a b
c
4
2
(b)
a b
bc
5
3
2 4
4
×
b
a
9
10
2
3 =
b
ac
3
2
5
4
(c)
d e
ef
4
3
2
×
e f
d
27
16
2 3
4 =
e f
d
9
4
2 2
2
(d)
d e
ef
16
7
2 4
2 ×
e f
d e
21
24
4 3
3 3 =
e f
d
2 4
(e)
h x
ky
9
4
3 2
2
×
k y
h y
5
12
2 4
2
=
hkx y
15
16
2
(f)
xy
abc
16
15
3
2
×
a bc
x yz
25
8
3
2
=
a y
cxz
10
3
2 2
6. (a)
a b
c
2
3
2
÷
abc
c
3
8 3
=
a b
c
2
3
2
×
c
abc
8
3
3
=
ac
16
9
58

1
(b)
d e
d e
18
14
4 3
2
÷
de
ef
27
21
5
2
=
d e
d e
18
14
4 3
2
×
ef
de
21
27
2
5
=
df 2
e2
(c)
a b
xy
14
6
3
÷
abc
x y
21
12 2 3
=
a b
xy
14
6
3
×
x y
abc
12
21
2 3
=
a xy
c
4
3
2 2
(d)
a x
bxy
81
16
3 3
÷
ax
b y
63
24
2
2 3
=
a x
bxy
81
16
3 3
×
b y
ax
24
63
2 3
2
=
a by
27
14
2 2
7. (a)
+
a b
a b
4( 3 )
– 3
×
+
a b
a b
3( – 3 )
25( 3 )
=
12
25
(b)
c d
e
7 – 28
2
×
e
c d
2 – 8
=
c d
e
7( – 4 )
2
×
e
c d
2( – 4 )
=
e
7
2
(c)
+
g h
f
3( )
10
÷
+
g h
f
8 8
5 3
=
+
g h
f
3( )
10
×
+
f
g h
5
8( )
3
=
f
3
16
2
(d)
+ +
j k
2( 5)
9
÷ (3j + 3k + 15)
=
+ +
j k
2( 5)
9
×
+ +
j k
1
3( 5)
=
2
27
8. (a)
a
3
5
+
a
1
4
=
+
a
12 5
20
=
a
17
20
(b)
b
1
2
+
b
3
4
–
b
1
6
=
+
b
6 9 – 2
12
=
b
13
12
(c)
c
2
7
–
d
1
7
=
d c
cd
2 –
7
(d)
ef
g
4
3
+
ef
g
–
ef
g
2
5
=
+
ef ef ef
g
20 15 – 6
15
=
ef
g
29
15
(e)
+
h j
k
2
+
h j
k
3 –
3
–
j h
k
–
5
=
+ +
h j h j j h
k
15( ) 10(3 – ) – 6( – )
30
=
+ + +
h j h j j h
k
15 15 30 – 10 – 6 6
30
=
h j
k
51 –
30
(f)
p q
r
2( – )
+
+
p q
r
3( 2 )
4
–
p q
r
5( – 4 )
6
=
+ +
p q p q p q
r
24( – ) 9( 2 ) – 10( – 4 )
12
=
+ + +
p q p q p q
r
24 – 24 9 18 – 10 40
12
=
+
p q
r
23 34
12
(g)
+
u
v
2 3
+
+
u
v
6
4 6
=
+
u
v
2 3
+
+
u
v
6
2(2 3)
=
+
u
v
2 3
+
+
u
v
3
2 3
=
+
u
v
4
2 3
(h)
+
z
x y
1
– 2
–
z
x y
2 – 3
2 – 4
+
z
x y
3 – 6
=
z
x y
1
– 2
+
–
z
x y
2 – 3
2( – 2 )
+
z
x y
3( – 2 )
=
+ +
z z z
x y
6( 1) – 3(2 – 3) 2
6( – 2 )
=
+ + +
z z z
x y
6 6 – 6 9 2
6( – 2 )
=
+
z
x y
2 15
6( – 2 )
9. (a)
a
2
+
a
5
2( – 1)
=
+
a a
a a
4( – 1) 5
2 ( – 1)
=
+
a a
a a
4 – 4 5
2 ( – 1)
=
a
a a
9 – 4
2 ( – 1)
59

1
(b)
+
d
2
3 1
–
+
d
1
5 3
=
+ +
+ +
d d
d d
2(5 3) – (3 1)
(3 1)(5 3)
=
+
+ +
d d
d d
10 6 – 3 – 1
(3 1)(5 3)
=
+
+ +
d
d d
7 5
(3 1)(5 3)
10. (a) a + x = b
a = b – x
(b) b – k = h
b = h + k
(c) c – d = e + f
c = d + e + f
(d) g – h – j = k2
g = h + j + k2
(e) q – p + r = s3
–p = –q – r + s3
p = q + r – s3
(f) 7k + h – q = 2q
3q = 7k + h
q =
+
k h
7
3
(g) 5pq – r = p2
– q
r = 5pq + q – p2
(h) 3ab + s = a2
b
s = a2
b – 3ab
(i) wx = 3y
w =
y
x
3
(j) 2xy = 3ak
x =
ak
y
3
2
(k) x2
y = 5k – 4
y =
k
x
5 – 4
2
(l) kz = p – q + k
z =
+
p q k
k
–
11. (a) ah = b – c + k3
a =
+
b c k
h
– 3
(b) 5by2
= x4
– y
b =
x y
y
–
5
4
2
(c) cx =
A
B
c =
A
Bx
(d) 3dk =
x
y
15
7
d =
x
ky
5
7
(e) 12mk =
+
ek
x y
3
5
2
3ek2
= 12mk(5x + y)
e =
+
mk x y
k
12 (5 )
3 2
=
+
m x y
k
4 (5 )
(f)
f
y
5
2
=
bk
x
3
4
f =
bk
x
3
4
×
y
2
5
=
bky
x
3
10
(g)
eg
t
=
+
t
n
5 4
g =
+
t
e n
(5 4)
2
(h) v = m(h + c)
h + c =
v
m
h =
v
m
– c
12. (a)
a
5
=
6
7
6a = 35
a =
35
6
= 5
5
6
(b)
b
7
2
= 3
6b = 7
b =
7
6
= 1
1
6
(c)
c
3
– 2
=
1
2
c – 2 = 6
c = 8
(d)
d
5
– 4
– 3 = 0
d
5
– 4
= 3
3(d – 4) = 5
d – 4 =
5
3
d = 4 +
5
3
= 5
2
3
60

1
(e)
e
9
5 – 2
+ 7 = 0
e
9
5 – 2
= –7
–7(5 – 2e) = 9
–35 + 14e = 9
14e = 44
e =
44
14
= 3
1
7
(f)
f
f
2
5 – 4
+
1
3
= 0
f
f
2
5 – 4
= –
1
3
6f = –(5f – 4)
= –5f + 4
11f = 4
f =
4
11
13. (a)
a
2
5
=
a
4
– 1
2(a – 1)= 4(5a)
2a – 2 = 20a
18a = –2
a = –
2
18
= –
1
9
(b)
b
7
2 – 1
=
b
3
– 4
7(b – 4) = 3(2b – 1)
7b – 28 = 6b – 3
b = 25
(c)
+
c
3
2
=
c
14
2 – 1
3(2c – 1) = 14(c + 2)
6c – 3 = 14c + 28
8c = –31
c = –
31
8
= –3
7
8
(d)
d
7
2 – 5
=
+
d
9
3 4
7(3d + 4) = 9(2d – 5)
21d + 28 = 18d – 45
3d = –73
d = –
73
3
= –24
1
3
(e)
+
e
3
1
+
+
e
1
2 1
= 0
+
e
3
1
= –
+
e
1
2 1
3(2e + 1) = –(e + 1)
6e + 3 = –e – 1
7e = –4
e = –
4
7
(f)
f
5
2 – 5
–
+
f
4
7 1
= 0
f
5
2 – 5
=
+
f
4
7 1
5(7f + 1) = 4(2f – 5)
35f + 5 = 8f – 20
27f = –25
f = –
25
27
14. (a)
x
4
+ 1
1
2
=
x
5
2
x
4
–
x
5
2
= –1
1
2
x
8 – 5
2
= –
3
2
x
3
2
= –
3
2
6 = –6x
x = –1
(b)
1
3
1
5y
– 3





 =
1
2
2 –
1
y






y
1
15
– 1= 1 –
y
1
2
y
1
15
+
y
1
2
= 2
+
y
2 15
30
= 2
y
17
30
= 2
17 = 2(30y)
60y = 17
y =
17
60
15.
x
5
2 – 7
–
x
6
– 7
= 0
x
5
2 – 7
=
x
6
– 7
5(x – 7) = 6(2x – 7)
5x – 35 = 12x – 42
7x = 7
x = 1
61

1
Intermediate
16. (a)
a b c
abc
(–3 )
27
2 3
4
=
a b c
abc
9
27
2 3
4
=
ab
c
3
2
3
(b)
d e
d e
(–3 )
9
2 4 3
2 5
=
d e
d e
–27
9
6 12
2 5
= –3d4
e7
(c)
fg h
f gh
(–4 )
–16
3 3
4 5
=
f g h
f gh
–64
–16
3 9 3
4 5
=
g
fh
4 8
2
(d)
j kl
jkl
(–9 )
(27 )
4 3
2
=
j k l
j k l
–729
729
12 3 3
2 2 2
= –j10
kl
17. (a)
a b
a ab
(2 – 3 )
6 – 9
2
2
=
a b
a a b
(2 – 3 )
3 (2 – 3 )
2
=
a b
a
2 – 3
3
(b)
+
+
c d x y
c x y
5 ( )
10 ( )
3
2
=
+
c d
x y
2( )
2
(c)
x e f
xy e f
15 ( – )
35 ( – )
3 2
2
=
x
y
3
7
2
(d)
+
+
g g
g g
– 6
– 9 14
2
2
=
+
g g
g g
( 3)( – 2)
( – 7)( – 2)
=
+
g
g
3
– 7
(e)
+ +
h h
h h
6 – 13 – 5
6 17 5
2
2
=
+
+ +
h h
h h
(3 1)(2 – 5)
(3 1)(2 5)
=
+
h
h
2 – 5
2 5
(f)
+
+
k k
k k
6 – 11 4
3 – 14
2
2
=
+
k k
k k
(4 – 3)( – 2)
(3 7)( – 2)
=
+
k
k
4 – 3
3 7
(g)
+
+
p q r
q r p
( ) –
( ) –
2 2
2 2
=
+ + +
+ + +
p q r p q r
q r p q r p
( )( – )
( )( – )
=
+
+
p q r
q r p
–
–
(h)
+
+ +
x y z
x xy xz
(3 ) – 4
15 5 10
2 2
2
=
+
+ +
x y z
x xy xz
(3 ) – (2 )
15 5 10
2 2
2
=
+ + +
+ +
x y z x y z
x x y z
(3 2 )(3 – 2 )
5 (3 2 )
=
+
x y z
x
3 – 2
5
(i)
+
+
xz yz
x xz xy yz
6 3
6 – 2 3 –
2
=
+
+
z x y
x x z y x z
3 (2 )
2 (3 – ) (3 – )
=
+
+
z x y
x y x z
3 (2 )
(2 )(3 – )
=
z
x z
3
3 –
(j)
+
+
x xz xy yz
x xy xz yz
– 2 – 2
– –
2
2
=
+
+ +
x x z y x z
x x y z x y
( – 2 ) ( – 2 )
( ) – ( )
=
+
+
x y x z
x z x y
( )( – 2 )
( – )( )
=
x z
x z
– 2
–
(k)
+
+
ac bc ad bd
cx cy dx dy
2 – 2 –
– 3 – 3
=
+ +
c a b d a b
c x y d x y
(2 ) – (2 )
( – 3 ) – ( – 3 )
=
+
c d a b
c d x y
( – )(2 )
( – )( – 3 )
=
+
a b
x y
2
– 3
18. (a)
a
b
2
×
c
a
3
4
×
a
c
8
9
=
a
b
4
3
(b)
d
ef
3 2
×
e
ef
6
21
2
×
f
de
28
3
2
=
d
e
8
62

1
(c)
h
2
2
×
k
1
3
÷
h
k
2
3
=
h
2
2
×
k
1
3
×
k
h
3
2
=
h k
3
3 2
(d)
m n
m
4
36
2 4
×
m
m n
24
8 2 3 ÷
m
mn
16
6 2
=
m n
m
4
36
2 4
×
m
m n
24
8 2 3
×
mn
m
6
16
2
=
n
8
3
(e)
p q
r
3
8
3 3
4
×
q r
p
6
5
2 3
5
÷
q
pr
9
10
2
=
p q
r
3
8
3 3
4
×
q r
p
6
5
2 3
5
×
pr
q
10
9 2
=
q
p
2
3
(f)
x y
az
2
7
2 3
3
÷
x z
a z
4
21
2
2 ×
a
xy
3
8
=
x y
az
2
7
2 3
3
×
a z
x z
21
4
2
2
×
a
xy
3
8
=
a y
xz
9
16
2 2
3
(g)
b
c
9
21
÷
3d
4e
×
16be
9a






=
b
c
3
7
÷
bd
a
4
3
=
b
c
3
7
×
a
bd
3
4
=
a
cd
9
28
(h)
x
y
3
4
÷
7x2
15z
÷
3y2
10z2






=
x
y
3
4
÷
7x2
15z
×
10z2
3y2






=
x
y
3
4
÷
x z
y
14
9
2
2
=
x
y
3
4
×
y
x z
9
14
2
2
=
y
xz
27
56
19. (a)
x x
ax a
–
–
5 4
÷
ax
ax x
–
2
=
x x
a x
( – 1)
( – 1)
4
÷
ax
x a
( – 1)
2
=
x
a
4
×
a
ax
– 1
=
x a
a
( – 1)
3
2
(b)
+
x
x 1
÷
x x
x x
– 2
– 2 – 3
2
2
=
+
x
x 1
÷
+
x x
x x
( – 2)
( – 3)( 1)
=
+
x
x 1
×
+
x x
x x
( – 3)( 1)
( – 2)
=
x
x
– 3
– 2
20. (a)
+
a
3
1
+
+
+ +
a
a a
4
( 1)( 2)
=
+ + +
+ +
a a
a a
3( 2) 4
( 1)( 2)
=
+ + +
+ +
a a
a a
3 6 4
( 1)( 2)
=
+
+ +
a
a a
4 10
( 1)( 2)
(b)
b
2
– 1
–
b
1
– 2
+
+
b
b b
3( 2)
( – 1)( – 2)
=
+ +
b b b
b b
2( – 2) – ( – 1) 3( 2)
( – 1)( – 2)
=
+ + +
b b b
b b
2 – 4 – 1 3 6
( – 1)( – 2)
=
+
b
b b
4 3
( – 1)( – 2)
(c)
c c
4
( – 2)( – 4)
–
c c
2
( – 2)( – 3)
=
c c
c c c
4( – 3) – 2( – 4)
( – 2)( – 3)( – 4)
=
+
c c
c c c
4 – 12 – 2 8
( – 2)( – 3)( – 4)
=
c
c c c
2 – 4
( – 2)( – 3)( – 4)
=
c
c c c
2( – 2)
( – 2)( – 3)( – 4)
=
c c
2
( – 3)( – 4)
(d)
+
d
d d
– 4
( 1)( – 5)
–
+
+ +
d
d d
5
( 1)( 3)
=
+ +
+ +
d d d d
d d d
( – 4)( 3) – ( 5)( – 5)
( 1)( 3)( – 5)
=
+
+ +
d d d d
d d d
( 3 – 4 – 12) – ( – 25)
( 1)( 3)( – 5)
2 2
=
+
+ +
d d d
d d d
– – 12 – 25
( 1)( 3)( – 5)
2 2
=
+ +
d
d d d
13 –
( 1)( 3)( – 5)
63

1
(e)
+
e
e f e f
( )( – 3 )
2
–
e f
e f
–
– 3
=
+
+
e e f e f
e f e f
– ( )( – )
( )( – 3 )
2
=
+
e e f
e f e f
– ( – )
( )( – 3 )
2 2 2
=
+
+
e e f
e f e f
–
( )( – 3 )
2 2 2
=
+
f
e f e f
( )( – 3 )
2
(f)
g
g
3
– 3
–
g
g – 9
2
=
g
g
3
– 3
–
+
g
g g
( 3)( – 3)
=
+
+
g g g
g g
3 ( 3) –
( 3)( – 3)
=
+
+
g g g
g g
3 9 –
( 3)( – 3)
2
=
+
+
g g
g g
3 8
( 3)( – 3)
2
(g)
h
h – 4
2
–
+
h
1
2
=
+
h
h h
( 2)( – 2)
–
+
h
1
2
=
+
h h
h h
– ( – 2)
( 2)( – 2)
=
+
+
h h
h h
– 2
( 2)( – 2)
=
+
h h
2
( 2)( – 2)
(h)
+
j
j
2
– 1
2
–
j
3
2( – 1)
=
+
+
j
j j
2
( 1)( – 1)
–
j
3
2( – 1)
=
+ +
+
j j
j j
2( 2) – 3( 1)
2( 1)( – 1)
=
+
+
j j
j j
2 4 – 3 – 3
2( 1)( – 1)
=
+
+
j
j j
– 1
2( 1)( – 1)
=
+
j
j j
–( – 1)
2( 1)( – 1)
= –
+
j
1
2( 1)
21. (a)
a
1
– 1
+
a
a
2
1 – 2
=
a
1
– 1
+
+
a
a a
2
(1 )(1 – )
=
a
1
– 1
–
+
a
a a
2
(1 )( – 1)
=
+
+
a a
a a
1 – 2
( – 1)(1 )
=
+
a
a a
1 –
( – 1)(1 )
= –
+ a
1
1
(b)
+
b
3
2
–
b
b
4 – 2
=
+
b
3
2
–
+
b
b b
(2 )(2 – )
=
+
b
3
2
+
+
b
b b
( 2)( – 2)
=
+
+
b b
b b
3( – 2)
( 2)( – 2)
=
+
+
b b
b b
3 – 6
( 2)( – 2)
=
+
b
b b
4 – 6
( 2)( – 2)
(c)
+
c
2
4
+
c
3
4 –
–
c
c – 16
2
=
+
c
2
4
–
c
3
– 4
–
+
c
c c
( 4)( – 4)
=
+
+
c c c
c c
2( – 4) – 3( 4) –
( 4)( – 4)
=
+
c c c
c c
2 – 8 – 3 – 12 –
( 4)( – 4)
=
+
c
c c
–2 – 20
( 4)( – 4)
= –
+
+
c
c c
2 20
( 4)( – 4)
(d)
+
d e
1
2 3
+
d
e d
4
9 – 4
2 2
–
e d
2
3 – 2
=
+
d e
1
2 3
–
d
d e
4
4 – 9
2 2
+
d e
2
2 – 3
=
+
d e
1
2 3
–
+
d
d e d e
4
(2 3 )(2 – 3 )
+
d e
2
2 – 3
=
+ +
+
d e d d e
d e d e
2 – 3 – 4 2(2 3 )
(2 3 )(2 – 3 )
=
+
+
d e
d e d e
2 3
(2 3 )(2 – 3 )
=
d e
1
2 – 3
64

1
22. (a)
+
a
a a
2
– 6
2
+
a
1
– 2
=
+
a
a a
2
( 3)( – 2)
+
a
1
– 2
=
+ +
+
a a
a a
2 3
( 3)( – 2)
=
+
+
a
a a
3 3
( 3)( – 2)
(b)
b
1
2( – 1)
+
+
+
b
b b
1
– 2
2
=
b
1
2( – 1)
+
+
+
b
b b
1
( – 1)( 2)
=
+ + +
+
b b
b b
2 2( 1)
2( – 1)( 2)
=
+ + +
+
b b
b b
2 2 2
2( – 1)( 2)
=
+
+
b
b b
3 4
2( – 1)( 2)
(c)
+
c c
4
2 – 3
2
–
+
c c
1
– 5 4
2
=
+
c c
4
( – 1)( 3)
–
c c
1
( – 1)( – 4)
=
+
+
c c
c c c
4( – 4) – ( 3)
( – 1)( – 4)( 3)
=
+
c c
c c c
4 – 16 – – 3
( – 1)( – 4)( 3)
=
+
c
c c c
3 – 19
( – 1)( – 4)( 3)
(d)
+
d
d d
3 – 2
– 3 2
2
–
d
d d
3 – 1
– 2
2
=
d
d d
3 – 2
( – 1)( – 2)
–
d
d d
3 – 1
( – 2)
=
d d d d
d d d
(3 – 2) – (3 – 1)( – 1)
( – 1)( – 2)
=
+
d d d d d
d d d
3 – 2 – (3 – 3 – 1)
( – 1)( – 2)
2 2
=
+ +
d d d d d
d d d
3 – 2 – 3 3 – 1
( – 1)( – 2)
2 2
=
d
d d d
2 – 1
( – 1)( – 2)
(e)
e
e f
4
–
+
+
e
e f
2
2
+
+
e ef f
1
– 2
2 2
=
e
e f
4
–
+
+
e
e f
2
2
+
+
e f e f
1
( – )( 2 )
=
+ + +
+
e e f e e f
e f e f
4 ( 2 ) 2 ( – ) 1
( – )( 2 )
=
+ + +
+
e ef e ef
e f e f
4 8 2 – 2 1
( – )( 2 )
2 2
=
+ +
+
e ef
e f e f
6 6 1
( – )( 2 )
2
(f)
+
g h
1
3 2
+
g h
1
2 – 3
–
g h gh
1
6 – 6 – 5
2 2
=
+
g h
1
3 2
+
g h
1
2 – 3
–
+
g h g h
1
(3 2 )(2 – 3 )
=
+ +
+
g h g h
g h g h
2 – 3 3 2 – 1
(3 2 )(2 – 3 )
=
+
g h
g h g h
5 – – 1
(3 2 )(2 – 3 )
23. (a)
+
a
2
1
2
=
+
a
4
2 1
(b)
+
b
c
1
4
2
1
2
=
+
b
c
8 2
24. (a) 2x –
8
x





 ÷ 1 –
2
x






=
x
x
2 – 8
2
÷
x
x
– 2
=
x
x
2( – 4)
2
÷
x
x
– 2
=
+
x x
x
2( 2)( – 2)
×
x
x – 2
= 2(x + 2)
(b)
1
x
–
1
y





 ÷
1
x2
–
1
y2






=
y x
xy
–
÷
y x
x y
–
2 2
2 2
=
y x
xy
–
×
x y
y x
–
2 2
2 2
=
y x
xy
–
×
+
x y
y x y x
( )( – )
2 2
=
+
xy
x y
25.
1
x3
–
1
x





 ÷
1
x2
–
1
x






=
x
x
1 – 2
3 ÷
x
x
1 –
2
=
+ x x
x
(1 )(1 – )
3
×
x
x
1 –
2
=
+ x
x
1
k = 1
65

1
26. (a) (a + p)y = q(2a – q)
ay + py = 2aq – q2
2aq – ay = py + q2
a(2q – y)= py + q2
a =
+
py q
q y
2 –
2
(c)
+
k m c
m
( )
=
x
4
k(m + c) =
m
x
4
m + c =
m
kx
4
c =
m
kx
4
– m
(c)
d
1
+ c =
b
d
b
d
–
d
1
= c
b
d
– 1
= c
b – 1 = cd
d =
b
c
– 1
(d) y =
+
bj k
j
7
7 – 4
y(7 – 4j) = 7bj + k
7y – 4jy = 7bj + k
7bj + 4jy = 7y – k
j(7b + 4y) = 7y – k
j =
7y – k
7b + 4y
(e)
+
x
k y
=
y
k
2
kx = y2
(k + y)
= ky2
+ y3
kx – ky2
= y3
k(x – y2
) = y3
k =
y
x y
–
3
2
(f)
3
5
=
+
n a
n b
– 4
7
3(n + 7b) = 5(n – 4a)
3n + 21b = 5n – 20a
2n = 20a + 21b
n =
+
a b
20 21
2
(g) kx + 4 =
x r
r
2 – 3
2 – 5
(kx + 4)(2r – 5) = 2x – 3r
2krx – 5kx + 8r – 20 = 2x – 3r
2krx + 11r = 5kx + 2x + 20
r(2kx + 11) = 5kx + 2x + 20
r =
+ +
+
kx x
kx
5 2 20
2 11
(h) k – 3ux =
uy
3
4
4k – 12ux = 3uy
12ux + 3uy = 4k
u(12x + 3y) = 4k
u =
+
k
x y
4
12 3
(i) x =
+ +
+
kw hx
bw xy
3 4 4
5 – 4 2
x(5bw – 4xy + 2) = 3kw + 4hx + 4
5bwx – 4x2
y + 2x = 3kw + 4hx + 4
5bwx – 3kw = 4x2
y + 4hx – 2x + 4
w(5bx – 3k) = 4x2
y + 4hx – 2x + 4
w =
+ +
x y hx x
bx k
4 4 – 2 4
5 – 3
2
(j)
x
1
+
y
3
2
=
z
4
5
y
3
2
=
z
4
5
–
x
1
=
x z
xz
4 – 5
5
15xz = 2y(4x – 5z)
y =
xz
x z
15
2(4 – 5 )
27. (a) ax2
+ bd2
+ c = 0
bd2
= –ax2
– c
d2
=
ax c
b
– –
2
= –
+
ax c
b
2
d = ± +
ax c
b
–
2
(b) k = hae
b e
2
–
2
2
k(b – e2
) = 2hae2
bk – e2
k = 2hae2
e2
k + 2hae2
= bk
e2
(k + 2ha) = bk
e2
=
+
bk
k ha
2
e = ±
+
bk
k ha
2
66

1
(c) y = +
f
f
2 –
2 3
2
2
y(2f2
+ 3) = 2 – f2
2f2
y + 3y = 2 – f2
2f2
y + f2
= 2 – 3y
f2
(2y + 1) = 2 – 3y
f2
=
+
y
y
2 – 3
2 1
f = ±
+
y
y
2 – 3
2 1
(d)
a
1
+
n
1
= y
n
1
= y –
a
1
=
ay
a
– 1
a = n (ay – 1)
n =
a
ay – 1
n =
a
ay
( – 1)
2
2
(e) x =
+
k t
k t
–
2 3
2 2
2 2
x2
=
+
k t
k t
–
2 3
2 2
2 2
x2
(2k2
+ 3t2
) = k2
– t2
2k2
x2
+ 3t2
x2
= k2
– t2
3t2
x2
+ t2
= k2
– 2k2
x2
t2
(3x2
+ 1) = k2
– 2k2
x2
t2
=
+
k k x
x
– 2
3 1
2 2 2
2
t = ±
k2
– 2k2
x2
3x2
+ 1
(f) k =
b x b
h
( – )
3
k3
=
b x b
h
( – )
hk3
= b(x – b)
x – b =
hk
b
3
x = b +
hk
b
3
28. u + v = m —(1)
u
1
+
v
1
=
f
1
—(2)
From (1),
v = m – u —(3)
u
1
+
m u
1
–
=
f
1
+
m u u
u m u
–
( – )
=
f
1
m
u m u
( – )
=
f
1
fm = u(m – u)
= mu – u2
u2
= mu – fm
= m(u – f)
m =
u
u f
–
2
29. y = p +
q
x
—(1)
z = p +
q
y
—(2)
z = p +
+
q
p
q
x
= p +
+
qx
px q
z – p =
+
qx
px q
(z – p)(px + q) = qx
pxz + qz – p2
x – pq = qx
p2
x + qx – pxz = qz – pq
x(p2
+ q – pz) = qz – pq
x =
+
qz pq
p q pz
–
–
2
30. (i)
a
1
+
b
2
+
c
3
=
d
4
b
2
=
d
4
–
c
3
–
a
1
=
ac ad cd
acd
4 – 3 –
2acd = b(4ac – 3ad – cd)
b =
acd
ac ad cd
2
4 – 3 –
(ii) When a = 6, c = 4, d =
1
2
,
b =
2(6)(4)
1
2
( )
4(6)(4) – 3(6)
1
2
( )– 4
1
2
( )
=
24
85
67

1
31. (i) x = y +
k y
gm
2
gmx = gmy + k2
y
= y(gm + k2
)
y =
+
gmx
gm k2
(ii) When x = 5, k = 9, g = 3 and m = 4,
y =
+
(3)(4)(5)
3(4) 92
=
20
31
32. (i) x a
–
2 2
= x + a
x2
– a2
= (x + a)2
= x2
+ 2ax + a2
2ax + 2a2
= 0
2a(x + a) = 0
a = 0 or x = –a
x = –a
(ii) When a =
3
4
,
a =
9
16
x = –
9
16
33. (i)
x
a
2
2
+
y
b
2
2
= 1
x
a
2
2
= 1 –
y
b
2
2
=
b y
b
–
2 2
2
x2
=
a
b
2
2 (b2
– y2
)
x = ±
a
b
b y
–
2 2
(ii) When y = 4, a = 2, b = 5,
x = ±
2
5
5 – 4
2 2
= ±1
1
5
34. (i) Q = mcq
m =
q
Q
c
(ii) When c = 4186, Q = 12 560, q = 3,
m =
12 560
4186(3)
= 1.00 (to 3 s.f.)
The mass of the water is 1.00 kg.
35. (i) P = 5000n2
– 8000
5000n2
= P + 8000
n2
=
+
P 8000
5000
n =
+
P 8000
5000
(ii) Given that P > 1 000 000,
5000n2
– 8000 > 1 000 000
5000n2
> 1 008 000
n2
> 201.6
When n = 14, n2
= 196 < 201.6.
When n = 15, n2
= 225 > 201.6.
The minimum number of employees is 15.
36. (i) T =
l l
l l
–
–
T 0
100 0
× 100
T
100
=
l l
l l
–
–
T 0
100 0
T
100
(l100 – l0) = lT – l0
lT =
T
100
(l100 – l0) + l0
(ii) When T = 80, l0 = 1.5, l100 = 13.5,
lT =
80
100
(13.5 – 1.5) + 1.5
= 11.1
The length of mercury thread is 11.1 cm.
37. (a) a =
+
a
3
2
a(a + 2) = 3
a2
+ 2a = 3
a2
+ 2a – 3 = 0
(a – 1)(a + 3) = 0
a = 1 or a = –3
(b) b – 2 =
b
9
– 2
(b – 2)2
= 9
b – 2 = 3 or b – 2 = –3
b = 5 b = –1
(c) c = 8 –
c
7
c2
= 8c – 7
c2
– 8c + 7 = 0
(c – 1)(c – 7) = 0
c = 1 or c = 7
68

1
(d)
d
d
6
2 – 1
= 2d
6d = 2d(2d – 1)
= 4d2
– 2d
4d2
– 8d = 0
4d(d – 2)= 0
d = 0 or d = 2
(e)
f
84
– 4
= 1 +
f
75
84f = f(f – 4) + 75(f – 4)
= f2
– 4f + 75f – 300
f2
– 13f – 300 = 0
(f – 25)(f + 12) = 0
f = 25 or f = –12
(f)
+
h
1
3
+
4
5
=
h
h
4 –
5(4 – h) + 4(h + 3)(4 – h) = 5h(h + 3)
20 – 5h + 4(4h – h2
+ 12 – 3h) = 5h2
+ 15h
20 – 5h + 16h – 4h2
+ 48 – 12h = 5h2
+ 15h
9h2
+ 16h – 68 = 0
(h – 2)(9h + 34) = 0
h = 2 or h = –
34
9
= –3
7
9
(g)
+
j
1
2
+
+
j
3
4
=
+
j
4
3
(j + 4)(j + 3) + 3(j + 2)(j + 3) = 4(j + 2)(j + 4)
j2
+ 3j + 4j + 12 + 3(j2
+ 3j + 2j + 6)
= 4(j2
+ 4j + 2j + 8)
j2
+ 7j + 12 + 3j2
+ 15j + 18 = 4j2
+ 24j + 32
2j = –2
j = –1
(h)
+
k
3
1
=
+
k
8
2
–
+
k
5
3
3(k + 2)(k + 3) = 8(k + 1)(k + 3) – 5(k + 1)(k + 2)
3(k2
+ 3k + 2k + 6)
= 8(k2
+ 3k + k + 3) – 5(k2
+ 2k + k + 2)
3k2
+ 15k + 18 = 8k2
+ 32k + 24 – 5k2
– 15k – 10
2k = 4
k = 2
38.
x
x – 1
+
+
x
x 1
= 3 +
x
1
1 – 2
x
x – 1
+
+
x
x 1
–
x
1
1 – 2
= 3
x
x – 1
+
+
x
x 1
+
x
1
– 1
2
= 3
x
x – 1
+
+
x
x 1
+
+
x x
1
( 1)( – 1)
= 3
+ + +
+
x x x x
x x
( 1) ( – 1) 1
( 1)( – 1)
= 3
x2
+ x + x2
– x + 1 = 3(x + 1)(x – 1)
2x2
+ 1 = 3(x2
– 1)
= 3x2
– 3
x2
= 4
x = 2 or x = –2
39. y =
+ x
3
1 2
—(1)
y =
+ x
5
3 4
—(2)
+ x
3
1 2
=
+ x
5
3 4
3(3 + 4x) = 5(1 + 2x)
9 + 12x = 5 + 10x
2x = –4
x = –2
Advanced
40. (a) +
a
a
15
25
n
n 3
=
a
3
5 3
(b)
a b
a b
49
7
n n
–1
2 3
= 7an – 3
bn – 3
41.
+
a b
a b
6
16
n n
n
5 – 2
4 – 4
=
3
8
an + 1
b2
h =
3
8
, k = 2, n = 8
42.
a
16
n–1
× bn + 3
÷
b
a
48 n
n
=
a
16
n–1 × bn + 3
×
a
b
48
n
n
=
ab
3
3
69

1
43.
a b
xy
16
7
3 4
4 ÷
ab
xy
4
21
2
3 ×
+
a
a
27
9
n
n
1
– 2
=
a b
xy
16
7
3 4
4 ×
xy
ab
21
4
3
2 ×
+
a
a
27
9
n
n
1
– 2
=
a b
y
36 5 2
44.
+
a b
1
2 3
=
+
b a
ab
1
2 3
=
+
ab
a b
3 2
45.
+
x y
x
2 5
3
=
+
y x
xy
x
2 5
3
=
+
x y
y
5 2
3
46. (i)
x
x
– 3
5 – 4
–
x
x
12 – 9
4 – 5
=
x
x
– 3
5 – 4
+
x
x
12 – 9
5 – 4
=
+
x x
x
– 3 12 – 9
5 – 4
=
x
x
9 – 8
5 – 4
(ii)
x
x
– 3
5 – 4
2
2
–
x
x
12 – 9
4 – 5
2
2
= 1
From (i),
x
x
9 – 8
5 – 4
2
2 = 1
9 – 8x2
= 5x2
– 4
13x2
= 13
x2
= 1
x = ±1
47. (i) F =
mv
r
2
Fr = mv2
v2
=
Fr
m
v = ±
Fr
m
(ii) When F = 30, m = 2, r = 3,
v = ±
30(3)
2
= ± 45
= ±6.71 (to 3 s.f.)
48. y =
x
1
– 1
–
+
x
1
2 1
–
x
1
2 – 3
When y = 0,
x
1
– 1
–
+
x
1
2 1
–
x
1
2 – 3
= 0
(2x + 1)(2x – 3) – (x – 1)(2x – 3) – (x – 1)(2x + 1) = 0
4x2
– 6x + 2x – 3 – (2x2
– 3x – 2x + 3) – (2x2
+ x – 2x – 1)
= 0
4x2
– 4x – 3 – 2x2
+ 5x – 3 – 2x2
+ x + 1 = 0
2x = 5
x =
5
2
= 2
1
2
The coordinates are 2
1
2
, 0





 .
New Trend
49. t = 2p
d
g
t
2π
=
d
g
t
4π
2
2
=
d
g
gt2
= 4p2
d
g =
d
t
4π2
2
50. (a) (i)
d e
df
25
49
3
÷
de
d f
15
21
2
3 2
=
d e
df
25
49
3
×
d f
de
21
15
3 2
2
=
d f
e
5
7
4
(ii)
c
2
– 1
–
c
3
– 2
=
c c
c c
2( – 2) – 3( – 1)
( – 1)( – 2)
=
+
c c
c c
2 – 4 – 3 3
( – 1)( – 2)
=
c
c c
– – 1
( – 1)( – 2)
= –
+
c
c c
1
( – 1)( – 2)
70

1
(b) 3 – e =
+
e
8
3
(3 + e)(3 – e) = 8
9 – e2
= 8
e2
= 1
e = 1 or e = –1
51.
2
x −6
+
5x
x −6
( )
2
=
2 x −6
( )+5x
x −6
( )
2
=
2x −12+5x
x −6
( )
2
=
7x −12
x −6
( )
2
52. (a) (i) When r = 3 and h = 5,
A = π(3) 52
− 32
= 37.7 (to 3 s.f.)
(ii) A = pr h r
–
2 2
A
r
π
= h r
–
2 2
A
r
π
2
2 2
= h2
– r2
h2
=
A
r
π
2
2 2
+ r2
=
+ π
π
A r
r
2 2 4
2 2
h = ±
+ π
π
A r
r
2 2 4
2 2
= ±
+ π
π
A r
r
2 2 4
(b)
u + 3v
( )2
− 4v2
u2
− 25v2
=
u + 3v + 2v
( ) u + 3v − 2v
( )
u + 5v
( ) u − 5v
( )
=
u + 5v
( ) u + v
( )
u + 5v
( ) u − 5v
( )
=
u + v
u − 5v
(c)
2g
2g − 3
+ 1 =
1
2 − 3g
4g − 3
2g − 3
=
1
2 − 3g
(4g – 3)(2 – 3g) = 2g – 3
8g – 12g2
– 6 + 9g = 2g – 3
12g2
−15g + 3 = 0
(12g – 3)(g – 1) = 0
g =
1
4
or g = 1
53.
5x
3x −2
( )
2
−
3
3x −2
=
5x − 3 3x −2
( )
3x −2
( )
2
=
5x −9x+6
3x −2
( )
2
=
6 − 4x
3x −2
( )
2
54. (i) When l = 0, k = 0.4, b = 12 and m = 2
a = 0.4[0(2) + 12(2)]
= 9.6
(ii) k(lm + bm) = a
lm + bm =
a
k
bm =
a
k
– lm
b =
a
km
– l
55. (a)
32x2
−50
8x2
+14x − 30
=
2 16x2
−25
( )
2 4x2
+ 7x −15
( )
=
4x+5
( ) 4x −5
( )
4x −5
( ) x+ 3
( )
=
4x+5
x+ 3
(b)
7
5 −2x
−
3
4 − x
=
7 4 − x
( )− 3 5 −2x
( )
5 −2x
( ) 4 − x
( )
=
28 − 7x −15+6x
5 −2x
( ) 4 − x
( )
=
13− x
5 −2x
( ) 4 − x
( )
56. (a)
a b
ab
3
8
2
3
÷
ac
abc
21
49
4
2
=
a b
ab
3
8
2
3
×
abc
ac
49
21
2
4
=
a
bc
7
8 2
(b) (i) V =
4π
3
(r3
+ s3
)
When r = 2.1 and s = 0.9,
V =
4π
3
(2.13
+ 0.93
)
= 41.8 (to 3 s.f.)
(ii) V =
4π
3
(r3
+ s3
)
r3
+ s3
=
3V
4π
r3
=
3V
4π
− s3
r =
3V
4π
− s3
3
71

1
(c)
3x − 7
4x +5
= 6
3x − 7 = 24x + 30
−37 = 21x
x = −
37
21
= −1
6
21
57. (a)
5m
3
÷
60m2
n
=
5m
3
×
n
60m2
=
n
36m
(b)
6x2
−20x −16
6x2
−5x −6
=
2 3x2
−10x −8
( )
3x +2
( ) 2x − 3
( )
=
2 3x+2
( ) x − 4
( )
3x +2
( ) 2x − 3
( )
=
2 x − 4
( )
2x − 3
58. (a) k =
+
m
m
2 – 1
4
k(m + 4) = 2m – 1
km + 4k = 2m – 1
2m – km = 4k + 1
m(2 – k) = 4k + 1
m =
+
k
k
4 1
2 –
(b)
+
b
3
2 1
+
b
6
2 – 1
=
+ +
+
b b
b b
3(2 – 1) 6(2 1)
(2 1)(2 – 1)
=
+ +
+
b b
b b
6 – 3 12 6
(2 1)(2 – 1)
=
+
+
b
b b
18 3
(2 1)(2 – 1)
(c)
9x2
− y2
3x2
+ xy
=
3x+ y
( ) 3x − y
( )
x 3x + y
( )
=
3x − y
x
72

1
Chapter 7 Relations and Functions
Basic
1. (a) No, the element a has two images.
(b) Yes.
(c) No, the element c has no image.
(d) Yes.
2. f(x) =
3
7
x + 4
f(−14) = 3
7
(−14) + 4 = −2
f(28) = 3
7
(28) + 4 = 16
f
7
8
⎛
⎝
⎜
⎞
⎠
⎟ =
3
7
7
8
⎛
⎝
⎜
⎞
⎠
⎟ + 4 = 4
3
8
f −
2
9
⎛
⎝
⎜
⎞
⎠
⎟ =
3
7
−
2
9
⎛
⎝
⎜
⎞
⎠
⎟ + 4 = 3
19
21
3. f(x) = 9 −4x
(i) f(4) = 9 – 4(4)
=
−7
(ii) f(−2) = 9 – 4(−2)
=
17
(iii) f(0) = 9 – 4(0)
= 9
(iv) f(1) + f(−5) = [9 – 4(1)] + [9 – 4(−5)]
= 34
4. g(x) = 3x – 13
(i) g(8) = 3(8) – 13
=
11
(ii) g(−6) = 3(−6) – 13
=
−31
(iii) g 4
1
3
⎛
⎝
⎜
⎞
⎠
⎟ = 3 4
1
3
⎛
⎝
⎜
⎞
⎠
⎟ – 13
= 0
(iv) g(7) + g(−3) = [3(7) – 13] + [3(−3) – 13]
=
8 + (−22)
=
−14
(v) g 1
1
3
⎛
⎝
⎜
⎞
⎠
⎟ −g −
2
3
⎛
⎝
⎜
⎞
⎠
⎟ = 3 1
1
3
⎛
⎝
⎜
⎞
⎠
⎟ − 13
⎡
⎣
⎢
⎤
⎦
⎥ − 3 −
2
3
⎛
⎝
⎜
⎞
⎠
⎟ − 13
⎡
⎣
⎢
⎤
⎦
⎥
= −9 – (−15)
=
6
Intermediate
5. h(x) = −3x + 12
(i) h(3a) – h(2a) = [−3(3a) + 12] – [−3(2a) + 12]
=
−9a+ 12 + 6a – 12
=
−3a
(ii) h(a) = 0
∴ −3a + 12 = 0
3a = 12
a = 4
(iii) h
2
3
a
⎛
⎝
⎜
⎞
⎠
⎟ + h(a) = −3
2
3
a
⎛
⎝
⎜
⎞
⎠
⎟ + 12
⎡
⎣
⎢
⎤
⎦
⎥ + [−3a + 12]
=
−2a + 12 – 3a + 12
=
−5a + 24
6. f(x) =
2
7
x + 3, g(x) = 1
5
x – 4
(a) (i) f(7) + g(35) =
2
7
7
( ) + 3
⎡
⎣
⎢
⎤
⎦
⎥ +
1
5
35
( ) − 4
⎡
⎣
⎢
⎤
⎦
⎥
=
2 + 3 + 7 – 4
=
8
(ii) f(−2) – g(−15) =
2
7
−2
( ) + 3
⎡
⎣
⎢
⎤
⎦
⎥ –
1
5
−15
( ) − 4
⎡
⎣
⎢
⎤
⎦
⎥
=
−
4
7
+ 3 + 3 + 4
= 9
3
7
(iii) 3f(2) – 2g(10) = 3
2
7
2
( ) + 3
⎡
⎣
⎢
⎤
⎦
⎥ − 2
1
5
10
( ) − 4
⎡
⎣
⎢
⎤
⎦
⎥
=
12
7
+ 9 – 4 + 8
=
14
5
7
(iv)
1
3
f(−7)–9g(0)=
1
3
2
7
−7
( ) + 3
⎡
⎣
⎢
⎤
⎦
⎥ −9
1
5
0
( ) − 4
⎡
⎣
⎢
⎤
⎦
⎥
=
−
2
3
+ 1 − 0 + 36
=
36
1
3
(b) (i) f(x) = g(x)
2
7
x + 3 =
1
5
x – 4
2
7
x –
1
5
x = −4 – 3
3
35
x = −7
x = −81
2
3
(ii) f(x) = 8
2
7
x + 3 = 8
2
7
x = 5
x = 17
1
2
73

1
7. f(x) = 12x – 1, g(x) = 9 – 5x
(i) f(x) = 23
12x – 1 = 23
12x = 24
x = 2
(ii) g(x) = 24
9 – 5x = 24
5x = −15
x = −3
(iii) g(x) = 2x
9 – 5x = 2x
7x = 9
x = 1
2
7
(iv) f(x) = −5x
12x – 1 = −5x
17x = 1
x =
1
17
(v) f(x) = g(x)
12x – 1 = 9 – 5x
17x = 10
x =
10
17
8. f(x) = 11x – 7, F(x) =
3
4
x + 3
(i) f(p) = 11p – 7
(ii) F 8 p +
1
2
⎛
⎝
⎜
⎞
⎠
⎟ =
3
4
8 p +
1
2
⎛
⎝
⎜
⎞
⎠
⎟ + 3
=
6p +
3
8
+ 3
=
6p + 3
3
8
(iii) f
3
11
p
⎛
⎝
⎜
⎞
⎠
⎟ + F(4p – 12)
=
11
3
11
p
⎛
⎝
⎜
⎞
⎠
⎟ – 7 +
3
4
4 p − 12
( ) + 3
⎡
⎣
⎢
⎤
⎦
⎥
= 3p – 7 + 3p – 9 + 3
=
6p – 13
74
Advanced
9. f(x) =
3x − 7
8
(i) f(−2) =
3 −2
( ) − 7
8
=
−1
5
8
f 2
5
6
⎛
⎝
⎜
⎞
⎠
⎟ =
3 2
5
6
⎛
⎝
⎜
⎞
⎠
⎟ − 7
8
= 3
16
(ii) f(x) = 5
3x − 7
8
= 5
3x – 7 = 40
3x = 47
x = 15
2
3

1
75
Chapter 8 Congruence and Similarity
Basic
1. A and D, B and H, C and F, E and I
2. (i) QP
(ii) PC
(iii) CA
(iv) QPC
(v) CAB
(vi) BCA
3. (i) AB = CD, BD = DB, AD = CB
(ii) ABD = CDB, ADB = CBD,
BAD = DCB
4. (i) PQ = QP, QS = PR, PS = QR
(ii) PQS = QPR, PSQ = QRP,
QPS = PQR
5. (i) AB = AC, BQ = CP, AQ = AP
(ii) ABQ = ACP, AQB = APC,
BAQ = CAP
6. (a) x = 40, y = 50, z = 50
(b) x = 44, y = 54, z = 82
(c) x = 6.75, y = 88
(d) x = 6.3
7. (a)
2
900 000
=
1
450 000
(b)
3
450 000
=
1
150 000
(c)
0.5
40 000
=
1
80 000
(d)
7.5
10 500 000
=
1
1 400 000
8. 4 cm represents 30 m
1 cm represents 7.5 m
Scale is 1 cm to 7.5 m
9. (i) Actual perimeter = 2(5 + 4)(15)
= 270 m
(ii) Actual area = (5 × 15) × (4 × 15)
= 4500 m2
10. 2 cm represent 3 km
1 cm represents 1.5 km
(a) 24 cm represent 36 km
(b) 10.5 cm represent 15.75 km
(c) 14.2 cm represent 21.3 km
(d) 2.6 cm represent 3.9 km
11. 1 cm represents 0.4 km
0.4 km is represented by 1 cm
(a) 800 m is represented by 2 cm
(b) 0.2 km is represented by 0.5 cm
(c) 3.6 km is represented by 9 cm
(d) 2 km 400 m is represented by 6 cm
12. (a) 2 cm represents 3 km
16 cm represents 24 km
(b) 1.2 cm represents 3 km
(c) 2.4 cm represents 9 km
(d) 0.5 cm represents 0.25 km
13. (a) 25 km is represented by 2 cm
480 km is represented by 38.4 cm
(b) 75 km is represented by 5 cm
480 km is represented by 32 cm
(c) 25 km is represented by 9 cm
480 km is represented by 172.8 cm
(d) 120 km is represented by 0.5 cm
14. 1 cm represent 5 km
17.6 cm represents 88 km
(a) 18 cm represent 4.5 km
(b) 16.5 cm represent 4.125 km
(c) 65 cm represent 16.25 km
(d) 7.4 cm represent 1.85 km
16. 0.5 km is represented by 1 cm
0.25 km2
is represented by 1 cm2
20 km2
1 cm2
represents 0.04 km2
(a) 5 cm2
represent 0.2 km2
(b) 18 cm2
represent 0.72 km2
(c) 75 cm2
represent 3 km2
(d) 124 cm2
represent 4.96 km2
18. 1 cm represents 75 km
(a) 12 cm represent 900 km
(b) 8 cm represent 600 km
(c) 20.5 cm represent 1537.5 km
(d) 22 cm represent 1650 km
19. (i) 1 cm represents 1.2 km
5.4 cm represent 6.48 km
(ii) 10 km 80 m is represented by 8.4 cm
(iii) 1 cm2
represents 1.44 km2
3.6 cm2
represent 5.184 km2
20. (i) 4 cm represent 3 km
10.5 cm represent 7.875 km
(ii)
1
75 000
(iii) 0.5625 km2
32.4 km2
is represented by 57.6 cm2

1 76
21. (i) Actual length = 2.6 × 1.6
= 4.16 m
Actual dimensions are 4.16 m by 4.16 m
(ii) Actual area = (2.6 × 1.6) × (1.8 × 1.6)
= 12.0 m2
(to 3 s.f.)
(iii) Actual total area = (6.0 × 1.6) × (6.0 × 1.6)
= 92.16
= 92 m2
(to the nearest m2
)
Intermediate
22. (a) c = 48, q = 92, z = 7
a = p = 180 – 92 – 48
= 40
a = 40, c = 48, p = 40, q = 92, z = 7
(b) a = 58, b = 10, q = 8.5, y = 32
(c) a = 39, p = 6, q = 66
r = 180 – 66 – 39
= 75
a = 39, p = 6, q = 66, r = 75
(d) b = 6.5, p = 6, r = 7
23. (a) b = 102, p = 73, q = 6, s = 7
(b) a = 11.5, c = 42, d = 62, x = 41, y = 11
(c) a = 7.6, b = 8.0, p = 92, r = 57
(d) a = 5.8, b = 7, s = 83, x = 7.2
24. (a)
B C
6 cm
10 cm
8 cm
A
8 cm
6 cm
Q P
R
10 cm
AB = PQ = 6 cm
BC = QR = 8 cm
CA = RP = 10 cm
ABC ≡ PQR
(b)
50°
65°
7 cm 7 cm
B Q
C R
A P
BAC = QPR = 50°
ABC = PQR = 65°
ACB = PRQ = 65°
BC = QR = 7 cm
ABC ≡ PQR
(c)
5 cm
40°
R Q
P
60°
70°
4.6 cm
B C
A
5 cm 4.6 cm
ACB = 180° – 60° – 70°
= 50°
Since the triangles do not have the same shape and
size, they are not congruent.
25. (i) ABP = 180° – 90° – 23°
= 67°
(ii) Using Pythagoras’ Theorem,
AP2
+ 52
= 132
AP2
= 132
– 52
= 144
AP = 12 cm
Area of ABC =
1
2
(10)(12)
= 60 cm2
26. BAC = EDF = 60°
ABC = DEF = 50°
ACB = DFE = 70°
Since all the corresponding angles are equal, ABC is
similar to DEF.
27.
PQ
XY
=
3.9
2.6
= 1.5
QR
YZ
=
13.9
9.1
= 1.53
PR
XZ
=
12.6
8.4
= 1.5
Since not all the ratios of the corresponding sides are equal,
PQR is not similar to XYZ.

1
77
28.
l
8
=
3.2
4
l =
3.2
4
× 8
= 6.4
Actual length is 6.4 m.
29. (i)
PS
AD
=
PQ
AB
PS
18
=
36
24
PS =
36
24
× 18
= 27
Width of PQRS is 27 cm.
(ii)
PQ
AB
=
QR
BC
PQ
24
=
36
18
PQ =
36
18
× 24
= 48
Length of PQRS is 48 cm.
30. (a)
h
8.4
=
1.2
2
h =
1.2
2
× 8.4
= 5.04
Height of the smaller mould is 5.04 cm.
(b)
l
7.6
=
2
1.2
l =
2
1.2
× 7.6
= 12.7 (to 3 s.f.)
Length of the base of the larger mould is 12.7 cm.
31. (a)
r
5.5
=
24
10
r =
24
10
× 5.5
= 13.2
Radius of the larger cone is 13.2 cm.
(b)
c
84
=
10
24
c =
10
24
× 84
= 35
Circumference of the smaller cone is 35 cm.
32.
+ x
7
7
=
18
8
7 + x = 15
3
4
x = 8
3
4
y
24
=
8
18
y =
8
18
× 24
= 10
2
3
x = 8
3
4
, y = 10
2
3
33.
+
x
x 3
=
3
4
4x = 3x + 9
x = 9
+
y
y 2.8
=
3
4
4y = 3y + 8.4
y = 8.4
x = 9, y = 8.4
34. (i) 5 cm represents 2 km
(ii) 4 km is represented by 6 cm
2.8 km is represented by 4.2 cm
35. (i) 16 km2
64 km2
(ii) 4 km2
64 km2
36. 1 cm represents 15 km
(a) 7 km is represented by 3 cm
(b) 35 km is represented by 4 cm
(d) 6 km is represented by 7 cm
37. (a) 2 cm represent 15 m
4 cm2
represent 225 m2
24 cm2
represent 1350 m2
(b) 4 cm represent 25 m
16 cm2
represent 625 m2
24 cm2
represent 937.5 m2
(c) 4 cm represent 600 m
16 cm2
represent 360 000 m2
24 cm2

1 78
(d) 1.5 cm represent 120 m
2.25 cm2
represent 14 400 m2
24 cm2
1 cm2
represents 0.25 km2
36 cm2
represent 9 km2
(a) 0.25 km is represented by 1 cm
0.0625 km2
9 km2
(b) 0.125 km is represented by 1 cm
0.015 625 km2
9 km2
0.5625 km2
9 km2
(d) 2 km is represented by 1 cm
4 km2
9 km2
39. 100 m2
is represented by 1 m2
10 m is represented by 1 m
40 m is represented by 4 m
40. 5 km is represented by 1 cm
25 km2
225 km2
Advanced
41. (a) False
(b) False
(c) True
(d) False
(e) False
(f) True
(g) False
(h) True
(i) True
(j) True
(k) False
(l) False
42. (a)
+
x 6
6
=
+
9 5
5
x + 6 =
14
5
× 6
x = 10
4
5
y
4
=
+
9 5
5
y =
14
5
× 4
= 11
1
5
x = 10
4
5
, y = 11
1
5
(b)
+
x 9
9
=
18
7
x + 9 =
18
7
× 9
x = 14
1
7
+
y 8
8
=
18
7
y + 8 =
18
7
× 8
y = 12
4
7
x = 14
1
7
, y = 12
4
7
(c)
x
12
=
10
18
x =
10
18
× 12
= 6
2
3
+
y 6
2
3
8
=
18
10
y + 6
2
3
=
18
10
× 8
y = 7
11
15
x = 6
2
3
, y = 7
11
15
(d)
+
x
x 5
=
8
12
12x = 8x + 40
4x = 40
x = 10
y
15
=
12
8
y =
12
8
× 15
= 22
1
2
x = 10, y = 22
1
2
43. (i)
+
CQ 6
6
=
7
4
CQ + 6 =
7
4
× 6
CQ = 4
1
2
cm

1
79
(ii)
CR
12
=
+
4
1
2
4
1
2
6
CR =
+
4
1
2
4
1
2
6
× 12
= 5
1
7
cm
44. Let 1 cm represent 100 m.
4 cm
3 cm
5 cm
A B
C
5 cm represents 500 m
AC = 500 m
45. (i) 1 cm represents 500 cm
Scale is 1 : 500
(ii) Perimeter = 2(2.4 + 4)(5)
= 64 m
Area = (2.4 × 5) × (4 × 5)
= 240 m2
46. (i)
PB
PQ
=
BA
QR
PB
18
=
16
24
PB =
16
24
× 18
= 12 cm
(ii)
PR
PA
=
QR
BA
PR
9
=
24
16
PR =
24
16
× 9
= 13.5 cm
BR = PB + PR
= 12 + 13.5
= 25.5 cm
47. (a) 1 cm represents 45 000 cm
n = 0.45
(b) Actual distance = 32.5 × 0.45
= 14.625 km
(c) 1 cm represents 450 m
1 cm2
represents 202 500 m2
1 cm2
represents 202.5 ha
Area on the map =
2227
202.5
= 11.0 cm2
(to 3 s.f.)

1
Chapter 9 Geometrical Transformation
Basic
1.
2
1
4
3
5
1
(2, 5) (6, 5)
x = 4
0 2 3 4 5 6
x
y
∴ The coordinates of the reflection of the point (2, 5) is
(6, 5).
2.
–4
–6 –2 –1
–3
–5 1
y = x
2 3 4 5 6
0
2
1
4
3
–2
–1
–4
–3
y
x
–5
–6
∴ p = −1, q = 3
3.
1 3 5 7
2 4 6 8
x
–1
–2
3
2
1
4
5
–3
0
y
(7, 3)
(2, 4)
(6, –2)
(4, 1)
(a) (7, 3)
(b) (6, −2)
4.
–1
–2
–3
3
2
4
6
5
7
8
1
2 3 5
1
0 4 6
x
y
N M
L
R
Q
P
5. Let the translation vector T be a
b
⎛
⎝
⎜
⎞
⎠
⎟ .
a
b
⎛
⎝
⎜
⎞
⎠
⎟ + 9
1
⎛
⎝
⎜
⎞
⎠
⎟ = 2
−3
⎛
⎝
⎜
⎞
⎠
⎟
a
b
⎛
⎝
⎜
⎞
⎠
⎟ =
2
−3
⎛
⎝
⎜
⎞
⎠
⎟ − 9
1
⎛
⎝
⎜
⎞
⎠
⎟
=
−7
−4
⎛
⎝
⎜
⎞
⎠
⎟
−7
−4
⎛
⎝
⎜
⎞
⎠
⎟ + Q =
−5
6
⎛
⎝
⎜
⎞
⎠
⎟
Q =
−5
6
⎛
⎝
⎜
⎞
⎠
⎟ − −7
−4
⎛
⎝
⎜
⎞
⎠
⎟
=
2
10
⎛
⎝
⎜
⎞
⎠
⎟
∴ The translation vector of T is
−7
−4
⎛
⎝
⎜
⎞
⎠
⎟ and Q(2, 10).
80

1
6.
6
2 3 4 5 7
1
–1
–2
–4 –3
y = 4
8
9
8
5
10
12
11
1
2
3
4
6
7
–1
–2
0
y
B(3, 10)
A(3, –2)
C(8, 3)
x + y = 6
x
(a) The coordinates of B are (3, 10).
(b) The coordinates of C are (8, 3).
7.
(d) y = –x – 1
(b) y = –x + 3
(c) y = –x + 9
x = 3
y = x + 3
(a) y = –x – 3
6
2 3 4 5 7
1
–1
–2
–4 –3 8
x
7
6
3
8
–1
1
2
4
5
–3
–4
0
y
–2
9
9
(a) y = −x – 3
(b) y = −x + 3
(c) y = −x + 9
(d) y = −x − 1
81
8.
(4, 6)
P
Q
R
A B
C
7
10
6
3
8
11
1
2
4
5
y
9
12
0 9
5 6 7 8 10
4
3
2
1 11
x
(ii) The centre of rotation is (4, 6). The angle of rotation
is 90° clockwise or 270° anticlockwise.
9. R4
represents (4 × 160°) − 360°
= 280° anticlockwise rotation about the origin.
R5
represents (5 × 160°) − 720°
= 80° anticlockwise rotation about the origin.
10.
y
6
2 3 4 5
1
–1
–2
–3 0
x
y = 3x + 4
y = −
1
3
x +
4
3
(0, 4)
4
1
1
2
3
5
6
7
–1
–2
(4, 0)
∴ The equation of the image of the line
y = 3x + 4 is y = −
1
3
x +
4
3
.

1 82
11.
y = x + 4
x + y = 4
0
5
4
7
6
2
3
–1
1
y
–4 –2 –1
–3 1 2 3 4 5 6
x
–2
(0, 2)
∴ The equation of the image of the line x + y = 4 is
y = x + 4.
12. Let the translation vector T be
p
q
⎛
⎝
⎜
⎞
⎠
⎟ .
p
q
⎛
⎝
⎜
⎞
⎠
⎟ +
1
3
⎛
⎝
⎜
⎞
⎠
⎟ = 9
7
⎛
⎝
⎜
⎞
⎠
⎟
p
q
⎛
⎝
⎜
⎞
⎠
⎟ = 9
7
⎛
⎝
⎜
⎞
⎠
⎟ −
1
3
⎛
⎝
⎜
⎞
⎠
⎟
=
8
4
⎛
⎝
⎜
⎞
⎠
⎟
(i)
8
4
⎛
⎝
⎜
⎞
⎠
⎟ + −2
−3
⎛
⎝
⎜
⎞
⎠
⎟ =
x
y
⎛
⎝
⎜
⎞
⎠
⎟
x
y
⎛
⎝
⎜
⎞
⎠
⎟ =
6
1
⎛
⎝
⎜
⎞
⎠
⎟
∴ x = 6, y = 1
(ii) 8
4
⎛
⎝
⎜
⎞
⎠
⎟ + h
k
⎛
⎝
⎜
⎞
⎠
⎟ =
4
6
⎛
⎝
⎜
⎞
⎠
⎟
h
k
⎛
⎝
⎜
⎞
⎠
⎟ = −4
2
⎛
⎝
⎜
⎞
⎠
⎟
h = −4, k = 2
13. (i) ZXX= 20°
XZ = XZ
∴ ZXX =
180° − 20°
2
= 80° (base ∠ of isos. r)
(ii) YZY = 20°
tan XZY =
7
4
XZY = tan−1 7
4
= 60.3° (to 1 d.p.)
∴ YZX = 60.3° − 20°
= 40.3°
14.
0
5
4
2
3
–1
1
y
–2 2 4 6
x
–2
Q
S
R
K
P
8
y = x – 1
(i) The line of reflection is y = x – 1.
(ii) The centre of rotation is (0, 2) and the angle of rotation
is 180°.
(iii) rK is mapped onto rQ by a translation
7
−2
⎛
⎝
⎜
⎞
⎠
⎟ .
(iv) A 90° anticlockwise rotation about (0, 0).

1
83
Revision Test B1
1. x2
– y2
= 28
(x + y)(x – y)= 28
4(x + y) = 28
x + y = 7
(2x + 2y)2
= 4(x + y)2
= 4(7)2
= 196
2. (a)
+
b
a a
4 – 1
3
2
×
+
+
a
b b
3
4 11 – 3
2
=
+
b
a a
4 – 1
( 3)
×
a
b b
3
(4 – 1)( 3)
+
+
=
+
a b
1
( 3)
(b) 4 – 4x +
x
3
= 0
4x – 4x2
+ 3 = 0
4x2
– 4x – 3 = 0
(2x – 3)(2x + 1) = 0
x =
3
2
or x = –
1
2
= 1
1
2
x = 1
1
2
or x = –
1
2
3. (i) (2p – q)(r + 5) = r(p – 1)
2pr + 10p – qr – 5q = pr – r
10p – 5q = qr – pr – r
= r(q – p – 1)
r =
p q
q p
10 – 5
– – 1
(ii) When p = 6, q = –3,
r =
10(6) – 5(–3)
–3 – 6 – 1
= –7
1
2
4. (i) x = –1 or x = 6
(ii) x =
+
–1 6
2
= 2
1
2
1
2
.
(iii) When x = 2
1
2
,
y = 2
1
2






2
– 5 2
1
2





 – 6
= –12
1
4
Minimum value of y is –12
1
4
when x = 2
1
2
.
5. h(x) = 5x +
4
5
h(−3) = 5(−3) +
4
5
= −14
1
5
h(4) = 5(4) +
4
5
= 20
4
5
h
7
25
⎛
⎝
⎜
⎞
⎠
⎟ = 5
7
25
⎛
⎝
⎜
⎞
⎠
⎟ +
4
5
= 2
1
5
h −
9
10
⎛
⎝
⎜
⎞
⎠
⎟ = 5 −
9
10
⎛
⎝
⎜
⎞
⎠
⎟ +
4
5
= −3
7
10
6. (i) nADO
(ii) nCOB
(iii) nADC
7.
+
x 8
8
=
+
21 7
7
x + 8 =
28
7
× 8
= 32
x = 24
y
6
=
+
21 7
7
y =
28
7
× 6
= 24
x = 24, y = 24

1 84
8.
6
2 3 4 5 7
1
–1
–2
–4 –3 8
9
8
5
10
12
11
1
2
3
4
6
7
–1
–2
0
y
(7, 9)
(–3, 9)
x = 2
x
The coordinates of the reflection of (7, 9) in x = 2 is
(−3, 9).
9. Let the translation vector T be a
b
⎛
⎝
⎜
⎞
⎠
⎟ .
4
6
⎛
⎝
⎜
⎞
⎠
⎟ =
2
3
⎛
⎝
⎜
⎞
⎠
⎟ +
a
b
⎛
⎝
⎜
⎞
⎠
⎟
a
b
⎛
⎝
⎜
⎞
⎠
⎟ =
4
6
⎛
⎝
⎜
⎞
⎠
⎟ −
2
3
⎛
⎝
⎜
⎞
⎠
⎟
=
2
3
⎛
⎝
⎜
⎞
⎠
⎟
8
−7
⎛
⎝
⎜
⎞
⎠
⎟ = A +
2
3
⎛
⎝
⎜
⎞
⎠
⎟
A =
8
−7
⎛
⎝
⎜
⎞
⎠
⎟ − 2
3
⎛
⎝
⎜
⎞
⎠
⎟
=
6
−10
⎛
⎝
⎜
⎞
⎠
⎟
The coordinates of A are (6, −10).
10. Let h be the height of the larger jar.
h
12
=
6
4
h =
6
4
× 12
= 18
Height of larger jar is 18 cm.
11. (i) 5 km is represented by 4 cm
(ii) 16 cm2
represents 25 km2
12 cm2
represents 18.75 km2
18.75 km2
= 187 500 ha
12. (i)
x
560
(ii)
x
560
–
+
x
560
1
= 0
560(x + 1) – 560x = 10x(x + 1)
560x + 560 – 560x = 10x2
+ 10x
10x2
+ 10x – 560 = 0
x2
+ x – 56 = 0 (shown)
(iii) (x + 8)(x – 7)= 0
x = –8 or x = 7
(rejected)
x
560
= 80
Original price of each casing is $7,
number of casings bought is 80.
13. (a) When x = 2, y = a,
a = 5 + 4(2) – 22
= 9
When x = 3, y = b,
b = 5 + 4(3) – 32
= 8
a = 9, b = 8
(b)
–1 0
6
8
9
7
4
5
y = 5 + 4x – x2
1
2
3
1 2 3 4 5
y
x
(c) (i) When x = 1.5, y = 8.8
(ii) When y = 6, x = 0.3 or x = 3.7

1
85
Revision Test B2
1. (a) 20.752
– 0.752
= (20.75 + 0.75)(20.75 – 0.75)
= (21.5)(20)
= 430
(b) 1597 × 1603
= (1600 – 3)(1600 + 3)
= 16002
– 32
= 2 560 000 – 9
= 2 559 991
2. (a)
p q
p pq q
–
5( 2 )
2 2
2 2
+ +
÷
+
+
q p pq
p q
– 2
25( )
2 2
=
+
+
p q p q
p q
( )( – )
5( )2
×
+
p q
p q
25( )
( – )2
=
p q
5
–
(b)
w
w
2 – 5
–
w
1
10 – 4
=
w
w
2 – 5
+
w
1
2(2 – 5)
=
+
w
w
2 1
2(2 – 5)
3. (i)
b ac
bc a
5 –
3 – 4
2
2
=
2
3
15b – 3ac2
= 6bc2
– 8a
3ac2
+ 6bc2
= 8a + 15b
c2
(3a + 6b) = 8a + 15b
c2
=
+
+
a b
a b
8 15
3 6
c = +
+
a b
a b
8 15
3 6
(ii) When a = 2, b = 1,
c =
+
+
8(2) 15(1)
3(2) 6(1)
= 1.61 (to 3 s.f.)
4. y = (3 – x)(2x + 3)
When y = 0,
x = 3 or x = –
3
2
= –1
1
2
A –1
1
2
, 0





, C(3, 0)
At maximum point,
x =
+
–1
1
2
3
2
=
3
4
When x =
3
4
,
y = 3 –
3
4





 2
3
4





 + 3






= 10
1
8
Coordinates of the maximum point are
3
4
, 10
1
8





 .
5. Let the lengths of the two squares be x cm and
(72 – x) cm.
x
4






2
+
72 – x
4






2
= 170
x
16
2
+
+
x x
5184 – 144
16
2
= 170
x2
+ 5184 – 144x + x2
= 2720
2x2
– 144x + 2464 = 0
x2
– 72x + 1232 = 0
(x – 28)(x – 44) = 0
x = 28 or x = 44
72 – x = 44 72 – x = 28
The length of each part is 28 cm and 44 cm respectively.
6. (i) f(x) =
2
5
x − 4
f(a) =
2
5
a – 4
(ii) F(x) = 8x + 3
F
1
8
−
1
2
a
⎛
⎝
⎜
⎞
⎠
⎟ = 8
1
8
−
1
2
a
⎛
⎝
⎜
⎞
⎠
⎟ + 3
= 1 – 4a + 3
= 4 – 4a
(iii) f(5a) + F(2a – 3) =
2
5
(5a) – 4 + 8(2a – 3) + 3
= 2a – 4 + 16a – 24 + 3
= 18a – 25
7. BAC = 180° – 90° – 45°
= 45°
Since the triangles do not have the same shape and size,
they are not congruent.

1 86
8.
BC
QR
=
AB
PQ
x
10
=
8
14
x =
8
14
× 10
= 5
5
7
PR
AC
=
PQ
AB
y
10
=
14
8
y =
14
8
× 10
= 17
1
2
x = 5
5
7
, y = 17
1
2
9. Let h be the height of the smaller rocket.
h
24
=
5
7
h =
5
7
× 24
= 17
1
7
Height of smaller rocket is 17
1
7
cm.
10.
x = 2
y = 2
A(–1, 6) A(5, 6)
A(–1, –2)
6
2 3 4 5
1
–1
–2
–4 –3
x
7
6
3
8
–1
1
2
4
5
–3
–4
0
y
–2
9
(i) The coordinates of A under the two reflections are
(−1, −2).
(ii) The point which remains invariant is (2, 2).
11. (i) 1 cm represents 0.75 km
(ii) 46 cm represent 34.5 km
(iii) 1 cm2
8 cm2
represent 4.5 km2
(iv) 3 cm2
represent 1.6875 km2
0.0625 km2
1.6875 km2
12. (a) When x = –2, y = a,
a = 3(–2)2
– 4(–2) – 30
= –10
When x = 1, y = b,
b = 3(1)2
– 4(1) – 30
= –31
a = –10, b = –31
(b)
–2
–3 –1 0
10
20
30
–10
–20
–30
1
y = 3x2
– 4x – 30
2 3 4 5
y
x
(c) (i) When x = 3.6, y = –5.5
(ii) When y = 0, x = 3.9 or x = –2.6
(iii) When y = –20, x = 2.6 or x = –1.3

1
87
Mid-Year Examination Specimen Paper A
Part I
1. Average speed =
1200m
6 minutes
=
1200 ÷1000
6 ÷ 60
= 12 km/h
2. (a) 5x(x – 3) = 0
x = 0 or x = 3
(b) 6y2
+ y – 1 = 0
(3y – 1)(2y + 1) = 0
y =
1
3
or y = –
1
2
3. 3x – y = 13 —(1)
x
3
–
y
4
= 1 —(2)
(1) × 3: 9x – 3y = 39 —(3)
(2) × 12: 4x – 3y = 12 —(4)
(3) – (4): 5x = 27
x = 5
2
5
Substitute x = 5
2
5
into (1):
3 5
2
5





 – y = 13
16
1
5
– y = 13
y = 16
1
5
– 13
y = 3
1
5
x = 5
2
5
, y = 3
1
5
4. (a) 40 – 10x2
= 10(4 – x2
)
= 10(2 + x)(2 – x)
(b) 2ac – 2bc – bd + ad
= 2c(a – b) + d(a – b)
= (a – b)(2c + d)
5. 3(a2
+ b2
)
= 3[(a + b)2
– 2ab]
= 3 189 – 2
78
6












= 489
6. (i) a2
– b2
= (a + b)(a – b)
(ii) 88.742
– 11.262
= (88.74 + 11.26)(88.74 – 11.26)
= (100)(77.48)
= 7748
7. (i) 3x – y2
= ax + b
3x – ax = y2
+ b
x(3 – a) = y2
+ b
x =
+
y b
a
3 –
2
(ii) When a = 5, b = 7, y = –1,
x =
+
(–1) 7
3 – 5
2
= –4
8. (2x – y)(x + 3y) – x(2x – 3y)
= 2x2
+ 6xy – xy – 3y2
– 2x2
+ 3xy
= 8xy – 3y2
9. (a)
a
3
–
a – 2
6
=
a a
2 – ( – 2)
6
=
+
a 2
6
(b)
m
5
–
mn
7
=
n
mn
5 – 7
(c)
+
p q
2
3 4
+
p
q p
3
16 – 9
2 2
–
p q
5
3 – 4
=
+
p q
2
3 4
–
+
p
p q p q
3
(3 4 )(3 – 4 )
–
p q
5
3 – 4
=
+
+
p q p p q
p q p q
2(3 – 4 ) – 3 – 5(3 4 )
(3 4 )(3 – 4 )
=
+
p q p p q
p q p q
6 – 8 – 3 – 15 – 20
(3 4 )(3 – 4 )
=
+
p q
p q p q
–12 – 28
(3 4 )(3 – 4 )
=
+
+
p q
q p q p
12 28
(4 3 )(4 – 3 )
10. 1 –
25
4x2





 ÷ 1 –
5
2x






=
x
x
4 – 25
4
2
2
÷
x
x
2 – 5
2
=
+
x x
x
(2 5)(2 – 5)
4 2
×
x
x
2
2 – 5
=
+
x
x
2 5
2

1 88
11.
x
1
3 –
+
x
1
1 – 2
= 6
1
3 – 4x






+
x x
x x
1 – 2 3 –
(3 – )(1 – 2 )
=
x
6
3 – 4
x
x x
4 – 3
(3 – )(1 – 2 )
=
x
6
3 – 4
(4 – 3x)(3 – 4x) = 6(3 – x)(1 – 2x)
12 – 16x – 9x + 12x2
= 6(3 – 6x – x + 2x2
)
12 – 25x + 12x2
= 18 – 42x + 12x2
17x = 6
x =
6
17
12. Let the cost of each bracelet and each pair of earrings be
$x and $y respectively.
3x + 6y = 1140 —(1)
7x + 9y = 1910 —(2)
(1) ×
3
2
:
9
2
x + 9y = 1710 —(3)
(2) – (3):
5
2
x = 200
x = 80
3(80) + 6y = 1140
240 + 6y = 1140
6y = 900
y = 150
Each bracelet costs $80 and each pair of earrings costs
$150.
13. R =
k
d2
When d = 2, R = 23,
23 =
k
22
k = 23 × 4
= 92
R =
d
92
2
When d = 2.3,
R =
92
2.32
= 17.4 (to 3 s.f.)
The resistance is 17.4 ohms.
14. On map A,
2 cm represent 5 km
4 cm2
represent 25 km2
72 cm2
represent 450 km2
On map B,
16 km2
450 km2
is represented
by 253
1
8
cm2
The forest is represented by an area of 253
1
8
cm2
on
map B.
15. (a) Yes
(b) No, the relation is not a function since the element 3
in the domain has two images, p and s in the codomain.
16. (i) ABD and BCD
(ii)
CD
CB
=
BD
AB
CD
a
=
x
c
CD =
ax
c
17. (a) ACT = 180° − 56° − 78° ( sum of CAT)
= 46°
DGO = 180° − 46° − 78° ( sum of OGD)
= 56°
C ↔ O
A ↔ G
T ↔ D
AT = GD = 9 cm
CA = OG = 12.2 cm
CT = OD = 10.4 cm
CAT  OGD
(b) RUN = 180° − 78° − 56° ( sum of RUN)
= 46°
EPI = 180° − 56° − 46° ( sum of PIE)
= 78°
R ↔ P
U ↔ E
N ↔ I
Since RU ≠ PE, UN ≠ EI and RN ≠ PI, RUN is not
congruent to PIE.
18. R3
represents (3 × 130°) − 360° = 30° anticlockwise about
the origin.
R5
represents (5 × 130°) − 360° = 290° anticlockwise about
the origin.

1
89
Part II
Section A
1.
+
x
x
– 1
2 3
= y + 4
x – 1 = (y + 4)(2x + 3)
= 2xy + 3y + 8x + 12
2xy + 7x = –3y – 13
x(2y + 7) = –(3y + 13)
x = –
+
+
y
y
3 13
2 7
2. (a) 24m2
– 13m – 2
= (8m + 1)(3m – 2)
(b) 2a2
– ap – 2ac + pc
= a(2a – p) – c(2a – p)
= (a – c)(2a – p)
(c) 64x2
– 25y2
– (8x – 5y)
= (8x + 5y)(8x – 5y) – (8x – 5y)
= (8x – 5y)(8x + 5y – 1)
3. (a) 3x – 4 – 7(3 – 2x) = 0
3x – 4 – 21 + 14x = 0
17x = 25
x =
25
17
= 1
8
17
(b) (8y – 5)2
= 98 – (y + 9)2
64y2
– 80y + 25 = 98 – y2
– 18y – 81
65y2
– 62y + 8 = 0
(13y – 2)(5y – 4) = 0
y =
2
13
or y =
4
5
4. x – 2y = 3 —(1)
6y – 3x = 4 —(2)
(1) × (–3): 6y – 3x = –9 —(3)
The two equations represent two parallel lines which do
not meet.
5. (i) 4 cm represents 5 km
Scale is 1 : 125 000
(ii) 9.4 cm represents 11.75 km
(iii) 1.5625 km2
64 km2
6. V = kr2
—(1)
1.96V = kR2
—(2)
(2) ÷ (1):
R
r
2
2 = 1.96
R2
= 1.96r2
R = 1.4r
The radius will increase by 40%.
Section B
7. (a)
x
x
9 – 15
9 – 25
2 =
+
x
x x
3(3 – 5)
(3 5)(3 – 5)
=
+
x
3
3 5
(b)
y y y
y
(3 – 2)( – 2) – 5
– 4
=
+
y y y y
y
3 – 6 – 2 4 – 5
– 4
2
=
+
y y
y
3 – 13 4
– 4
2
=
y y
y
(3 – 1)( – 4)
– 4
= 3y – 1
x2
+ (x + 2)2
= 1460
x2
+ x2
+ 4x + 4 = 1460
2x2
+ 4x – 1456 = 0
x2
+ 2x – 728 = 0
(x + 28)(x – 26) = 0
x = –28 or x = 26
x + 2 = 28
9. (a) p = a + bq
When q =
1
6
, p = 6,
a +
1
6
b = 6 —(1)
When q =
1
3
, p = 10,
a +
1
3
b = 10 —(2)
(1) × 6: 6a + b = 36 —(3)
(2) × 3: 3a + b = 30 —(4)
(b) (1) – (2): 3a = 6
a = 2
3(2) + b = 30
6 + b = 30
b = 24
(c) p = 2 + 24q
(i) When q = 2,
p = 2 + 24(2)
= 50
(ii) When p = 0,
0 = 2 + 24q
24q = –2
q = –
1
12

1 90
10. (i)
x
480
(ii)
x
480
– 2
(iii)
x
480
– 2
–
x
480
= 8
480x – 480(x – 2) = 8x(x – 2)
480x – 480x + 960 = 8x2
– 16x
8x2
– 16x – 960 = 0
x2
– 2x – 120 = 0 (shown)
(iv) (x – 12)(x + 10) = 0
x = 12 or x = –10
x
480
= 40
Mr Lim’s car used 40 l to travel 480 km.
11. (a) When x = –1, y = a,
a = 5 – (–1) – (1)2
= 5
When x = 2, y = b,
b = 5 – 2 – 22
= –1
a = 5, b = –1
(b)
–3 –2 –1
–1
1
2
3
4
5
1
y = 5 – x – x2
0 2
x
y
(c) (i) The equation of the line of symmetry of the graph
is x = –0.5.
(ii) Greatest value of y = 5.25.
(iii) When y = 4, x = 0.6 or x = –1.6.

1
91
Mid-Year Examination Specimen Paper B
Part I
1. 3x – 4y = 9 —(1)
4x + 5y = 43 —(2)
(1) × 4: 12x – 16y = 36 —(3)
(2) × 3: 12x + 15y = 129 —(4)
(4) – (3): 31y = 93
y = 3
3x – 4(3) = 9
3x – 12 = 9
3x = 21
x = 7
x = 7, y = 3
2. (a) (x + 2)(x + 3) – x(x – 3)
= x2
+ 3x + 2x + 6 – x2
+ 3x
= 8x + 6
(b) (2x + y)(x – y) – 2x(x – 2y)
= 2x2
– 2xy + xy – y2
– 2x2
+ 4xy
= 3xy – y2
3. (a) 12p3
– 3pq2
= 3p(4p2
– q2
)
= 3p(2p + q)(2p – q)
(b) 6ax + 3bx – 6ay – 3by
= 3(2ax + bx + 2ay – by)
= 3[x(2a + b) – y(2a + b)]
= 3(2a + b)(x – y)
4. 6x –
x
6
= 5
6x2
– 6 = 5x
6x2
– 5x – 6 = 0
(2x – 3)(3x + 2) = 0
x =
3
2
or x = –
2
3
= 1
1
2
5. a2
– b2
= 72
(a + b)(a – b) = 72
6(a – b) = 72
a – b = 12
b – a = –12
3b – 3a = –36
6. (a)
x
4
– 3
–
x
5
=
x x
x x
4 – 5( – 3)
( – 3)
=
+
x x
x x
4 – 5 15
( – 3)
=
x
x x
15 –
( – 3)
(b)
x
y2
– xy
+
y
x2
– xy





 ÷
+
x y
xy
=
x
y(y – x)
+
y
x(x – y)





 ×
+
xy
x y
=
y x
xy x y
–
( – )
2 2
×
+
xy
x y
=
+
x y y x
xy x y
( )( – )
( – )
×
+
xy
x y
= –1
7. x –
y
2
7
=
y
a
3
5
+ 2
35ax – 10ay = 21y + 70a
10ay + 21y = 35ax – 70a
y(10a + 21) = 35ax – 70a
y =
+
ax a
a
35 – 70
10 21
8. (i) y = ax2
+ bx + 5
When x = 1, y = 0,
0 = a(1)2
+ b(1) + 5
a + b = –5 —(1)
When x = 3, y = 2,
2 = a(3)2
+ b(3) + 5
9a + 3b = –3
3a + b = –1 —(2)
(ii) (2) – (1): 2a = 4
a = 2
2 + b = –5
b = –7
Equation of the curve is y = 2x2
– 7x + 5
9. (i) (2x + 1)(x – 1) = 90
2x2
– 2x + x – 1 = 90
2x2
– x – 91 = 0 (shown)
(ii) (x – 7)(2x + 13) = 0
x = 7 or x = –
13
2
= –6
1
2
(iii) Perimeter = 2[2(7) + 1 + 7 – 1]
= 42 cm
1 cm2
represents 0.04 km2
40 cm2
represent 1.6 km2
0.0025 km2
1.6 km2

1 92
11. (a) y = kx3
When x = 3, y = 108,
108 = k(3)3
= 27k
k = 4
y = 4x3
(b) y = k(x + 4)
When x = 1, y = 10,
10 = k(1 + 4)
= 5k
k = 2
y = 2(x + 4)
When x = 2, y = a,
a = 2(2 + 4)
= 12
When x = b, y = 5,
5 = 2(b + 4)
5
2
= b + 4
b = –1
1
2
a = 12, b = –1
1
2
12. (i)
–2 –1 0
2
1
3
4
5
6
–1
2
1
y =
1
2
x + 2
2y + 3x = 12
4
3 5
y
x
(ii) x = 2, y = 3
13. f(x) = 5x –
2
3
f(3) = 5(3) –
2
3
= 14
1
3
f(−5) = 5(−5) –
2
3
=
−25
2
3
f
2
5
⎛
⎝
⎜
⎞
⎠
⎟ = 5
2
5
⎛
⎝
⎜
⎞
⎠
⎟ –
2
3
= 1
1
3
f −
3
5
⎛
⎝
⎜
⎞
⎠
⎟ = 5 −
3
5
⎛
⎝
⎜
⎞
⎠
⎟ –
2
3
=
−3
2
3
14. AP = PB = BR = RC = CQ = QA
/PAQ = /APQ = /AQP = 60°
/BPR = /PBR = /PRB = 60°
/RQC = /QRC = /QCR = 60°
/PRQ = /RPQ = /PQR = 60°
Since APQ and RPQ are equilateral triangles with
sides of equal length, APQ is congruent to RPQ.
15.
+
x 9
9
=
+
5 7
7
x + 9 =
12
7
× 9
x = 6
3
7
y
6
=
+
5 7
7
y =
12
7
× 6
= 10
2
7
x = 6
3
7
, y = 10
2
7
16. (a) Length of PQ = Length of PQ
= 4 units
x-coordinate of Q = 4 + 4
= 8
The coordinates of Q are (8, 2).
k = 8
(b) Since (2, 3.5) is 1.5 units away from P, its image will
be 1.5 units away from P i.e. (5.5, 2)
(c) Since (7, 2) is 1 unit away from Q, the coordinates
of the original point will be 1 unit away from Q
i.e. (2, 5).
Part II
Section A
1. (a) (i) y = k(2x + 1)2
When x = 2, y = 75,
75 = k[2(2) + 1]2
= 25k
k = 3
y = 3(2x + 1)2
(ii) When x = 3,
y = 3[2(3) + 1]2
= 147

1
93
(b) (i) F =
k
R
3
When R = 125, F = 4,
4 =
k
125
3
=
k
5
k = 20
F =
R
20
3
(ii) F
O
R
3
2. (a) (i) y =
2x
x + 3
y2
=
2x
x + 3
y2
(x + 3) = 2x
2x − xy2
= 3y2
x(2 − y2
) = 3y2
x =
3y2
2 – y2
(ii) When y = 1,
x =
3 1
( )
2
2 – 1
( )
2
= 3
(b)
3
4x – 2y
+
5
y – 2x
=
3
2y – 4x
+
2(5)
2(y – 2x)
=
7
2y – 4x
3. (i) A(–1, 0), B(5, 0)
(ii) x =
+
–1 5
2
= 2
When x = 2,
y = 2(2 + 1)(2 – 5)
= –18
Coordinates of minimum point are (2, –18).
4. (i)
x
80
h
(ii)
x
80
– 3
h
(iii)
x
80
– 3
–
x
80
=
80
60
x
1
– 3
–
x
1
=
1
60
60x – 60(x – 3) = x(x – 3)
60x – 60x + 180 = x2
– 3x
x2
– 3x – 180 = 0 (shown)
(iv) (x – 15)(x + 12) = 0
x
80
= 5
1
3
Time taken is 5 h 20 min.
Section B
5. (a) (a – b)2
= 87
a2
– 2ab + b2
= 87
a2
+ b2
– 2(7.5) = 87
a2
+ b2
– 15 = 87
a2
+ b2
= 102
3a2
+ 3b2
= 306
(b) xy + 2x – 3y = 6
xy + 2x – 3y – 6 = 0
x(y + 2) – 3(y + 2)= 0
(x – 3)(y + 2) = 0
x = 3 or y = –2
6. Let the original fraction be
x
y
.
x
y
– 1
– 1
=
1
6
—(1)
+
+
x
y
3
3
=
1
2
—(2)
From (1),
6x – 6 = y – 1
6x – y = 5 —(3)
From (2),
2x + 6 = y + 3
2x – y = –3 —(4)
(3) – (4): 4x = 8
x = 2
2(2) – y = –3
4 – y = –3
y = 7
The fraction is
2
7
.

1 94
7. (i) 0.25 km is represented by 1 cm
The length of the line representing the coastline is
24 cm.
(ii) 1 cm2
60 cm2
represent 3.75 km2
The actual area of the marine park is 3.75 km2
.
8. (i)
x x
x x
( 2)( – 1)
( 1)( – 2)
+
+
=
10
7
7(x + 2)(x – 1) = 10(x + 1)(x – 2)
7(x2
– x + 2x – 2) = 10(x2
– 2x + x – 2)
7x2
+ 7x – 14= 10x2
– 10x – 20
3x2
– 17x – 6 = 0
(x – 6)(3x + 1) = 0
x = 6 or x = –
1
3
(rejected)
(ii) Perimeter of A = 2(6 + 2 + 6 – 1)
= 26 cm
Perimeter of B = 2(6 + 1 + 6 – 2)
= 22 cm
Perimeter of A : Perimeter of B
= 26 : 22
= 13 : 11
9. (a) When x = –1
1
2
, y = a,
a = –1
1
2
3 – 2 –1
1
2












= – 9
When x = 2, y = b,
b = 2[3 – 2(2)]
= –2
a = –9, b = –2
(b)
–2 –1 0
1
–1
–2
–3
–4
–5
–6
–7
–8
–9
–10
–11
–12
–13
–14
1 2 3
x
y
y = x(3 – 2x)
(c) (i) The equation of the line of symmetry of the graph
is x = 0.75.
(ii) When x = 2.2, y = –3.1
(iii) 2x(3 – 3x) = –13
x(3 – 2x) = –
13
2
When y = –
13
2
, x = 2.7 or x = –1.2

1
95
Chapter 10 Pythagoras’ Theorem
Basic
1. (a) Using Pythagoras’ Theorem,
a2
= 11.92
+ 6.82
= 187.85
a = 187.85
= 13.7 (to 3 s.f.)
(b)
x cm
12.4 cm
b cm
4.8 cm
7.4 cm
Using Pythagoras’ Theorem,
x2
+ 4.82
= 12.42
x2
= 130.72
x = 130.72
b2
= 130.72 + (7.4 + 4.8)2
= 279.56
b = 279.56
= 16.7 (to 3 s.f.)
(3a)2
+ (2a)2
= 18.92
9a2
+ 4a2
= 357.21
13a2
= 357.21
a2
= 27.47 (to 4 s.f.)
a = 27.47
= 5.24 (to 3 s.f.)
(b) Using Pythagoras’ Theorem,
(3b + 4b + 3b)2
+ 16.32
= 29.62
(10b)2
= 29.62
– 16.32
100b2
= 610.47
b2
= 6.1047
b = 6.1047
= 2.47 (to 3 s.f.)
3. Using Pythagoras’ Theorem,
a2
= 52
+ 122
= 169
a = 169
= 13
b2
+ 122
= 212
b2
= 212
– 122
= 297
b = 297
= 17.2 (to 3 s.f.)
a = 13, b = 17.2
4. Let the length of the square be x cm.
x2
= 350
x = 350
Length of diagonal = +
350 350
= 700
= 26.5 cm (to 3 s.f.)
(x + 1)2
+ (4x)2
= (4x + 1)2
x2
+ 2x + 1 + 16x2
= 16x2
+ 8x + 1
x2
– 6x = 0
x(x – 6) = 0
x = 0 (rejected) or x = 6
6. Let the length of the ladder be x m.
x2
= 3.22
+ 0.82
= 10.88
x = 3.30 (to 3 s.f.)
The length of the ladder is 3.30 m.
7. Let the vertical height of the cone be h cm.
h2
+ 82
= 122
h2
= 122
– 82
= 80
h = 80
= 8.94 (to 3 s.f.)
The vertical height of the cone is 8.94 cm.

1 96
8. Let the length of the diagonal be x m.
x2
= 142
+ 122
= 340
x = 340
= 18.4 (to 3 s.f.)
The length of the fence is 18.4 m.
9. Let the distance between the tips of the hands be x m.
x2
= 3.052
+ 3.852
= 24.125
x = 24.125
= 4.91 (to 3 s.f.)
The distance between the tips of the hands is 4.91 m.
10.
1.6 m
3 m
3 m
x m
14 m
x2
= 142
+ 1.62
= 198.56
x = 198.56
= 14.1 (to 3 s.f.)
The distance between the top of the two posts is 14.1 m.
d
2






2
+ 92
= 182
d
2






2
= 182
– 92
= 243
d
2
= 243
d = 2 243
= 31.2 (to 3 s.f.)
12. (a) AC2
= 322
= 1024
AB2
+ BC2
= 242
+ 282
= 1360
Since AC2
≠ AB2
+ BC2
,
ABC is not a right-angled triangle.
(b) DF2
= 852
= 7225
DE2
+ EF2
= 132
+ 842
= 7225
Since DF2
= DE2
+ EF2
,
DEF is a right-angled triangle with /DEF = 90°.
(c) HI2
= 6.52
= 42.252
GH2
+ GI2
= 3.32
+ 5.62
= 42.25
Since HI2
= GH2
+ GI2
,
GHI is a right-angled triangle with /HGI = 90°.
(d) KL2
= 2
3
17






2
= 4
213
289
JK2
+ JL2
=
12
17






2
+ 22
= 4
144
289
Since KL2
≠ JK2
+ JL2
,
JKL is not a right-angled triangle.
Intermediate
13. (a)
x cm
x cm
7.5 cm
a cm
14.8 cm
x2
+ x2
= 14.82
2x2
= 219.04
x2
= 109.52
x = 109.52
= 10.47 (to 4 s.f.)
a2
= 10.472
+ (7.5 + 10.47)2
= 432.2 (to 4 s.f.)
a = 432.2
= 20.8 (to 3 s.f.)

1
97
(b)
x cm
b cm
17.6 cm
8.9 cm
x2
= 8.92
+ 17.62
= 388.97
x = 388.97
Area of triangle =
1
2
× base × height
1
2
× 388.97 × b =
1
2
× 17.6 × 8.9
b =
×
17.6 8.9
388.97
= 7.94 (to 3 s.f.)
(c)
x cm
c cm
24.9 cm
15.6 cm
13.8 cm
(x + 13.8)2
+ 15.62
= 24.92
(x + 13.8)2
= 376.65
x + 13.8 = 376.65
x = 376.65 – 13.8
= 5.607 (to 4 s.f.)
c2
= 15.62
+ 5.6072
= 274.8 (to 4 s.f.)
c = 274.8
= 16.6 (to 3 s.f.)
a2
= 82
+ 92
= 145
a = 145
= 12.0 (to 3 s.f.)
b2
= 162
+ 92
= 337
b = 337
= 18.4 (to 3 s.f.)
a = 12.0, b = 18.4
15. (i) Using Pythagoras’ Theorem,
QR2
+ 8.52
= 12.32
QR2
= 12.32
– 8.52
= 79.04
QR = 79.04
= 8.89 cm (to 3 s.f.)
PS2
+ 12.32
= 17.82
PS2
= 17.82
– 12.32
= 165.55
PS = 165.55
= 12.9 cm (to 3 s.f.)
(iii) Area of trapezium PQRS =
1
2
(8.5 + 17.8) 79.04
= 117 cm2
(to 3 s.f.)
16. Area of ABC =
1
2
× AB × 14
180 = 7AB
AB =
180
7
cm
AC2
=
180
7






2
+ 142
= 857.2 (to 4 s.f.)
AC = 857.2
= 29.3 cm (to 3 s.f.)

1 98
BK2
+ 72
= 122
BK2
= 122
– 72
= 95
BK = 95
= 9.746 cm (to 4 s.f.)
BC = 2(9.746)
= 19.49 cm (to 4 s.f.)
(2x + 3)2
= 19.492
+ 82
= 444
2x + 3 = 444
= 21.07 (to 4 s.f.)
2x = 18.07
x = 9.04 (to 3 s.f.)
18.
17 cm 17 cm
8 cm 8 cm
h cm
A
B C
h2
+ 82
= 172
h2
= 172
– 82
= 225
h = 225
= 15
Area of ABC =
1
2
(16)(15)
= 120 cm2
19.
14 cm
x cm
27 cm
x2
= 142
+ 272
= 925
x = 925
= 30.41 (to 4 s.f.)
Perimeter = 4(30.41)
= 122 cm (to 3 s.f.)
PQ2
= (28 – 11)2
+ (28 – 9)2
= 172
+ 192
= 650
Area of PQRS = PQ2
= 650 cm2
BD2
= 122
+ 52
= 169
BD = 13 cm
(AD + 5)2
+ 122
= 152
(AD + 5)2
= 152
– 122
= 81
AD + 5 = 9
AD = 4 cm
(ii) Area of ABC =
1
2
(12)(9)
= 54 cm2
(iii) Let the shortest distance from C to AB be h cm.
Area of ABC =
1
2
× 15 × h
54 =
15
2
h
h = 54 ×
2
15
= 7.2 cm
XB2
+ 1.32
= 52
XB2
= 52
– 1.32
= 23.31
XB = 23.31
= 4.828 cm (to 4 s.f.)
XY = 2(4.828)
= 9.66 cm (to 3 s.f.)
23. P(–2, –1), T(6, 5)
PT2
= 82
+ 62
= 100
PT = 10
The player has to run 10 units.

1
99
24. Let the height of the LCD screen be h inches.
h2
+ 48.52
= 552
h2
= 552
– 48.52
= 672.75
h = 672.75
= 25.9 (to 3 s.f.)
Since h > 24, the box will not fit the LCD screen.
25.
h q
20
p
16
h2
+ 162
= 202
h2
= 202
– 162
= 144
p2
+ 144 = q2
(x + 2)2
+ x2
= (x + 4)2
x2
+ 4x + 4 + x2
= x2
+ 8x + 16
x2
– 4x – 12 = 0
(x – 6)(x + 2) = 0
Perimeter = 2(x + 2 + x)
= 4x + 4
= 4(6) + 4
= 28 m
(2x)2
+ (4x – 1)2
= (4x + 1)2
4x2
+ 16x2
– 8x + 1 = 16x2
+ 8x + 1
4x2
– 16x = 0
4x(x – 4) = 0
x = 0 or x = 4
(rejected)
Cross-sectional area of sandwich =
1
2
(8)(15)
= 60 cm2
28.
x km 240 km
140 km
x2
= 1402
+ 2402
= 77 200
x = 77 200
= 278 (to 3 s.f.)
The distance from the starting point is 278 km.
29. D
A
B
C
10 km
8 km
AB = 40 ×
12
60
= 8 km
BC = 15 km
CD = 60 ×
10
60
= 10 km
DA2
= 152
+ 22
= 229
DA = 229
= 15.1 km (to 3 s.f.)
The shortest distance is 15.1 km.

1 100
Advanced
30. A
E
B
D
C
h Y
YC2
+
h
2






2
= h2
YC2
+
h
4
2
= h2
YC2
= h2
–
h
4
2
=
h
3
4
2
YC =
3
4
h
=
3
2
h
AC2
=
h
2






2
+ h +
3
2
h






2
= 0.25h2
+ 3.482h2
= 3.732h2
AC = h
3.732 2
= 1.93h units (to 3 s.f.)
31. Let the length of the diagonal of the base be x m.
x2
= 32
+ 42
= 25
x = 25
= 5
PQ2
= 52
+ 122
= 169
PQ = 169
= 13 m
32. (i) Let the radii of P, Q, and R be p, q and r respectively.
Area of P =
1
2
pp2
578 =
π
2
p2
p2
=
1156
π
Area of Q =
1
2
pq2
128 =
1
2
pq2
q2
=
256
π
Area of R =
1
2
pr2
x =
1
2
pr2
r2
=
x
2
π
(2q)2
+ (2r)2
= (2p)2
4q2
+ 4r2
= 4p2
q2
+ r2
= p2
256
π
+
x
2
π
=
1156
π
2x = 1156 – 256
= 900
x = 450
(ii) Since q2
=
256
π
and r2
=
900
π
,
then q =
256
π
and r =
900
π
.

AB
BC
=
256
π
900
π
=
16
π
30
π
=
8
15

1
101
33. (i) At x-axis, y = 0
3x + 15 = 0
x = −5
At y-axis, x = 0
y + 15 = 0
y = −15
The coordinates of A are (−5, 0) and B are (0, −15).
AB2
= 52
+ 152
= 250
AB = 250
= 15.8 units (to 3 s.f.)
The length of the line joining A to B is 15.8 units.
34. (i) BC = 23x − 2 − (3x − 2) − (5x + 1) − (6x −7)
= 23x − 3x − 5x − 6x − 2 + 2 − 1 + 7
= (9x + 6) cm
(ii) Since BC = 2AD,
9x + 6 = 2(5x + 1)
9x + 6 = 10x + 2
x = 4
Perimeter of trapezium = 23x − 2
= 23(4) − 2
= 90 cm
(iii) BX + CY = BC − AD
= 9(4) + 6 − [5(4) + 1]
= 42 − 21
= 21
Since 5BX = 2CY,
BX
CY
=
2
5
BX =
21
7
× 2
= 6
AB = 3(4) − 2
= 10
AX2
= 102
− 62
= 64
AX = 64
= 8 cm
Area of trapezium =
1
2
× 8 × (21 + 42)
= 252 cm2

1
Chapter 11 Trigonometric Ratios
Basic
1. (a) (i) AC
(ii) AB
(iii) BC
(b) (i) YZ
(ii) XZ
(iii) XY
2. (a) (i) sin X =
4
5
(ii) cos X =
3
5
(iii) tan X = 1
1
3
(iv) sin Y =
3
5
(v) cos Y =
4
5
(vi) tan Y =
3
4
(b) (i) sin X =
12
13
(ii) cos X =
5
13
(iii) tan X = 2
2
5
(iv) sin Y =
5
13
(v) cos Y =
12
13
(vi) tan Y =
5
12
(c) (i) sin X =
4
5
(ii) cos X =
3
5
(iii) tan X = 1
1
3
(iv) sin Y =
3
5
(v) cos Y =
4
5
(vi) tan Y =
3
4
(d) (i) sin X =
24
25
(ii) cos X =
7
25
(iii) tan X = 3
3
7
(iv) sin Y =
7
25
(v) cos Y =
24
25
(vi) tan Y =
7
24
3. (a) (i) sin X =
b
c
(ii) cos X =
a
c
(iii) tan X =
b
a
(iv) sin Y =
a
c
(v) cos Y =
b
c
(vi) tan Y =
a
b
(b) (i) sin X =
q
p
(ii) cos X =
r
p
(iii) tan X =
q
r
(iv) sin Y =
r
p
(v) cos Y =
q
p
(vi) tan Y =
r
q
4. (a) cos 27° + cos 54° = 1.479 (to 3 d.p.)
(b) 5 cos 51° + 2 sin 16° = 3.698 (to 3 d.p.)
(c) 7 tan 20° – 5 sin 13° = 1.423 (to 3 d.p.)
(d) 14 sin 43° – 6 cos 7° = 3.593 (to 3 d.p.)
(e) 12 cos 13° × 12 tan 49° = 161.407 (to 3 d.p.)
(f) 9 cos 41° – 4 tan 12° = 5.942 (to 3 d.p.)
5. (a) sin x = 0.4
x = 23.6° (to 1 d.p.)
(b) cos x = 0.4
x = 66.4° (to 1 d.p.)
(c) tan x = 0.3
x = 16.7° (to 1 d.p.)
(d) sin x = 0.45
x = 26.7° (to 1 d.p.)
(e) cos x = 0.74
x = 42.3° (to 1 d.p.)
(f) tan x = 1.34
x = 53.3° (to 1 d.p.)
(g) sin x = 0.453
x = 26.9° (to 1 d.p.)
(h) cos x = 0.973
x = 13.3° (to 1 d.p.)
(i) tan x = 0.354
x = 19.5° (to 1 d.p.)
102

1
(j) tan x = 1
x = 45°
6. (a) sin 34° =
a
15
a = 15 sin 34°
= 8.39 (to 3 s.f.)
cos 34° =
b
15
b = 15 cos 34°
= 12.4 (to 3 s.f.)
a = 8.39, b = 12.4
(b) tan 64° =
c
12
c = 12 tan 34°
= 24.6 (to 3 s.f.)
cos 64° =
d
12
d =
12
cos 64°
= 27.4 (to 3 s.f.)
c = 24.6, d = 27.4
(c) tan 51.7° =
e
7.53
e =
7.53
tan 51.7°
= 5.95 (to 3 s.f.)
sin 51.7° =
f
7.53
f = 7.53
sin 51.7°
= 9.60 (to 3 s.f.)
e = 5.95, f = 9.60
(d) cos 31.9° =
g
71.6
g = 71.6
cos 31.9°
= 84.3 (to 3 s.f.)
tan 31.9° =
h
71.6
h = 71.6 tan 31.9°
= 44.6 (to 3 s.f.)
g = 84.3, h = 44.6
7. (a) tan a° =
5.5
7.6
a° = 35.9° (to 1 d.p.)
a = 35.9
b2
= 5.52
+ 7.62
= 88.01
b = 9.38 (to 3 s.f.)
a = 35.9, b = 9.38
(b) cos c° =
24.3
35.7
c° = 47.1° (to 1 d.p.)
c = 47.1
24.32
+ d2
= 35.72
d2
= 684
d = 26.2 (to 3 s.f)
c = 47.1, d = 26.2
8. tan /QPR =
32
43
/QPR = 36.7° (to 1 d.p.)
9. (i) sin 21.6° =
SF
86.5
SF = 86.5 sin 21.6°
= 31.8 m (to 3 s.f.)
(ii) cos 21.6° =
FH
86.5
FH = 86.5 cos 21.6°
= 80.4 m (to 3 s.f.)
10. tan 38° =
BE
45
BE =
45
tan 38°
= 57.6 m (to 3 s.f.)
The distance between the enemy and the foot of the
observatory is 57.6 m.
Intermediate
11. (a)
2 sin 26°
3 cos17°
= 0.306 (to 3 d.p.)
(b)
(tan 45°)2
tan 10°
= 5.671 (to 3 d.p.)
(c)
° + °
°
sin 30 cos 40
tan 50
= 1.062 (to 3 d.p.)
(d) cos19°
tan 22° – sin 58°
= –2.129 (to 3 d.p.)
(e)
° °
° × °
sin 20 – cos 61
tan 47 sin 91
= –0.133 (to 3 d.p.)
(f)
° °
° ÷ °
cos 63 – sin 2
tan 54 tan 3
= 0.016 (to 3 d.p.)
12. (a) tan 27.7° =
a
18.1
2
a =
18.1
2 tan 27.7°
= 17.2 (to 3 s.f.)
sin 27.7° =
b
18.1
b =
18.1
sin 27.7°
= 38.9 (to 3 s.f.)
a = 17.2, b = 38.9
103

1
(b) sin 29° =
c
15.4
c = 15.4 sin 29°
= 7.47 (to 3 s.f.)
sin 32° =
d
15.4
d = 15.4 sin 32°
= 8.16 (to 3 s.f.)
cos 32° =
e
15.4
e = 15.4 cos 32°
= 13.1 (to 3 s.f.)
13. (i) Area of BCD =
1
2
(12)(AB)
45 = 6AB
AB =
45
6
= 7.5 cm
(AD + 12)2
+ 7.52
= 192
(AD + 12)2
= 304.75
AD + 12 = 304.75
AD = 304.75 – 12
= 5.46 cm (to 3 s.f.)
(iii) tan /BDA =
AB
AD
=
7.5
304.75 – 12
= 1.374 (to 4 s.f.)
/BDA = 53.95° (to 2 d.p.)
/BDC = 180° – 53.95°
= 126.0° (to 1 d.p.)
AP2
= 82
+ 52
= 89
AP = 89
= 9.43 cm (to 3 s.f.)
(ii) tan /APC =
8
5
/APC = 57.99° (to 2 d.p.)
/APQ = 180° – 57.99°
= 122.0° (to 1 d.p.)
(iii) /BRP = 360° – 2(57.99°) – 90°
= 154.0° (to 1 d.p.)
15. (i) sin 65° =
BQ
7.6
BQ = 7.6 sin 65°
= 6.887 (to 4 s.f.)
= 6.89 cm (to 3 s.f.)
PQ2
+ 6.8872
= 8.72
PQ2
= 28.24 (to 4 s.f.)
PQ = 5.314 (to 4 s.f.)
= 5.31 cm (to 3 s.f.)
(iii) Using Pythagoras’ Theorem,
(AP + 5.314)2
+ 6.8872
= 10.22
(AP + 5.314)2
= 56.59
AP + 5.134 = 7.523 (to 4 s.f.)
AP = 2.21 cm (to 3 s.f.)
(iv) sin /BPQ =
6.887
8.7
/BPQ = 52.34° (to 2 d.p.)
/APB = 180° – 52.34°
= 127.7° (to 1 d.p.)
16. (a)
p cm
h cm
11.7 cm
12.5 cm
a cm
64°
73°
q cm
sin 73° =
h
11.7
h = 11.7 sin 73°
= 11.18 (to 4 s.f.)
sin 64° =
a
11.18
a =
11.18
sin 64°
= 12.4 (to 3 s.f.)
cos 73° =
p
11.7
p = 11.7 cos 73°
= 3.420 (to 4 s.f.)
tan 64° =
q
11.18
q =
11.18
tan 64°
= 5.457 (to 4 s.f.)
b = 3.420 + 5.457 + 12.5
= 21.4 (to 3 s.f.)
a = 12.4, b = 21.4
104

1
(b)
e cm
d cm
w cm
7.5 cm
6.3 cm
c°
78°
3.3 cm
tan 78° =
d
3.3
d = 3.3 tan 78°
= 15.5 (to 3 s.f.)
cos 78° =
w
3.3
w =
3.3
cos 78°
= 15.87 (to 4 s.f.)
cos c° =
7.5
15.87
c° = 61.80° (to 2 d.p.)
c = 61.8 (to 1 d.p.)
tan 61.80° =
e
7.5
e = 7.5 tan 61.80°
= 14.0 (to 3 s.f.)
c = 61.8, d = 15.5, e = 14.0
(c)
16.9 cm
f° 52°
g°
9.6 cm
h cm w cm
sin f° =
9.6
16.9
f° = 34.61° (to 2 d.p.)
f = 34.6 (to 1 d.p.)
g° = 52° – 34.61°
= 17.4° (to 1 d.p.)
g = 17.4
tan 52° =
w
9.6
w =
9.6
tan 52°
= 7.500 (to 4 s.f.)
(h + 7.500)2
+ 9.62
= 16.92
(h + 7.500)2
= 193.45
h + 7.500 = 193.45
h = 193.45 – 7.500
= 6.41 (to 3 s.f.)
f = 34.6, g = 17.4, h = 6.41
QT2
+ 8.62
= 11.32
QT2
= 53.73
QT = 53.73
= 7.330 cm (to 4 s.f.)
9.822
+ (10.2 + 7.330)2
= PS2
PS2
= 403.3
PS = 403.3
= 20.08 (to 4 s.f.)
= 20.1 cm (to 3 s.f.)
(ii) cos /SPQ =
9.8
20.08
/SPQ = 60.8° (to 1 d.p.)
18.
12.8 cm
58°
w cm
h cm
Q
P R
sin 58° =
h
12.8
h = 12.8 sin 58°
= 10.85 (to 4 s.f.)
cos 58° =
w
12.8
w = 12.8 cos 58°
= 6.782 (to 4 s.f.)
Area of PQR =
1
2
(2 × 6.782)(10.85)
= 73.6 cm2
(to 3 s.f.)
105

1
19.
68°
h cm
z cm
24 cm 24 cm
z cm
cos 68° =
z
24
z =
24
cos 68°
= 64.06 (to 4 s.f.)
Perimeter = 48 + 2(64.06)
= 176 cm (to 3 s.f.)
tan 68° =
h
24
h = 24 tan 68°
= 59.40 (to 4 s.f.)
Area =
1
2
(48)(59.40)
= 1430 cm2
(to 3 s.f.)
20.
a cm
b cm
8.9 cm
12.6 cm
42.8°
Q
P
R
tan 42.8° =
a
8.9
a =
8.9
tan 42.8°
= 9.611 (to 4 s.f.)
b2
+ 8.92
= 12.62
b2
= 79.55
b = 79.55
= 8.919 (to 4 s.f.)
Area of PQR =
1
2
(9.611 + 8.919)(8.9)
= 82.5 cm2
(to 3 s.f.)
21. Let the distance from the boat to the foot of the cliff be
d m.
tan 26° =
d
55
d =
55
tan 26°
= 113 (to 3 s.f.)
Thedistancefromtheboattothefootofthecliffis113m.
22. (i) cos 54° =
PQ
1.8
PQ =
1.8
cos 54°
= 3.062 (to 4 s.f.)
= 3.06 m (to 3 s.f.)
(ii) tan 54° =
QN
1.8
QN = 1.8 tan 54°
= 2.477 (to 4 s.f.)
= 2.48 m (to 3 s.f.)
(iii) Q′N = 2.477 – 0.8
= 1.677 (to 4 s.f.)
= 1.68 m (to 3 s.f.)
(iv) sin /NP′Q′ =
1.677
3.062
/NP′Q′ = 33.2° (to 1 d.p.)
23. (i) sin 47° =
KH
240
KH = 240 sin 47°
= 176 m (to 3 s.f.)
(ii) Assume that the string is taut.
(BC + 5.2)2
+ 18.32
= 242
(BC + 5.2)2
= 241.11
BC + 5.2 = 241.11
BC = 241.11 – 5.2
= 10.32 (to 4 s.f.)
= 10.3 m (to 3 s.f.)
(ii) tan /BMC =
10.32
18.3
/BMC = 29.43° (to 2 d.p.)
= 29.4° (to 1 d.p.)
(iii) cos /AMC =
18.3
24
/AMC = 40.31° (to 2 d.p.)
/AMB = 40.31° – 29.43°
= 10.9° (to 1 d.p.)
106

1
25.
16°
32°
14.5 m
a m
b m
Q
P
K
tan 32° =
a
14.5
a = 14.5 tan 32°
= 9.060 (to 4 s.f.)
tan 16° =
b
14.5
b = 14.5 tan 16°
= 4.157 (to 4 s.f.)
PQ = 9.060 + 4.157
= 13.2
The height of the monument is 13.2 m.
26.
5 cm
42 cm
L
A B
H
O
OH = 42 – 5
= 37 cm
cos /AOL =
37
42
/AOL = 28.24° (to 2 d.p.)
/AOB = 2(28.24°)
= 56.5° (to 1 d.p.)
27. (i)
q
28°
18 m
28 m
h m
d m
1st
storey
2nd
storey
3rd
storey
w m
sin 28° =
h
18
h = 18 sin 28°
= 8.450 (to 4 s.f.)
= 8.45 (to 3 s.f.)
The height of the first storey is 8.45 m.
(ii) cos 28° =
w
18
w = 18 cos 28°
= 15.89 (to 4 s.f.)
= 15.9 (to 3 s.f.)
(d + 8.450)2
+ 15.892
= 282
(d + 8.450)2
= 531.4
d + 8.450 = 531.4
d = 531.4 – 8.450
= 14.6 (to 3 s.f.)
The height of the second storey is 14.6 m.
(iii) cos (q + 28°) =
15.89
28
q + 28° = 55.41° (to 2 d.p.)
q = 27.4° (to 1 d.p.)
107

1
Advanced
28.
q
+ x
1
2
1
x
(a) 2 sin q =
+
x
x
2
1 2
(b) 3 cos q =
+ x
3
1 2
29.
45°
135°
y
x
O
a
P
3√2
a
a2
+ a2
= (3 2 )2
= 18
2a2
= 18
a2
= 9
a = ±3
The coordinates of P are (–3, 3).
30.
P R
28.5°
30 m
Q
tan 28.5° =
PQ
30
PQ = 30 tan 28.5°
= 16.3 m
The width of the river is 16.3 m.
31. tan 34° =
TA
AB
AB =
TA
tan 34°
—(1)
tan 26° =
+
TA
AB 25
AB tan 26° + 25 tan 26° = TA —(2)
tan 26°
tan 34°





 TA + 25 tan 26°= TA
TA –
tan 26°
tan 34°





 TA = 25 tan 26°
1 –
tan 26°
tan 34°





 TA = 25 tan 26°
TA =
25 tan 26°
1 –
tan 26°
tan 34°
= 44.0 m (to 3 s.f.)
The height of the office tower is 44.0 m.
32. (i) tan 56° =
PQ
250
PQ = 250 tan 56°
= 370.6 (to 4 s.f.)
= 371 m (to 3 s.f.)
P is 371 m above the parade ground.
(ii) tan 46° =
PQ
250
P′Q = 250 tan 46°
= 258.8 (to 4 s.f.)
= 259 m (to 3 s.f.)
PP′ = 370.6 – 258.8
= 111.7 m (to 4 s.f.)
Speed of descent =
111.7
45
= 2.484 m/s (to 4 s.f.)
Time taken to descend from P to Q
=
370.6
2.484
= 149 s (to 3 s.f.)
108

1
Chapter 12 Volume and Surface Area of
Pyramids, Cones and Spheres
Basic
1. (a) Volume of pyramid =
1
3
× 162
× 27
= 2304 cm3
(b) Volume of pyramid =
1
3
×
1
2
× 12 × 9





 × 20
= 360 cm3
(c) Volume of pyramid =
1
3
× 9 × 5 × 3
= 45 m3
2. Volume of pyramid =
1
3
× 8 × h
42 =
8
3
h
h = 15.75
The height of the figurine is 15.75 cm.
1
3
× 8 × 3 × h
86 = 8h
h = 10.75
The height of the pyramid is 10.75 m.
1
3
×
1
2
× 12 × 5





 × h
160 = 10h
h = 16
The height of the pyramid is 16 m.
5. Total surface area = 162
+ 4 ×
1
2
× 16 × 17
= 800 m2
6. V =
1
3
pr2
h
(a) When r = 8 and V = 320,
320 =
1
3
p(8)2
h
h =
960
64π
= 4.77 (to 3 s.f.)
(b) When r = 10.6 and V = 342.8,
342.8 =
1
3
p(10.6)2
h
h =
1028.4
112.36π
= 2.91 (to 3 s.f.)
(c) When h = 6 and V = 254,
254 =
1
3
pr2
(6)
r2
= 762
6π
r =
762
6π
= 6.36 (to 3 s.f.)
(d) When h = 11 and V = 695,
695 =
1
3
pr2
(11)
r2
=
2085
11π
r =
2085
11π
= 7.77 (to 3 s.f.)
Radius, r cm Height, h cm Volume, V cm3
(a) 8 4.77 320
(b) 10.6 2.91 342.8
(c) 6.36 6 254
(d) 7.77 11 695
7. (a) Volume of cone =
1
3
p(6)2
(8)
= 302 cm3
(to 3 s.f.)
Total surface area of cone = p(6)2
+ p(6)(10)
= 302 cm2
(to 3 s.f.)
(b) Volume of cone =
1
3
p(12)2
(28.8)
= 4340 cm3
(to 3 s.f.)
Total surface area of cone = p(12)2
+ p(12)(31.2)
= 1630 cm2
(to 3 s.f.)
8. (a) Volume of sphere =
4
3
p(5.8)3
= 817 cm3
(to 3 s.f.)
(b) Volume of sphere =
4
3
p(12.6)3
= 8380 m3
(to 3 s.f.)
4
3
p
24.2
2






3
= 7420 cm3
(to 3 s.f.)
4
3
p
6.25
2






3
= 128 mm3
(to 3 s.f.)
109

1
4
3
pr3
34 =
4
3
pr3
r3
=
51
2π
r =
51
2π
3
= 2.009 (to 4 s.f.)
= 2.01 cm (to 3 s.f.)
Surface area of sphere = 4p(2.009)2
= 50.8 cm2
(to 3 s.f.)
4
3
pr3
68.2 =
4
3
pr3
r3
=
51.15
π
r =
51.15
π
3
= 2.534 (to 4 s.f.)
= 2.53 m (to 3 s.f.)
Surface area of sphere = 4p(2.534)2
= 80.7 m2
(to 3 s.f.)
11. Surface area of sphere = 4p(8)2
= 256p m2
Cost of painting =
256π
8
× 8.5
= $854.51 (to 2 d.p.)
Intermediate
12. Let the height and the slant height of the pyramid be h cm
and l cm respectively.
Total surface area of pyramid = 82
+ 4 ×
1
2
(8)l
144 = 64 + 16l
16l = 80
l = 5
42
+ h2
= 52
16 + h2
= 25
h2
= 9
h = 3
Volume of pyramid =
1
3
× 82
×3
= 64 cm3
13. (i) Let the radius of the base be r m.
2pr = 8.5
r =
4.25
π
= 1.352 (to 4 s.f)
Volume of rice =
1
3
p(1.352)2
(1.2)
= 2.29 (to 3 s.f.)
= 2.3 m3
(to 2 s.f.)
(ii) Number of bags =
2.29
0.5
= 4.59 (to 3 s.f.)
≈ 5
Assume that the space between the grains of rice is
negligible.
14. Volume of crew cabin
=
1
3
p
75
2






2
(92) –
1
3
p
27
2






2
(92 – 59)
= 129 000 cm3
(to 3 s.f.)
15. (i) Let the radius of the base be r cm.
2pr = 88
r =
44
π
= 14.00 (to 4 s.f.)
Curved surface area of cone = p
44
π





 (15)
= 660 cm2
(ii) Total surface area of cone
= 660 + p(14.00)2
= 1276 cm2
(to the nearest integer)
16. (i) Curved surface area of cone = p(x – 5)(x + 5)
75p = p(x2
– 25)
75 = x2
– 25
x2
= 100
x = 10
(ii) Base radius = 5 cm
Slant height = 15 cm
Height = 15 – 5
2 2
= 200
Volume of cone =
1
3
p(5)2
( 200 )
= 370 cm3
(to 3 s.f.)
110

1
17. (i) Volume of solid =
2
3
ph3
–
2
3
p h
2






3
=
2
3
ph3
–
1
12
ph3
=
7
12
ph3
(ii) Total surface area of solid
= 2ph2
+ πh2
– π
h
2






2





 + 2p
h
2






2
= 2ph2
+ ph2
–
1
4
ph2
+
1
2
ph2
=
13
4
ph2
18. Volume of plastic =
4
3
p(4)3
–
4
3
p(3.6)3
= 72.7 cm3
(to 3 s.f.)
19. Volume of steel
= 100 ×
4
3
π
16
2






3
–
4
3
π
16
2
– 0.8






3






= 58 100 cm3
(to 3 s.f.)
20. Amount of space = 63
–
4
3
p
6
2






3
= 103 cm3
(to 3 s.f.)
21. (a) Total surface area of hemisphere = 2pr2
+ pr2
374 = 3pr2
r2
=
374
3π
r =
374
3π
= 6.3 cm (to 1 d.p.)
Volume of hemisphere =
2
3
p 374
3π








3
= 523.6 cm3
(to 1 d.p.)
(b) Total surface area of hemisphere = 2pr2
+ pr2
1058.4 = 3pr2
r2
=
352.8
π
r =
352.8
π
= 10.6 m (to 1 d.p.)
Volume of hemisphere =
2
3
p 352.8
π








3
= 2492.5 m3
(to 1 d.p.)
22. (i) Volume of sphere =
4
3
p
x + 2
2






3
972p =
4
3
p x + 2
2






3
x + 2
2






3
= 729
+
x 2
2
= 9
x + 2 = 18
x = 16
(ii) Surface area of sphere = 4p
18
2






2
= 1020 cm2
(to 3 s.f.)
23. Volume of glass
= volume of prism + volume of pyramid
=
1
2
× 3.6 × 4.8





 (6) +
1
3
1
2
× 3.6 × 4.8





 (12)
= 86.4 m3
24. Volume of hemisphere =
2
3
p(4)3
=
128
3
p cm3
Volume of model =
37
4
×
128
3
p
= 1240 cm3
(to 3 s.f.)
25. (i) Capacity of container =
1
3
p(21)2
(21)
= 9698 cm3
(to 4 s.f.)
= 9.70 l (to 3 s.f.)
(ii) Mass of container = 9698 × 1.5
= 14 540 g (to 4 s.f.)
= 15 kg (to the nearest kg)
Advanced
26. (i) Volume of iron =
1
3
p(1)2
(0.5)
=
π
6
= 0.524 m3
(to 3 s.f.)
Volume of lead = p(2)2
(3) –
π
6
= 12p –
π
6
=
71π
6
= 37.2 m3
(to 3 s.f.)
111

1
(ii) Let the denisty of lead be ρ g/m3
.
Original mass of cylinder = p(2)2
(3)ρ
= 12pρ g
New mass of cylinder =
π
6
2
3
ρ
⎛
⎝
⎜
⎞
⎠
⎟ + 71π
6
(ρ)
=
π
9
ρ +
71π
6
ρ
=
215π
18
ρ g
Percentage reduction in mass
=
ρ ρ
ρ
π
π
π
12 –
215
18
12
× 100%
=
25
54
%
27.
20 cm
18 cm
B
C
O
D
A
X
AC2
= 202
+ 182
= 724
AC = 724 cm
tan 50° =
AX
OX
OX =
AX
tan 50°
=
1
2
724
tan 50°
Volume of pyramid =
1
3
(20 × 18)
1
2
724
tan 50°








= 1350 cm3
(to 3 s.f.)
New Trend
h2
+ 82
= 172
h2
= 225
h = 225
= 15
The height of the cone is 15 cm. (shown)
(b) Volume of solid
= volume of cone + volume of hemisphere
=
1
3
π(8)2
(15) +
1
2
4
3
π 8
( )
3






= 2080 cm3
(to 3 s.f.)
29. Total surface area of solid =
1
2
(4πx)2
+ 2πx(3x) + πx2
= 2πx2
+ 6πx2
+ πx2
= 9πx2
Total surface area of solid = 2 × surface area of cone
9πx2
= 2(πxl + πx2
)
7πx2
= 2πxl
l =
7πx2
2πx
=
7x
2
30. (i) Let the height of the pyramid be h cm.
h2
+ 152
= 392
h2
= 1296
h = 1296
= 36
Volume of solid = (30)(30)(70) +
1
3
(30)2
(36)
= 73 800 cm3
(ii) Volume of spherical candle =
1
10
× 73 800
4
3
πr3
= 7380
r3
=
7380× 3
4π
r =
7380× 3
4π
3
= 12.078 cm (to 5 s.f.)
= 12.1 cm (to 3 s.f.)
(shown)
(iii) Volume of cuboid
= 4(12.078) × 2(12.078) × 2(12.078)
= 28 191 cm3
(to 5 s.f.)
Volume of empty space = 28 191 − 2(7380)
= 13 400 cm3
(to 3 s.f.)
112

1
31. Total surface area = π(4r)2
+2(2πr)(3r) +
1
2
[4π(4r)2
]
= 16πr2
+ 12πr2
+ 32πr2
= 60πr2
cm2
x2
= (15 − 9)2
+ 162
x2
= 292
x = 292
= 17.088 (to 5 s.f.)
= 17.09 cm (to 4 s.f.) (shown)
(ii) Let the slant height of the cone with radius 9 cm
be l cm.
l2
= (40 − 16)2
+ 92
l2
= 657
l = 657
= 25.63 cm (to 2 d.p.)
Total surface area of vase
= π(15)(17.088 + 25.63) − π(9)(25.63) + π(15)2
= 1995 cm2
(to the nearest whole number)
113

1
Chapter 13 Symmetry
Basic
1. (a) (i) The figure has 1 line of symmetry.
(ii) The figure has rotational symmetry of order 1
i.e. no rotational symmetry.
(b) (i) The figure has 1 line of symmetry.
(c) (i) The figure has 2 lines of symmetry.
(ii) The figure has rotational symmetry of order 2.
(d) (i) The figure has 0 lines of symmetry, i.e. no line
symmetry.
(e) (i) The figure has 1 line of symmetry.
(f) (i) The figure has 4 lines of symmetry.
(g) (i) The figure has 1 line of symmetry.
2. (a) y
6 7
2 3 4 5
1
0
x
4
1
2
3
5
6
7
Line of symmetry: x = 4
(b)
y
6 7
2 3 4 5
1
0
x
4
1
2
3
5
6
7
Line of symmetry: y = 3.5
3. (a) The figure has rotational symmetry of order 1 i.e. no
rotational symmetry.
(b) The figure has rotational symmetry of order 5.
(c) The figure has rotational symmetry of order 2.
(d) The figure has rotational symmetry of order 1 i.e. no
rotational symmetry.
(e) The figure has rotational symmetry of order 4.
(f) The figure has rotational symmetry of order 8.
(g) The figure has rotational symmetry of order 2.
4. (i) The letters with line symmetry are O, E, H and I.
(ii) The letters with rotational symmetry are O, S, H and I.
5. (a) False
(b) False
(c) True
(d) True
(e) False
(f) True
(g) True
(h) False
(i) True
(j) False
(k) False
(l) False
6. (a) An equilateral triangle has 3 lines of symmetry.
(b)
A
B
7. (i) y
3 4
–1
–3 1 2
–2
–4 0
x
4
1
–2
2
–1
3
5
6
7
R
Q
P
y = 3
(ii) The equation of the line of symmetry is y = 3.
114

1
8. (a) The letters with a vertical line of symmetry are M, U,
I and A.
(b) The letters with horizontal line of symmetry are I and
C.
(c) The letter I has two lines of symmetry.
(d) The letters S and L are not symmetrical.
(e)
9. (a) The figure has rotational symmetry of order 4.
(b) The figure has rotational symmetry of order 3.
(c) The figure has rotational symmetry of order 5.
(d) The figure has infinite rotational symmetry.
(e) The figure has rotational symmetry of order 8.
(f) The figure has infinite rotational symmetry.
(g) The figure has rotational symmetry of order 2.
10. (i) There are infinite planes of symmetry.
(ii) There is 1 axis of rotational symmetry.
(iii) The pencil has infinite rotational symmetry.
115

1 116
Revision Test C1
AB2
+ BC2
= AC2
52
+ BC2
= 132
BC2
= 132
– 52
= 144
BC = 12 cm
Area of ABC =
1
2
(5)(12)
= 30 cm2
PR2
= PQ2
+ QR2
= 32
+ 32
= 18
RS2
= PR2
+ PS2
l = 18 + 32
= 27
3. (i) tan 62° =
AB
46
AB = 46 tan 62°
= 86.5 m (to 3 s.f.)
Height of building is 86.5 m
(ii) tan 64° =
+
AB BC
46
AB + BC = 46 tan 64°
BC = 46 tan 64° – 46 tan 62°
= 7.80 m (to 3 s.f.)
Height of flag pole is 7.80 m
4. Volume of sphere =
4
3
p(13.5)3
Volume of cone =
1
3
p(4.5)2
(6)
Number of cones =
π
π(4.5)
4
3
(13.5)
1
3
(6)
3
2
= 81
5. (a) (i) Using Pythagoras’ Theorem,
PS2
+ SR2
= PR2
PS2
+ 52
= 132
PS2
= 132
– 52
= 144
PS = 12 cm
PQ2
= PS2
+ QS2
= 144 + 92
= 225
PQ = 15 cm
(iii) Area of PQS =
1
2
(9)(12)
= 54 cm2
(b)
1
2
(13)(QT) =
1
2
(14)(12)
QT =
×
14 12
13
= 12
12
13
cm (shown)
6. (a) (i) sin 55°=
PT
6
PT = 6 sin 55°
= 4.91 cm (to 3 s.f.)
(ii) cos 55° =
UR
6
UR = 6 cos 55°
= 3.44 cm (to 3 s.f.)
(iii) tan /PST =
6 sin 55°
3
/PST = 58.6° (to 1 d.p.)
(iv) cos /PST =
PS
3
PS =
PST
3
cos
= 5.76 cm (to 3 s.f.)
(b) Using Pythagoras’ Theorem,
TR2
+ PT2
= PR2
TR2
+ (6 sin 55°)2
= 12.02
TR2
= 12.02
– (6 sin 55°)2
TR = 10.95 cm (to 4 s.f.)
TU = TR – UR
= 10.95 – 6 cos 55°
= 7.509 cm (to 4 s.f.)
Area of PQUT = (7.509)(6 sin 55°)
= 36.9 cm2
(to 3 s.f.)
7. (i) In ABC,
cos q =
12
24
=
1
2
q = 60°
AC2
+ 122
= 242
AC2
= 432
AC = 432 cm
In ACD,
sin 60° =
AD
432
AD = 432 sin 60°
= 18 cm

1
117
In ADE,
cos 60° =
ED
18
ED = 18 cos 60°
= 9 cm
(ii) tan 60° =
AE
9
AE = 9 tan 60°
= 15.6 cm (to 3 s.f.)
8. (i)
+
x
x 5
=
+
8
8 6
=
8
14
14x = 8x + 40
6x = 40
x = 6
2
3
y
4
=
+
8 6
8
y =
14
8
× 4
= 7
x = 6
2
3
, y = 7
(ii) AP2
+ PQ2
= 11
2
3






2
+ 72
= 185
1
9
AQ2
= 142
= 196
Since AP2
+ PQ2
≠ AQ2
,
APQ is not a right-angled triangle.
9. (i) Volume =
2
3
p(5)3
+ p(5)2
(15) +
1
3
p(5)2
(12)
=
1675
3
p
= 1750 cm3
(to 3 s.f.)
(ii) Let the slant height of the cone be l cm.
l = 122
+ 52
= 13
Cost = 1.4[2p(5)2
+ 2p(5)(15) + p(5)(13)]
= 371p
= $1165.53 (to 2 d.p.)
10.
Lines of symmetry: 3
Order of rotational symmetry: 3
11.
A hemisphere has only one axis of rotational symmetry.

1 118
Revision Test C2
1. 2 tan q + 3 cos q
= 2
3
4





 + 3
4
5






= 3
9
10
x2
+ 23.42
= 32.72
x2
= 32.72
– 23.42
= 521.73
x = 22.84 m (to 4 s.f.)
Perimeter = 2(23.4 + 22.84)
= 92.5 m (to 3 s.f.)
(ii) Area = (23.4)(22.84)
= 534 m2
(to 3 s.f.)
x2
+ x2
= 34.22
2x2
= 34.22
x2
= 584.82
x = 24.18 cm (to 4 s.f.)
Length of ribbon = 4(24.18)
= 96.7 cm (to 3 s.f.)
PN2
+ NR2
= PR2
102
+ NR2
= 262
NR2
= 262
– 102
= 576
NR = 24 cm
(ii) sin /QRP =
10
26
=
5
13
/QRP = 22.6° (to 1 d.p.)
(iii) cos 34° =
PQ
10
PQ =
10
cos 34°
= 12.1 cm (to 3 s.f.)
(iv) tan 34° =
NQ
10
NQ = 10 tan 34°
= 6.75 cm (to 3 s.f.)
(v) Area of PQR =
1
2
(10 tan 34° + 24)(10)
= 154 cm2
5. tan 34° =
d
123
1
d1 =
123
tan 34°
tan 49° =
d
123
2
d2 =
123
tan 49°
d = d1 + d2
=
123
tan 34°
+
123
tan 49°
= 289 m (to 3 s.f.)
AC2
= AB2
+ BC2
= 112
+ 152
= 346
AC = 346 cm
1
2
346
( )(KB) =
1
2
(11)(15)
KB =
×
11 15
346
= 8.87 cm (to 3 s.f.)
(ii) cos /KBC =
KB
15
=
8.870
15
/KBC = 53.7° (to 1 d.p.)
7. (i) Total volume = 100 ×
4
3
p(1.2)3
+ 2000
= 2720 cm3
(to 3 s.f.)
(ii) Number of cups =
2724
π(4)2
(8)
= 6.77
≈ 7 (round up to the nearest integer)
8. (i) Volume =
1
3
(1.8 × 1.6)(1.1)
= 1.1 m3
(to the nearest 0.1 m3
)
VB2
= 1.12
+ 1.62
= 3.77
VB = 3.77 m
AC2
= 1.62
+ 1.82
= 5.8
VC2
= 5.8 + 1.12
= 7.01
VC = 7.01 m

1
119
VD2
= 1.12
+ 1.82
= 4.45
VD = 4.45 m
Sum of lengths = 3.77 + 7.01 + 4.45
= 6.70 m (to 3 s.f.)
9. (i) Capacity =
1
3
p(3.5)2
(2.1) + p(3.5)2
(4.9) + (1)(1)(1.5)
+
1
3
(1)2
(0.9)
= 217 m3
(to 3 s.f.)
(ii) Total area = p(3.5) +
3.5 2.1
2 2
+ 2p(3.5)(4.9)
+ p(3.5)2
– (1)2
+ 4(1)(1.5)
+ 4
1
2





 (1) +
0.5 0.9
2 2
= 198 m2
(to 3 s.f.)
10. (a) T, I and A
(b) I and E
(c) I
(d) R, S, N, G, L
(e)
11.
A square pyramid has only one axis of rotational symmetry
and rotational symmetry of order 4.

1 120
Chapter 14 Sets
Basic
1. (a) Yes, because it is clear if a pupil has no siblings.
(b) No, because a bag may be considered nice by some
but not to others.
(c) No, because a singer may be considered attractive to
some, but not others.
(d) No, because a song may be well-liked by some, but
not others.
(e) Yes, because it is clear whether a teacher teaches Art.
(f) No, because a move may be considered funny to some,
but not others.
2. (a) A  B′ = {a, b, c, x, y, m, n}
(b) A′  B′ = {m, n}
(c) A  B′ = {a, b, c}
3. (a) A  B′ = {1, 2, 3, 4, 5, 7, 8}
(b) A′  B′ = {4, 8}
(c) A  B′ = {1, 2, 7}
4. (a) A′  B′
A B

(b) A  B′
A B

5. (a) T
(b) T
(c) T
(d) F
(e) F
(f) T
(g) T
(h) T
(i) F
(j) F
6. (a) T
(b) T
(c) F
(d) T
(e) T
(f) F
(g) T
(h) F
7.  = {x : x is an integer, 1  x  14} = {1, 2, 3, …, 13, 14}
P = {x : x is a prime number} = {2, 3, 5, 7, 11, 13}
Q = {x : x is a factor of 12} = {1, 2, 3, 4, 6, 12}
(a) P
2
3
10
1
6
9
8
5 7
11 13
14
12
4
Q

(b) (i) P  Q′ = {2, 3, 5, 7, 8, 9, 10, 11, 13, 14}
(ii) P′  Q′ = {8, 9, 10, 14}
8. (a) A
5
15
10
6
3
9
8
13
4
B

(b) (i) (A  B)′ = {3, 4, 6, 8, 9, 13}
(ii) A′  B = {6, 9}
9.  = {x : x is an integer, 1  x  12} = {1, 2, 3, …, 11, 12}
A = {x : x is a prime number} = {2, 3, 5, 7, 11}
B = {x : x is a multiple of 3} = {3, 6, 9, 12}
(i) A
4
9
1
6
2
8
3
5
7
10
11 12
B

(ii) A  B′ = {2, 5, 7, 11}
10. (a) B  A′
(b) B′

1
121
Intermediate
11. (a) A  B = {e, x}
(b) A  C′ = {a, b, e, d, x, h, m, y, z}
(c) B  A′ = {e, x, h, m, n, k, y, z}
(d) B′  C′ = {a, b, y, z}
(e) A  B  C = {e, x, d, k, n}
12.  = {polygons}
A = {quadrilaterals}
B = {regular polygons}
(a) square or rhombus
(b) rectangle or parallelogram
13.  = {x : x is an integer, 12  x  39}
= {12, 13, 14, …, 37, 38, 39}
A = {x : x is a multiple of 5} = {15, 20, 25, 30, 35}
B = {x : x is a perfect square} = {16, 25, 36}
C = {x : x is odd} = {13, 15, 17, …, 35, 37, 39}
(a) A  B = {25}
(b) A  C = {15, 25, 35}
(c) B  C = {13, 15, 16, 17, 19, 21, 23, 25, 27, 29, 31,
33, 35, 36, 37, 39}
14.  = {x : x is an integer}
A = {x : x 4}
B = {x : −1 x  10 }
C = {x : x 8}
(a) A  B = {x : 4 x  10 }
(b) B  C = {x : −1 x 8}
(c) A′  B = {x : −1 x  4 }
(d) A′  C = {x : x  4 }
15.  = {x : x is an integer, 0  x 25}
= {0, 1, 2, 3, …, 23, 24}
B = {x : x is divisible by 5} = {0, 5, 10, 15, 20}
C = {x : x is prime and x  19}
= {2, 3, 5, 7, 11, 13, 17, 19}
16.  = {x : x is an integer, 0 x  13}
= {1, 2, 3, …, 11, 12, 13}
A = {x : 2x 9}
B = {x : (x − 2)(x − 5) = 0}
C = {x : x is prime}
(a) C = {5, 6, 7, 8, 9, 10, 11, 12, 13}
(b) C = {2, 5}
(c) C = {1, 3, 5, 7, 11, 13}
A  C = {5, 7, 11, 13}
17. c
18.  = {x : x is whole number and x  20}
A = {2, 4, 6, 8, 10, 12}
B = {1, 4, 9, 16}
(a) A  B′ = {2, 6, 8, 10, 12}
(b) A′  B = {1, 9, 16}
(c) A′  B′ = {3, 5, 7, 11, 13, 14, 15, 17, 18, 19, 20}
(d) A′  B′ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,
14, 15, 16, 17, 18, 19, 20}
19.  = {(x, y) : x and y are integers}
P = {(x, y) : 0 x  3 and 0  y 6}
Q = {(x, y) : 2  x 8 and 5  y  9}
P  Q = {(x, y) : 2  x  3 and 5  y 6}
x = 2, 3 and y = 5
∴ P  Q = {(2, 5), (3, 5)}
20.  = {a, b, c, d, e, f, g}
A = {a, c, f, g}
B = {a, c, g}
C = {b, c, e, f}
(i) (A  B)′ = {b, d, e, f}
(ii) A  C′ = {a, c, d, f, g}
21.  = {all triangles}
A = {isosceles triangles}
B = {equilateral triangles}
C = {right-angled triangles}
(a) A  B = A
(b) B  C = ∅
(c) A  B = B
22. (a) A = B
(b)
P Q
R

23.  = {x : x is an integer}
A = {x : 20 x  32}
B = {x : 24  x  37}
(a) A  B = {x : 24  x  32 }
= {24, 25, 26, 27, 28, 29, 30, 31, 32}
(b) A  B = {x : 20 x  37}
= {21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32,
33, 34, 35, 36, 37}
24.  = {x : x is an integer, 4  x  22} = {4, 5, 6, …, 22}
A = {x : x is a multiple of 5} = {5, 10, 15, 20}
B = {x : x is a prime number} = {5, 7, 11, 13, 17, 19}
C = {x : x is a factor of 30} = {5, 6, 10, 15}
(a) A  C = {5, 6, 10, 15, 20}
(b) B  C = {5}

1 122
25.  = {6, 8, 10, 12, 13, 14, 15, 16, 18, 20, 21}
B = {x : 2x 33} = {6, 8, 10, 12, 13, 14, 15, 16}
A  B = {6, 8, 10, 12, 13, 14, 15, 16, 18, 21}
26.  = {x : x is a natural number, 2  x  15}
= {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
B = {x : x is even} = {2, 4, 6, 8, 10, 12, 14}
A′  B = {2, 4, 8, 10, 14}
27.  = {x : x is a positive integer}
A = {x : 7 3x 28} = {3, 4, 5, 6, 7, 8, 9}
B = {x : 3 2x + 1 25 } = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
C = {x : 1
x
2
 9}
= {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18}
28. (a) A′  B = B
(b) A  B′ = B′
29.  = {x : x is a positive integer and 20  x  90}
A = {x : x is a multiple of 3}
= {21, 24, 27, 30, 33, 36, ..., 90}
B = {x : x is a perfect square} = {25, 36, 49, 64, 81}
C = {x : unit digit of x is 1} = {21, 31, 41, 51, 61, 71, 81}
(i) A  B = {36, 81}
(ii) A  C = {21, 51, 81}
30.  = {x : x is a positive integer and 0  x  24}
A = {x : x is a prime number}
= {2, 3, 5, 7, 11, 13, 17, 19, 23}
B = {x : 12 3x 37} = {5, 6, 7, 8, 9, 10, 11, 12}
A  B = {5, 7, 11}
31. (a) P  Q = P
(b) Q  P′ = ∅
32. (a) A  B = A
(b) A  B = B
33.  = {integers}
A = {factors of 4} = {1, 2, 4}
B = {factors of 6} = {1, 2, 3, 6}
C = {factors of 12} = {1, 2, 3, 4, 6, 12}
D = {factors of 9} = {1, 3, 9}
(a) A  B = {1, 2, 3, 4, 6}
(b) B  C = {1, 2, 3, 6}
(c) C  D = {1, 3}
Advanced
34.  = {polygons}
A = {polygons with all sides equal}
B = {polygons with all angles equal}
C = {triangles}
D = {quadrilaterals}
(a) A  C = equilateral triangle
(b) A  D = rhombus
(c) B  D = square or rectangle
35.
B C
A

36.  = {x : x is an integer less than 22}
A = {x : x is a prime number less than 20}
= {2, 3, 5, 7, 11, 13, 17, 19}
B = {x : a x b}
For A  B = ∅,
8 x 10 or 14 x 16
∴ a = 8, b = 10 or a = 14, b = 16.
37. A = {(x, y) : x + y = 4}
B = {(x, y) : x = 2}
C = {(x, y) : y = 2x}
(a) A  B = {(x, y) : x = 2, y = 2} = {(2, 2)}
(b) B  C = {(x, y) : x = 2, y = 4} = {(2, 4)}
(c) A  C = {(x, y) : x + y = 4, y = 2x}
= (x, y) : x = 1
1
3
, y = 2






= 1
1
3
, 2
2
3













1
123
New Trend
38.  = {x : x is an integer, 30 x  40}
= {31, 32, 33, …, 39, 40}
A = {x : x is a multiple of 3} = {33, 36, 39}
B = {x : 2x – 4 73} = {31, 32, 33, 34, 35, 36, 37, 38}
(i) A′  B = {31, 32, 34, 35, 37, 38}
(ii) A
33
36
39
40
34
31
37
35
32
38
B

(iii)
A B

39.  = {x : x is an integer, 0  x 12}
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
A = {x : x(x − 5) = 0} = {0, 5}
B = {x :
1
3
x − 1 3
1
3
}
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
(a) A
0
5
1
6
2 3
9
8
7
11
10
4
B

(b) (i) A  B = {0, 5}
(ii) A  B′ = {0, 5}
40.  = {1, 2, 3, 4, 5, …,19}
(i) A = {x : x is prime} = {2, 3, 5, 7, 11, 13, 17, 19}
(ii) C = {x : x is a factor of 12} = {1, 2, 3, 4, 6, 12}
(iii) B = {3, 6, 9, 12, 15, 18}
C′ = {5, 7, 8, 9, 10, 11}
∴ B  C′ = {9}
(iv) A  C = {1, 2, 3, 4, 5, 6, 7, 11, 12, 13, 17, 19}
∴ (A  C)′ = {8, 9, 10, 14, 15, 16, 18}
41.  = {x : x is a positive integer, 5 3x  28}
= {2, 3, 4, 5, 6, 7, 8, 9}
A = {x : x is a multiple of 3} = {3, 6, 9}
B = {x : x is divisible by 2} = {2, 4, 6, 8}
(i)
A B

6
2
5
8
3
9
4
7
(ii) (A  B)′ = {5, 7}
n(A  B)′ = 2
(iii) A  B′ = {3, 9}
42. R S
C
B A


1 124
Chapter 15 Probability of Single Events
Basic
1. (a) {A1, A2, C, E, H, I, M1, M2, S, T1, T2}
(b) (i) Probability of obtaining the letter ‘A’ =
2
11
(ii) Probability of obtaining the letter ‘H’ =
1
11
(iii) Probability of obtaining a vowel =
4
11
2. (a) {HH, HT, TH, TT}
(b) (i) Probability of obtaining two tails =
1
4
(ii) Probability of obtaining a head and a tail =
2
4
=
1
2
3. (i) Probability of getting an odd number =
3
6
=
1
2
(ii) Probability of getting a number less than 4 =
3
6
=
1
2
(iii) Probability of getting a ‘5’ or a ‘6’ =
2
6
=
1
3
(iv) Probability of getting a number which is not ‘6’ =
5
6
4. (i) Probability of drawing a number that is a multiple
of 3 =
5
8
(ii) Probability of drawing a prime number =
2
8
=
1
4
(iii) Probability of drawing a number whose digits have a
sum that is divisible by 2 =
3
8
5. (i) Probability of drawing a King =
4
52
=
1
13
(ii) Probability of drawing the King of diamonds =
1
52
(iii) Probability of drawing a heart =
13
52
=
1
4
(iv) Probability of drawing a picture card =
12
52
=
3
13
6. (i) Number of white pearls = 50 – 24 – 15 = 11
Probability of selecting a white pearl =
11
50
(ii) Probability that the pearl selected is not green
=
+
24 11
50
=
35
50
=
7
10
(iii) Probability of selecting a pink pearl = 0
7. (i) Probability that the month is December =
1
12
(ii) Probability that the month begins with the letter J
=
3
12
=
1
4
(iii) Probability that the month has exactly 30 days
=
4
12
=
1
3
8. (a) (i) Probability that the customer wins $88 cash
=
1
8
(ii) Probability that the customer wins a $10 shopping
voucher =
3
8
(iii) Probability that the customer wins a packet of
dried scallops = 0
(b) A pair of movie tickets and a can of abalone
9. (i) Angle corresponding to the sector representing beans
= 360° – 150° – 90° – 50°
= 70°
Probability that the student prefers beans =
70°
360°
=
7
36
(ii) Probability that the student prefers broccoli or carrots
=
° + °
°
90 50
360
=
140°
360°
=
7
18

1
125
10. (i) Probability that a bag selected has a mass of exactly
1 kg = 1 –
1
40
–
1
160
=
31
32
(ii) Number of bags each with a mass of less than 1 kg
=
1
160
× 8000
= 50
Intermediate
11. (i) Number of cards remaining = 13
Probability of drawing the Jack of diamonds =
1
13
(ii) Probability of drawing a King, a Queen or a Jack
=
12
13
(iii) Probability of drawing the ace of hearts or the King
of hearts =
2
13
(iv) Probability of drawing a joker = 0
12. (i) Number of slots = 37
Probability that the ball lands in the slot numbered 13
=
1
37
(ii) Prime numbers from 0 to 37: 2, 3, 5, 7, 11, 13, 17, 19,
23, 29, 31
Probability that the ball lands in the slot numbered
with a prime number =
11
37
(iii) Probability that the ball lands in the slot numbered
with a number less than 19 =
19
37
(iv) Probability that the ball lands in the slot numbered
with an odd number =
18
37
13. (i) Number of two-digit numbers = 90
Two-digit numbers greater than 87: 88, 89, 90, 91,
92, 93, 94, 95, 96, 97, 98, 99
Probability that the number generated is greater
than 87 =
12
90
=
2
15
(ii) Two-digit numbers less than 23: 10, 11, 12, 13, 14,
15, 16, 17, 18, 19, 20, 21, 22
Probability that the number generated is less than 23
=
13
90
(iii) Two-digit numbers divisible by 4: 12, 16, 20, 24, 28,
32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84,
88, 92, 96
Probability that the number generated is divisible by 4
=
22
90
=
11
45
(iv) Number of two-digit numbers between 55 and 72
inclusive = 18
Probability that the number is between 55 and 72
inclusive =
18
90
=
1
5
14. (i) Number of cards = 16
Probability of selecting a vowel =
7
16
(ii) Probability of selecting a letter which appears in the
word ‘SCIENCE’ =
6
16
=
3
8
(iii) Probability of selecting a letter which appears in the
word ‘SMART’ =
7
16
(iv) Probability of selecting a letter which appears in the
word ‘DUG’ = 0
15. (a) Number of cards = 11
(i) Probability that the card shows the letter ‘P’
=
1
11
(ii) Probability that the card shows the letter ‘E’
=
3
11
(iii) Probability that the card shows a vowel or a
consonant = 1
(b) Number of cards = 10
(i) Probability that the card shows the letter ‘P’
=
1
10
(ii) Probability that the card shows the letter ‘E’
=
2
10
=
1
5
(iii) Probability that the card shows a vowel =
3
10

1 126
16. (a) Number of students = 210
(i) Probability of selecting a Secondary 1 student
=
+
22 38
210
=
60
210
=
2
7
(ii) Probability of selecting a girl
=
+ + +
38 25 35 22
210
=
120
210
=
4
7
(iii) Probability of selecting an upper secondary
student =
+ + +
25 35 24 22
210
=
106
210
=
53
105
(iv) Probability of selecting a Secondary 2 student
who is a boy =
19
210
(b) (i) Probability of selecting a Secondary 3 student
who is a girl =
38
215
(ii) Probability of selecting a Secondary 2 student or
a Secondary 4 student =
+ + +
21 25 24 22
215
=
92
215
17. Probability that it is labelled Gold = 1 –
1
5
–
1
4
=
11
20
Total number of boxes = 55 ÷
11
20
= 100
18. (i) Number of medical staff =
1
5
× 30 – 2
= 4
(ii) Number of footballers = 30 – 4 – 2
= 24
Number of midfielders =
3
8
× 24
= 9
Number of goalkeepers =
1
3
× 9
= 3
Number of forwards = 24 – 3 – 7 – 9
= 5
Probability of selecting a forward from the contingent
=
5
30
=
1
6
19. (a) (i) Probability of selecting a vowel =
1
7
(ii) Probability of selecting a card that bears the letter
C =
3
7
(b)
+ x
3
7
=
1
7
21 = 7 + x
x = 14
20. (a) (i) Probability that the mark is less than 44
=
8
15
(ii) Probability that the mark is not a prime number
=
14
15
(iii) Probability that the mark is divisible by 11
=
3
15
=
1
5
(b) Probability that the student obtained the badge
=
9
15
=
3
5
(c) Probability that the mark was 39 =
2
6
=
1
3

1
127
21. (i) Experimental probability of obtaining a ‘1’ =
2
20
=
1
10
Experimental probability of obtaining a ‘2’ =
4
20
=
1
5
4
20
=
1
5
3
20
4
20
=
1
5
3
20
(ii) No. As the number of rolls increases, the experimental
probability of an outcome occurring tends towards the
theoretical probability of the outcome happening






i.e.
1
6
.
22. (i)
+
x
x
35
=
1
6
6x = 35 + x
5x = 35
x = 7
(ii) Probability of selecting a sports car =
+
+ +
35 5
35 7 5
=
40
47
23.
+ +
+ + + +
x
x x
12 2
36 12 2 2
= 0.3
+
+
x
x
14
3 50
= 0.3
x + 14 = 0.9x + 15
0.1x = 1
x = 10
24. (i)
+
x
x
18
=
3
5
5x = 54 + 3x
2x = 54
x = 27
(ii) Probability of selecting a pink sweet
=
+ + +
15
18 27 10 15
=
15
70
=
3
14
Advanced
25. (i) Number of elements of S = 50
Integers in S that are not divisible by 2 or 3: 1, 5, 7,
11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49
Probability that the element is not divisible by 2 or 3
=
17
50
(ii) Number of elements that contain the digit ‘2’ at least
once: 2, 12, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 32,
42
Probability that the element contains the digit ‘2’ at
least once =
14
50
=
7
25
26. (i)
Smoke Do not smoke Total
Male 18 42 60
Female 8 32 40
Total 26 74 100
(ii) Probability that a randomly selected smoker is male
=
18
26
=
9
13
(iii) The respondents of this online survey may not be a
good representation of the country’s population.
New Trend
27. (a) Probability of selecting a red chip =
10
24
=
5
12
(b) Let x be the number of extra green chips added,
6+ x
24 + x
=
1
3
18 + 3x = 24 + x
2x = 6
x = 3
∴3 green chips must be placed in the bag so that the
probability of choosing a green chip would be
1
3
.

1 128
Chapter 16 Statistical Diagrams
Basic
1. (i) Most common length = 7 cm
(ii) Length of longest fish = 9 cm
(iii) Percentage of fish which have lengths of more
than 6 cm =
10
20
× 100%
= 50%
2. (i)
0 1 2 3 4 5 6
(ii) Most common number of universities = 3
(iii) Probability that the student has not applied to a
university =
3
20
3. (i) Total number of people = 29
(ii) Most common duration = 20 minutes
(iii) Percentage of people who take less than half an hour
=
17
29
× 100%
= 58.6% (to 3 s.f.)
4. Stem Leaf
1
2
3
5
1
0
6
4
5
8
7 8 9
Key: 1 | 5 means 15 ohms
5. (i) Take two points on the line and draw dotted lines to
form the right-angled triangle.
Vertical change (or rise) = 7.6 – 3.8
= 3.8
= 4
Gradient = −
rise
run
=
−
3.8
4
=
−0.950 (to 3 s.f.)
(ii) y-intercept = 9.5
∴ The equation of the line of best fit is
y = –0.950x + 9.5.
(iii) Extrapolating the line of best fit, we see that 1900
people will visit the gallery 8 years after its opening.
(iv) It would be unreliable since year 8 lies outside of the
range between year 0 and year 6.
6. (a) The data shows strong, negative correlation.
(b) The data shows strong, positive correlation.
(c) The data shows no correlation.
7. (a) The data shows strong, positive correlation.
y
8
10
14
18
4
6
2
12
16
20
8
5 6 7
2 3 4
1
0 10 12
9 11
x
(b) The data shows strong, negative correlation.
y
8
10
14
18
4
6
2
12
16
20
22
24
26
16
10 12 14
4 6 8
2
0 20 24
18 22
x

1
129
(c) The data shows no correlation.
y
8
10
14
18
4
6
2
12
16
20
8
5 6 7
2 3 4
1
0 10 12
9 11
x
8. (i)
5 10 15 20 25
1
2
3
4
5
6
7
8
9
10
11
Number of cases, n
Frequency
0
(ii) Number of days = 3 + 1
= 4
9. (a) The 4th
day had the greatest number of employees
report sick. 35 workers reported sick.
(b) The 10th
day had the least number of employees report
sick. 13 workers reported sick.
(c) The number of employees who reported sick was more
than 30 on the 4th
and 8th
day.
10.
2
1 3 4
1
2
3
4
5
6
7
8
9
10
Distance (x km)
Frequency
density
0
6
5 7 8 9 10 11 12
11. Since the class intervals are unequal, the histogram is to
be drawn using either height of rectangle or frequency
density.
Weekly
earnings ($)
Class width Frequency
Rectangle’s
height
180  x 185 5 1×standard 4 4 ÷ 1 = 4
185  x 190 5 1×standard 6 6 ÷ 1 = 6
190  x 200 10 2×standard 8 8 ÷ 2 = 4
200  x 210 10 2×standard 18 18 ÷ 2 = 9
210  x 225 15 3×standard 18 18 ÷ 3 = 6
225  x 230 5 1×standard 6 6 ÷ 1 = 6
230  x 235 5 1×standard 8 8 ÷ 1 = 8
Weekly earnings ($)
185
180 190
1
2
3
4
5
6
7
8
9
10
Height
of
rectangle
0
200 210 225 230 235

1 130
12. (a) Total number of cars
= 60 + 56 + 86 + 150 + 60 + 105 + 60 + 48
= 625
(b) Since the class intervals are unequal, the histogram
is to be drawn using either height of rectangle or
frequency density.
Class
interval
Rectangle’s
height
5 – 24 20 2 × standard 60 60 ÷ 2 = 30
25 − 59 35 3.5 × standard 56 56 ÷ 3.5 = 16
60 − 79 20 2 × standard 86 86 ÷ 3 = 43
80 – 104 25 2.5 × standard 150 150 ÷ 2.5 = 60
105 – 114 10 1 × standard 60 60 ÷ 1 = 22
115 − 129 15 1.5 × standard 105 105 ÷ 1.5 = 70
130 − 149 20 2 × standard 60 60 ÷ 2 = 30
150 − 189 40 4 × standard 48 48 ÷ 4 = 12
Length of stay (min)
25
5
50
30
10
60
40
20
70
Height
of
rectangle
0
60 80 105
115
130 150 190
Intermediate
13. (i) Total number of children who participated in the
survey = 23
(ii) Greatest number of children in a family = 6 + 1
= 7
(iii) Average number of children in a family
=
× + × + × + × + × + ×
5 1 8 2 4 3 3 4 2 5 1 7
23
=
62
23
= 2.70 (to 3 s.f.)
(iv) Number of children with fewer than 2 siblings = 13
+ k
13
23
=
13
25
23 + k = 25
k = 2
14. (i) Fraction of people in Group 1 =
2
12
=
1
6
Fraction of people in Group 2 =
11
12
(ii) Group 1 consists of healthy human beings because a
large proportion of the people do not have to undergo
a blood test.
15. (i) Total number of boys = 56
(ii) Most common mass = 63 kg
(iii) Number of boys who have to gain mass = 18
Number of boys who have to lose mass = 16
∴ Ratio is 18 : 16 = 9 : 8
16. (i)
Stem Leaf
1
1
2
2
3
3
4
0
7
1
5
3
5
0
3
8
2
6
6
4
4
4
6
6
6 7 9
Key: 1 | 0 means 10
(ii) The most common number of smartphone applications
downloaded last month is 26.
(iii) Percentage of people =
9
20
× 100%
= 45%
17. (i) Publishing House A: 46 hours
Publishing House B: 48 hours
(ii) Publishing House A:
10
17
× 100% = 58.8% (to 3 s.f.)
Publishing House B:
10
18
× 100% = 55.6% (to 3 s.f.)
18. (i)
Leaves for Factory A Stem Leaves for Factory B
8 7
8
5
3
8
7
5
3
3
7
4
4
2
3
2
0
0
1
0
43
44
55
66
77
88
3
5
3
1
2
2
7
6
7
5
3
6
7
5
5
9
9
6
9
6
Key: 43 | 3 means 433 hours
(ii) Factory A produces longer-lasting light bulbs as there
are more light bulbs with durations of more than
770 hours.

1
131
19. (i)
Leaves for test scores
before the remedial
Stem
Leaves for test scores
after the remedial
9 8 7
8
7
8
7
6
5
5
6
6
1
4
3
6
3
0
2
0
3
4
5
6
7
8
9
0
2
1
0
0
6
3
3
2
0
7
8
2
2
9
5
9
7 9
Key: 3 | 9 means 39
(ii) Yes, it is effective because the test scores after the
remedial are generally higher than those before the
remedial.
20. (a)
20
10
30
40
50
60
70
4 7
3
1 2
0 5 8
6
Hours
Distance
travelled
(km)
(b) The line of best fit is drawn passing through as many
points as possible and as close as possible to all the
other points.
(c) Using the line of best fit on the scatter diagram, the
hiker travels 48 km in 6.5 hours.
(d) Take two points on the line and draw dotted lines to
Vertical change (or rise) = 58 – 2.5
= 55.5
Horizontal change (or run) = 7.9 – 0
= 7.9
Gradient =
55.5
7.9
=
7 (to nearest whole number)
y-intercept = 2.5
∴ The equation of the line of best fit is y = 7x + 2.5.
(e) The data displays strong, positive correlation.
21. (i)
y
100
50
150
200
250
300
160
150
130 140
0 170
x
Height (cm)
Pocket
money
(cents)
(ii) The data displays no correlation.
(iii) Since there is no correlation between the heights of
pupils and the amount of pocket money they receive,
we cannot use the graph to predict the amount of
pocket money that a child of height 147 cm will
receive.

1 132
22. (i)
Age of patient, x years Frequency
10  x 20 85
20  x 30 117
30  x 40 38
40  x 50 24
50  x 60 18
60  x 70 16
Total frequency 300
(ii) Percentage of patients who are at least 50 years old
=
+
18 16
300
× 100%
= 11.3% (to 3 s.f.)
(iii) No. The actual ages of the patients in the interval
20  x 30 are not known, so it is incorrect for Priya
to assume that al the patients in this interval are
+
20 30
2
= 25 years old.
23. (i)
2
14
40 50 60 70 80 90 100 110
4
16
6
18
8
20
10
22
26
12
24
28
Number of cases, n
Number
of
days
0
(ii) No, the most number of cases occur in the interval
70  n 80, but it is not correct to take the mid-value
of this interval.
24. (i)
2
20 30 40 50 60 70 80 90
4
6
8
10
12
Number of staff, x
Number
of
events
0
(ii) Number of events = 45% × 40
= 18
∴ p = 60
25. (i)
Age of crew, x years Frequency Frequency density
25  x 30 2 0.4
30  x 35 4 0.8
35  x 45 17 1.7
45  x 50 8 1.6
50  x 55 6 1.2
55  x 60 3 0.6
0.2
1.4
25 30 35 45 50 55 60
0.4
1.6
0.6
1.8
0.8
1.0
1.2
Age of crew, x years
Number
of
crew
0
(ii) Number of crew = 0.85 × 40
= 34
∴p = 35
26. (i) pH values, x Tally Frequency
6.5  x 7.0 //// 4
7.0  x 7.5 /// 3
7.5  x 8.0 //// /// 8
8.0  x 8.5 //// /// 8
8.5  x 9.0 // 2
9.0  x 9.5 //// 5
Total frequency 30

1
133
(ii)
1
7
6.5 7.0 7.5 8.0 8.5 9.0 9.5
2
8
3
4
5
6
pH values, x
Frequency
0
(iii) There are many distinct values in the set of data. Using
a histogram for grouped data would be more suitable.
(iv) Percentage of the types which are alkaline
=
26
30
× 100%
= 86.7% (to 3 s.f.)
27. (a)
5
40 50 60 70 80 90 100 110
10
15
20
25
30
Mass (kg)
Number
of
members
0
(b)
Mass, (x kg) Mid-value Frequency
40 x  50 45 7
50 x  60 55 10
60 x  70 65 14
70 x  80 75 27
80 x  90 85 12
90 x  100 95 6
100 x  110 105 4
The points to be plotted are (35, 0), (45, 7), (55, 10),
(65, 14), (75, 27), (85, 12), (95, 6), (105, 4) and (115, 0).
y
15
10
5
25
20
30
105
95
85
75
45
35 65
55
0 115
x
Mass (kg)
Frequency
28. (a) Since the class intervals are unequal, the histogram
is to be drawn using either height of rectangle or
frequency density.
Class
interval
Rectangle’s
height
10 − 29 20 2×standard 32 32 ÷ 2 = 16
30 − 39 10 1×standard 38 38 ÷ 1 = 38
40 − 49 10 2×standard 64 64 ÷ 1 = 64
50 − 59 10 2×standard 35 35 ÷ 1 = 35
60 − 69 10 1×standard 22 22 ÷ 1 = 22
70 − 99 30 3×standard 9 9 ÷ 3 = 3
y
20
10
30
40
50
60
70
70
50 60
40
10 30
0 100
x
Mass (kg)
Height
of
rectangle

1 134
(b) The points to be plotted are (5, 0), (20, 16), (35, 38),
(45, 64), (55, 35), (65, 22), (85, 9) and (105, 0).
y
20
10
30
40
50
60
70
70
50 60
40
10 30
0 100
x
Mass (kg)
Frequency
29. (a)
Height, (x cm) Number of plants
0 x  20 0.4 × 20 = 8
20 x  30 11
30 x  40 9
40 x  45 8
45 x  50 2.2 × 5 = 11
50 x  60 2.1 × 10 = 21
60 x  70 1.2 × 10 = 12
(b) Number of plants = 8 + 11 + 9 + 8 + 11 + 21 + 12
= 80
(c) The points to be plotted are (0, 0), (10, 0.4), (25, 1.1),
(35, 0.9), (42.5, 1.6), (47.5, 2.2),
(55, 2.1), (65, 1.2) and (70, 0).
2
2.5
1.5
1
0.5
40 70
30
10 20
0 50 60
Height (x cm)
Frequency
density Advanced
30. (i)
50 100 150 200 250
20
2
40
34
66
69
60
195
80
100
120
140
160
180
200
PSI value, x, in 2012
0
50 100 150 200 250
20
1
40
32
60
190
80
100
120
140
160
180
200
PSI value, x, in 2013
0
48
94
(ii) The measures taken have been effective in improving
the air quality as the PSI values in 2013 are generally
lower than those in 2012.

1
135
Chapter 17 Averages of Statistical Data
Basic
1. (a) 11, 11, 12, 13, 16
Mean =
+ + + +
11 11 12 13 16
5
= 12.6
Median = 12
Mode = 11
(b) 11, 12, 18, 18, 20, 20, 20, 24, 29, 41
Mean =
+ + + +
+ + + + +
11 12 18 18 20
20 20 24 29 41
10
= 21.3
Median =
+
20 20
2
= 20
Mode = 20
(c) 10.5, 12.6, 12.6, 13.5, 14.3, 15.3, 16.0, 16.4
Mean =
+ + +
+ + + +
10.5 12.6 12.6 13.5
14.3 15.3 16.0 16.4
8
= 13.9
Median =
+
13.5 14.3
2
= 13.9
Mode = 12.6
(d) 7, 8.1, 8.1, 8.1, 9.4, 9.4, 9.6, 10.4, 10.5, 11, 11.7
Mean =
+ + + + +
+ + + + +
7 8.1 8.1 8.1 9.4 9.4
9.6 10.4 10.5 11 11.7
11
= 9.39 (to 3 s.f.)
Median = 9.4
Mode = 8.1
2. 35, 36, 38, 38, 38, 39, 39, 40, 42, 43, 45, 45, 45, 45, 47
(i) Mean =
+ + + + + + +
+ + + + + + +
35 36 38 38 38 39 39 40
42 43 45 45 45 45 47
15
= 41
(ii) Mode = 45
(iii) Median = 40
3. Mean =
+ + + + + + + x
3 7 13 14 16 19 20
8
=
+ x
92
8
Median =
+
14 16
2
= 15
Since mean = median,
+ x
92
8
= 15
92 + x = 120
x = 28
4. (i) Total number of seeds = 100 × 5
= 500
(ii) Number of seeds that germinated
= 30 × 1 + 25 × 2 + 20 × 3 + 10 × 4 + 5 × 5
= 205
Fraction of seeds that germinated =
205
500
=
41
100
(iii) Mean =
× + × + ×
+ × + × + ×
10 0 30 1 25 2
20 3 10 4 5 5
100
= 2.05
Median = 2
Mode = 1
5. (i)
Number of countries Frequency
0 7
1 9
2 7
3 4
4 2
5 1
Total frequency 30
(ii) Mean =
× + × + ×
+ × + × + ×
7 0 9 1 7 2
4 3 2 4 1 5
30
= 1.6
Median = 1
Mode = 1
Intermediate
6. Let the eighth number be x.
1, 2, 2, 4, x, 7, 8, 13
Median =
+ x
4
2
4.5 =
+ x
4
2
9 = 4 + x
x = 5
The eighth number is 5.
Mode = 2

1 136
7. Sum of the set of 12 nunbers = 12 × 5
= 60
Sum of the set of 8 numbers = 8a
Mean of combined set of 20 numbers =
+ a
60 8
20
8 =
+ a
60 8
20
160 = 60 + 8a
8a = 100
a =
100
8
= 12.5
8. (a) (i) Modal profit = $3 million
(ii) Median profit = $2 million
(b) Mean profit
=
× + × + × + × + ×
2 0 6 1 8 2 10 3 4 4
30
=
$2.27 million (to 3 s.f.)
Raj is incorrect.
9. (a) (i) 12 + 9 + x + 6 + y + 7 = 49
x + y + 34 = 49
x + y = 15 (shown)
(ii) Mean =
× + × + ×
+ × + × + ×
x
y
12 1 9 2 3
6 4 5 7 6
49
3
2
49
=
+ +
x y
96 3 5
49
149 = 96 + 3x + 5y
3x + 5y = 53 (shown)
(iii) x + y = 15 —(1)
3x + 5y = 53 —(2)
(1) × 3: 3x + 3y = 45 —(3)
(2) – (3): 2y = 8
y = 4
x + 4 = 15
x = 11
x = 11, y = 4
(b) (i) Mode = 1
(ii) Median = 3
(c) Let the number shown on the die be n.
Mean =
× + × + ×
+ × + × + × + n
12 1 9 2 11 3
6 4 4 5 7 6
50
3 =
+ n
149
50
150 = 149 + n
n = 1
The number shown on the die is 1.
10. Initial sum of eye pressure = 30 × 12.4
= 372 mm Hg
New sum of eye pressure = 30 × 12.6
= 378 mm Hg
Nora’s actual eye presssure = 8 + (378 – 372)
= 14 mm Hg
11. 62.0, 62.0, 62.6, 63.1, 63.7, 64.2, 64.3, 64.7, 65.1, 65.2,
65.2, 65.2, 65.5, 65.9, 66.8, 67.1, 67.4, 68.2
(a) (i) Mean =
62.0 62.0 62.6 63.1 63.7
64.2 64.3 64.7 65.1 65.2
65.2 65.2 65.5 65.9 66.8
67.1 67.4 68.2
18
+ + + +
+ + + + +
+ + + + +
+ + +
= 64.9 s
(ii) Mode = 65.2 s
(iii) Median =
+
65.1 65.2
2
= 65.15 s
(b) Percentage =
100
62.0
100
68.2
× 100%
= 110%
12. 1.6, 1.7, 1.8, 1.8, 1.8, 1.8, 1.9, 1.9, 1.9, 2.0, 2.0
(a) (i) Modal height = 1.8 m
(ii) Median height = 1.8 m
(iii) Mean height =
1.6 1.7 1.8 1.8 1.8 1.8
1.9 1.9 1.9 2.0 2.0
11
+ + + + +
+ + + + +
=
20.2
11
= 1.84 m (to 3 s.f.)
(b) Sum of heights of the first 11 boys = 20.2 m
Sum of heights of the 12 boys = 12 × 1.85
= 22.2 m
Height of the 12th
boy = 22.2 – 20.2
= 2.0 m
13. (a) 1 + 4 + 8 + x + 9 + y + 2 = 40
x + y + 24 = 40
x + y = 16 —(1)
× + × + × + ×
+ × + × + ×
x
y
1 0 4 1 8 2 3
9 4 5 2 6
40
= 3.2
+ +
x y
3 5 68
40
= 3.2
3x + 5y + 68 = 128
3x + 5y = 60 —(2)
(1) × 3: 3x + 3y = 48 —(3)
(2) – (3): 2y = 12
y = 6
x + 6 = 16
x = 10
x = 10, y = 6

1
137
(b) (i) Largest possible value of x = 8
(ii) Mean number of fillings
=
× + × + × + ×
+ × + × + ×
1 0 4 1 8 2 8 3
9 4 8 5 2 6
40
= 3.3
14. (i) Total number of pages = 1 + 3 + 10 + 7 + 4 + 3 + 2
= 30
(ii) Number of pages with fewer than 3 errors
= 1 + 3 + 10
= 14
Percentage of pages with fewer than 3 errors
=
14
30
× 100%
= 46.7% (to 3 s.f.)
(iii) Mode = 2
(iv) Mean =
× + × + × + ×
+ × + × + ×
1 0 3 1 10 2 7 3
4 4 3 5 2 6
30
= 2.9
15. (i) p = 7, q = 4, r = 4, s = 3, t = 1
(ii) Mean =
× + × + × + ×
+ × + × + ×
7 0 6 1 4 2 5 3
4 4 3 5 1 6
30
= 2.2
Median = 2
Mode = 0
(iii) Percentage of students who consume at least 5 servings
of fruit and vegetables on a typical weekday
=
4
30
× 100%
= 13.3% (to 3 s.f.)
Most of the students do not consume at least
5 servings of fruit and vegetables.
16. (i) Total number of days = 3 + 5 + 8 + 7 + 10 + 6 + 1
= 40
(ii) Mean number of security cameras sold
=
× + × + × + ×
+ × + × + ×
3 32 5 57 8 82 7 107
10 132 6 157 1 182
40
= 105.75
(iii) Median = 107
Mode = 132
The median gives a better comparison.
17. (i) Modal number of emergency calls received in
December = 49
(ii) Median number of emergency calls received in
October = 26
Median number of emergency calls received in
December = 37
Mean number of emergency calls received in October
=
+ + + + + +

4 4 4 41 44 45
31
= 23.4 (to 3 s.f.)
Mean number of emergency calls received in
December =
+ + + + + +

8 10 13 49 49 49
31
= 34.1 (to 3 s.f.)
(iii) More emergency calls were received per day in
December than in October.
18. (i) Mean mass ≈
× + × + ×
+ × + × + ×
32 20 38 35 64 45
35 55 22 65 9 85
200
= 44.85 kg
(ii) Probability that the steel bar requires another
transportation vehicle =
9
200
19. (a) Mean amount of medical claims
=
+ + + + + +

150 44 225 77 55 136
20
=
$70.40
(b)
Amount of medical claims, $m Frequency
0  m 50 8
50  m 100 7
100  m 150 3
150  m 200 1
200  m 250 1
Total frequency 20
(c) (i)
50 100 150 200 250
1
2
3
4
5
6
7
8
Amount of medical claims, $m
Frequency
0

1 138
(ii) Estimate for the mean amount of medical claims
=
× + × + ×
+ × + ×
8 25 7 75 3 125
1 175 1 225
20
=
$75
(d) There is a difference of $4.60 in the answers in (a)
and (c)(ii). The mean amount calculated in (a) is the
exact value as it is based on the individual values, but
the mean amount calculated in (c)(ii) is an estimate
as the mid-values of each interval are used.
20. Total number of vehicles along Section A = 50
The median average speed along Section A lies in the
interval 60  v 70.
Total number of vehicles along Section B = 49
The median average speed along Section B lies in the
interval 70  v 80.
As the actual data in these intervals is not known, it is
incorrect for Ethan to obtain the median average speed
along Section A by taking
+
60 70
2
= 65 km/h or to
obtain the median average speed along Section B by
taking
+
70 80
2
= 75 km/h.
Advanced
21. Total mass of the children
= 15 + 15 + 11 + 13 + 9 + 20 + 15 + a + 13 + 18
= 129 + a
Mean mass of the children =
+ a
129
10
Arrangement of the masses without a:
9, 11, 13, 13, 15, 15, 15, 18, 20
X X X X Y Z Z Z Z Z
Case 1: a lies at one of the points labelled X.
Median = 14
+ a
129
10
= 14 – 0.4
129 + a = 136
a = 7
Case 2: a lies at the point labelled Y.
Median =
+
a 15
2
+ a
129
10
=
+
a 15
2
– 0.4
129 + a = 5a + 75 – 4
4a = 58
a = 14.5
Case 3: a lies at one of the points labelled Z.
Median = 15
+ a
129
10
= 15 – 0.4
129 + a = 146
a = 17
a = 7 or a = 14.5 or a = 17
New Trend
22. Let the numbers be x, y, 60 and 60, such that x y.
Since the median is 56,
+
y 60
2
= 56
y + 60 = 112
y = 52
Since the mean is 54,
+ + +
x 52 60 60
4
= 54
x + 172 = 216
x = 44
The four numbers are 44, 52, 60 and 60.
23. (a) Difference = 100 − (−210)
= 310°C
(b) Mean boiling points =
2856+100+ −195.79
( )
3
=
2760.21
3
= 920.07°C (to 2 d.p.)
Mean melting points =
1064.18+0+ −210
( )
3
=
854.18
3
= 284.73°C (to 2 d.p.)

1
139
Revision Test D1
1. (a) A  B = {2, 4, 6, 8, 10, 12, 14}
(b) A  B = {10, 12}
(c) A  B = {8, 14}
2. (i) Most common number of pencils = 7
(ii) Mean =
× + × + × + × + ×
+ × + × + × + ×
2 2 3 3 5 4 4 5 4 6
6 7 4 8 1 9 1 10
30
= 5.67 (to 3 s.f.)
(iii) Probability =
20
30
=
2
3
3. (i) Mean =
+ + + +

7 8 12 74
40
= 33 minutes
(ii) Fraction of students =
11
40
(iii) No. The median time is
+
29 32
2
= 30.5 minutes.
4. (i)
Stem Leaf
1
1
1
1
1
0
2
4
6
9
0
2
4
6
0
2
4
6
0
2
4
6
0
2
4
6
1
2
5
6
1
2
5
7
1
2
5
7
1
2
5
1
2
5
1
3
5
1
3 3 3 3 3 3 3 3 3 3 3 3
Key: 1 | 0 means 10 points
Note that a frequency table would be a more
appropriate statistical diagram as compared to a
stem-and-leaf diagram.
(ii) Mode = 13
Median = 13
5. (a)
40
50
70
30
20
10
60
80
100
90
16
12 14
4
2 6 8 10
0 18
(c) (i) 6 hours
(ii) 18 hours
(d) Since 100 marks lies outside of the range, the result
obtained in (c)(ii) is not reliable.
(e) Take two points on the line and draw dotted lines to
= 60
Horizontal change (or run) = 14.4 − 0
= 14.4
Gradient =
rise
run
=
60
14.4
= 4.17 (to 3 s.f.)
The equation of the line of best fit is y = 4.17x + 25.
(f) The data displays a strong, positive correlation.
6. Total height of the 9 players = 9 × 1.8
= 16.2 m
Mean height of 3 reserve players =
×
16.2 – 6 1.82
3
= 1.76 m
7. (a) Mean =
× + × + ×
+ × + × + ×
+ + + + +
x
x
5 1 8 2 5 3
4 4 5 2 6
5 8 5 4 2
3.5 =
+
+
x
x
68 4
24
84 + 3.5x = 68 + 4x
0.5x = 16
x = 32
(b) x = 3, 4, 5, 6, 7, 8, 9, 10, 11
(c) x = 7
(d) Median =
+
1 2
2
= 1.5
(e) We do not know the exact number of books read by
the last student in the category ‘ 6’
8. (i)
+
+ + + +
x
x y x y
2
2 2 3
=
1
5
5x + 10 = 3x + 4y + 2
2x – 4y = –8
x = 2y – 4 —(1)
+
+ +
x y
x y
2 3
3 4 2
=
24
35
70x + 105y = 72x + 96y + 48
2x – 9y = –48 —(2)

1 140
(ii) Substitute (1) into (2):
2(2y – 4) – 9y = –48
4y – 8 – 9y = –48
5y = 40
y = 8
x = 2(8) – 4
= 12
x = 12, y = 8
(iii) Probability =
+
+ + +
2(12) 3(8)
3(12) 4(8) 2 2
=
48
72
=
2
3
9. (i)
Number of cases, x Frequency
20  x 40 1
40  x 60 11
60  x 80 18
80  x 100 14
100  x 120 5
120  x 140 1
Total frequency 60
(ii)
2
20 40 60 80 100 120 140
4
6
8
10
14
12
16
18
Number of cases, x
Frequency
0
(iii) Median =
+
72 74
2
= 73
Mode = 48
(iv) Estimate for the mean
=
× + × + ×
+ × + × + ×
1 30 11 50 18 70
14 90 5 110 1 130
50
= 75.6
(v) There is a difference of 0.5 in the actual mean and
the estimated mean. The mean calculated in (iv) is an
estimate as the mid-values of each interval were used.
10. (i) Estimate for the mean profit
=
× + × + ×
+ × + ×
6 2.5 11 7.5 18 12.5
12 17.5 3 22.5
50
= $12 million
(ii) Profit = $120 000 × 125
= $15 million
Percentage of number of years =
15
50
× 100%
= 30%

1
141
Revision Test D2
1. (a) A  B = {8, 10, 12, 14, 16, 20}
(b) A  B = {3, 5, 11, 13}
(c) A  B = {0, 1, 7, 9, 15, 17, 19}
(d) A  B = {0, 1, 3, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15,
16, 17, 19, 20}
2. (i) Probability that is divisible by 2 =
5
10
=
1
2
(ii) Probability that it is divisible by 5 =
2
10
=
1
5
(iii) Probability that is divisible by 4 =
2
10
=
1
5
3. (a)
Number of mobile devices Frequency
0 5
1 4
2 8
3 5
4 2
5 1
Total frequency 25
(b)
0 1 2 3 4 5
2
4
6
8
0
Frequency
Number of mobile devices
(c) (i) Median number of mobile devices owned = 2
(ii) Modal number of mobile devices owned = 2
(iii) Mean number of mobile devices owned
=
× + × + ×
+ × + × + ×
5 0 4 1 8 2
5 3 2 4 1 5
25
= 1.92
(d) (i) Probability that he owns 2 mobile devices =
8
25
(ii) Probability that he owns at least 3 mobile devices
=
+ +
5 2 1
25
=
8
25
4. (i) Total number of students = 57
(ii) Mass of lightest school bag = 3.0 kg
(iii) Most common mass = 6.3 kg
(iv) Percentage of bags that were considered ‘overweight’
=
21
57
× 100%
= 36.8% (to 3 s.f.)
5. (i)
Stem Leaf
0
0
0
0
0
0
4
5
6
7
8
9
4
5
6
7
8
9
4
5
6
7
8
9
4
5
6
7
8
9
4
5
6
7
8
4
5
6
8
4
5
6
4
6
4
6
Key: 0 | 4 means 4
Note that a frequency table would be a more
appropriate statistical diagram as compared to a
stem-and-leaf diagram.
(ii) Percentage of patients =
10
40
× 100%
= 25%
6. (a)
4
5
7
3
2
1
6
8
16
12 14
4
2 6 8 10
0 18 20
Height
(cm)
Time (days)
(c) (i) 5 days
(ii) 18 days
(d) The result obtained in (c)(ii) is not reliable since the
height of 7.2 cm lies outside of the range.

1 142
(e) Take two points on the line and draw dotted lines to
Vertical change (or rise) = 6.6 – 1.6
= 5
Horizontal change (or run) = 16 – 3
= 13
Gradient =
rise
run
=
5
13
= 0.385 (to 3 s.f.)
The equation of the line of best fit is
y = 0.385x + 0.4
(f) The data displays a strong, positive correlation.
7. (i) Modal score = 78
(ii) Number of points
= 9 × 79 – (78 + 85 + 64 + 97 + 68 + 78 + 73 + 77)
= 91
8. (a) Modal class is 48 x  52
(b) Estimate of the mean number of hours worked
=
× + × + × + ×
8 42 11 46 14 50 9 54
42
= 48.3 h
(c) (i) Probability that he worked more than 52 hours
=
9
42
=
3
14
(ii) Probability that he worked not more than 44 hours
=
8
42
=
4
21
9. (a) x = 13
(b) x = 21
(c) Mean number of fish caught
=
× + × + × + × + ×
+ + + +
4 1 14 2 7 3 21 4 3 5
4 14 7 21 3
= 3.10 (to 3 s.f.)
10. (i) x + 6y – 9 + 2x + 4y – 4 + 3x + 2y + 2 + 5x + 2y + 9
+ 9x + 4y – 3 = 26 × 5
20x + 18y = 135 —(1)
(ii) 8x + 5y + 2 + 5x + 7y + 6x + 4y + 7 + x + 6y – 3
= 39 × 4
20x + 22y = 150 —(2)
10x + 11y = 75
(iii) (2) – (1): 4y = 15
y = 3
3
4
Substitute y = 3
3
4
into (1):
20x + 18 3
3
4





 = 135
20x = 67
1
2
x = 3
3
8
x = 3
3
8
, y = 3
3
4
(iv) Mean =
× + ×
5 26 4 39
9
= 31
7
9
(v) When x = 3
3
8
, y = 3
3
4
,
x + 6y – 9 = 5
5
8
2x + 4y – 4 = 17
3
4
3x + 2y + 2 = 19
5
8
5x + 2y + 9 = 33
3
8
9x + 4y –3 = 42
3
8
8x + 5y + 2 = 47
3
4
5x + 7y = 43
1
8
6x + 4y + 7 = 42
1
4
x + 6y – 3 = 22
7
8
Probability that the number is greater than 30 =
5
9

1
143
End-of-Year Examination Specimen Paper A
Part I
1. 5x + 3y = 23 —(1)
7y – x = –35 —(2)
From (2),
x = 7y + 35 —(3)
5(7y + 35) + 3y = 23
35y + 175 + 3y = 23
38y = –152
y = –4
x = 7(–4) + 35
= 7
x = 7, y = –4
2. (i) a2
– b2
= (a + b)(a – b)
(ii) 20x = 4022
– 3982
= (402 + 398)(402 – 398)
= (800)(4)
= 3200
x = 160
3. 2x4
y – 18x2
y3
= 2x2
y(x2
– 9y2
)
= 2x2
y(x + 3y)(x – 3y)
4.
+
x
x
6 – 3
2 7
=
+
x
x
3 – 2
5
(6x – 3)(x + 5) = (3x – 2)(2x + 7)
6x2
+ 30x – 3x – 15 = 6x2
+ 21x – 4x – 14
10x = 1
x =
1
10
5. (i) y =
+
x
x
5 3
– 5
xy – 5y = 5x + 3
xy – 5x = 3 + 5y
x(y – 5) = 3 + 5y
x =
+ y
y
3 5
– 5
(ii) When y = 1,
x =
+
3 5(1)
1 – 5
= –2
6.
+
p q
p q
5
2 –
=
9
5
5p + 25q = 18p – 9q
13p = 34q
p
q
=
34
13
p
q
2
2
=
1156
169
p
q
338 2
2
= 2312
7. (i) y = k(x + 1)2
When x = 1,
y = k(1 + 1)2
= 4k
When x = 2,
y = k(2 + 1)2
= 9k
9k – 4k = 20
5k = 20
k = 4
y = 4(x + 1)2
(ii) When x = 3,
y = 4(3 + 1)2
= 64
(iii) y
x2
+ 2x + 1
O
8. (i) A(4, 0)
(ii) x =
+
0 4
2
= 2
(iii) When x = 2,
y = 2(2 – 4)
= –4
Minimum value of y is –4 when x = 2
9. Let the length of the rhombus be x cm.
x2
= 52
+ 122
= 169
x = 13
Perimeter = 4(13)
= 52 cm

1 144
10. (i) k =
9
6
= 1.5
(ii)
+
BP 8
8
= 1.5
BP + 8 = 12
BP = 4 cm
(iii)
+
AC
AC 5
=
6
9
9AC = 6AC + 30
3AC = 30
AC = 10 cm
8 km is represented by 13
1
3
cm
1
3
(15w)(18)
826 = 90w
w = 9.18 (to 3 s.f.)
13.
1
3
p(6)2
(3) =
4
3
pr3
r3
= 27
r = 3
14. (i) Probability that the player does note win anything
=
9
20
(ii) Probability that the player wins either a key chain or
a can of soft drink =
11
20
(iii) Probability that the player wins a soft toy = 0
15. (i) Estimate for the mean height
=
× + × + × + ×
8 125 13 135 12 145 7 155
40
= 139.5 cm
(ii) Fraction of plants = 1 –
8
40
=
4
5
16. Let the translation vector T be
a
b
⎛
⎝
⎜
⎞
⎠
⎟ .
7
5
⎛
⎝
⎜
⎞
⎠
⎟ =
3
4
⎛
⎝
⎜
⎞
⎠
⎟ +
a
b
⎛
⎝
⎜
⎞
⎠
⎟
a
b
⎛
⎝
⎜
⎞
⎠
⎟ = 7
5
⎛
⎝
⎜
⎞
⎠
⎟ −
3
4
⎛
⎝
⎜
⎞
⎠
⎟
= 4
1
⎛
⎝
⎜
⎞
⎠
⎟
The column vector representing the translation vector
T is
4
1
⎛
⎝
⎜
⎞
⎠
⎟ .
−4
2
⎛
⎝
⎜
⎞
⎠
⎟ = Q + 4
1
⎛
⎝
⎜
⎞
⎠
⎟
Q =
−4
2
⎛
⎝
⎜
⎞
⎠
⎟ −
4
1
⎛
⎝
⎜
⎞
⎠
⎟
= −8
1
⎛
⎝
⎜
⎞
⎠
⎟
The coordinates of Q are (−8, 1).
17. The object shown is a regular hexagon.
A regular hexagon has a rotational symmetry of order 6.
Part II
Section A
1. x
x
x
3
–
3
1
–
1
3
=
x
x
9 –
3 –
2
=
+ x x
x
(3 )(3 – )
3 –
= 3 + x
2. (i) x2
+ y2
= 2xy + 64
(ii) x2
– 2xy + y2
= 64
(x – y)2
= 64
x – y = ±8
Difference is 8
3. c = at3
+
b
t2
When t = 1, c = 74,
74 = a(1)3
+
b
12
a + b = 74 —(1)
When t = 2, c = 34,
34 = a(2)3
+
b
22
8a +
1
4
b = 34
32a + b = 136 —(2)
(2) – (1): 31a = 62
a = 2

1
145
2 + b = 74
b = 72
c = 2t3
+
t
72
2
When t = 3,
c = 2(3)3
+
72
32
= 62
4. f(x) = 4x – 6
f 2
1
8
⎛
⎝
⎜
⎞
⎠
⎟ = 4 2
1
8
⎛
⎝
⎜
⎞
⎠
⎟ – 6
= 2
1
2
f −
1
2
⎛
⎝
⎜
⎞
⎠
⎟ = 4 −
1
2
⎛
⎝
⎜
⎞
⎠
⎟ – 6
= −8
5. (i) tan /ACB =
34
43
/ACB = 38.3° (to 1 d.p.)
AC2
= 342
+ 432
= 3005
AC = 3005 cm
AS = ( 3005 – 250 ) cm
Area of APQS = ( 3005 – 250 )2
= 1520 cm2
(to 3 s.f.)
6. (a)
2
4
6
8
2
0 4 6 8
y = 2
2y = x + 4
x + y = 8
A
B C
x
y
(b) A(4, 4), B(0, 2)
(c) C(6, 2)
(d) Area of ABC =
1
2
(6)(2)
= 6 units2
Section B
7. (a) n = 3
(b) (i) F =
k
R
3
When R = 125, F = 4,
4 =
k
125
3
=
k
5
k = 20
F =
R
20
3
(ii) When R = 512,
F =
20
512
3
= 2.5
8. (a) QLM and MRQ
(b) (i) Using Pythagoras’ Theorem,
RS2
= 72
+ 52
= 74
RS = 74
= 8.60 cm (to 3 s.f.)
(ii) sin 50° =
PQ
5
PQ =
5
sin 50°
= 6.53 cm (to 3 s.f.)
(iii) tan 50° =
PL
5
PL =
5
tan 50°
PS =
5
tan 50°
+ 4 + 7
= 15.2 cm (to 3 s.f.)
(iv) tan /MSR =
5
7
/MSR = 35.5° (to 1 d.p.)
(v) Area of PQRS =
1
2
5
tan 50°
+ 4 + 7 + 4





 (5)
= 48.0 cm2
(to 3 s.f.)
9. (i)
Stem Leaf
2
3
4
5
6
7
8
5
6
1
1
2
9
8
9
1
3
3
1
5
2
8
3 4 6 6 9
Key: 2 | 5 means 25

1 146
(ii) Modal number of points = 41
(iii) Median number of points =
+
44 46
2
= 45
(iv) Mean number of points =

+ + +
25 28 89
20
= 51.6
(v) Fraction of clubs =
5
20
=
1
4
(vi) Percentage of clubs =
2
20
× 100%
= 10%
10. (a) When x = –1, y = a,
a = (–1 + 2)(–1 – 4)
= –5
When x = 3, y = b,
b = (3 + 2)(3 – 4)
= –5
a = –5, b = –5
(b)
–2
–3 –1 0
2
4
6
–2
–4
–6
–8
1 2 3 4 5
y = (x + 2)(x – 4)
y
x
(c) (i) When x = 1
1
2
, y = –2.8
(ii) Least value of y = –9
(iii) When y = 3, x = 4.45 or x = –2.45

1
147
End-of-Year Examination Specimen Paper B
Part I
1. y2
– x2
= (y + x)(y – x)
= (–4)(–8)
= 32
2. (a) (x + 7)(x – 3) – 5(x – 3)
= x2
– 3x + 7x – 21 – 5x + 15
= x2
– x – 6
(b) 4y2
– 3(y – 2)(y + 3) – 7
= 4y2
– 3(y2
+ 3y – 2y – 6) – 7
= 4y2
– 3y2
– 3y + 18 – 7
= y2
– 3y + 11
3. (a) 18x2
– 3x – 6
= 3(6x2
– x – 2)
= 3(2x + 1)(3x – 2)
(b) 2x2
– xy – 15y2
= (x – 3y)(2x + 5y)
4. 6x – 4 =
x
2
6x2
– 4x = 2
6x2
– 4x – 2 = 0
3x2
– 2x – 1 = 0
(x – 1)(3x + 1) = 0
x = 1 or x = –
1
3
5. (a)
x
2
3
–
x
x
– 3
10 2
=
x x
x
20 – 3( – 3)
30 2
=
+
x x
x
20 – 3 9
30 2
=
+
x
x
17 9
30 2
(b)
x y
2
3 –
–
y x
5
2 – 6
=
x y
2
3 –
+
x y
5
6 – 2
=
+
x y
4 5
6 – 2
=
x y
9
6 – 2
6. x(y + 2) – 3(y + 2) = 0
(x – 3)(y + 2) = 0
x = 3 or y = –2
7. x =
+
y
y y
3 – 1
(2 1)(2 – 1)
2
x(4y2
– 1) = 3y2
– 1
4xy2
– x = 3y2
– 1
4xy2
– 3y2
= x – 1
y2
(4x – 3) = x – 1
y2
=
x
x
– 1
4 – 3
y = ±
x
x
– 1
4 – 3
8. x –
y
3
= 4
1
3
—(1)
0.5x – 0.25y = 2 —(2)
(1) × 3: 3x – y = 13 —(3)
(2) × 4: 2x – y = 8 —(4)
(3) – (4): x = 5
2(5) – y = 8
10 – y = 8
y = 2
x = 5, y = 2
9.
+
+ +
x
x x
2 3
2 3
=
5
7
14x + 21 = 15x + 15
x = 6
There are 6 $50-vouchers.
10. f(x) = 13 – 4x
f(−2) = 13 – 4(−2)
= 21
11.
DE
8
=
+
6 8
6
DE =
14
6
× 8
= 18
2
3
cm
12. (i)
3
5
h = 36 minutes
Percentage of students =
9
36
× 100%
= 25%
(ii) Modal time taken = 29 minutes
Angle =
4
36
× 360°
= 40°

1 148
13. 2.0, 2.5, 2.5, 3.0, 3.5, 3.5, 3.5, 4.0, 4.0, 4.0, 4.5, 4.5, 4.5,
4.5, 4.5
(i) Mode = 4.5 kg
(ii) Median = 4.0 kg
(iii) Mean
=
+ + + + +
2.0 2(2.5) 3.0 3(3.5) 3(4.0) 5(4.5)
15
= 3.67 kg (to 3 s.f.)
14. (i) 0.25 km is represented by 1 cm
(ii) 1 cm2
40 cm2
represents 2.5 km2
(iii) 0.5 km is represented by 1 cm
0.25 km2
2.5 km2
PQ2
= 252
+ 342
= 1781
PQ = 42.2 m (to 3 s.f.)
16. Estimate for the mean lifespan
=
× + × + × + × + ×
45 10 28 30 19 50 6 70 2 90
100
= 28.4 days
17. (i) (A  B) = {o, p, q, r, s, u, v}
(ii) A  C = {o, p, q, r, s, t, u}
Part II
Section A
h2
+ 2.42
= 8.52
h2
= 8.52
– 2.42
= 66.49
h = 66.49
= 8.15 m (to 3 s.f.)
The ladder reaches 8.15 m up the wall.
2. (a) (i) Gradient =
4
2
= 2
(ii) y = 2x + 2
(b)
x
y = –2x + 6
y
6
4
2
–2
K
–4 –2 2 4 6
0
(c) K(1, 4)
3. (i) h = kt2
When t = 5, h = 200,
200 = k(5)2
= 25k
k = 8
h = 8t2
When t = 7,
h = 8(7)2
= 392
It falls 392 m in 7 seconds.
(ii) When h = 1250,
1250 = 8t2
t2
= 156.25
t = ±12.5
It takes 12.5 seconds.
4.
A
25
24
?
(a) 2 cos A = 2
24
25






= 1
23
25
(b) tan (90° – A) =
24
7
= 3
3
7
(c) 2 cos (90° – A) + 4 tan A = 2
7
25





 + 4
7
24






= 1
109
150
5. (i) Total volume =
1
3
p(7)2
(18) +
2
3
p(7)3
= 522
2
3
p cm3
(ii) Area to be painted pink
= 2p(7)2
= 98p cm2
Area to be painted brown
= p(7)( +
18 7
2 2
)
= 7 373 p cm2
Area to be painted pink : Area to be painted brown
= 98p : 7 373 p
= 1 : 1.4 (to 1 d.p.)
n = 1.4

1
149
Section B
6. (i) C = a +
b
n
When n = 300, C = 8.5,
8.5 = a +
b
300
300a + b = 2550 —(1)
When n = 700, C = 4.5,
4.5 = a +
b
700
700a + b = 3150 —(2)
(ii) (2) – (1): 400a = 600
a = 1
1
2
Substitute a = 1
1
2
into (1):
300 1
1
2





 + b = 2550
450 + b = 2550
b = 2100
a = 1
1
2
, b = 2100
(iii) C = 1
1
2
+
n
2100
When n = 200,
C = 1
1
2
+
2100
200
= 12
The cost of each book is $12.
(iv) When C = 5.7,
5.7 = 1
1
2
+
n
2100
n
2100
= 4.2
n = 500
500 copies are printed.
7. (a) (i) Mode = 2
(ii) Median = 3
(iii) Mean =
× + × + ×
+ × + × + ×
7 1 9 2 6 3
4 4 5 5 8 6
39
=
132
39
= 3.38 (to 3 s.f.)
(b) Number shown = 40 × 3.45 – 132
= 6
(c) Number shown = 6
8. (i) Total surface area
= (10)(10) + 4 ×
1
2
(10)( 13 – 5
2 2
)
= 340 cm2
(ii) Let the height of the pyramid be h cm.
h2
+ 52
= 122
h2
= 122
– 52
= 119
h = 119 cm
Volume =
1
3
(10)(10)( 119 )
= 364 cm3
(to 3 s.f.)
9. (a)
x –4 –3 –2 –1 –
1
2
0
1
2
1 2
y –9 0 5 6 5 3 0 –4 –15
(b)
–3
–4 –2 –1
–2
2
4
6
–4
–6
–8
–10
–12
–14
1
0 2
x
y
y = (1 – 2x)(3 + x)
(c) (i) Greatest value of y = 6.1
(ii) When x = –2.3, y = 3.9
(iii) When y = –3, x = 0.9 or x = –3.4

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