Table of Contents
Answers
Chapter 1
Pre-test
1 a
d
2 a
3 a
4 a
b
c
5 a
6 a
b
c
6
7
8
9
200
b 0.05
c 2.3
430
e 62.9
f 1 380 000
21
b 30
c 3.8
d 10
12 cm
b 7m
c 32 cm
C = 12.6 m, A = 12.6 m2
C = 37.7 cm, A = 113 cm2
C = 24.5 km, A = 47.8 km2
5
b 2.2
c 5
Surface area = 76 cm2, Volume = 40 cm3
Surface area = 24 m2, Volume = 8 m3
Surface area = 100.5 cm2, Volume = 75.4 cm3
a $5258.88
b $43.82
c 164 MB
1:54.27 means 1 min and 54.27 s or 114 270 ms.
nearest millisecond
8.64 × 107
10 0.00024
11 171
12 109
13 11.574
14 a 9002 kB
b 9.002 MB
c No, at least two emails are needed, sending the 1.2 MB
picture separately.
15 a
Exercise 1A
1 a
c
e
g
2 a
c
e
3
× 24
day
4
× 60
hour
min
÷ 60
× 1000
× 1000
× 1000
÷ 1000
× 1000 × 1000 × 1000
ms
s
÷ 60
GB
millimetres
megatonnes
milligrams
decimetre
8 gigabytes
3 nanoseconds
× 60
÷ 24
TB
5 a
c
e
g
i
k
m
o
q
s
b
d
f
h
b
d
milliseconds
kilometres
megagrams
nanometre
$40 000
2 megatonnes
40 kilobytes
MB
÷ 1000 ÷ 1000
0.002
5000
1
3.2 × 104, 3.2 × 1010
0.2
60
1000
2 × 103
2 × 1021
103, 106, 109
ms
÷ 1000 ÷ 1000 ÷ 1000
× 1000
kB
B
÷ 1000
b
d
f
h
j
l
n
p
r
t
5000
2
3 × 106
0.06
5000
500 000
106
0.035
109
40
ns
b
millisecond
microsecond
nanosescond
minute
hour
day
week
month
year
century
millennium
1000
1 000 000
100 000 0000
i
0.016
i
0.00027
0.000011574
0.000001653
0.000000381
3.2 × 10–8
3.1709… × 10–10
3.1709… × 10–11
millisecond
microsecond
nanosescond
second
minute
hour
day
week
month
century
millennium
3.1536 × 1010
3.1536 × 1013
3.1536 × 1016
31 536 000
525 600
8760
365
52.142857
12
0.01
0.001
16 a 213Po, 216Po, O, Ba, Zn, Ar, Na, Au, Cr, H, U, C, Ca
b i 4.4 × 1010 ii 441.75
c 13.48 days
d Answers will vary.
749
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Exercise 1B
1
2
3
4
Some examples are 3.35, 3.37, 3.40 and 3.42.
a 347 cm
b 3m
6.65
a i 1 cm
ii 44.5 cm to 45.5 cm
b i 0.1 mm
ii 6.75 mm to 6.85 mm
c i 1m
ii 11.5 m to 12.5 m
d i 0.1 kg
ii 15.55 kg to 15.65 kg
e i 0.1 g
ii 56.75 g to 56.85 g
f i 1m
ii 9.5 m to 10.5 m
g i 1h
ii 672.5 h to 673.5 h
h i 0.01 m
ii 9.835 m to 9.845 m
i i 0.01 km
ii 12.335 km to 12.345 km
j i 0.001 km
ii 0.9865 km to 0.9875 km
5 a 4.5 cm to 5.5 cm
b 7.5 cm to 8.5 cm
c 77.5 mm to 78.5 mm
d 4.5 ns to 5.5 ns
e 1.5 km to 2.5 km
f 34.15 cm to 34.25 cm
g 3.85 kg to 3.95 kg
h 19.35 kg to 19.45 kg
i 457.85 t to 457.95 t
j 18.645 m to 18.655 m
k 7.875 km to 7.885 km
l 5.045 s to 5.055 s
6 a $4450 to $4550
b $4495 to $4505
c $4499.50 to $4500.50
7 a 30 m
b 145 g
c 4.6 km
d 9.0 km
e 990 g
f 990 g (nearest whole)
8 a 149.5 cm to 150.5 cm
b 145 cm to 155 cm
c 149.95 cm to 150.05 cm
9 a 24.5 cm to 25.5 cm
b 245 cm
c 255 cm
10 a 9.15 cm
b 9.25 cm
c 36.6 cm to 37 cm
d 83.7225 cm2 to 85.5625 cm2
11 a 9.195 cm
b 9.205 cm
c 36.78 cm to 36.82 cm
d 84.548025 cm2 to 84.7832025 cm2
e Increasing the level of accuracy lowers the difference
between the upper and lower limits of any subsequent
working.
12 a different rounding (level of accuracy being used)
ody used to the nearest kg, Jacinta used to the nearest
b C
100 g and Lachlan used to the nearest 10 g.
c yes
13 a d istances on rural outback properties, distances between
towns, length of wires and pipes along roadways
b building plans, measuring carpet and wood
c g iving medicine at home to children, paint mixtures,
chemical mixtures by students
d buying paint, filling and costing filling a pool
14 a 1.79%
b 5.6%
c 0.56%
d 0.056%
e 0.28%
f 0.06%
g 0.12%
h 0.07%
Exercise 1C
1 a
b
c
55
11
77
d 2
e
8 =2 2 f
50 = 5 2
2 a x 2 + y 2 = z 2 b 2a 2 = b 2 c 2x 2 = c 2
3 a 5 cm
b 11.18 m
c 16.55 km
d 1.81 mm
e 0.43 km
f 77.10 cm
4 a 4.58 m
b 7.94 m
c 0.63 m
d 1.11 cm
e 14.60 cm f 0.09 cm
5 a i 34 ii 6.16
b i 80 (or 4 5 )
ii 16.61
c i 10 ii 7.68
d i 68 (or 2 17 )
ii 12.21
6 a no
b yes
c no
d no
e yes
f yes
7 8.3 cm
8 a 2.86 m
b 2.11 cm c 26.38 m
d 4.59 cm
e 0.58 km f 3.65 km
9 a 13.19 mm b 13.62 m c 4.53 cm
d 2.61 m
e 12.27 km f 5.23 cm
10 a 2 13
b 4 2
c
181
11 a i 22.4
ii 24.5
b Student’s own investigation.
5
cm
12
2
13 a 4 5 cm by 2 5 cm
b 3 10 cm by 10 cm
c
100
100
cm by 10
cm
101
101
14 a i 5.41 m ii 4.61 m
iv 8.70 m v 8.91 m
b 7.91 m
15 Student’s own research.
iii 5.70 m
vi 6.44 m
Exercise 1D
1 a πr 2
b
θ
× πr 2
360
c s2
d l×b
750
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
1
xy, where x and y are the diagonals
2
f
1
h( a + b )
2
i bh
2 a
d
g
j
k
l
3 a
d
g
j
4 a
d
g
30 cm2
5000 cm2
230 cm2
10 000 000 mm2
2 200 000 cm2
0.000 145 km2
25 cm2
0.03 mm2
1472 m2
2.36 km2
2.88 cm
1.05 m
1.26 cm
1
xy
2
1
bh
2
1 2
j
πr
2
k 1 πr 2
4
b 2.98 m2
e 5 000 000 m2
h 53 700 mm2
c 0.205 km2
f 100 m2
i 2700 m2
g
h
Let x be the base of the triangle.
1
1
A = ( b − x ) × h + xh + xh
2
2
A = bh − xh + xh
b
54.60 m2
c 1.82 km2
2
153.94 m
f 75 cm2
2
0.05 mm
i 0.17 km2
2
1.12 m
l 3.97 cm2
14.35 m
c 1.44 km
1.91 mm
f 8.89 m
0.52 m
i 5753.63 km
25
5 a 9π cm2, 28.27 cm2
b
π m2 , 39.27 m2
2
49
26
c
π m2 , 51.31 m2
d
π m2 , 9.08 m2
3
9
f
7
π mm2 , 2.75 mm2
8
7 a
25
π + 25 cm2 , 34.82 cm2
8
c
289
104 2
π+
m , 8.70 m2
200
25
d
(3969 − 441π )
mm2 , 103.34 mm2
25
8 a
9 a
10 a
c
11 a a =
1
AC bisects BD, hence DE = EB = y .
2
1
1
1
1
A= × x × y + × x × y
2
2
2
2
1
1
A = xy + xy
4
4
1
A = xy
2
c Consider the following trapezium.
b
b 49 m2
h
a
1
A = bh + ( a − b )h
2
c
1
1
A = bh + ah − bh
2
2
1
1
A = h( b + a )
2
2
b 14 bags
b 1: 2
b 200 000 m2
d 2.5 acres
2A
−b
h
triangle
B
E
Let x = AC and y = BD.
49
99 2
π−
m , 0.52 m2
200
400
66 m2
1:3
100 ha
0.4 ha
C
A
e 81π + 324 km2, 578.47 km2
f
A = bh
D
b
e
h
k
b
e
h
e 21π km2, 65.97 km2
6 43.2 m2
Answers
e
1
A = h( a + b )
2
b i 3
1
3
ii 4.7
iii 0
13 a 63.7%
b 78.5%
c 50%
d 53.9%
Exercise 1E
1 a
12 a
h
b
b
751
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
10 33.5 m2
11 a 6x 2
 1 2
c π d  +
2 
c
12 a 6π
b 2(ab + ac + bc)
1
πdh + dh
2
1
b 5 π
2
b 7.71 m
13 a 0.79 m
14 1 cm
15 a 4πr 2
2 a
d
1 2
1
πr + 2rh + πrh
2
2
b 2x (x + 2y )
c 2rh + πr (h + r )
d 2rh +
θ
πr ( h + r )
180
Exercise 1F
1 a
4 cm
10 cm
2 a
b
1
bh
2
b πr 2
29 cm b
c πrl
221 m
c
109 cm
3 a
2 cm
2 cm
c
b
3 cm
2 cm
4 cm
1 cm
2 cm
2
3 a 90 cm
d 920 cm2
4 a 8.64 cm2
d 872 m2
5 24.03 m2
6 3880 cm2
7 a 121.3 cm2
c 236.5 m2
8 a 66.2
d 207.3
9 a 144.5 cm2
c 1192.7 cm2
b
e
b
e
47.82 mm2
502.91 m2
96 mm2
4.74 m2
c
f
c
f
111.3 cm2
168.89 m2
836.6 m2
43.99 m2
b
d
b
e
b
d
10.2 m2
2437.8 cm2
17.9
2308.7
851.3 m2
4170.8 m2
c 243.12
f 65.0
c
3 cm
752
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
b
b
b
b
b
b
b
0.82 m2
105 cm2
5.18 cm2
126.7 cm2
25.0 cm
17.8 cm
66.6 m2
b 63 cm2
e 502.8 mm2
c 435.90 km2
c 0.16 m2
c 1960.35 mm2
(
πr ( r + ) = πr r + r 2 + h 2
1 4 cm3
2 15 m3
3 a 10 m3
c 163.3 cm2
f 76.6 m2
5
6
7
)
14 Substitute h = r into the equation given in Question 13.
πr (r + r 2 + h2 ) = πr (r + r 2 + r 2 ) = πr (r + 2r 2 ) = πr (r + 2r )
8
9
= πr 2 (1 + 2 ) as required
15 182.3 cm
2
16 a 4 26 cm
b 306.57 cm2
c 4 2 cm
d 20.199 cm
e 260.53 cm2
f 85%
1 a 80 cm3
b 32 m3
c 108 mm3
3
3
2 a 2000 mm
b 200 000 cm
c 15 000 000 m3
3
3
d 5.7 cm
e 0.0283 km
f 0.762 m3
3
3
g 130 000 cm
h 1000 m
i 2094 mm3
j 2700 mL
k 0.342 ML
l 0.035 kL
m 5720 kL
n 74.25 L
o 18 440 L
3 a 40 cm3
b 10 500 m3
c 259.7 mm3
3
3
4 a 785.40 m
b 18.85 cm
c 1583.36 m3
3
3
5 a 12 cm
b 1570.8 m
c 2.448 mm3
3
3
6 a 30 km
b 196 cm
c 30 m3
3
3
d 10 cm
e 0.0021 m
f 4752.51 cm3
3
3
g 0.157 m
h 1357.168 cm
i 24 m3
7 1000
8 480 L
9 a 379.33 cm3
b 223.17 m3
c 6.81 m3
3
3
d 716.46 mm
e 142.36 cm
f 42.85 cm3
3
3
10 a 27 cm
b 3 3m
11 21.1 cm2
12 0.5 cm
13 The student must use the perpendicular height of the oblique
prism instead of 5.
θ
πr 2 h
360
15 yes; 69.3 m3
16 a
1
2
b 5.8 m3
c
e 8 cm3
f 0.336 mm3
b 9.38 mm3
e 0.12 m3
c 25 132.74 m3
f 523.60 cm3
b 276 cm3
e 10.35 m3
c 48 m3
f 70.79 m3
c 50 km3
2
3
Wood wasted = volume of cylinder − volume of cone
Wood wasted = πr2h −
1 2
πr h
3
2 2
πr h
3
2
Wood wasted = of the volume of cylinder
3
Wood wasted =
Exercise 1G
14 V =
8 3
cm
3
a 0.82 m3
d 25.13 m3
47 mL
a 282.74 m3
d 56.88 mm3
11.11 cm
d
1
3
58 mm
3
8
cm3
3
b 585 m3
b
4 a 4 cm3
13 Slant height = r 2 + h2 ,
so
Exercise 1H
Answers
4 a 593.76 mm2
5 a 64 m2
6 a 62.83 m2
7 a 10.44 cm
8 a 25.5 cm
9 a 18.9 cm
10 a 6.3 m
11 hat B
12 a 105 cm2
d 299.4 m2
10 a i V =
1 2
x h
3
ii V =
1 2
πx h
12
b
π
4
11 a 3.7 cm
3V
3V
ii r =
πh
πr 2
12 a Similar triangles are formed, so corresponding sides are in
the same ratio.
b i h=
b
1 2
π(r h1 − r22 h2 )
3 1
c i 18.3 cm3
ii 14.7 cm3
Exercise 1I
1 a 314.16
d 33.51
2 r=
3
3V
4π
4 a
1
2
3
b 60.82
e 91.95
c 3.14
f 1436.76
A
4π
b
1
8
c
1
4
5 a 50.27 cm2, 33.51 cm3
b 3.14 m2, 0.52 m3
c 18145.84 mm2, 229 847.30 mm3
d 1017.88 cm2, 3053.63 cm3
e 2.66 km2, 0.41 km3
f 5.81 m2, 1.32 m3
753
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
6 a 113.10 cm2, 113.10 cm3
b 201.06 m2, 268.08 m3
c 688.13 m2, 1697.40 m3
d 15.71 mm2, 5.85 mm3
e 21.99 m2, 9.70 m3
f 15.21 km2, 5.58 km3
7 a i 1.53 cm
ii 3.50 cm
iii 0.50 km
b i 0.89 m
ii 3.09 cm
iii 0.18 mm
8 a 113.10 cm3
b 5654.9 cm3
c 21 345.1 cm3
9 11.5 cm
10 52%
11 a 32.72 cm3
b 67.02 cm3
c 0.52 m3
2
12 1570.8 cm
13 a 4 m
b 234.6 m3
2
14 a 235.62 m
b 5.94 cm2
c 138.23 mm2
2
2
d 94.25 m
e 27.14 m
f 26.85 cm2
3
3
15 a 5.24 m
b 942.48 m
c 10.09 cm3
3
3
d 1273.39 cm
e 4.76 m
f 0.74 cm3
3
3 16 a i 523.60 cm
ii 4188.79 cm
iii 14 137.17 cm3
b 61.2 cm
17 a 5 cm
b 5 5 cm
c 332.7 cm2
A
4π
18 a r =
b r=3
19 a 4 times
3V
4π
b 8 times
4
20 V = × πr 3
3
Substitute
d
into r, giving
2
3
3
4π
ii
iv 6 units2
b i 4πr
3
iii 1
36 π
v 80.6%
ii x = 3
2
4π
r
3
2
 4π  3
iii 6   r 2
 3 
c Proof required. Example:
4 πr 2
2
 4π  3 2
6
r
 
 3 
=
2π
1
2
3 3 (4 π ) 3
d They are the same.
1 6
2 1.3 m
3 As the sphere touches the top, bottom and curved surface, the
height of the cylinder is 2r and the radius of the base is r. So
the curved surface area = 2 × π × r × h and h = 2r, therefore
this equals 4πr 2, which is equal to the surface area of the sphere.
4 h = 4r
5 (4 − π)r 2
6 2 :1
7 67%
Multiple-choice questions
1 D
5 C
9 E
2 E
6 A
10 C
1
=
2π 3
1
83
1
× 63
=3
π as required.
6
3 A
7 B
11 D
4 D
8 D
12 E
Short-answer questions
1 a
d
g
j
2 a
b
c
3 a
23 cm
b 2.7 cm2
8372 mL
e 0.63825 m2
6
h 777 600
1 000 000
k 125
5.5 m to 6.5 m
8.85 g to 8.95 g
12.045 min to 12.055 min
32 m
b 28.6 m
7
m
4 a
π
d 3
4
V = × π 
2
3
4 πd 3 1 πd 3
V= ×
= ×
3
8
3
2
1
V = πd 3
6
4
21 h = r
3
22 a i
Challenges
2 600 000 cm3
3 000 000 cm2
1000
0.089
c 20.4 cm
b 15.60 m2
5 a
b
65
6 a 13.02 m2
b
d 78.54 cm2
e
7 a 4.8 m
b
8 a i 236 m2
b i 184 cm2
c i 1407.43 cm2
d i 360 cm2
e i 201.06 m2
f i 282.74 cm2
175
cm
b
9 a
3π
10 a 18 cm
b 3 61 cm
c 2305.8 cm2
11 12 m
12 a i 414.25 cm2
b i 124 m2
c i 19.67 mm2
c
f
i
l
8.31
216 m2
100.43 m2
25.48 m
ii 240 m3
ii 120 cm3
ii 4021.24 cm3
ii 400 cm3
ii 268.08 m3
ii 314.16 cm3
c 38.5 m2
f 46.69 m2
17.6 cm
ii 535.62 cm3
ii 88 m3
ii 6.11 mm3
754
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
i 117.27 cm2
i 104 cm2
i 25.73 cm2
4950π cm3
ii
ii
ii
b
84.94 cm3
75 cm3
9.67 cm3
1035π cm2
8
9
5 a yes
e no
6 a 2 3
Extended-response questions
1 a 72 m3
2 a 100 m
d 36%
17
90
b yes
f yes
b 3 5
2
11
c no
g yes
c 2 6
5
12
d no
h no
d 4 3
e 5 3
f
10 5
g 7 2
h 3 10
i
j
6 10
k 9 2
l 4 5
b 6 5
c 16 3
d 6 7
f
20 5
g
5
h
7
j
6
4
k
5
5
l
11
6
b
4 a
b
c 138.7 m2
37 m
b 50 2 m c 5000 m2
e athlete A, 0.01 seconds
d 6 L, $120
8 2
7 a 6 2
e 21 2
Chapter 2
i
1
9
4
j 13
25
b 32 × 43
5×5×5 b 4×4×3
m×m×m×m
y7
b b3
e 27
i
2 a
3 a
c
4 a
c −25
b 25
f
n 2 2
o 2 2
p 2 17
7
q 3 3
2
r
4 6
s
2 3
3
t 3 3
2
8 a
2 2
3
b
2 3
7
c
3 2
5
d
11
5
e
10
3
f
21
12
g
13
4
h
14
5
i
5
3
j
3 3
2
k
d 1
e 10d 7e3
g 8g3h12
h
b
e
b
b
d
5 a
d
6 a
7 a
c
8 a
b
c
d
9 a
5
rational
rational
4a − 2b
−2x − 6
−4x + 28
rational
rational
x
3x + 12
−2x − 10
x 2 + 4x + 3
x 2 + 4x − 5
2x 2 − 5x − 12
x 2 − 4x + 4
i $2205 ii $3258
d −25
1
16
k 2
c 2a 2b
h −
g
1
2
l 4
d 7×x×x×x×y×y
c a15
3s 2
f
2
i 1
c irrational
f irrational
c −10ab + a
irrational
non-recurring
4
b
16
f
irrational b
rational
e
irrational h
–1.24
–2
–1
–√2
5
l
2 2
14
19
12
b
32
c
50
d
e
45
f
108
g
128
h
700
i
810
j
125
k
245
l
363
9 a
10 a 15 3
b 13 7
c 19 5
11 a 4 2 m
b 2 30 cm c 4 15 mm
27
d 31 3
12 a radius = 2 6 cm, diameter = 4 6 cm
b radius = 3 6 m, diameter = 6 6 m
c radius = 8 2 m, diameter = 16 2 m
13 a 2 5 cm
b 15
d
14
Exercise 2A
1 a
c
2 a
e
3 a
d
g
d
m 3 11
Pre-test
1 a 25
6
2
c
Answers
13 a
b
c
14 a
11 m
b 3 5 m
c
e
f
11 mm
145 mm
2 21 cm
72 = 36 × 2 (i.e. 36 is highest factor of 72)
=6 2
9
16
rational
rational
irrational
b
d
c
g
c
f
root
rational
25
36
rational
irrational
√5
0.18
2
0
2
5
1 57
d 4
h 36
15 a 9, 25, 225 b 15 2
16 a Draw triangle with shorter side lengths of 1 cm and 3 cm.
b Draw triangle with shorter side lengths of 2 cm and 5 cm.
c
d
2
1
√22
√6
2√3
3
π
1
1
4
2
1
4
17 Check with your teacher.
755
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Exercise 2B
1 a
e
2 a
e
3 a
b
4 a
b
yes
yes
6x
17a
13 a 6 3 − 3 2 , unlike surds
b
f
b
f
e 11 5
6 a
no
yes
−5x
t
d
h
d
h
b 8 2 + 2 5 , unlike surds
yes
yes
−2b
−2y
c 5 2 − 6 5 , unlike surds
d 10 10 + 10 3 , unlike surds
4 3
i 5 3 ii −3 3 iii 17 3
10 2
i 5 2 ii −16 2 iii 0
5 a 6 5
i
c
g
c
g
no
no
7y
−5x
−2 21
f 4 5 − 6 6 , unlike surds
b 3 3
c 4 2
d 3 2
f
g 6 10
h 5 2
k − 13
l −7 30
j
3 +5 2
e 20 2 + 30 3 , unlike surds
3
−6 11
2
b 5 2
c 4 3
d
5
f
12 3
g 8 11
h 3 2
5 6
j
5
k 32 2
l 20 2
b 9 6
8 a 13 2
d 5 5+6 7
c 2 5− 7
e
6 −3 2
f
9 a
9 + 18 2
5 3
6
d
7
6
g
13 5
18
2 3 + 11 5 − 5 2
h 11 − 9 3
g 9 3+2 2
i
j
b
−9 2 + 15 5
7 5
12
− 2
e
10
h
−7 3
30
10 a 4 3 + 2 5 cm
c
c
2
30
f
13 3
14
i
−11 10
24
b 14 2 cm
10 + 3 2 cm
d 2 10 + 4 5 cm
e 4 3 + 30 cm
11 a
f 12 3 cm
h
29 6
28
6 6
35
l
29 5
42
c
3
2
f
−7 7
15
j
8 3
k
e
5
12
1 a
15
= 5 3
b
42
= 6
7
c
6 × 5 = 30 d
11 × 2 = 22
c
26
2 a
15
b
d
35
e − 30
f
− 30
g
66
h
6
i
70
3 a
10
b
6
c − 3
d
5
e
3
f
10
g
5
h − 13
i
− 5
b
10
f 9
c
30
g 7 2
d 3
h 2 11
j
k 4 6
l 10
b 21 2
c 12 14
d −50 3
e −18 3
f
15 5
g 42 6
h −60 10
i
j
42 2
k 24 30
l 216 7
5
c
2
2 5
f
3
b
14 − 10
4 a
21
e 5
i
338 = 13 2
12 a 5 3 − 6 3 + 3 = 0
c 6 2 −8 2+2 2 =0
b
6+2 6 −3 6 =0
d 2 2 −3 2+ 2 =0
e 4 5 −7 5+3 5 =0
3 2 −6 3 −5 2+6 3+2 2 =0
3 6
5 a 10 3
−20 10
6 a 2 2
d
7 a
20 = 2 5
b 3 72 = 18 2 ,
f
g − 2
2 3
3
Exercise 2C
e 7 2
i
−3 2
4
b
d −2 2 + 5
e 4 3
f 0
g −3 2 − 3 10 h −2 5 + 3 15
7 a
d
7 2
15
i 0
b 3 6 + 7 11
c 4 5 −7 2
14 a
−4
13
21
5 2
b 3 6
e
6 + 15 −1
3 7
c − 55 − 65
d −2 15 − 2 21
e 6 26 − 3 22
f
g 30 2 + 15 30
h −12 3 + 12 2
i
42 + 63 2 j
90 3 − 24 10
k −16 + 24 10
l
42 2 + 30
20 − 20 2
756
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
g 2
9 a 2 6
3
cm2
10 a
4
b 18
e 3 3 +4
c −75
f − 10 + 5
h 8 2
b
30
i
4 a − 2 −4
d 7+3 3
2 −6
c 6
b 2 6 cm
6 × 6 = 6 × 6 = 36 = 6
11 a
b − 8 × 8 = − 8 × 8 = − 64 = −8
c − 5 × − 5 = + 5 × 5 = 25 = 5
12 a Simplify each surd before multiplying.
b Allows for the multiplication of smaller surds, which is
simpler.
c i 3 2 ×3 3 =9 6
ii 2 6 × 2 5 = 4 30
iii 5 2 × 3 5 = 15 10
iv 3 6 × 5 3 = 45 2
v 6 2 × 4 3 = 24 6
vii −12 3 ×−2 7 = 24 21
viii 7 2 × 10 3 = 70 6
ix 12 2 × 12 5 = 144 10
b 2
2
e
5
c −9
b 375 3
c 162 3
d 25
e 9
f
h −108 2
i 720
j
g −120 5
27 2
k
2
o −96 15
1
d −
5
14 a 54 2
m 100 3
5
q
3 3
128 2
14 7
n 144
r
f 3
9 6
2
b
21
11
e
125
h
0
b
0
e
x 2 + 5x + 6
x2 + x − 12
6x2 − 11x − 10
x2 − 16
25x 2 − 36
4x 2 − 4x + 1
− 10
13
147
0
2 3
c
f
i
c
f
b
d
f
h
j
l
f
−13 − 2 2
i
c
f
i
c
f
i
l
c
f
i
b
23 − 8 7
h
b
e
h
b
e
h
k
b
e
h
27 + 12 2
2 6 − 118
23 3 − 46
14 − 6 5
15 + 4 11
9 + 2 14
32 + 2 247
7
6
3
30 − 9 10
21+ 2 3
35 − 13 10
43 − 19 2
10 − 4 6
28 + 10 3
13 − 2 22
40 + 2 391
19
4
6
118 + 28 10
25 + 32 5
18 7 − 65
24 5 − 89
23 + 8 7
54 − 14 5
13 − 2 30
60 − 2 899
13
−6
−4
139 + 24 21
c 195 + 30 30
d 176 − 64 6
f
g 66 + 36 2
h 140 + 60 5
107 − 40 6
97
163
0
7 + 4 3 cm2
2 m2
b 17
e 26
h −33
87 − 12 42
c 41
f 10
i −40
15 6 − 5 2 − 18 + 2 3 mm2
d 5 + 6 5 m2
l 81
81 3
p
25
Exercise 2D
1 a
d
g
2 a
d
3 a
c
e
g
i
k
6 5 − 13
c 4+ 6
e 207 − 36 33
i
9 a
d
g
10 a
b
c
vi 6 3 ×−10 5 =−60 15
13 a 3
g
5 a
d
g
6 a
d
g
j
7 a
d
g
8 a
b 2 5 −3
e 5+ 7
Answers
8 a 28
d 2− 6
6 6
12
162
0
6 3
x 2 − 4x − 5
2x2 − 9x − 5
6x 2 − 17x − 28
4x 2 − 9
x2 + 4x + 4
9x 2 − 42x + 49
e
7 6 −7 2 − 3 +1 2
cm
2
f 81−30 2 mm2
11 a −5
b 7
12 a 11 21− 26 3
c 128
b 2 5 + 30
c 5 35 + 31 7
d 19 7 + 2
e
f
2 −2 6
13 Yes. Possible example:
b
14 a 19 − 2 6
c 2 15 − 85
d
e 30 − 10 2
f
g 4 3 − 14
h
10 + 3 5
a = 12 , b = 3
16
10 3 − 37
0
47 2 − 10 30 + 11
Exercise 2E
1 a 1
b 1
e −2
2 a
3
f −9
b
5
1
2
g 6
c 10
e
f
d
g
5
3
3
h
c
3
7
7
i
1
2
h −1
d −
7
13
13
757
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
3 a 0.377… b 2.886… c 16.31…
Notice that all pairs of numbers are equal.
c 4+ 2
12 + 3 2
−3 3 − 3
ii
14
2
2
2
b
7
7
c
3 11
11
d
4 5
5
5 3
3
f
4 2
g
15
3
h
14
7
iii 2 2 + 6 iv
6
3
b
35
7
c
66
11
d
10
5
14 a 5 3 − 5
2
21
3
f
42
7
g
30
3
h
34
2
6 a
4 14
7
b
5 6
3
c
d
3 42
7
e
7 30
10
7 a
4 21
15
b
d
2 2
5
g
9 2
2
4 a
e
5 a
e
7− 3
l
14 − 2
6
n
14 + 2 2
o 10 − 4 5
6
3
c
35
3
e
2 5
15
f
h
3 7
2
d
g
j
2 5 − 15
5
e
35 + 14
7
f
30 − 21
3
2 3 + 42
6
h
5 2 +2 5
10
i
30 − 5 2
5
8 3 − 15 2
6
k
3 2 +2 5
2
l
6 5 +5 6
2
5 3
cm2
3
b
2 2
m
3
c
10 + 15
mm2
10
6 5 +5 2
b
10
c
9 7 − 14 3
21
d
5 3 −2 2
6
e
2 2 +5 3
12
f
9 5 +4 3
30
g
−2 14
15
h
6 30 + 4 6
9
i
3 10 − 2 42
9
11 Does not change because
21 + 7 a
7
d 1− 3a
4 10a + 5 2
10
b
x
x
is equal to 1.
30 + 5a
5
e 1− 5a
h
6a + 2
2
b a −b b
a a +a b
q
a−b
a−b
p
c
2 3 +3 2
10 a
6
g
k
j
6 − 10
2
12 a
2 11 + 2 2
9
2 105
15
3 7 + 35
7
9 a
i
f
3+ 6
3
s a − ab
a−b
f
1− 7 a
i
2 14 a + 7 2
14
13 a i 14
ii 2
iii 47
b E ach question is a difference of two squares, and each
answer is an integer.
r
a + b − 2 ab
a−b
a b +b a
a−b
t
Exercise 2F
1 a 34
d 6x 3
2
b 76
e 15y 4
c 83
f 8a3b2
x
4
3
2
1
0
2x
24 = 16
23 = 8
22 = 4
21 = 2
20 = 1
3 a 22 × 23 = 2 × 2 × 2 × 2 × 2
b
= 25
x5 x × x × x × x × x
=
x3
x×x×x
= x2
c ( a2 )3 = a × a × a × a × a × a × a d (2 x )0 × 2 x 0 = 1 × 2
=2
= a6
4 a a9
d 14m5
g
p3
5
m
5 a
e
i
150x5y6
x3
y5
2xy3
m 5b3
q −x
3
6 a x10
e 64t
b x5
e 6s7
h
j 6x3y3
2 3 + 6a
c
6
−3 − 3 3
2
h −14 − 7 5
g −12 − 4 10
m 6+ 6
5
c 3 5 +6
42 + 7 7
29
3 10
2
2 5 −2 2
)
7
f
e
10
9
(
− 6 + 2 30
b 2 3 +2
d −4 − 4 2
b
8 a
d i
6
c7
6
k 15a3b6
n
b
f
j
12r7s6
a
d5
3r 2s
n 4st
2
r −y
2
b t6
f 4u4
c b6
f 2t16
i
9 2
s
25
l 18v9w 2
o
c
g
k
20m8n10
q3
j
2p 2
o
1 2
v
4
c 4a6
g 27r9
d b4
h m6
l 2m4x
p
1
a
2
d 5y15
h 81p16
758
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
a4
b6
j
x9
y12
k
x 4 y6
z8
l
u16 w8
v8
e
6
a2
f
m
27 f 6
125 g3
n
9a 4 b 2
4 p2 q 6
o
a3 t 9
27 g12
p
256 p8 q12
81r 4
i
2x
3
j
8
5
x8
xy 2
b
f
b
e
3
3
x 2y 2
m
c
g
c
f
1
−5
x 6n 8
r 4s7
m
5
3s 3
9 x 8 y2
2
h 2y4
7 a
e
8 a
d
g
j 27m7n14
m 2m6n3
9 a −27
k −45a8b5
n 21y3z 2
b −27
d 1
h 3
i 2a 2b 2
16 3
f
3
o 1
c 81
l
p −6m 2n7
d −81
a30
b15
11 a 13
b 18
c 81
d 64
e 1
f 1
g 9
h 8
12 Billy has not included the minus sign inside the brackets; i.e.
has applied it only afterwards. Need (−2)4 not −24.
13 a 3
b 4
c 1
d 3
e 4
f −1
14 a 9
b 2
c 162
d −18
10 a x12
15 a ± 2
16 a
b
c
d
b a105
b 5
c
c 2
d
7
2
x = 2, y = 4 or x = 4, y = 2 or x = 16, y = 1
x = 8, y = 2 or x = 4, y = 3 or x = 64, y = 1 or x = 2, y = 6
x = 9, y = 2 or x = 3, y = 4 or x = 81, y = 1
x = 1, y = any real number, but in this situation y = any
positive integer.
Exercise 2G
1 a 2−2, 2−3, 2−4
c 3x −1, 3x −2, 3x −3
1
32
1
3 a b
a
2 a
4 a
1
x5
3a 2
e
b3
q3r
3 p2
5 a x2
b
1
52
b x −1, x −2, x −3
5
42
d
−2
33
2
m4
d
3
y7
b ab
b
1
a4
c
f
4m3
n3
10 y5 z
g
x2
3z 3
h 4 2
x y
3u2 w7
8v6
c 4m7
2b 3
5c5 d 2
d 3b 5
4 b4 a3
3
2
c
b4
5h 3 g 3
2
3
d 3
y
d2 f 5
5e4
b 2y 3
j
k
e 2b4d 3
f 3m 2n 4
g
6 a x
b a3
f
1
xy
b
8 a
a
b3
p
i
q2r
f5
m 5
g
e
b
j
f
j
n
1
64 m 6
5
x4
h12
81
4 y2
a3
q
p5
x
2y
1
r 6s2
g
−10
m6
h
−18
a10
k
1
2
l
3b 2
4
o
1
2d 2
p
5
6t 2
c
g
2
x 21
81
x 20
k 7j 8
c
6
a5 b2
a2
b2
4m
k
7 n3
g
o
wx 5
2
b
16 p4
9q 2
c 54x 7y10
d 4a 8b 3
e
324r 11
s
f
g a 2b18
h
m14
n8
i
b
1
64
c
9 a a7b 2
1
25
1
e
9
9
i
4
11 0.0041 cm
10 a
f 1
j
−64
125
4
d6
−8
h 15
x
d
l 2t 6
18b4
a2
m2
h
n
4r 3 s 7
l
3
d
p
5c5 d 4
4
d
−5
81
2 y14
x3
27 x
2y
2
49
g 98
h −48
1
16
l 100
k
3
7
y
b
b
ii
iii
2
5
2x
a
13 The negative index should be applied to x only, not to 2:
2
2x−2 = 2
x
12 a i
c
i
7d 2
10
4
n
3f2
16
x4
9
e 8
t
y4
i
16
7 a
12
x
Answers
i
l
h
14 a
5
6
d −
7
12
b
5
18
c
1
3
e
71
48
f
106
9
 1 x
x
15 Proof:   = (2−1) = 2−1 × x = 2− x
2
16 a
e
i
m
−2
−2
0
−2
b
f
j
n
−5
−3
0
1
c
g
k
o
−3
−3
1
−2
d
h
l
p
−1
−4
2
2
759
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Exercise 2H
1 a 3
b 4
e 3
f 2
2 a 103
b 107
4
3 a 4.3 × 10
b
d 1.001 × 104
e
g 3 × 10−5
h
4 a 3120
c 710 500
e 59 500
g 10 120
i 210 500 000
k 2 350 000 000
5 a 0.0045
c 0.0003085
e 0.000092
g 0.0001002
i 0.98
k 0.000000000003285
6 a 6.24 × 103
b
d 4.24 × 105
e
g 7.25 × 104
h
j 9.09 × 105
k
7 a 2.42 × 10−3
b
d 7.87 × 10−3
e
g 6.40 × 10−6
h
j 7.01 × 10−7
k
8 a 2.4 × 104
b
d 4.88 × 103
e
g 9.8 × 10−6
h
9 a 7.7 × 106 km2
c 7.4 × 109 km
e 1.675 × 10−27 kg
10 a 2.85 × 10−3
b
d 6.38 × 10−3
e
g 1.80 × 10−3
h
11 328 min
12 38 is larger than 10.
13 a 2.1 × 104
b
d 1.79 × 10−4
e
g 1 × 107
h
j 3.1 × 10−14
k
14 a 9 × 104
b
d 1.44 × 10−8
e
g 2.25 × 10−6
h
j 1.275 × 10−4
k
m 8 × 10−1
n
15 a i 9 × 1017 J
iii 2.7 × 1015 J
c 3
d 2
g 1
h 3
c 10−6
d 10−3
7.12 × 105
c 9.012 × 105
−4
7.8 × 10
f 1.01 × 10−3
−2
3.00401 × 10
b 54 293
d 8 213 000
f 800 200
h 9 990 000
j 55 000
l 1 237 000 000 000
b 0.0272
d 0.00783
f 0.265
h 0.000006235
j 0.000000000545
l 0.000000875
5.73 × 105
c 3.02 × 104
4
1.01 × 10
f 3.50 × 107
5
3.56 × 10
i 1.10 × 108
6
4.56 × 10
l 9.83 × 109
−2
1.88 × 10
c 1.25 × 10−4
−4
7.08 × 10
f 1.14 × 10−1
−5
7.89 × 10
i 1.30 × 10−4
−9
9.89 × 10
l 5.00 × 10−4
6
5.71 × 10
c 7.0 × 108
−3
1.9 × 10
f 7.05 × 10−4
−1
3.571 × 10
i 5.00 × 10−5
6
b 2.5 × 10
d 1 × 10−2 cm
f 9.5 × 10−13 g
1.55 × 10−3
c 4.41 × 10−8
7
8.00 × 10
f 3.63 × 108
15
3.42 × 10
i 8.31 × 10−2
3.94 × 109
2 × 103
6 × 106
2.103 × 10−4
8 × 109
4 × 104
1.25 × 107
1.8 × 10−1
4 × 10−14
ii 2.34 × 1021 J
iv 9 × 1011 J
c
f
i
l
c
f
i
l
o
6.004 × 101
7 × 10−1
4 × 10−3
9.164 × 10−21
6.4 × 109
6.25 × 10−12
1 × 10−5
2 × 102
2.5 × 104
b i 1.11 × 108 kg ii 4.2 × 10−1 kg
iii 9.69 × 10−13 kg iv 1.89 × 10−19 kg
c 5.4 × 1041 J
Exercise 2I
1 a
d
2 a
e
i
3 a
3, 2
5, 5
3
2
2
1.91, 1.91
b
e
b
f
j
b
2, 3
4, 4
5
3
3
1.58, 1.58
c
f
c
g
k
c
3, 5
5, 5
11
5
2
1.43, 1.43
d 25
h 4
l 10
1
1
2
3
4 a 29 2
b 35 3
c x5
d b4
1
1 7
1
e 22 a 2
f
43 t 3
5
5 a 7x 2
d
2 1
5 p3r 3
7
b 6n 3
e
4 2
2a 3 b 3
6 a
5
e
3
h 72
i
43
2
b
7
8
c
3
9
f
3
49
g
5
7 a 6
3
b 3
e 2
i
3 5
2g 4 h 4
3
1
3
8 a 4
j
1
10
b 8
1
e
8
i 125
9 a a2
6
e s7
10 a 5s2
e x
i 4ab 4
5
7
11 a method B
b i 32
m
4
6
d
10
11
8
h
7
81
c 4
f 5
1
g
3
1
k
20
d 7
1
h
2
l
1
10
c 216
d 32
f
1
9
1
g
16
h
1
125
j
1
16
k
1
81
l
1
100
b m3
c x
y9
b 3t 2
f b4
j 6m 2n
n
2x
3
1
d b2
1
1
f
1
h 88 m 2
c 3y 3
f
g 52
1
1 2
g 10 5 t 5
g 1
c 2t 2
g t3
k 2x 2y 3
o
2
x2
ii 216
iii 128
1
32
vii
v 625
vi
27
3125
12 It equals 2 since 26 = 64.
13 a i −3
ii −10
iii −2
b i no
ii yes
iii yes
c y is a real number when n is odd, for x < 0.
2
h a
b
d 5t 4
h m2
l 7r 3 t 2
p 10x
iv 81
viii
81
10000
iv −3
iv no
760
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
1 a
b
2 a
c
e
3 a
d
4 a
e
i
5 a
e
i
i 4
ii 8
i 3
ii 5
16, 32, 64, 128, 256, 512
64, 256, 1024, 4096
216, 1296
32
b 53
7
2
e 36
3
b 3
3
f 3
4
j 5
−2
b −2
−5
f 3
3
4
3
b
6 a
3
2
1
1
e
f
2
3
i −2
7 a 1
b i 2
c i 3 min
8 a
1
2
3
4
15
i
4
e
iii
iii
b
d
16
iv 32
6
81, 243, 729, 2187, 6561
125, 625, 3125
c 35
c
g
k
c
g
2
4
4
−2
2
3
2
1
g
5
c
d
h
l
d
h
2
3
3
−4
6
3
2
1
h
4
d
3
2
3
2
j −4
k −
ii 32
ii 8 min
iii 260
iii 10 min
iv 21440
b 1
c 3
d 1
f 2
g 9
h
k 4
l −3
2
j
−
11
2
l −
6
7
9 Prefer 1 cent doubled every second for 30 seconds because
receive 229 cents, which is more than 1 million dollars.
10 a i 1
ii 1
iii 1
b No solutions. If a = 1, then a x = 1 for all values of x.
2
3
d
11 a 2
b 1
c
3
4
5
3
1
g
e
f 1
h −
4
10
2
3
12 a i 0.25
ii 0.125
iii 0.001
iv 0.0016
b i 5−2
ii 2− 4
iii 2−1
iv 5−4
13 a −4
b −6
c −5
1
3
e −1.5
f −
d
2
4
14 a 1
b −1
c 8
d −3
2
2
e −5
f −2
g 3
h
3
2
i
1
5
j 2
k 0
Exercise 2K
1 a $50
d $55.13
2
, 0.98 c 4.52 kg
100
3 a growth
b growth
c decay
d decay
e growth
f decay
4 a A = amount of money at any time, n = number of years of
investment
A = 200 000 × 1.17n
2 a 4.9 kg
b $1050
c $52.50
e $1276.28
l −2
b
b A = house value at any time, n = number of years since
initial valuation
A = 530 000 × 0.95n
c A = car value at any time, n = number of years since purchase
A = 14 200 × 0.97n
d A = population at any time, n = number of years since
initial census
A = 172 500 × 1.15n
e A = litres in tank at any time, n = number of hours elapsed
A = 1200 × 0.9n
f A = cell size at any time, n = number of minutes elapsed
A = 0.01 × 2n
g A = size of oil spill at any time, n = number of minutes elapsed
A = 2 × 1.05n
h A = mass of substance at any time, n = number of hours
elapsed
A = 30 × 0.92n
5 a 1.1
b i $665 500
ii $1 296 871.23
iii $3 363 749.97
c After 7.3 years.
6 a 300 000
b i $216 750
ii $96 173.13
iii $42 672.53
c 3.1 years
7 a V = 15 000 × 0.94t
b i 12 459 L ii 9727 L
c 769.53 L d 55.0 h
8 a V = 50 000 × 1.11n
b i $75 903.52 ii $403 115.58
c 6.65 years
9 a 3000
b i 3000
ii 20 280
iii 243 220
c 10 h 11 min
10 a D = 10 × 0.875t, where t = number of 10 000 km travelled
b 100 000
c yes
11 a T = 90 × 0.92t
b i 79.4°C ii 76.2°C
c 3.2 minutes = 3 minutes 12 seconds
12 a i $1610.51 ii $2143.59 iii $4177.25
b i $1645.31 ii $2218.18 iii $4453.92
13 a $2805.10 b $2835.25 c $2837.47
14 a i 90 g
ii 72.9 g
iii 53.1 g
b 66 years
15 a 60 L
b 22.8 min
16 0.7%
Answers
Exercise 2J
761
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Challenges
1 3n
2 a 5
1
5
4 a −8
9 a 3
4
b  
9
e −2
x
b 22− a
f −3
2
3
l 0
h
b V = 3000 × 0.82t
1 a 36 15 + 3 45 = 36 15 + 9 5 cm2
6
b 360 3 + 144 15 + 90 + 36 5 cm3
xy ( x − y )
c
d
2 a
b
c
d
8 12 + 8 2
Multiple-choice questions
2
5
8
11
d 6
3
g
2
k −4
Extended-response questions
5 length = 10 2 cm, breadth = 10 cm
1 C
4 E
7 D
10 D
c 1
i 4
j 3
10 a V = 800 × 1.07t
3
−3 − 2 + 7 3
7
x−y
7 a
b
xy
b 2
3
6
9
12
D
A
C
C
B
D
B
D
4 3 +1
i 10 000 cm2 ii 1.6%
A = 10 000 × 1.065n
i $11 342.25 ii $13 700.87
11.1 years
i $14 591 ii V = 14 591 × 0.97t
iii $12 917; profit of $2917
Chapter 3
Short-answer questions
1 a 2 6
e
2
7
b 6 2
f
2 a 4+7 3
e 2 30
2 2
3
c 30 2
d 12 6
g 5 7
h 2 5
5
b 2 5 +2 7 c 5 2
d 4 3 +2 2
f
h
−12 5
g 2 5
3 a 2 6 + 4 2 b 12 5 − 6 c 12 3 − 4
e 6
6
4 a
6
e 3 2
4
5 a
d
e
g 16 + 6 7
h 56 − 16 6
b 5 2
c 3 6
d 2 14
f
5 2
8
b 6
3 y4
2x3
e
1
1 3
10 2 x 2
7 a 5
d 6 5
f −9
25
y6
6 a 212
7
3
1
6t 3
1
b x3
f
1
1
2 3 a3 b 3
b 4
g
10 + 2 2 h 4 6 − 3
3
2
c
20 x 3
y4
f
27
4 b8
5
2
c m3
d a5
g
3
72
c
1
2
h
Pre-test
1 a i 14
18
b i
25
ii 25
7
ii
25
1
3 9
2 0, 1 in 5, 39%, 0.4, , 0.62, 71%, , , 1
2
4 10
1
8
5
e
8
4 a 11
3 a
1
2
1
g
4
ii
2
11
iii
4
11
vi
8
11
f
1
11
b i
1
2
7
8
b
iv
7
11
v
3
11
5 a
7
16
b
9
16
6 a
c
d
1
4
Roll 1
4
43
1
1
1
d
e
f
7
10
5
8 a i 3210
ii 4 024 000 iii 0.00759
iv 0.0000981
b i 3.08 × 10−4
ii 7.18 × 10−6
iii 5.68 × 106
iv 1.20 × 108
iii 11
7
iii
25
Roll 2
1
2
3
4
1
2
3
4
5
2
3
4
5
6
3
4
5
6
7
4
5
6
7
8
b 16
c i
1
16
ii
3
16
iii
3
8
762
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
v
13
16
vi
3
16
2
5
d
7
A
not A
B
2
8
10
not B
5
2
7
7
10
17
Toss 2
Outcome
8 a
Toss 1
T
(H, T)
13 a
H
(T, H)
e
T
(T, T)
14 a
c 8, as
iv
1
4
iii
3
4
v
1
2
vi 1
1
4
1
2
Exercise 3A
b { H, T }
1
e
2
v
c yes
1
4
b
1
6
c
d
3
8
e
2
3
f 0
3 a
1
7
b
2
7
c
5
7
d
3
7
4 a
3
10
b
2
5
c
3
5
d
1
2
5 a
1
10
b
1
2
c
1
2
d
1
2
2
5
6 a 0.09
7 a 0.62
1
8 a
50
1
2
5
d
24
6
10 a
25
1
5
b 0.43
b 0.03
3
b
10
f
b
3
8
e 1
b
1
50
1
5
iii
1
20
v
1
20
41
of 20 is closest to 8.
100
1
1
b
c
13
52
f
4
13
g
12
13
d
1
2
h
9
13
1
25
24
25
ii
8
25
iii
16
25
vi
17
25
vii 1
iv
viii
9
25
17
25
d Yes. Hint: Try changing 10, 10, 10 to 14, 14, 14. Also try
changing 10, 10, 10 to (10 + x), (10 + x), (10 + x).
f 1
2 a
e
ii
41
of 10 is closest to 4.
100
2
13
c i
1 a 2
1
d
2
4
25
7
15
b 15; any multiple of 15 is a possibility as 3 and 5 must be
factors.
15 a 625π
b i 25π
ii 200π
iii 400π
b 4
ii
f
12 a 59
(H, H)
1
4
2
25
1
10
b
H
T
c i
iv 0
b 4, as
H
9 a
7
10
11 a i
e
Answers
5
8
iv
1
4
3
10
c 0.47
c 0.97
49
c
50
Exercise 3B
1 a
c
A
B
A
B
A
B
A
B
b
d
A
B
A
B
A
B
A
B
g
c
1
4
d 0.91
d 0.38
e
g
f 0
c
f
h
21
25
763
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
2 a ∅
e E∩F
3 a no
4 a
b A∩B
f W∪Z
b yes
A
c A∪B
d ∅
g A∪B∪C h A∩B∩C
c no
3
5
9
8
b i A ∩ B = {2, 5, 8}
A
11
ii
F
iii 1
2
B′
1
3
4
5
5
10
2
3 17
13
19 23
7
10
C
O P
L
iii
1
5
iv 1
b i
9
13
ii
D
C M
E E T
M
6
13
F
8 a
7
9
iii
ii 5
2
9
iii
8
1
2
iv
v
9
9
9
A′
B
2
6
8
B′
5
3
8
7
9
16
ii 6
vi 8
9
ii
16
4
13
3
13
v
B
3
3
6
B′
4
1
5
7
4
11
A
A′
B
2
7
B′
2
1
3
4
8
12
iii 5
vii 13
5
iii
16
13 3
14 a 1 − a
15
A
b a+b
9
c 0
B
A ∩ B′ A ∩ B A′ ∩ B
A
b i 2
v 7
1
c i
8
iv
A′
1
5
B
ii
10
13
b
5
c i
N
A
25 10 5
b i 25
iii
M
A
R
Y
12 a
N
ii 5
1
ii
5
7
10
d nothing
iv
iii
b 10, 12
29
6
2
5
c a, c, e
ii 3
10 a 4
11 a
5
7 a
A′
4
c i 2
25 10 10
b i 25
2
c i
5
A
B
ii A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
b i A ∩ B = {2, 13}
ii A ∪ B = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}
6 a
2
6
B
7
1
2
4
3
7
3
ii
c i
10
10
d No, since A ∩ B ≠ 0.
c i
1
B
b
2
7 10
5
A
B
1 4
5 a
9 a
A′ ∩ B′
16 a
M
iv 3
w
(A ∪ B)′
E
c n s
v
m
764
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
1
3
17 a
ii
2
3
iii
1
6
iv
2
3
v
1
3
L
5
S
1
2
2
2
I
1
13
20
1
16 a
4
7
d
100
ii
1
3
iii
13
15
L
2x
7
2 2
x y
1
15
iv
L = Own State
S = Interstate
I = Overseas
18 − 2x
S
I
c i
5
19
ii 10
1
ii
19
b No, A ∩ B ≠ 0
2 a 0.8
3 0.05
4 a i 13
1
b i
4
4
c
13
b 0.8
ii 4
3
ii
4
3
d
52
5 a i {3, 6, 9, 12, 15, 18}
1
b i
20
7
c
20
6 a
7 a
8 a
9 a
10 a
1
8
0.1
0.3
3
8
4
13
49
e
52
11 a 0.4
b
b
b
b
b
5
24
0.2
0.1
5
32
4
13
10
f
13
b 0.45
1 a i 2
ii 9
b
2 a i 7
ii 10
3 a
7
38
iii
iv
35
38
v
25
38
Exercise 3C
1 a i {4, 5, 6} ii {2, 4, 6}
3
5
33
c
500
7
f
500
iii {2, 4, 5, 6}
2
c
3
c 0.7
e
1
3
d 1
1
iii
52
ii {2, 3, 5, 7, 11, 13, 17, 19}
13
ii
20
1
2
3
ii
13
9
13
iv
1
3
b i
14
17
ii
4
17
iii
4
7
iv
2
7
c i
3
4
ii
5
8
iii
5
7
iv
5
6
d i
7
16
ii
1
8
iii
1
4
iv
2
7
5 a i
7
18
ii
1
9
iii
1
5
iv
2
7
b i
4
9
ii
1
9
iii
1
5
iv
1
4
c i
8
17
ii
7
17
iii
7
10
iv
7
8
d i
3
4
ii
1
4
iii
2
3
iv
1
3
B′
b
10
g
13
7
13
1
5
V
3
A
A′
9
6
15
4
1
5
13
7
20
3
5
c
7 a
25
h
26
12 Because P (A ∩ B ) = 0 for mutually exclusive events.
7
12
3
7
B
d
c
iii
6 a
7
13
2
9
7
b
10
b
4 a i
iv {4, 6}
iii 1
c
f
Exercise 3D
3
b i 4
9
20
71
b
500
1
e
25
d
1
1
b 1
3
c i
5
18 a
13 a P (A) < P (A ∩ B )
b P (A) + P (B ) < P (A ∪ B )
14 P (A ∪ B ∪ C ) = P (A ) + P (B ) + P (C ) − P (A ∩ B ) −
P (A ∩ C ) − P (B ∩ C ) + P (A ∩ B ∩ C )
3
1
3
15 a
b
c
10
4
20
Answers
b i
d
9
13
d
1
4
P
2
6
4
b 4
c
2
5
765
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
8 a
A
14
ii
C
6
1st
1
2nd
A
A′
C
6
9
15
C′
14
1
15
20
10
30
3
10
b i
ii
7
15
2
5
c
d
3
10
9 a
A
A′
B
2
2
4
B′
3
1
4
5
3
8
i 1
b
ii
2
5
B
3
13
16
B′
5
6
11
8
19
27
3
ii
8
1
b
13
3
iii
16
1
c
4
1
13
d
1
1
b
11 a
3
2
12 P (A | B) = P (B | A) = 0 as P (A ∩ B) = 0
1
b
13 a 1
5
14 a P (A ∩ B) = P (A) × P (B | A)
15 a 329
174
b
329
24
d
155
31
e
231
(O, D)
(G, D)
O
(D, O)
X
(G, O)
G
(D, G)
(O, G)
X
b i 9
ii
d i 0
ii
2 a 9
3 a
2nd roll
5
9
2
3
b 6
iii
4
9
iv
8
9
v
2
9
iii
1
3
iv 1
v
1
3
2
3
4
1
(1, 1)
(2, 1)
(3, 1)
(4, 1)
2
(1, 2)
(2, 2)
(3, 2)
(4, 2)
3
(1, 3)
(2, 3)
(3, 3)
(4, 3)
4
(1, 4)
(2, 4)
(3, 4)
(4, 4)
c
1
16
1
d i
4
4 a
ii
5
8
iii
13
16
1st toss
2nd toss
1
2
H
T
H
(H, H)
(T, H)
T
(H, T)
(T, T)
1
c
4
b 4
1
d i
2
ii
3
4
e 250
5 a
1st
S
E
T
S
×
(E, S)
(T, S)
E
(S, E)
×
(T, E)
T
(S, T)
(E, T)
×
b 0.18
2nd
18
31
Exercise 3E
1
b 16
81
c
329
f
ii 6
1
c i
3
b i
1
6
ii
2
3
iii
2
3
iv
1
3
v 1
6 a
1st
1 a i
1st
2nd
G
X
1
2
iii
A′
10 a
O
D
1st roll
A
i 6
D
9
L
E
V
E
L
D
O
G
L
×
(E, L)
(V, L)
(E, L)
(L, L)
D
(D, D)
(O, D)
(G, D)
E
(L, E)
×
(V, E)
(E, E)
(L, E)
O
(D, O)
(O, O)
(G, O)
G
(D, G)
(O, G)
(G, G)
2nd
V
(L, V)
(E, V)
×
(E, V)
(L, V)
E
(L, E)
(E, E)
(V, E)
×
(L, E)
L
(L, L)
(E, L)
(V, L)
(E, L)
×
766
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
13 a
ii 12
3
ii
5
iii 12
3
iii
5
1st
1
e
5
7 a
Die 1
b 36
c i 2
e
8 a
1
6
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
ii 6
ii
1
6
iii
35
36
iv
1
12
1
. Min’s guess is wrong.
6
2nd
b 21
9 a i 100
1
b i
10
19
100
1
10 a i
4
b
20
12.5
22.5
5
7.5
10
15
25
12.5
15
20
30
20
22.5
25
30
40
O
L
D
C
(O, C)
(L, C)
(D, C)
O
(O, O)
(L, O)
(D, O)
L
(O, L)
(L, L)
(D, L)
L
(O, L)
(L, L)
(D, L)
E
(O, E)
(L, E)
(D, E)
G
(O, G)
(L, G)
(D, G)
E
(O, E)
(L, E)
(D, E)
1
c
7
ii 90
1
ii
10
2
i
5
without
30
1
i
15
1
c
18
b 16
c i 1
e
1 a i
2
5
ii
3
5
c i
1
4
ii
3
4
5
8
3
8
3
16
2
iii
15
M, F
3 5 15
× =
8 8 64
3
8
M
F, M
5 × 3 = 15
8 8 64
5
8
F
F, F
2
7
M
M, M
5
7
F
M, F
3 5
×
8 7
= 15
56
3
7
M
F, M
5 3
×
8 7
= 15
56
4
7
F
F, F
5 4
×
8 7
5
= 14
F
d
3
8
3 2
×
8 7
3
= 28
3
4
b
1
2
5 × 5 = 25
8 8 64
M
Counter
Outcome
Probability
1
4
yellow
(A, yellow)
1 1 1
× =
2 4 8
3
4
orange
(A, orange)
1 3 3
× =
2 4 8
3
4
yellow
(B, yellow)
1 3 3
× =
2 4 8
1
4
orange
(B, orange)
1 1 1
× =
2 4 8
A
d without
4
iv
15
3
5
3× 3 = 9
8 8 64
F
F
1
4
1
2
ii
M, M
5
8
Box
2
iii
3
c with
2
5
b i
M
c
1
15
iv
M
b
4
5
5
8
1
ii
10
b with
ii
3
8
3
8
3 a
ii
iii 8
1
iii
4
7
16
5
8
iii
ii 8
1
ii
8
1
d i
16
2 a
c
11 a
12 a
10
7.5
Exercise 3F
iii 15
1st
b
5
5
10
2nd
Die 2
d i
2.5
2.5
Answers
b 20
c i 8
2
d i
5
B
e
1
2
767
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
4 a
1st toss
2nd toss
1
4 1
4
1
1
4
1
4
1
4
1
4 1
4
2
1
4
1
4
1
4
1
4
1
4 1
4
3
1
4
1
4
1
4
1
4 1
4
4
1
4
1
4
Outcome
1
(1, 1)
2
3
(1, 2)
(1, 3)
4
1
(1, 4)
(2, 1)
2
3
(2, 2)
(2, 3)
d i
2
3
(3, 2)
(3, 3)
4
1
(3, 4)
(4, 1)
2
3
(4, 2)
(4, 3)
4
(4, 4)
ii
1
4
iii
5
8
5 a
Outcome
3
5
2
3
1
3
R
W
(R, R)
2 3 2
× =
3 5 5
(R, W)
2 2 4
× =
3 5 15
4
5
R
(W, R)
1 4 4
× =
3 5 15
1
5
W
(W, W)
1 1 1
× =
3 5 15
W
4
b i
15
2
ii
5
8
iii
15
2
c i
9
6 a
4
ii
9
4
iii
9
3
7
4
7
2
7
iii
4
7
iv
3
7
b i
9
49
ii
16
49
iii
24
49
iv
25
49
7 a
Probability
(M, M)
3 1 1
× =
7 3 7
2
3
F
(M, F)
3 2 2
× =
7 3 7
F
4
9
G
B
G
G
3
9
4
8
5
8
G
(B, B, G)
B
(B, G, B)
3
8
G
(B, G, G)
5
8
B
(G, B, B)
3
8 6
8
G
(G, B, G)
B
(G, G, B)
2
8
G
(G, G, G)
b i P (B, B, G) + P (B, G, B) + P (G, B, B)
= 3 × P (B, B, G)
6 5 4
=3×
× ×
10 9 8
1
=
2
1
2
1
2
(F, M)
4 1 2
× =
7 2 7
(F, F)
4 1 2
× =
7 2 7
white (Falcon, white)
1 3 3
× =
2 4 8
1
4
silver
(Falcon, silver)
1 1 1
× =
2 4 8
2
3
white (Commodore, white)
1 2 1
× =
2 3 3
1
3
red
1 1 1
× =
2 3 6
Commodore
3
8
7
24
(Commodore, red)
ii
1
6
iii
17
24
v
5
6
vi
1
3
Outcome
1
2
1
2
i
Probability
3
4
Falcon
1
4
1
2
R
(R, R)
1
2
W
(R, W)
1
2
R
(W, R)
1
2
W
(W, W)
R
M
1
2
B
9 a
M
M
5
9
6
9
4
10
iv
Outcome
F
Outcome
(B, B, B)
B
6
10
b i
1
3
1
2
B
4
8
Probability
R
2
5
ii
ii P (at least 1 girl) = 1 – P (B, B, B)
6 5 4
=− × ×
10 9 8
5
=
6
8 a
Outcome
1
ii
4
1
16
1
7
(2, 4)
(3, 1)
4
1
b 16
1
c i
16
a i
W
ii
1
2
iii
3
4
iv
3
4
768
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
1
2
1
2
2
3
W
(R, W)
1 2 1
× =
2 3 3
2
3
R
(W, R)
1 2 1
× =
2 3 3
1
3
W
(W, W)
1 1 1
× =
2 3 6
2
ii
3
5
iii
6
1
5
4
5
U
(U, U)
1 1 1
× =
5 9 45
d
8
9
N
(U, N)
1 8 8
× =
5 9 45
15 a i
2
9
U
(N, U)
4 2 8
× =
5 9 45
7
9
N
(N, N)
4 7 28
× =
5 9 45
1 a i
iii 0.83
iii 0.8555
2 a i
R
R
B
3
4
1
2
3
5
R
B
B
1
2
0
R
(R, R, R)
1
B
(R, R, B)
1
3
R
(R, B, R)
B
(R, B, B)
2
3
ii
1
2
1
10
ii
b i 1
ii
3
10
1
3
R
(B, R, R)
2
3 2
3
B
(B, R, B)
R
(B, B, R)
1
3
B
(B, B, B)
2
5
iii 0
(A, G)
1 3 1
× =
3 4 4
1
2
P
(B, P)
1 1 1
× =
3 2 6
1
2
G
(B, G)
1 1 1
× =
3 2 6
3
4
P
(C, P)
1 3 1
× =
3 4 4
1
4
G
(C, G)
1 1 1
× =
3 4 12
1
6
iii
1
4
1
2
1
c
2
ii
Probability
0
1
10
1
10
1
5
1
10
1
5
1
5
1
10
3
10
ii
b no
3 a with
4 a
iv
9
10
v
9
10
1
3
c no
b without
A
1
B
2
2
3
3
b i
8
1
ii
2
c not independent
5 a
A
B
2
i
G
1
2
b yes
Outcome
1
4
3
4
Exercise 3G
44
45
13 a
1 1 1
× =
3 4 12
7
1
ii
8
8
b $87.50 to player A, $12.50 to player B
c i A $68.75, B $31.25
ii A $50, B $50
iii A $81.25, B $18.75
iv A $34.375, B $65.625
4
7
b
(A, P)
B
b 6
1
c i
12
1
9
ii 0.11
ii 0.0965
P
C
Probability
iii
1
3
1
3
Outcome
16
45
ii
3
7
2
5
1
3
5
iv
6
N
1
45
Outcome Probability
1
4
A
U
c 62.2%
11 a i 0.17
b i 0.1445
12 a
14 a
4
5
ii
b
i
1 1 1
× =
2 3 6
W
1
5
10 a i
Probability
(R, R)
R
R
1
6
i
Outcome
1
3
Answers
b
2
1
1
2
2
ii
3
3
c independent
b i
769
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
6 a i
3 1
,
4 2
ii not independent
6
b i
1 1
,
4 4
ii independent
c i
1 1
,
3 3
ii independent
d i
2
,0
7
ii not independent
3
5
1
8
12
9 true
7
Multiple-choice questions
1
1
7 a P(A) = , P (A | B ) = , independent
2
2
1 A
6 B
3
1
, P (A | B ) = , not independent
b P (A) =
10
4
c P (A) =
8
2 B
7 D
1 a
1
8
5
8
5
2 a
8
3 a i
T′
8
2
G
G′
15
i
17
b no
9 a
1
32
1
10 a
216
b
10
7
0
7
15
2
17
7
ii
17
31
32
1
b
216
c
5
63
64
2
5
1
10
3
b i
5
4 a
C
1
b
2
c
3
8
c
5
8
iii
1
5
H
6
5
b
1
d
36
C
C′
H
6
5
11
H′
12
13
25
18
18
36
d 2
c 13
d i
c 0.144
4
7
3
c
4
1
4
1
v
20
17
ii
20
ii
iv
1
6
c
2
d
3
ii
5
36
6 a i 13
6
13
ii 4
3
b i
4
1
ii
52
4
13
7 a 0.5
2
8 a
5
10
13
b 0.5
1
b
5
5 a 6
b 1
5 A
10 E
13
Challenges
b 0.192
1
4
1
2
1
b
2
12
31
32
1
c
72
4 E
9 A
e
4
iii
5
2
11 False; P (A | B ) = 0 but P (A) = .
9
b 22
c 49
12 a 6
13 a 0.24
b 0.76
5
14
6
1 a 0.16
2 0.593 75
7
3 a
8
1
4 a
12
b
d
2
T
3 C
8 C
Short-answer questions
5
3
, P (A | B ) =
, not independent
12
20
1
1
d P (A) = , P (A | B ) = , independent
9
9
8 a
T
G
7
1
13 983 816
c
iii
1
2
b
iii 1
d
770
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
iii
iii
Extended-response questions
1
5
1 a 3
7
b i
15
c
1
2
1st
2nd
H
A
P
P
Y
H
(H, H)
(A, H)
(P, H)
(P, H)
(Y, H)
E
(H, E)
(A, E)
(P, E)
(P, E)
(Y, E)
Y
(H, Y)
(A, Y)
(P, Y)
(P, Y)
(Y, Y)
d i
1
15
ii
R
R′
S
3
1
4
S′
3
8
11
6
9
15
1
2
3
4
ii
2 a
1st
b 15
c i
1
15
ii
11 a
2
15
iii
1st
2nd
1
4 1
4
1
1
4
1
4
1
4
1
4
1
4 1
4
2
1
4
1
4
1
4
1
4 1
4
3
1
4
1
4
1
4
1
4 1
4
4
1
4
1
4
b i
1
16
ii
12
2
5
3
5
2
5
13 a 0.12
a
Answers
4
5
ii
11
11
b No, P (A | B ) ≠ P (A )
1
1
c i
ii
2
4
d Yes, P (A | B) = P (A )
10 a
9 a i
13
15
2nd
Total
1
2
2
3
3
4
4
1
5
3
2
3
4
5
4
1
6
4
2
3
5
6
4
1
7
5
2
3
6
7
4
8
1
4
b i
1
9
R
S
W
R
(R, R)
(S, R)
(W, R)
S
(R, S)
(S, S)
(W, S)
W
(R, W)
(S, W)
(W, W)
1
3
5
d
9
ii
c 4
e
iv
4
9
iv
16
21
Outcome
1
6
R
1
3
1
2
2
7
2
7
1
3
S
1
6
1
2
3
7
1
3
W
1
3
i
iii 0
5
9
iii
1
21
ii
10
21
1
3
R
(R, R)
S
(R, S)
W
(R, W)
R
(S, R)
S
(S, S)
W
R
(S, W)
(W, R)
S
(W, S)
W (W, W)
iii
6
7
iv 1
Chapter 4
1
4
F
3
4
M
1
2
F
1
2
M
F
Pre-test
M
b
3
10
b 0.58
c
3
10
d
3
5
e
7
10
1 a
b
c
2 a
b
c
i mean = 5.9
ii median = 6
iii modes = 2, 6, 9
i mean = 44.6 ii median = 41 iii mode = 41
i mean = 0.7
ii median = 0.65 iii mode = 0.3
6
i 19
ii 23
30
d 10%
771
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
c 82
d 32.5%
c 111, 139 are most frequent.
b
Class interval
Frequency
Percentage
frequency
80 −84
8
16%
85 −89
23
46%
90 −94
13
26%
Exercise 4A
1 Answers will vary and should be discussed in class.
2 a E
b F
c A
d B
e H
f D
g G
h C
3 a B
b E
c C
d D
e F
f A
4 C
5 D
6 a numerical and discrete
c categorical and nominal
e categorical and ordinal
7 D
95 − 100
Total
12%
100%
4 a
Percentage
Class interval
Frequency
frequency
0−4
5
25%
5−9
9
45%
b numerical and discrete
d numerical and continuous
10 − 15
Total
6
30%
20
100%
b My interpretation:
Histogram of wins
10
Frequency
8 C
9 D
10 a T he first 30 students are likely to be keen, conscientious
students.
b T he students in the highest Maths class are probably good at
mathematics and they probably like it too.
c S ome classes might be bigger than others, so the number
of students chosen from each class should be related to the
number of students in that class.
11 Answers will vary and should be discussed in class.
12 Answers will vary and should be discussed in class.
13 Answers will vary and should be discussed in class.
14 Answers will vary and should be discussed in class.
15 Answers will vary and should be discussed in class.
16 Answers will vary and should be discussed in class.
6
50
50
8
40
6
30
4
20
2
10
2.5
Percentage frequency
3 a 8
b 40
4 a 15
b 123 g
d 47
5 Q1 = 4, Q3 = 12, IQR = 8
7.5 12.5
Wins
c
Stem
Leaf
0
01344556778999
1
012235
d 7.5
Exercise 4B
1 a 10
b 1.4
2 a numerical
c categorical (ordinal)
3 a
Class interval
5 a
c 1
d 1
e 90%
b categorical (nominal)
d numerical
Frequency
Type of transport
Percentage
frequency
Frequency
Percentage
frequency
Car
16
40%
Train
6
15%
Tram
8
20%
Walking
5
12.5%
5%
0 −9
2
20%
Bicycle
2
10 −19
1
10%
Bus
3
7.5%
40
100%
20 −29
5
50%
30 − 40
2
20%
10
100%
Total
Total
b
6 a
c
d
7 a
b
i 6 ii car iii 40% iv 17.5% v 42.5%
symmetrical b negatively skewed
positively skewed
symmetrical
i 34.3
ii 38
iii 39
i 19.4
ii 20
iii no mode
772
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Mass
0
1
2
3
Nick’s goal scoring
b
0
1
2
3
4
Jack’s goal scoring
5
Sensor A
frequency
Sensor B
frequency
Sensor C
frequency
0 −2
21
12
6
3 −5
0
1
11
6 −8
0
1
3
9 −11
0
1
1
12 −14
0
0
0
15 −17
0
2
0
18 −20
0
2
0
21 −23
0
1
0
24 − 27
0
1
0
21
21
21
Total
b
Frequency
25
Sensor A
20
15
10
10 −14
3
6%
15 −19
6
12%
20 −24
16
32%
25 −29
21
42%
30 − 35
4
8%
50
100%
b 50
c 32%
d At least 25 g but less than 30 g.
e 42%
f 94%
11 a
Section
Frequency
Percentage frequency
Strings
21
52.5%
Woodwind
8
20%
Brass
7
17.5%
Percussion
4
10%
40
100%
Total
b 40
c 52.5%
d 47.5%
e 9.3%
f 65.6%
12 Eight students scored between 20 and 30 and there are
32 students all together, so this class interval makes up 25%
of the class.
13 No discrete information; only intervals are given, not
individual values.
14 3 ≤ a ≤ 7, 0 ≤ b ≤ 4, c = 9
15 a
5
0
Bill ($)
Frequency
0−
2
2
40−
1
3
8.1%
80−
12
15
40.5%
6
120−
18
33
89.2%
4
160−
3
36
97.3%
2
200−240
1
37
100%
14
10
8
0
1.5 4.5 7.5 10.5 13.5 16.5 19.5 22.5 25.5
12
Sensor C
10
8
6
4
2
100
40
80
30
60
10
20
1.5 4.5 7.5 10.5 13.5 16.5 19.5 22.5 25.5
ii very sensitive iii moderately sensitive
20
40
0
c i insensitive
5.4%
b 15
c
Percentage cummulative
frequency
Sensor B
12
Frequency
Percentage
cumulative
frequency
Cumulative
frequency
1.5 4.5 7.5 10.5 13.5 16.5 19.5 22.5 25.5
Frequency
Percentage
frequency
Frequency
Total
c Well-spread performance.
d Irregular performance, positively skewed.
9 a
Answers
10 a
8a
d i $130
e $180
0
40
80
120
Bill ($)
160
200
240
ii $100
iii $150
f approx. 20%
773
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Exercise 4C
1 a Min, lower quartile (Q1), median (Q2), upper quartile (Q3), max
bRange is max − min; IQR is Q3 − Q1. Range is the spread of
all the data; IQR is the spread of the middle 50% of data.
c A n outlier is a data point (element) outside the vicinity of the
rest of the data.
d W
hen the data point is greater than Q3 + 1.5 × IQR or less
than Q1 − 1.5 × IQR.
2 a 0, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 4, 8
b 2
c i 1
ii 3
d 2
e −2, 6
f Yes, the data value 8 is an outlier.
3 a i 10.5
ii 7.5
iii 12
b 4.5
c 0.75, 18.75
d Yes, the data value 0 is an outlier.
4 a m
in = 3, Q1 = 4, median = 8, Q3 = 10, max = 13, range = 10,
IQR = 6
b m
in = 10, Q1 = 10.5, median = 14, Q3 = 15.5, max = 18,
range = 8, IQR = 5
c min = 1.2, Q1 = 1.85, median = 2.4, Q3 = 3.05,
max = 3.4, range = 2.2, IQR = 1.2
d m
in = 41, Q1 = 53, median = 60.5, Q3 = 65, max = 68,
range = 27, IQR = 12
5 a median = 8, mean = 8
b median = 4.5, mean = 4.2
c median = 4, mean = 4.3
6 a min = 0, max = 17
b median = 13
c Q1 = 10, Q3 = 15
d IQR = 5
e 0
f Road may have been closed that day.
7 a i min = 4, max = 14
ii 7.5
iii Q1 = 5, Q3 = 9
iv IQR = 4
v no outliers
b i min = 16, max = 31
ii 25
iii Q1 = 21, Q3 = 27
iv IQR = 6
v no outliers
8 a i min = 25, max = 128
ii 47
iii Q1 = 38, Q3 = 52.5
iv IQR = 14.5
v Yes; 128
vi 51.25
b Median, as it is not affected dramatically by the outlier.
c A more advanced calculator was used.
9 a no outliers
b Outlier is 2.
c Outliers are 103, 182.
d Outliers are 2, 8.
10 a IQR = 12
b no outliers
c 24
d 22
11 1, 2, 3
12 a Increases by 5.
b It is doubled.
c It is divided by 10.
13 a It stays the same.
b It doubles.
c It is reduced by a scale factor of 10.
14 Answers may vary. Examples:
a 3, 4, 5, 6, 7
b 2, 4, 6, 6, 6
c 7, 7, 7, 10, 10
15 It is not greatly affected by outliers.
Exercise 4D
1 a
e
2 a
d
3 a
b
f
b
e
15
10
4
20
c 25
g 10
c 18
5
20
2
It is.
d 20
1 2 3 4 5 6 7 8
b
5 6 7 8 9 10 11 12 13 14 15 16 17
4 a i Q1 = 4, Q3 = 7; outlier is 13
ii
2
4
6
8
10
12
14
1.8
2.0
2.2
b i Q1 = 1.6, Q3 = 1.9; outlier is 1.1
ii
1.0
1.2
1.4
1.6
c i Q1 = 19, Q3 = 23; outliers are 11 and 31
ii
10 12 14 16 18 20 22 24 26 28 30 32
d i Q1 = 0.03, Q3 = 0.05; no outliers
ii
0.02
0.04
0.03
0.06
0.05
0.07
5 a
0 2 4 6 8 10 12 14 16 18 20
b
36
44
40
48
52
60
64
c
0
1
2
3
4
5
70
80
90
100
110
120
d
130
774
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
13 14 15 16 17 18 19 20
Lemur weights
8 a
b
c
d
9 a
10 a
They have the same median and upper quartile.
B
i 4
ii 5
Set B is more spread out.
A
b B
c B
Box plot of Set 1, Set 2
Set 2
Set 1
0 2 4 6 8 10 12 14 16
Spelling errors
b Yes, examiner 2 found more errors.
11 Answers may vary. Examples:
a i, ii Class results have a smaller spread in the top 25%, and
bottom 25% performed better.
iii State results have a larger IQR.
b The class did not have other results close to 0 but the
school did.
Exercise 4E
1 a larger
b smaller
2 a B
b A
3 a A
b The data values in A are spread farther from the mean than
the data values in B.
4 a Gum Heights
b Gum Heights
5 a mean = 6, σ = 2
b mean = 3.6, σ = 2.3
c mean = 8, σ = 3.5
d mean = 32.5, σ = 3.3
6 a mean = 2.7, σ = 0.9
b mean = 14.5, σ = 6.0
7 a The outer suburb has more data values in the higher range.
b There is less spread. Data values are closer to the mean.
c Students at outer-suburb schools may live some distance
from the school.
8 a false
b true
c true
Answers
Box plot of lemur weights
mean = 2, σ = 0.9
mean = 5.25, σ = 0.7
no
b no
c yes
Yes, one of the deviations would be calculated using the
outlier.
11 a No; standard deviation reflects the spread of the data values
from the mean, not the size of the data values.
b No. As for part a.
12 The IQRs would be the same, making the data more
comparable.
13 a i 85.16
ii 53.16
iii 101.16
iv 37.16 v 117.16
vi 21.16
b i 66%
ii 96%
iii 100%
c i Student’s own research required.
ii One SD from the mean = 68%
Two SDs from the mean = 95%
Three SDs from the mean = 99.7%
Close to answers found.
9 a
b
10 a
d
Exercise 4F
1 a linear
b no trend c non-linear d linear
2 a i 28°C
ii 33°C
iii 33°C
iv 35°C
b 36°C
c i 12 noon to 1 p.m.
ii 3−4 p.m.
d Temperature is increasing from 8 a.m. to 3 p.m. in a
generally linear way. At 3 p.m. the temperature starts
to drop.
3 a
Population
Same minimum of 1.
B
i 5
ii 10
Data points for B are more evenly spread than those for A.
Q1 = 14.6, Q2 = 15.3, Q3 = 15.8
19.7 kg
1000
900
800
700
600
500
400
0
1998 2000 2002 2004 2006 2008 2010 2012
Year
b Generally linear in a positive direction.
c i 500
ii 950
4 a
Price ($)
6 a
b
c
d
7 a
b
c
1.45
1.4
1.35
1.3
1.25
1.2
0
J
F M A M J J A S O N D
Month
775
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
100
90
80
70
0
1994 1996 1998 2000 2002 2004
Year
Sales ($’000)
b T he pass rate for the examination has increased marginally
over the 10 years, with a peak in 2001.
c 2001
d 11%
6 a linear
b i $650 000
ii $750 000
7 a i $6000
ii $4000
b 1
c
20
Southbank
18
16
14
12
10
City Central
8
6
4
2
0
Jul Aug Sept Oct Nov Dec
Month
d i The sales trend for City Central for the 6 months is fairly
constant.
ii Sales for Southbank peaked in August before taking a
downturn.
e about $5000
8 a i 5.84 km ii 1.7 km
b i Blue Crest slowly gets closer to the machine.
ii Green Tail starts near the machine and gets farther from it.
c 8.30 p.m.
9 a T he yearly temperature is cyclical and January is the next
month after December and both are in the same season.
b no
cNorthern Hemisphere, as the seasons are opposite; June is
summer.
10 a Increases continually, rising more rapidly as the years progress.
b Compound interest due to exponential growth.
11 a G
raphs may vary, but it should decrease from room
temperature to the temperature of the fridge.
b N
o. Drink cannot cool to a temperature lower than that of the
internal environment of the fridge.
12 a
Innings
1
Score
2
3
4
5 6
7
8
5 10 52 103 75 21
26 38
Moving
26 32 23 20 26
average
39 44 41
9 10
33
0
40 36
b
Number of runs
Pass rate (%)
b T he share price generally increased until it peaked in June
and then continually decreased to a yearly low in November
before trending upwards again in the final month.
c $0.21
5 a
120
100
80
60
40
20
0
Score
Moving
average
0
2
4
6
8
Innings number
10
12
c Innings number.
i The score fluctuates wildly.
ii The graph is fairly constant with small increases and
decreases.
dThe moving average graph follows the trend of the score
graph but the fluctuations are much less significant.
Exercise 4G
1 a unlikely
d likely
2 a i y
b likely
e likely
c unlikely
f likely
Scatter plot
12
10
8
6
4
2
0
2
4
6
8
10
x
ii y generally increases as x increases.
b i y
Scatter plot
10
8
6
4
2
0
0.0
0.5
1.0
1.5
2.0
2.5
x
ii y generally decreases as x increases.
776
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
6 a
c
7 a
c
8 a
Scatter plot
1.8
1.7
1.6
1.5
1.4
1.3
1.2
1.1
1.0
0.9
0
b positive
4 a y
1
2
3
4
c strong
5
6
7
x
8
d (8, 1.0)
Scatter plot
4
b
d
b
ii
none
positive
yes
i yes
weak negative
strong positive
decrease
car H
Answers
y
Scatter plot
Average diameter (cm)
3 a
8.5
E
8.0
7.5
B
7.0
C
A
6.5
D
6.0
20
3
25
30
35
Fertiliser (grams per week)
40
b D
c Yes, although small sample size does lead to doubt.
9 a
Scatter plot
2
1
5.0
7.5
b negative
5 a negative
10.0
c strong
y
12.5
15.0
x
No. of words
0
d (14, 4)
Scatter plot
24
22
20
18
16
14
12
10
8
2250
2000
1750
1500
1250
1000
750
500
250
0
1
2
3
4
5
No. of photos
6
7
b negative, weak correlation
10 a
Scatter plot
4.5
x
Volume (dB)
1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9
b positive
y
Scatter plot
150
4.0
3.5
3.0
2.5
2.0
1.5
140
130
0
120
110
100
0
2
4
6
8
10
x
b negative
c As Distance increases, Volume decreases.
11 a i weak, negative correlation
Scatter plot
c none
Scatter plot
26
24
22
20
18
16
14
12
10
8
Incidence of crime
y
200 400 600 800 1000120014001600
Distance (m)
35
30
25
20
15
5
15.0 17.5 20.0 22.5 25.0 27.5 30.0 32.5
x
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
10
15
20
25
No. of police
30
777
Cambridge University Press
b
d
5 a
Scatter plot
26
25
24
23
22
21
20
19
positive correlation
c see part a
All answers are approximate.
i 3.2
ii 0.9
iii 1.8
≈ 4.5
b ≈6
c ≈ 0.5
bSurvey 1, as this shows an increase in the number of police
has seen a decrease in the incidence of crime.
12 The positive correlation shows that as height increases, the
ability to play tennis increases.
13 Each axis needs a better scale. All data lie between 6 and 8 hours
sleep and shows only a minimum change in exam marks.
14 a i students I, I
ii students G, S
b i students H, C
ii students B, N
c students G, S d students B, I, N, R
e no
Exercise 4H
1 a y
b
y
x
x
c y
d
y
x
x
1
7
2 a y = x + 2
2
Growth (cm)
b i 42
c i 30
7 a, b
7.5 10.0 12.5 15.0 17.5 20.0 22.5 25.0 27.5
No. of police
ii 72
ii 100
Scatter plot
90
80
70
60
50
40
30
20
400
Max. heart rate (b.p.m.)
5
4
3
2
160
140
120
100
2
3
4
5
30
40
50
Age (years)
6
7
60
70
Scatter plot
200
180
160
140
120
100
20
1
20
220
1
0
800
180
Experiment 2
6
600
700
Rainfall (mm)
200
10
23
4
28
14
b i
ii
5
5
4 a y
Scatter plot
500
c i ≈ 25 cm ii ≈ 85 cm
d i ≈ 520 mm ii ≈ 720 mm
8 a y = 5x − 5
b 85 cm
c 21 kg
9 a Data does not appear to have any correlation.
b Too few data points.
10 a Too few data points to determine a correlation.
bThe data points suggest that the line of best fit is not linear.
11 a i 50
ii 110
b It is possible to obtain scores of greater than 100%.
12 a Experiment 1
Scatter plot
2
17
b y=− x+
3
3
3 a i 17 ii
0
iv 7.4
d ≈ 50
3
6 a y = x + 18
5
Max. heart rate (b.p.m.)
Incidence of crime
ii no correlation
x
30
40
50 60
Age (years)
70
80
90
778
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
i ≈ 140
ii ≈ 125
i ≈ 25
ii ≈ 22
experiment 2
Student’s own research required.
C a, c
Answers
b
c
d
e
Scatter plot
y
8
7
Exercise 4I
6
1 a
b
2 a
b
3 A
i 12
ii 3.26
i 7
ii 2
There is no linear correlation.
The correlation shown is not a linear shape.
a, c
Scatter plot
5
y
1
4
3
2
x
0
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6
8
b y = −3.45129x + 7.41988
4 a y = −3.54774x + 43.0398
5 a, c
7
6
d i −16.7
b $32 397
ii −34.0
c 8 years
Scatter plot
5
160
4
140
2
1
x
0
0
1
2
3
4
5
6
7
8
Number of jackets
3
120
100
80
60
40
b y = 0.554762x + 3.45357
d i 7.3
ii 10.1
B a, c
Scatter plot
y
20
0
0
5
10
15 20 25 30
Temperature (°C)
35
40
b y = –1.72461x + 190.569
d i 139
ii 130
iii 113
6 a, b
Scatter plot
4.0
3.5
3.0
35
2.0
30
1.5
1.0
0.5
x
0.0
0
5
10
15
20
b y = −0.077703x + 4.21014
d i 3.7
ii 3.3
25
30
Number of breakdowns
2.5
25
20
15
10
5
0
0
20
40
60
80 100 120 140 160
–5
Number of copies (× 1000)
779
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Record (m)
120
8
50
6
4
25
2
110
2.5 7.5 12.5 17.5 22.5
Number of hours of TV
100
c It is positively skewed.
d 90
80
Stem
70
1970 1975 1980 1985 1990 1995 2000
Year
b i 139 m ii 170 m
cNo, records are not likely to continue to increase at this rate.
8 A ll deviations are used in the calculation of the least squares
regression.
9 Line A, as it has been affected by the outlier.
10 Student’s own research required.
Challenges
1
2
3
4
5
66 kg
88%
19
1.1
a larger by 3
b larger by 3
c no change
d no change
e no change
6 y = x2 − 3x + 5
7 5.8 ≤ a < 6.2
1
012346
2
4
2 a i 10 ii Q1 = 2.5, Q3 = 5.5 iii 3 iv 12
v
2
6
4
10
8
12
b i 17
ii Q1 = 15, Q3 = 24
iii 9 iv none
v
0.5
3 a false
4 a y
20
15
30
25
1.0
2.0
1.5
b true
2.5
c true
3.0
d true
Scatter plot
25
20
3 D
7 D
4 C
8 E
15
10
5
1
Short-answer questions
b negative
1 a
Class interval
Frequency
0 −4
2
5 −9
7
43.75
10 −14
5
31.25
15 −19
1
6.25
20 − 25
135667889
c i 2.4
ii Q1 = 2.1, Q3 = 2.6
iii 0.50 iv 0.7
v
2 E
6 C
10 B
Total
e 8.5 hours
Leaf
0
10
Multiple-choice questions
1 B
5 A
9 C
% Frequency
Scatter plot
b
Frequency
c 232 000 copies
d N
o, the percentage of breakdowns is still very low.
7 a
1
16
Percentage frequency
12.5
6.25
100
2
c weak
3
7
5 a y= x+
5
5
b i 3.8
ii 7.4
2
2
ii 17
c i 2
3
3
6 a non-linear 7 a mean = 7, σ = 2.3
8 a The Eagles
c The Eagles
9 y = −3.75x + 25.65
3
4
5
x
d (3, 5)
b
b
b
d
linear
mean = 4, σ = 2.8
The Eagles
The Eagles
780
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
1 a i 14
b i no outliers
c
8 a i x = –4, y = 4
iii
ii 41
ii no outliers
ii 1
y
y=x+4
(0, 4)
Tree 2
10
20
30 40 50 60 70
Number of flying foxes
80
x
O
(–4, 0)
Tree 1
90
d More flying foxes regularly take refuge in tree 1 than in
tree 2, for which the spread is much greater.
2 a positive correlation
b i x = 3, y = −2
2
3
ii
iii
y
y = 23 x − 2
Scatter plot
1800
Number of shoppers
Answers
Extended-response questions
(3, 0)
x
O
(0, –2)
1600
1400
1200
1000
c i No x-intercept, y = 4
iii
y
800
ii 0
600
25.0
27.5
30.0
32.5
35.0
Maximum daily temperature (°C)
y=4
37.5
(0, 4)
b y = 74.5585x − 1010.06
c i 779
ii 33.7°C
x
O
Chapter 5
Pre-test
1 a 2x + 1
b 5(x − 1)
2 a 7x
1
( x + 4)
3
b −5ab
3 a 8y
4 a 3x + 3
5 a 4m
7
6 a
6
4
c
7
b −21x
b −4x + 4
b 3
7
b
12
19
d
72
4
7
3
g
4
7 a 4
21
20
49
h
9
b 7
d 2x − 3
e
b −2
b y = −x + 5
9 a 3
10 a y = 2x + 1
e
1
3
c x+ x = x
2
2
f 3x − 7
c −5x 2
c 15a 2
c −10x + 2x 2
2x
c a
d
3
f
c 10
c 1
Exercise 5A
1 C
2 D
3 a 7
1
e
2
4 a yes
5 a 9
6 a 10a
e 4ab
i −3m2n
m 3a + 7b
q 5st − s 2t
7 a 12ab
e 30ht
i 8a 2b 4
b 1
f
b
b
b
f
j
n
r
b
f
j
c
1
g
−
5
yes
c
−8
c
15d
c
9t
g
−0.7a 2b k
8jk − 7j
o
3x 3y 4 + 2xy 2
25ab
c
30bl
g
24p3q
k
−4
2
7
no
−8
0
9b
2gh + 5
ab 2 + 10a 2b
d −9
h −7
3
−6ad
12s 2t
−18h5i 5
d −10hm
h −21b 2d 5
l 63m 2pr
d
d
h
l
p
−9
5xy
−st 2
12xy − 3y
2mn − m 2n
781
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
782
Chapter 1
mx
n 3ab
q 2b
r −3x
o −a
3
s −y
2
p − ab
4
t −a
2
5x + 5
b 2x + 8
3x − 15
d −20 − 5b
−2y + 6
f −7a − 7c
6m + 18
h 4m − 12n + 20
−2p + 6q + 4
j 2x 2 + 10x
2
6a − 24a
l −12x 2 + 16xy
2
15y + 3yz − 24y
n 36g − 18g 2 − 45gh
2
−8ab + 14a − 20a
p 14y 2 − 14y 3 − 28y
3
2
−6a + 3a + 3a
r −5t 4 − 6t 3 − 2t
5x + 23
b 10a + 26 c 21y + 3
d 15m + 6
10
f 11t − 1
g 3x 2 + 15x h 15z − 7
−11d 3
j 9q 4 − 9q 3
3(x − 3)
b 4(x − 2)
c 10(y + 2)
6(y + 5)
e x (x + 7)
f 2a (a + 4)
5x (x − 1)
h 9y (y − 7)
i xy (1 − y )
x 2y (1 − 4y )
k 8a 2(b + 5)
l ab(7a + 1)
−5t (t + 1)
n −6mn (1 + 3n)
o −y(y + 8z)
−3a(ab + 2b + 1)
−32
b 7
c 61
d 12
13
g −7
e −1
f
h 1
5
2
5
8 a
c
e
g
i
k
m
o
q
9 a
e
i
10 a
d
g
j
m
p
11 a
2x 2 + 6x
b x 2 − 5x
P = 4x − 4, A = x 2 − 2x − 4
P = 4x + 2, A = 3x − 1
P = 4x + 14, A = 7x + 12
(−2)(−2) = 4, negative signs cancel
a 2 > 0 ∴ −a 2 < 0
(−2)3 = (−2)(−2)(−2) = −8
true
b false, 1 − 2 ≠ 2 − 1
1 2
c true
d false, ≠
2 1
e true
f false, 3 − (2 − 1) ≠ (3 − 2) − 1
g true
h false, 8 ÷ (4 ÷ 2) ≠ (8 ÷ 4) ÷ 2
12 a
13 a
b
c
14 a
b
c
15 a
16 a x + y or x + y
2
2
b It could refer to either of the above, depending on
interpretation.
c ‘Half of the sum of a and b ’ or ‘a plus b all divided by 2’.


π
π
17 a P = 4 +  x + 2, A = 1 +  x 2 + x


2
4
Exercise 5B
1 a 11
8
2
2 a
3
3 a 12
b 3
8
b 3
7a
b 6
4 a 5x
b 4x
e 5x
f −2x
1
i −
2p
j
1
24
7t
c −
4 xy
c
d 3
c 14
c a
4
g −9b
2
d − bc
8x 2a
d 2x
d 1
3a
h −2y
3x
y
4
9st
b a−5
k −
c 3x − 9
6b
7
d 1 − 3y
e 1 + 6b
f 1 − 3x
g 3−t
h x−4
i x+2
j 3 − 2x
k a−1
l 1 + 2a
3
5 a x+2
−
6 a x −1
2x
4
d
9
b
g 2
7 a 3
d
3
4
g −5
3
8 a 3a + 14
21
1
−
6a
e
9
b 3
4
3
2
h
5
e
x+4
5x
l
c −4
e 5
f
5a
2
h 15
i
−
c
f
i
1
2
18
5
1
25
1
−
3
b 4a + 3
c 3 − 15b
8
10
3
a
+
8
7
a
− 27
f
g
4a
9a
27 − 14 y
j
k −12 − 2 x
18 y
3x
d 4x + 6
15
16
− 3b
h
4b
b 7 x + 11
12
c −x − 7
4
d 2 x + 15
9
e 4x + 7
6
f
8x + 3
10
g 7x + 2
24
h 5x − 1
5
i
5 − 3x
14
i
4 b − 21
14 b
9 a 9 x + 23
20
10 a
7 x + 22
( x + 1)( x + 4 )
b
7 x − 13
( x − 7 )( x + 2 )
c
3x − 1
( x − 3 )( x + 5 )
d
x − 18
( x + 3 )( x − 4 )
l


π
π
b P = 6 +  x − 6, A = 3 −  x 2 − 3 x


4
2
e
−21
(2 x − 1)( x − 4 )
f
14 x − 26
( x − 5 )(3 x − 4 )
g
h


c P = 2 π x , A = 1 + π  x 2

2
41− 7 x
(2 x − 1)( x + 7 )
3 x + 17
( x − 3 )(3 x + 4 )
i
14 − 17 x
(3 x − 2 )(1− x )
−27 − 2 x
6x
782
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
11 a i a 2
ii x 2
b i 2a − 3
a2
2
ii a + a − 4
a2
iii 3 x + 14
4x2
12 The 2 in the second numerator needs to be subtracted; i.e. x − 2 .
6
13 a −(3 − 2x ) = −3 + 2x (since −1 × −2x = 2x )
2
x −1
b i
iv −
ii
7
3
14 a −1
3x − x2
d
( x − 2 )2
iii x + 3
7−x
2x
3− x
v − 12
x
vi
7
2
b 5a + 2
a2
c
3x + 5
( x + 1)2
f
yz − xz − xy
xyz
21x − 9 x 2
e
14( x − 3 )2
15 a 2
13 a $214
b $582
c i 1
ii 10.5
iii 21
14 a 41 L
b 90 s = 1 min 30 s
c 250 s = 4 min 10 s
15 x = 9. Method 2 is better; expanding the brackets is
unnecessary, given that 2 is a factor of 8.
a
c 5
16 a 5 − a
b
6
a
c−b
d 2a + 1
e 3a + 1
f
a
a
a
17 a 6
b 4
c −15
d 20
e 3
f 6
g 1
h −26
i −10
18 a a =
c
b+1
b a=
d a=b
b
b+1
Answers
Number and Algebra
1
c−b
bc
a=
b−c
c a=
e a = −b
f
b 1
Exercise 5D
Exercise 5C
1
2
3
4
a
a
a
a
b
b
b
b
no
false
true
8
c
c
c
c
5
e
2
f
11
4
g −1
3
h − 11
6
i −4
j
3
2
k −9
2
l −4
3
m −2
n 7
o −2
p
b 9
23
c
2
g 1
d −5
6
h 2
k −9
l 5
no
true
false
5
5 a 1
e
i
m
6 a
e
i
m
7 a
c
2
9
f
−
3
11
4
j 7
19
n 23
10
b 13
−5
f 6
−9
j 8
20
n 15
x + 3 = 7, x = 4
x − 4 = 5, x = 9
c
g
k
o
b
d
yes
false
true
−3
d yes
d 4
11
9
13
14
9 a 1
10 17 cm
11 17 and 18
12 24 km
h
b 6
b x≥5
−22
d 4
16
h 4
6
l −7
−9
x + 8 = 5, x = −3
15 − x = 22, x = −7
2
5
i
c 2
x
c x≥4
g 3x + 8 = 23, x = 5
h 2x − 5 = x − 3, x = 2
8 a 1
b 0
c −17
7
27
28
d
e
f
2
23
5
g
x
4
5
f 2(x − 5) = −15, x = −
e 2x + 5 = 13, x = 4
1 a 3, 6, 10 (Answers may vary.)
b −4, −3, −2 (Answers may vary.)
c 5, 6, 7 (Answers may vary.)
d −8.5, −8.4, −8.3 (Answers may vary.)
2 a B
b C
c A
3 11, 12 or 13 rabbits
4 a x≥1
b x<7
c x≤4
d x > −9
e −2 < x ≤ 1 f 8 < x ≤ 11 g −9 < x < −7
h 1.5 ≤ x ≤ 2.5
i −1 ≤ x < 1
5 a x<4
94
11
d 25
x
4
d x ≤ 10
x
10
e x≤2
5
2
x
2
f x>3
x
3
g x>6
6
x
h x≤6
6
x
783
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
784
Chapter 1
i x < −18
d −3 ≤ x ≤2
2
x
–18
x
j x > 32
–1.5
k x≤
e −3 ≤ x ≤ 7
3
x
32
10
9
x
f −4 ≤ x ≤ −2
x
l x < −3
8
–4
x
–0.375
e x < −8
7 a x>6
1
16
b x<2
c x ≤ −5
d x ≤ −7
f x≥4
g x ≥ −10
h x < −21
b x≤2
c x<
f
x<
8 a 2x + 7 < 12, x <
b 4−
1
3
e x≥
x
≥ −2, x ≤ 12
2
d x + (x + 2) ≤ 24, x ≤ 11
27
29
f
x<
1
2
11 a An infinite number of whole numbers (i.e. all the ones greater
than 8).
b Only one as 3 is the only whole number.
12 a x ≥ a + 3
10
h 1≤ x <
c x < 1 − 7 or x < a − 7
a
a
b x < 2 − 4a
13 a −4 ≤ x < 5 b −9.5 < x ≤ −7 c x = 10
14 a 3 ≤ x ≤ 9
x
3
9
12
15
4
b x<
c x≤1
1 a y = −2x + 5, m = −2, b = 5
b y = 2x − 3, m = 2, b = −3
c y = x − 7, m = 1, b = −7
d y=−
2x 3
2
3
− , m=− , b=−
5 5
5
5
2 a i 3
ii 6
iii
21
2
b i 2
ii 6
iii
8
3
3 a i
b
c ii
d
e v
f
4 a yes
b
c no
d
e yes
f
5 a m = 5, b = −3
iv
iii
vi
yes
no
no
y
x
y = 5x − 3
3
c −9 ≤ x < −7
(0.6, 0)
x
–9
19
5
Exercise 5E
b −7 ≤ x ≤ 3
–7
x
15
4
1
15 a x ≥ 23
e (x − 6) + (x − 4) + (x − 2) + x ≤ 148, x ≤ 40
9 a i C < $1.30
ii C > $2.30
b i less than 9 min
ii 16 min or more
11
10 a x < −5
b x≥
c x ≥ 11
4
29
d x ≤ 14
5
11
d x ≥ 10
11
4
5
2
c 3(x + 1) ≥ 2, x ≥ −
5
2
–2
g 11 ≤ x ≤ 12
x
6 a x ≥−2
5
x
7
3
–3
10
9
e x≤
2
–7
O
(0, –3)
x
784
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
b m = 2, b = 3
g m=
y
4
, b = −2
3
y
y = 2x + 3
(0, 3)
(–1.5, 0)
y = 43 x − 2
x
O
O
(0, –2)
c m = −2, b = −1
y
(1.5, 0)
7
h m=− ,b=6
2
y
y = –2x − 1
(0, 6)
(–0.5, 0)
y = – 72 x + 6
x
O
(0, –1)
x
Answers
Number and Algebra
O
( , 0)
x
12
7
d m = −1, b = 2
i m = 0.5, b = −0.5
y
y
y = –x + 2
y = 0.5x − 0.5
(0, 2)
O
x
(2, 0)
(0, −0.5)
e m = 1, b = −4
O
(1, 0)
x
j m = −1, b = 1
y
y
y=x−4
y=1−x
O
x
(4, 0)
(0, 1)
O
(1, 0)
x
(0, –4)
3
f m= − ,b=1
2
y
2
k m= ,b=3
3
y
y = – 32 x + 1
( 23 , 0)
(0, 1)
O
(0, 3)
x
(–4.5, 0)
O
y = 23 x + 3
x
785
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
786
Chapter 1
l m = −0.2, b = 0.4
4
e m = , b = −3
3
y
y
y = 43 x − 3
y = 0.4 − 0.2x
(0, 0.4)
x
O (2, 0)
O
(0, –3)
6 a m = −3, b = 12
1
3
f m = −1, b = −
y
x
( 94 , 0)
y
(0, 12)
y = –x –
y = –3x + 12
1
3
(– , 0)
1
3
x
O (4, 0)
b m = −5, b =
(0, – )
1
3
5
2
g m = −4, b = −8
y
y
x
(0.5, 0)
y = –5x +
x
(–2, 0) O
(0, 2.5)
O
x
O
y = –4x – 8
(0, –8)
5
2
c m = 1, b = −7
1
1
h m=− ,b=
2
4
y
O
(7, 0)
(0, –0.25)
x
y
O
x
(0.5, 0)
y = 12 x − 14
y=x−7
(0, –7)
7 a x = 2, y = −6
d m = 1, b = −2
y
y
y=x−2
(2, 0)
O
(0, –2)
(2, 0)
O
x
y = 3x − 6
x
(0, –6)
786
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Number and Algebra
g x = 4, y = 6
Answers
b x = −2, y = 4
y
y
y = – 32 x + 6
y = 2x + 4
(0, 4)
(0, 6)
(–2, 0)
x
O
(4, 0)
x
O
c x = −2.5, y = 10
h x = 5, y = 2
y
y
y = 4x + 10
y = – 25 x + 2
(0, 2)
(0, 10)
(–2.5, 0)
x
O
O
x
(5, 0)
i x = −8, y = 6
4
d x = , y = −4
3
y
y
(0, 6)
y = 3x − 4
O
(43 , 0)
(0, –4)
x
y = 34 x + 6
x
O
(–8, 0)
j x = 5, y = 2.5
y
e x = 3.5, y = 7
y
y = – 12 x +
y = –2x + 7
(0, 7)
O
5
2
(0, 2.5)
x
O
(3.5, 0)
x
(5, 0)
7
7
k x= ,y=
3
4
f x = 8, y = 4
y
y
y = – 34x + 74
y = – 12 x + 4
(0, 1.75)
(0, 4)
O
O
(8, 0)
x
x
7
3
,0
787
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
788
Chapter 1
y
e
l x = −6, y = 12
5
y
y=0
2
y = 5x +
(0, 2.4)
(0, 0)
12
5
x
(–6, 0) O
f
8 a
x
O
y
y
x=0
x
O (0, 0)
y = –4
x
O
(0, –4)
g
y
(1, 4)
y
b
y = 4x
y=1
x
O (0, 0)
(0, 1)
x
O
h
y
y
c
y = –3x
x=2
(2, 0)
O
(0, 0)
x
O
x
(1, –3)
d
y
i
y = – 13 x
5
x=–2
(–2.5, 0)
y
O
x
(0, 0)
O
x
(3, –1)
788
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Number and Algebra
11 a
12 a
13 a
b
c
14 a
y
(2, 5)
O
y = 52 x
x
(0, 0)
k
b
c
y
d
(0, 0)
O
15 a
x
b
(1, –1)
$7 per hour
b P = 7t
$0.05/km b C = 0.05k c C = 1200 + 0.05k
m = 25
The cyclist started 30 km from home.
(0, 30)
1
y = x + , gradient = 1
2
y = 0.5x + 1.5, y-intercept = 1.5
y = −3x + 7, gradient = −3
1
1
y = x − 2, gradient =
2
2
gradient = 3 , y-intercept = 7
a
a
gradient = a, y-intercept = −b
Answers
j
a
c gradient = − , y-intercept = 3
b
b
l
y
b
17 a 12 sq. units
b 9 sq. units
4
d
x
O
d
b
16 a d
a
121
sq. units
5
c −
c
32
sq. units
3
e
a
b
121
sq. units
4
Exercise 5F
1 a y = 4x − 10
9 a C = 2n + 10
b $C
c y = −x − 7
35
30
25
C = 2n + 10
20
15
10
(0, 10)
5
0
2 a 2
d −4
(10, 30)
1 2 3 4 5 6 7 8 9 10
n ( kg)
V (L)
(0, 90)
b 3
e −3
c 0
f infinite
1
4
e 0
f 0
5
2
g −1
i −1
j infinite
k
3 a
c i $28
ii 23.5 kg
10 a V = 90 − 1.5t
b
90
80
70
60
50
40
30
20
10
b y = −x + 3
1
11
d y= x+
2
2
b 2
y=x+3
y = −3x + 4
y = 2x + 4
y = −2x + 12
y = 3x + 5
1
3
c y= x−
2
2
4 a
d
5 a
d
6 a
c
d 3
3
2
b
e
b
e
b
y=x−2
y=4
y = 4x − 5
y = −3x − 4
y = −2x + 4
d
y = −2x − 2
5
2
3
l −
2
y = 3x + 6
y = −7x − 10
y=x−4
y = −3x − 2
h
c
f
c
f
2
3
2
mBC =
3
Yes; they are collinear.
A = 500t + 15 000
$15 000
4 years more; i.e. 10 years from investment
$21 250
7 a mAB =
V = 90 − 1.5t
b
(60, 0)
0
5 10 15 20 25 30 35 40 45 50 55 60
c i 82.5 L
ii 60 hours
t (hours)
c
8 a
b
c
d
789
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
790
Chapter 1
9 a
$C
90
80
70
60
50
40
30
20
10
C = 10t + 20
(0, 20)
0
1 2 3 4 5 6 7 8 9
t (hours)
b C = 10t + 20
c i $10 per hour
ii $20 up-front fee
10 a i V = 4t
ii V = 3t
iii V = t + 1 iv V = 1.5t + 2
b 1 L, 2 L
c Initially, the container has b litres and it is losing 1 litre
per minute.
5
b m=
= −1
11 a m = −5 = −1
−
5
5
c It doesn’t matter which pair of points is (x1, y1) and which
is (x2, y2).
d x2 − x1 and x1 − x2 have the same value but opposite sign.
y2 − y1
is changed to
Likewise for y2 − y1 and y1 − y2. If
x 2 − x1
y1 − y2
, both the numerator and denominator will change
x1 − x 2
−3
4
sign; e.g.
becomes
. These are equal to each other.
4
−3
13
4
4 x 13
b y= −
+
c y = − 4x +
3
3
3
3
3
d The results from parts b and c are the same (when
simplified). So it doesn’t matter which point on the line is
used in the formula y − y1 = m(x − x1).
12 a −
1
2
= 0.02
ii
= 0.04
13 a i
50
50
b i y = 0.02x + 1.5
ii y = 0.04x + 1.5
c The archer needs m to be between 0.02 and 0.04 to hit the
target.
5 a (1, 6.5)
b (1.5, 2.5) c (−0.5, 1)
d (−1, 4.5) e (1, −1.5) f (−3.5, 3)
g (−3, −0.5) h (2, 2.5)
i (−7, 10.5)
6 B and C are both 5 units away from (2, 3).
7 a a = 3, b = 5
b a = −4, b = 5
c a = −2, b = 2
d a = 11, b = 2
8 a 3, 7
b −1, 3
c −1, 9
d −6, 0
9 a 1478 m
b 739 m
10 a (−0.5, 1)
b (−0.5, 1)
c These are the same. The order of the points doesn’t matter
(x1 + x2 = x2 + x1).
d 5
e 5
f The order of the points doesn’t matter (x − y) 2 = (y − x) 2.
11 a = −4, 0
y
y=3
(–4, 3)
(0, 3)
d = √20
x
O
(–2, –1)
1 
12 a  , 2
2
 16 
d  2, 
 5
 1 4
 4 8
b − ,  c  , 
 3 3
 3 3
 3 
e  − , 1
 4 
f

 0,
8
5 
13 a C (x1, 0)
c
e
g
i
b D (x2, 0)
x2 − x1
d yes
f G (0, y2)
F (0, y1)
y2 − y1
h yes
AB 2 = AE 2 + BE 2
d 2 = (x2 − x1)2 + (y2 − y1)2
d = ( x2 − x1 )2 + ( y2 − y1 )2
Exercise 5G
 9
d  3, 
 2
1 a 4
b 5
c
41
2 a 4
b 4
c
32 = 4 2 d (0, −3)
3 a d = 20 = 2 5 , M = (2, 5 )
j In the diagram, x2 − x1 will be positive. x1 − x2 will equal value
but opposite sign; e.g. if x2 − x1 = 3, then x1 − x2 = − 3. When
they are squared they have equal value and equal sign.
14 a
( x − 7 )2 + y2
b d = 97 , M = (2, 3.5)
b
c d = 41, M = (−1, 1.5)
c i 721 m ii 707 m
iii 721 m
iv 762 m
d x=2
e The distance will be a minimum when the dotted line joining
Sarah to the fence is perpendicular to the fence (i.e. when it
has gradient −1). The closest point is (2, 5).
d d = 37 , M = (−1, −1.5)
4 a
d
g
29
65
101
b
e
h
58
37
193
c
37
f 15
i
37
( x − 7 )2 + ( x + 3 )2
790
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Exercise 5H
8
c −3
d
7
4
1
1
8
9
−
b
d
c −
2
3
7
4
5
b 4
c y = 5x + 4
parallel
b parallel
c neither d neither
perpendicular f perpendicular g parallel
parallel
i perpendicular j perpendicular
y=x+4
b y = −x − 6
c y = −4x − 1
4
1
2
y= x−6
e y=− x+7 f y=− x+6
5
2
3
1
3
3
y= x−2
h y=− x+5 i y=− x−5
4
2
4
7
y = x + 31
2
x=6
b x=0
c y = 11
d y = 8.4
4
2
y=3
f y = −3
g x=
h x= −
11
3
1 a 4
2 a
3 a
4 a
e
h
5 a
d
g
j
6 a
e
Exercise 5I
b −7
2
7 a y= x+5
3
1 a yes
e no
b yes
f no
y
y = 2x − 4
(3.5, 3)
3
c y= − x+1
2
x
1

b  , − 2
2

y
y = 2x − 3
x
O
y = –2
7
28
x+
5
5
1
10
d y= − x−
7
7
c (2, 4)
y
2
5
9 The second line has equation y = − x − . It cuts the x-axis
3
3
5
at x = − .
2
9
10 a 14
b −2
c 5
d
7
11 a m
b −
a
b
c −
1
m
d
1
3
b AB is parallel to CD, BD is parallel to AC; i.e. opposite
sides are parallel.
c parallelogram
3
4
14 a i
iii 0
ii −
4
3
ii −3
iii 1
x=2
y=4
(2, 4)
b
a
12 a y = 2x + d − 2c
b y = mx + d − mc
c y=x+d−c
c
1
d y= − x+d+
m
m
13 a i 1
y=3
O
d y = 7x + 20
b y=
d no
h no
7 
2 a  , 3
2 
5
54
b y= − x+
7
7
2
16
c y= x+
3
3
3
8 a y= − x+5
2
c no
g yes
Answers
Number and Algebra
iv −
b Right-angled triangle (AB is perpendicular to BC ).
c 20
1
15 y = − x + 4, x-intercept = 8
2
x
O
d (−1, 0)
y
x = –1
y=0
(–1, 0)
O
x
791
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
792
Chapter 1
e (1, 4)
3 a i Joe’s: $60, Paul’s: $150
ii Joe’s: $0.20 per km, Paul’s: $0.10 per km
iii Joe’s: C = 0.2k + 60, Paul’s: C = 0.1k + 150
iv 900 km
b Joe’s Car Rental
c Paul’s Motor Mart
4 a (2, 10)
b (1, −5)
c (−3, 3)
d (13, −2)
e (3, 1)
f (2, 1)
g (1, 4)
h (1, 3)
5 a (2, 7)
b (2, 5)
c (3, 1)
d (2, 1)
e (1, 1)
f (1, 1)
g (5, 1)
h (10, 4)
i (1, 2)
j (9, 2)
6 a i E = 20t ii E = 15t + 45
b t = 9, E = 180
c i 9 hours
ii $180
7 a i V = 62 000 − 5000t
ii V = 40 000 − 3000t
b t = 11, V = 7000
c i 11 years
ii $7000
8 18 years
9 197 600 m2
y
(1, 4)
y=x+3
x
O
y = – 23 x +
14
3
f (1, −4)
y
y = 3x − 7
y = 2x − 6
x
O
(1, –4)
10 a no
e no
11 a −4
g No intersection (lines are parallel).
 k 2k 
12 a  , 
3 3 
y
y = 3x − 3
y = 3x + 9
b
b2
, y=
a−b
a−b
a
−a
, y=
c x=
1+ b
1+ b
13 a x =
x
e x=
h No intersection (lines are parallel).
d yes
h no
c 12
(a − b)
1
, y=
a − 2b
a − 2b
 −2 k − 1 −2 k − 4 
d 
,

 3
3 
b x=
−b
a
, y=
a+b
a+b
d x=
b
b2
, y=
b−a
b−a
2c
c (1 − b )
, y=
a( b + 1)
b+1
f
x=
c
c
g
b
d
c
g
k
c
0
d 0
add
d add
subtract
h subtract
6x − 9y = 12
20x − 30y = 40
(4, 2)
d (2, 2)
(2, −1)
h (2, 2)
(2, 1)
l (−1, 2)
(2, 1)
d (4, −3)
2
g x = ab , y = a b
2
2
a +b
a +b
y
y=x
y=x+4
Exercise 5J
x
O
1 a 0
b
2 a subtract
b
e add
f
3 a 4x − 6y = 8
c 8x − 12y = 16
4 a (2, 5)
b
e (1, 1)
f
i (1, 2)
j
5 a (1, 1)
b
 1 17 
− ,

 5 5
y
(–0.2, 3.4)
x
O
y = 3x + 4
c yes
g yes
k k
b  ,− 
2 2
c (−1 − k, −2 − k)
O
i
b no
f yes
3
b
2
y = –2x + 3
1 
e  , 1
2 
6 a (4, −3)
d (2, 2)
0
subtract
add
(2, 3)
(2, 1)
(2, 1)
(4, 2)
 1 1
− , − 
 2 2
b (1, 1)
c (3, 4)
f
1

e  , −1
2

f

1
−3, 

3
792
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
7 799 and 834
8 $0.60
9 A = $15, C = $11
10 Should have been (1) − (2) to eliminate y : −2y − (−2y) = 0.
The correct solution is (1, −1).
1

 2 2
 13 1 
b  ,
c − , 
11 a  , − 1

a

 a b
 3 3b 


d a+b, a−b
 2a
2b 


e  c , c 
a+b a+b
12 The two lines are parallel; they have the same gradient.
2
2
−
x −1 x +1
b
2
1
−
2x − 3 x + 2
c
3
2
−
3x + 1 2x − 1
d
e
1
1
+
x+3 x −4
f
3
2
+
3x − 1 x + 2
1
3
−
7(2 x − 1) 7(4 − x )
13 a
c
e
g
i
A(2, 0)
E (3, 3)
B(2, 2)
1: 2
d
f
h
j
x=2
y=x
AB = 2, BC = 4
yes
b
e
h
b
yes
no
yes
C
Exercise 5L
1 a no
d no
g no
2 a B
3 x ≥ −1, y ≤ 4
c
f
i
c
yes
no
yes
A
y
x≥1
y≤4
(–1, 0)
(0, 4)
x
O
Exercise 5K
1 a x + y = 16, x − y = 2; 7 and 9
b x + y = 30, x − y = 10; 10 and 20
c x + y = 7, 2x + y = 12; 5 and 2
d 2x + 3y = 11, 4x − 3y = 13; 4 and 1
2 7 cm × 21 cm
3 Nikki is 16, Travis is 8.
4 Cam is 33, Lara is 30.
5 Bolts cost $0.10, washers cost $0.30.
6 There were 2500 adults and 2500 children.
7 Thickshakes cost $5, juices cost $3.
8 There are 36 ducks and 6 sheep.
9 $6.15 (mangoes cost $1.10, apples cost $0.65)
10 43
11 70
12 1 hour and 40 minutes
Answers
Number and Algebra
4 a y≥x+4
y
4
–4
y≥x+4
x
O
b y < 3x − 6
y
1
of an hour
7
14 200 m
15 a Student’s own research required.
b
13
y < 3x − 6
O
y
x
2
–6
8
6
c y > 2x − 8
C
y
4
y > 2x − 8
2
D
O
–1
–1
1 2 3 4 5 6 7 8
O
x
4
x
–8
793
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
794
Chapter 1
d y ≤ 3x − 5
h y > −3x + 6
y
y
y ≤ 3x − 5
O
5
3
6
x
O
–5
x
2
y > 6 − 3x
e y < −4x + 2
y
i y ≤ −x
y
y < –4x + 2
2
O
x
1
2
x
O
f y ≤ 2x + 7
y ≤ –x
y
7
j x>3
y
y ≤ 2x + 7
– 72
x
O
x>3
(3, 0)
O
x
g y < 4x
y
k x < −2
y
4
O
x
1
x < –2
y < 4x
(–2, 0) O
x
794
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Number and Algebra
l y≥2
Answers
d
y
y
y≥2
2
(9, 0)
x
O
O
x
(0, –6)
e
5 a
y
y
(0, 5)
3
(–2.5, 0)
O
9
x
x
O
f
b
y
y
1.5
x
O
(0, 1.5)
(–3, 0)
O
–3
–3
x
g
c
y
y
4
O
2
x
(–5, 0)
O
x
(0, –2)
795
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
796
Chapter 1
h
d
y
y
O
x
O
–9
1
x
5
(0, –3)
–4
e
6 a yes
b no
7 a no
b yes
8 a y≤x+3
3
c y<− x−3
2
9 a
y
c no
d yes
c no
d no
b y ≥ −2x + 2
2
d y> x−2
5
4
2
O
y
3
x
4
(6, –2)
4
(4, 0)
2
O
4
x
f
y
10
(3.33, 6.67)
b
2.5
y
–5
(0, 3)
x
10
g
4
O 1
O
y
x
9
(4, 3)
2
c
O
–8
y
h
O
2
3
x
6
y
x
6
(0, –2)
2.5
–5
O
(1, 3)
2
x
796
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Number and Algebra
f
y
y
Answers
10 a
y≥0
y ≥ 12 x + 3
(0, 1)
(0, 0) O (2, 0)
y≤
– 12 x
x≥0
x
(–5.5, 14.5)
(4, 5)
+1
(–1.43, 2.29)
y ≤ –x + 9
b
x
O
y ≥ –3x – 2
y
y ≥ 2x − 4
11 a y ≥ 0, y < 2x + 4, y ≤ −x + 7
(0, 0)
O
(2, 0)
x≥0
1
b y > − x + 6, y ≤ x + 3, x < 8
2
x
(0, –4)
12 a 1
y≤0
b 4
c 22
d
81
20
115
578
ii
6
15
b Answers may vary; e.g. x > 0, x < 3, y > 0, y < 2.
13 a i
c
y
Challenges
(0, 15)
y ≥ 14 x + 4
x≥0
O
y ≤ – 54 x + 15
d
1 0.75 km
6
2
8
3 a The gradient from (2, 12) to (−2, 0) = the gradient from
(−2, 0) to (−5, −9) = −3.
b The gradient from (a, 2b) to (2a, b) = the gradient from
b
(2a, b) to (−a, 4b) = − .
a
(4, 5)
(0, 4)
x
y
y > – 25 x + 2
(–2.5, 3)
3
5
and the gradient of AB is − . So
5
3
ABC is a right-angled triangle, as AC is perpendicular to AB.
4 The gradient of AC is
(2, 3)
(2, 1.2)
x
O
5
y<3
6
7
e
y
x≤0
y<x+7
7
(–5.4, 1.6)
–7
O
–3 –2
8
x
2x + 3y ≥ –6
Can also show that side lengths satisfy Pythagoras’ theorem.
4840
9680
The missiles are travelling at
km/h and
km/h.
9
9
The distance between the two points and (2, 5) is 5 units.
The diagonals have equations x = 0 and y = 3. These lines are
perpendicular and intersect at the midpoint (0, 3) of the diagonals.
It is not a square, since the angles at the corners are not 90°. In
particular, AB is not perpendicular to BC (mAB ≠ mBC ).
x = 2, y = −3, z = −1
Multiple-choice questions
1 E
5 D
9 C
13 A
2
6
10
14
D
B
D
A
3
7
11
15
B
C
E
D
4
8
12
16
C
A
B
C
797
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
798
Chapter 1
Short-answer questions
1 a 5xy + 6x
d 3b + 21
d
3
x
2
e −2m 2 + 12m f x + 2
b 12a 2b
c
6 − 7a
14
b
5a + 18
6a
7 x + 26
30
d
11− x
( x + 1)( x − 3 )
3 a 3x − 1
b
2
x+2
4 a x = −3
b x=−
2 a
c
y
5 a x<1
3
4
b x ≥ −4
6 a x>5
b x ≥ 10
7 a V = 2 − 0.4t
8 a
y
O
3
4
c
e
1
5
c −1 < x ≤ 3
c x=
c x > −3
b 1.4 L
c
(3, 0)
5 min
y
d x=2
y = 2x
2
7
d t ≤ 3.5 min
d x≤
x
O
x
x=5
(1, 2)
(0, 0)
x
O
f
y
y = 3x − 9
(0, –9)
(0, 0)
O
b
x
(1, –5)
y
g
y
(0, 5)
(0, 4)
(2.5, 0)
O
c
x
O
y = 5 − 2x
x
(8, 0)
y
y
h
(0, 3)
(0, 3)
O
x
y=3
O
(8, 0)
x
798
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
1
9 a y= x+3
2
b y=−
Extended-response questions
3
15
x+
2
2
1 a i h = 4t + 25
ii h = 6t + 16
b 16 cm
c Shrub B is growing at a faster rate because its gradient
is greater.
d h (cm)
c y = 2x − 3
10 a m = −
3
5
b y=−
3
34
x+
5
5
11 a M = (4, 8), d = 52 = 2 13
 11 
b M =  , 1 , d =
2 
90
80
70
60
50
40
30
20
10
61


c M =  1, − 5  , d =
2 2
18 = 3 2
12 a y = 3x − 2
b y = −1
1
c y=− x+5
d y = 3x − 1
2
13 a a = 7
b b = −8
c c = 0 or 4
14 a (−3, −1)
b (−8, −21)
15 a (−3, −1)
b (0, 2)
16 A regular popcorn costs $4 and a small drink costs $2.50.
17 a
y
O
(12, 88)
(12, 73)
0
2 4 6 8 10 12
t (months)
e after 4.5 months
f i 1.24 m
ii 26.25 months
iii between 8.75 and 11.25 months
2 a A (0, 0), B (8, 6), C (20, 0)
y
x
4
3
Answers
Number and Algebra
–4
B (8, 6)
D (14, 3)
b
y
A (0, 0)
C (20, 0)
x
8
3
x
O
–4
3
d i y= x
4
1
ii y = − x + 10 iii y = 0
2
4
80
f y=− x+
3
3
y
Semester review 1
2
–6
c The drink station is at (14, 3).
3
1
e y > 0, y < x, y < − x + 10
4
2
18 The point of intersection is (4, 0).
O
b 43.4 km
4
x
Chapter 1: Measurement
Multiple-choice questions
1 B
2 D
3 B
4 A
5 D
799
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Chapter 1
800
Short-answer questions
8 a
1 a
b
c
d
e
2 a
b
c
d
3 a
b
c
4 a
b
c
5 a
b
c
6 a
23 000 mm
8000 ms
7.8 × 109 ns
0.008 Mt
2.3 × 106 TB
6.5 s to 7.5 s
8.985 g to 8.995 g
699.5 km to 700.5 km
695 km to 705 km
4.25 m and 4.35 m
6.75 m to 6.85 m
22 m ≤ P < 22.4 m and 28.6875 m2 ≤ A < 29.7975 m2
36 cm, 52 cm2
1.3 m, 0.1 m2
220 mm, 2100 mm2
188.5 m2, 197.9 m3
50.3 cm2, 23.7 cm3
6.8 m2, 1.3 m3
1.8 cm
b 58.8 cm2
7
27
cm
π
1
5
9 a x=3
3
1 a 753.98 cm
b 206.02 cm
c 17 cm
d 1.79 cm
Chapter 2: Indices and surds
1 a V = 80 000(1.08)n
b i $86 400
c 11.91 years
3 E
4 E
1 a 3 6
1 C
c 3 6
f
48 3
i
10 2
7
d
10
e 21
g
3
h
5
3
2 a 7 5− 7 b 0
4 a
3 2
2
5 a 24x10y 2
2
5
b
b 3a 2b 2
6 a i 37 200
b i 7.30 × 10−5
7 a i
b i
1
10 2
6
3 B
ii
ii
c
c
c 4
1 a
x
w
A
v
B
r
g
u
e
f
a5
5 y3
ii 4.73 × 109
1
72 x3
5
20
3
3
iii 4 x 5
iv 15 2
iii
4
s
c
q
n
j
k
l
m
iii
9
26
iv
19
26
B
B′
A
3
1
4
A′
4
4
8
7
5
12
ii 4
iii 5
iv 8
1
ii
12
b 0.37
7
iii
12
iv
1
2
3
4
1
2
3
4
5
2
3
4
5
6
3
4
5
6
7
4
5
6
7
8
3
4
b 16
c i
ii 0.0000049
5C
y z
t
3
5
ii
26
26
c No, A ∩ B ≠ ∅
2 a
d 59 + 24 6
d
o
p
h
1
c i
4
3 a 0.18
4 a
2x2
i
a
b i
2 5 −5
5
3b 2
4D
Short-answer questions
c − 2 −4
3 a 2 15 − 4 3 b 11 5 − 62
1
2
ii $108 839
d 6% per year
2 E
b i 3
b 20 3
d −
Multiple-choice questions
5C
Short-answer questions
3
2
1
2
Chapter 3: Probability
Multiple-choice questions
2 D
c x=
d
Extended-response question
d
3
c 3
b x=2
b
Extended-response question
1 B
1
16
b
3
16
5 a
73 or ( 4 7 )3
5
8
ii
A
4
iii
1
5
B
4
2
2
800
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
c i 14
ii 50%
iii 20−24 days, those that maybe catch public transport
to work or school each week day.
2 a
2
2
ii
3
3
c Yes they are because P(A | B) = P(A).
b i
Extended-response question
1
0
Outcome
200 g 400 g
5
7
250 g
3
7
200 g 450 g
4
7
250 g
200 g
5
8
2
450 g
3 a
250 g
3
28
a i
9
14
15
ii
28
c
Balance ($)
3
8
b
2
7
5
10
15
20
25
b
500 g
4
3
5
6
7
8
Answers
Number and Algebra
9
2500
2000
1500
1000
J F M A M J J A S O N D
Months
5
iii
14
b Balance fluctuated throughout the year but ended up with
more money after 12 months.
c May and June
d increase of $500
4 a, c
3
5
y
6
5
4
3
2
Chapter 4: Single variable and bivariate
statistics
Multiple-choice questions
1 E
2 B
3 C
5 A
4 B
0
Short-answer questions
b
d
5 a
b
1 a
Frequency
Percentage
frequency
0−4
2
10%
5−9
4
20%
10−14
4
20%
15−19
3
15%
20− 24
6
30%
25−30
1
5%
Total
20
100%
1 a y = 1.50487x + 17.2287
b 41 cm
30
6
Frequency
5
4
20
3
10
2
1
0
0
2.5
7.5
12.5 17.5 22.5 27.5
Number of days
positive
i ≈ 3.2
ii ≈ 11.5
i under 40 years
ii over 40 years
Over 40 years: mean = 11, standard deviation (σ) = 7.0
Under 40 years: mean = 24.1, standard deviation (σ) = 12.2
Extended-response question
Percentage frequency (%)
b
Class
interval
x
1 2 3 4 5 6 7 8 9 10 11 12 13
Chapter 5: Expressions, equations and linear
relationships
Multiple-choice questions
1 A
2 C
3 D
4 E
5C
Short-answer questions
1 a 3 − 2x
b 20
2 a i x = −4
ii x = 2
c
3a − 8
4a
iii x = 13
d
9x − 2
( x + 2 )( x − 3 )
iv x = 2
801
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
802
Chapter 1
b i x ≤ 6,
ii x < 3,
c
x
3 4 5 6 7
x
0 1 2 3 4
y
–3
y
(1, 1)
1
O
Extended-response question
1 a x = 3, y = 4
b, c
y
2
x
1
O
–2
x
O
4
ii m = − , b = 2
3
3 a i m = 3, b = −2
y
x
3
(3, –2)
–2
Intersecting
region
y = 4x − 8
8.5
(3, 4)
b i
ii
y
y
O
O
x
3
3
O
–6
x
5
2
17
3
x
3x + 2y = 17
–8
d 167 500 m2
iii
iv
y
y
Chapter 6
O
3
O
x
1
(1, –2)
–2
b y=
4 a y = −x + 3
x
1 a
b
c
d
e
f
8
9
x−
5
5
5 a a = −3
b a = −4
c a = 1 or a = 7
d a = −4
6 a x = −3, y = −7
b x = −2, y = −4
c x = −1, y = 4
d x = 3, y = −5
7 A hot dog costs $3.50 and a can of soft drink costs $2.
8 a
b
y
Pre-test
2
3
y
4
3
O
O
x
3
2
–4.5
3
x
5
6
All side lengths and angles are equal.
Two side lengths and two angles are equal.
All side lengths and angles are different.
One angle of the triangle is 90°.
Opposite pairs of sides are parallel and of equal length.
Opposite pairs of sides are parallel and of equal length and
all angles are 90°.
g All sides are of equal length and all angles are 90°.
h One pair of opposite sides are parallel.
i All sides are of equal length, opposite pairs of sides are parallel.
j Two pairs of adjacent sides with the same length.
a b, j
b f, i
a 60°
b 55°
c 190°
d 40°
e 40°
f 60°
g 90°
h 130°
i 70°
a supplementary
b complementary
c neither
d supplementary
a 120
b 60
c 120
d 60
triangle, quadrilateral, pentagon, hexagon, heptagon, octagon,
nonagon, decagon
802
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
7 a 105
d 8.8
b 69
e
25
2
c
8
3
f 6
Exercise 6A
1 triangle, quadrilateral, pentagon, hexagon, heptagon, octagon,
nonagon, decagon
2 a false
b true
c true
d true
e false
f false
g true
h false
i true
3 a a = 110 (angles on a straight line), b = 70 (vertically
opposite)
b a = 140 (angles in a revolution)
c a = 19 (angles in a right angle)
d a = 113 (cointerior angles on parallel lines), b = 67
(alternate angles on parallel lines), c = 67 (vertically
opposite to b)
e a = 81 (equal angles are opposite sides of equal length),
b = 18 (angle sum of a triangle)
f a = 17 (angle sum of a triangle), b = 102 (angles on a
straight line)
g a = 106 (cointerior angles on parallel lines), b = 74
(opposite angles in a parallelogram)
h a = 17 (angle sum of a triangle), b = 17 (angles in a right
angle)
i a = 90 (vertically opposite), b = 60 (angle sum of a triangle)
4 a 72
b 60
c 56
5 a 60
b 60
c 110
d 80
e 10
f 20
g 109
h 28
i 23
j 121
k 71
l 60
6 a 50 (angle sum of a quadrilateral)
b 95 (angle sum of a quadrilateral)
c 125 (angle sum of a pentagon)
d 30 (angle sum of a pentagon)
e 45 (angle sum of a hexagon)
f 15 (angle sum of a quadrilateral)
7 a 108°
b 135°
c 144°
8 a 95
b 113
c 85
d 106
e 147
f 292
9 a 176.4°
b 3.6°
10 a 12
b 20
c 48
11 x = 36, y = 144
12 115
13 a Expand the brackets.
b n = S + 360
180
S
360
180
(
n
−
2
)
c I=
=
d E = 180 − I =
n
n°
n
14 a ∠BCA = 180° − a° − b° (angles in a triangle)
b c = 180 − ∠BCA = a + b (angles on a straight line)
15 a alternate angles (BA || CD)
b ∠ABC + ∠BCD = 180° (cointerior angles on parallel
lines), so a + b + c = 180.
c Angle sum of a triangle is 180°.
16 ∠ACB = ∠DCE (vertically opposite), so ∠CAB = ∠CBA
= ∠CDE = ∠CED (isosceles), since ∠CAB = ∠CED
(alternate), AB || DE.
17 Answers may vary.
18 a 15° (alternate angles in parallel lines)
b 315° (angle sum in an octagon)
19 ∠AMB ≡ ∠ABM = 60° ( AMB is equilateral)
∠AMB + ∠BMC = 180° (angles on a straight line)
∴∠BMC = 120°
∠MCB ≡ ∠MBC = 30° (angle sum of isosceles MCB)
∴∠ABC = 60° + 30° = 90°
20 Let ∠AOB = x and ∠COD = y. 2x + 2y = 180° (angles on
a straight line), so ∠BOD = x + y = 90°.
Answers
Number and Algebra
Exercise 6B
1 a
d
2 a
d
3 a
SAS
b SSS
c AAS
SAS
e RHS
f RHS
5
b 4
c 3
5
e 2
f 2
In ABC and DEF,
AB = DE (given)
∠ABC = ∠DEF (given)
BC = EF (given)
∴ ABC ≡ DEF (SAS test)
b In FED and CBA,
∠FED = ∠ABC = 90° (given)
FD = AC (given)
EF = BC (given)
∴FED ≡ CBA (RHS test)
c In ABC and DEF,
AC = DF (given)
BC = EF (given)
AB = DE (given)
∴ ABC ≡ DEF (SSS test)
d In DEF and ABC,
∠EDF = ∠BAC (given)
∠DFE = ∠ACB (given)
EF = BC (given)
∴ DEF ≡ ABC (AAS test)
4 a x = 7.3, y = 5.2
b x = 12, y = 11
c a = 2.6, b = 2.4
d x = 16, y = 9
5 a In ADC and CBA,
∠DAC = ∠ACB (alternate angles, AD || BC )
∠DCA = ∠CAB (alternate angles, AB || DC )
AC is common
∴ ADC ≡ CBA (AAS test)
803
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
804
Chapter 1
b In ABD and CDB,
∠ABD = ∠BDC (given)
∠ADB = ∠DBC (given)
BD is common
∴ ABD ≡ CDB (AAS test)
c In ABC and EDC,
∠BAC = ∠CED (alternate angles, AB || DE )
∠ABC = ∠CDE (alternate angles, AB || DE )
BC = CD
∴ ABC ≡ EDC (AAS test)
d In ABD and CBD,
AD = DC (given)
∠ADB = ∠CDB (given)
BD is common
∴ ABD ≡ CBD (SAS test)
e In OAB and OCD,
OA = OC (equal radii)
OB = OD (equal radii)
AB = CD (given)
∴ OAB ≡ OCD (SSS test)
f In ADC and ABC,
∠ADC = ∠ABC = 90° (given)
AC is common
DC = BC (given)
∴ ADC ≡ ABC (RHS test)
6 a In AOB and COB,
AO = CO (equal radii)
∠AOB = ∠COB (given)
OB is common
∴ AOB ≡ COB (SAS test)
b AB and BC are matching sides in congruent triangles,
∴ AB = BC.
c AB = 10 mm
7 a In ABC and EDC,
BC = DC (given)
∠ACB = ∠DCE (vertically opposite angles)
AC = EC (given)
∴ ABC ≡ EDC (SAS test)
b AB and DE are matching sides in congruent triangles,
∴ AB = DE.
c ∠ABC = ∠CDE (matching angles in congruent triangles)
∠ABC and ∠CDE are alternate angles, ∴ AB || DE.
d DE = 5 cm
8 a In ABD and CDB,
AB = CD (given)
AD = BC (given)
BD is common
∴ ABD ≡ CDB (SSS test)
b ∠DBC = ∠BDA are matching angles in congruent
triangles, ∴ ∠ DBC = ∠ BDA.
c ∠ADB and ∠DBC are equal alternate angles, ∴ AD || BC.
9 a In ABC and EDC,
BC = DC (given)
∠ BCA = ∠ ECD (vertically opposite angles)
AC = EC (given)
∴ ABC ≡ EDC (SAS test)
∠ ABC = ∠ CDE (matching angles in congruent triangles)
and they are alternate.
∴ AB  DE
b ABC is straight.
∴ ∠ ABO = ∠ OBC = 90°
In ABO and CBO,
∠ ABO = ∠ OBC (see above)
OA = OC (equal radii)
OB is common
∴ ABO ≡ CBO (RHS test)
∴ AB = BC (matching sides in congruent triangles)
∴ OB bisects AC.
c In ACD and CAB,
CD = AB (given)
AD = BC (given)
AC is common
∴ ACD ≡ CAB (SSS test)
∴ ∠ DAC = ∠ ACB (matching angles in congruent
triangles) and they are alternate.
∴ AD  BC
d AE = AB (given)
∴ AEB is isosceles.
∠ ABE = ∠ AEB (equal angles are opposite equal sides)
In AED and ABC,
AE = AB (given)
∠ ABE = ∠ AEB (see above)
DE = BC (given)
∴ AED ≡ ABC (SAS test)
∴ AD = AC (matching sides in congruent triangles)
e In OAD and OBC,
OD = OC (given)
∠ DOA = ∠ COB (vertically opposite angles)
OA = OB (given)
∴ OAD ≡ OBC (SAS test)
∴ ∠ OAD = ∠ OBC (matching angles in congruent triangles)
f In ADC and ABC,
AC is common
∠ DAC = ∠ BAC (given)
AD = AB (given)
∴ ADC ≡ ABC (SAS test)
∴ ∠ ACD = ∠ ACB (matching angles in congruent triangles)
and they are supplementary (since BD is straight).
∴ ∠ ACD = ∠ ACB = 90°
∴ AC ⊥ BD
10 a In ABD and ACD:
∠BDA = ∠CDA = 90° ( AD ⊥ BC )
804
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
AB = AC (given)
AD is common
∴ ABD ≡ ACD (RHS test)
b They are matching angles in congruent triangles.
c If two sides of a triangle are equal, then the angles
opposite those sides are also equal.
11 In line 2 ∠ABC should be ∠ACD.
In line 4 AB should be AD.
In line 5 RHS should be AAS.
12 a In OAM and OBM,
OA = OB (equal radii)
OM is common
AM = BM (M is the midpoint of AB)
∴ OAM ≡ OBM (SSS test)
∴ ∠OMA = ∠OMB (matching angles in congruent
triangles) and they are supplementary (since AB is straight).
∴ ∠OMA = ∠OMB = 90°
∴ OM ⊥ AB
b In OAC and OBC,
OA = OB (equal radii)
AC = BC (equal radii)
OC is common
∴ OAC ≡ OBC (SSS test)
∴ ∠AOC = ∠BOC (matching angles in congruent
triangles)
c ∠CAB = ∠CBA= x (equal angles are opposite sides of
equal length)
x
∴ ∠EAB = ∠DBA =
2
∴ AF = BF (sides of equal length are opposite equal
angles in AFB)
b AE = CE (corresponding sides), BE = DE (corresponding
sides).
6 a AB = CB (given), AD = CD (given), BD is common.
Therefore, ABD ≡ CDB (SSS).
b ∠ABD = ∠ADB = ∠CBD = ∠CDB (equal angles in
congruent isosceles triangles). Therefore, BD bisects
∠ABC and ∠CDA.
7 a AE = CE (given), BE = DE (given), ∠AEB = ∠CED
(vertically opposite angles). Therefore, ABE ≡ CDE
(SAS).
b ∠ABE = ∠CDE (corresponding angles), ∠BAE
= ∠DCE (corresponding angles). Therefore, AB ||
DC (alternate angles are equal). ∠ADE = ∠CBE
(corresponding angles),
∠DAE = ∠BCE (corresponding angles). Therefore,
AD || BC (alternate angles are equal).
8 a AD = CB (given), ∠DAC = ∠BCA (alternate angles), AC
is common. Therefore, ABC ≡ CDA (SAS).
b ∠BAC = ∠DCA (corresponding angles), therefore AB ||
DC (alternate angles are equal).
9 a ABE ≡ CBE ≡ ADE ≡ CDE (SAS)
b ∠ABE = ∠CDE (corresponding angles), ∠BAE = ∠DCE
(corresponding angles), therefore AB || CD.
∠ADE = ∠CBE (corresponding angles), ∠DAE = ∠BCE
(corresponding angles), therefore AD || CB.
Also, AB = AD = CB = CD (corresponding sides).
Therefore, ABCE is a rhombus.
10 a
D
C
E
A
Exercise 6C
1 a
c
2 a
c
d
3 a
b
4 a
b
5 a
rectangle
b parallelogram
square
d rhombus
rectangle, square
b rectangle, square
parallelogram, rhombus, rectangle, square
rhombus, square
e rhombus, square
A trapezium does not have both pairs of opposite sides
parallel.
A kite does not have two pairs of opposite sides parallel.
∠BAC = ∠DCA (alternate angles), ∠BCA = ∠DAC
(alternate angles), AC is common. Therefore, ABC ≡
CDA (AAS).
As ABC ≡ CDA, AD = CB, AB = CD (corresponding
sides).
∠ABE = ∠CDE (alternate angles, AB || DC )
∠EAB = ∠ECD (alternate angles, AB || DC )
AB = DC (opposite sides of a parallelogram)
∴ ABE ≡ CDE (AAS test)
Answers
Number and Algebra
B
∠CAB = ∠ACD and ∠CAD = ∠ACB (alternate angles).
So ∠ACB = ∠ACD.
CDE ≡ CBE (SAS)
So ∠CED = ∠CEB = 90°.
b From part a, ∠ECD = ∠ECB.
11 D
A
C
B
As ABCD is a parallelogram ∠BDC = ∠DBA and
∠DBC = ∠BDA.
So CBD ≡ ADB (AAS).
So ∠BAD = ∠DCB = 90°.
Similarly, ∠ADC = ∠CBA = 90°.
805
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
806
Chapter 1
12 D
C
E
A
B
First, prove AED ≡ BEC (SAS).
Hence, corresponding angles in the isosceles triangles are equal
and CED ≡ BEA (SAS).
Hence, corresponding angles in the isosceles triangles are equal.
So ∠ADC = ∠DCB = ∠CBA = ∠BAC, which sum to 360°.
Therefore, all angles are 90° and ABCD is a rectangle.
13
D
G
11 a true
b true
c false
d false
e false
f false
g false
h false
i true
j true
12 Yes, the missing angle in the first triangle is 20° and the
missing angle in the second triangle is 75°. So all three angles
are equal.
13 a x : y = 2 : 6 = 1 : 3
b i 4
ii 36
c i 8
ii 216
d
Cube
C
H
Length
Small
2
4
8
6
36
216
1: 3
1: 9
1 : 216
ii 1 : 16 or 1 : 42
e i 1: 4
A
iii 1 : 64 or 1 : 4
f i 1 : k2
B
E
First, prove all four corner triangles are congruent (SAS).
So EF = FG = GH = HE, so EFGH is a rhombus.
14 The simplest way to do this is to prove that the midpoint of one
diagonal is the same as the midpoint of the other diagonal.
Exercise 6D
1 a Yes, both squares have all angles 90° and all sides of
equal length.
b 3
c 15 cm
8
5
2 a 2
b
3 a A
b ∠C
c FD
c
4
3
3
2
c
3
2
d ABC ||| EFD
4 a ABCDE ||| FGHIJ
3
d
cm
2
b
AB DE
=
FG IJ
4
e
cm
3
EF GH
=
AB CD
5 a ABCD ||| EFGH
b
4
3
6 a 1.2
d 12 m
c
d 3.75
7 1.7 m
8 a 1.6
9 a 2
10 a BC
c 1
d
b 12.5
c 4.8
e 11.5
f 14.5
b
b
b
d
62.5 cm
1
c 1.875
ABC ||| EDC
4.5
Volume
Large
Scale factor
(fraction)
F
Area
3
ii 1 : k3
Exercise 6E
1 a
c
2 a
b
c
d
e
3 a
b
c
d
4 a
b
c
e 10.5 m
E
b ∠C
AB
d ABC ||| DEF
∠D (alternate angles, AB || DE )
∠A (alternate angles, AB || DE )
∠ECD
CA
ABC ||| EDC
sides about equal angles are in proportion
matching angles are equal
sides about equal angles are in proportion
three pairs of sides are in proportion
∠ABC = ∠DEF = 65°, ∠BAC = ∠EDF = 70°. Therefore,
ABC ||| DEF (matching angles are equal).
DE 2
EF 6
= = 2,
= = 2 (ratio of corresponding sides),
AB 1
BC 3
∠ABC = ∠DEF = 120°. Therefore, ABC ||| DEF (sides
about equal angles are in proportion).
DF 10
DE 8
= = 2,
= = 2 (ratio of corresponding sides),
CA 5
CB 4
∠ABC = ∠FED = 90°. Therefore, ABC ||| FED
(hypotenuse and side are in proportion).
d
AB 28
BC 16
AC 32
= = 4,
= = 4,
= = 4. Therefore,
DE 7
EF 4
DF 8
ABC ||| DEF (three pairs of sides are in proportion).
d 4.3
3
2
d a = 4, b = 15
5 a
b 19.5
c 2.2
e x = 0.16, y = 0.325 f a = 43.2, b = 18
806
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Number and Algebra
c Adding the two equations: AB 2 + AD 2 = CB × BD +
CD × BD = BD(CB + CD ) = BD × BD = BD 2
Exercise 6F
1 a–e
chord
4
cm
b
3
8 a ∠ACB = ∠DCE (common), ∠BAC = ∠EDC = 90°.
Therefore, BAC ||| EDC (matching angles are equal).
b 1.25 m
9 1.90 m
10 4.5 m
11 a Yes, matching angles are equal for both.
b 20 m
c 20 m
d Less working is required for Jenny’s triangles.
12 The missing angle in the smaller triangle is 47°, and the
missing angle in the larger triangle is 91°. Therefore, the two
triangles are similar (matching angles are equal).
13 a ∠AOD = ∠BOC (common), ∠OAD = ∠OBC
(corresponding angles), ∠ODA = ∠OCB (corresponding
angles). So OAD ||| OBC (matching angles are equal).
OC 3
= = 3 (ratio of corresponding sides), therefore
OD 1
OB = 3OA.
b ∠ABC = ∠EDC (alternate angles), ∠BAC = ∠DEC
(alternate angles), ∠ACB = ∠ECD (vertically opposite).
So ABC ||| EDC (matching angles are equal).
7
AE 7
BD 7
= and AE = AC.
= , therefore
5
AC 5
BC 5
14 a ∠BAD = ∠BCA = 90°, ∠ABD = ∠CBA (common). So
ABD ||| CBA (matching angles are equal). Therefore
AB BD
=
, AB 2 = CB × BD.
CB AB
b ∠BAD = ∠ACD = 90°, ∠ADB = ∠CDA (common). So
ABD ||| CAD (matching angles are equal). Therefore
AD BD
=
, AD 2 = CD × BD.
CD AD
minor
sector
centre
AB 3
EB 2
=
= 0.4,
= = 0.4 (ratio of corresponding
d
CB 7.5
DB 5
sides), ∠ABE = ∠CBD (vertically opposite angles).
Therefore, AEB ||| CDB (sides about equal angles are
in proportion).
7 a ∠EDC = ∠ADB (common), ∠CED = ∠BAD = 90°.
Therefore, EDC ||| ADB (matching angles are equal).
Answers
6 a ∠ABC = ∠EDC (alternate angles), ∠BAC = ∠DEC
(alternate angles), ∠ACB = ∠ECD (vertically opposite
angles). Therefore, ABC ||| EDC (matching angles are
equal).
b ∠ABE = ∠ACD (corresponding angles), ∠AEB =
∠ADC (corresponding angles), ∠BAE = ∠CAD
(common). Therefore, ABE ||| ACD (matching angles
are equal).
c ∠DBC = ∠AEC (given), ∠BCD = ∠ECA (common).
Therefore, BCD ||| ECA (matching angles are equal).
radius
major sector
2 a 55°
b 90°
c 75°
d 140°
3 a 85° each
b ∠AOB = ∠COD (equal chords subtend equal angles at
the centre)
c 0.9 cm each
d OE = OF (equal chords subtend equal angles at the
centre)
4 a 1 cm each
b 52° each
c AM = BM and ∠AOM = ∠BOM
5 a ∠DOC = 70° (equal chords subtend equal angles at the
centre)
b OE = 7.2 cm (equal chords subtend equal angles at the
centre)
c XZ = 4 cm and ∠XOZ = 51°
6 The perpendicular bisectors of two different chords of a circle
intersect at the centre of the circle.
7 a 3.5 m
b 9m
c 90°
d 90°
8 a 140°
b 40°
c 19°
d 72°
e 30°
f 54°
9 6m
10 3 + 128 mm = 3 + 8 2 mm
11 a Triangles are congruent (SSS), so angles at the centre of
the circle are corresponding, and therefore equal.
b Triangles are congruent (SAS), so chords are
corresponding sides, and therefore equal.
12 a Triangles are congruent (SSS), so the angles formed by the
chord and radius are corresponding, and therefore equal.
Since these angles are also supplementary, they must be
90°.
b Triangles are congruent (SAS), so the angles formed by the
chord and radius are corresponding, and therefore equal.
Since these angles are also supplementary, they must be 90°.
807
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
808
Chapter 1
13
A
O
B
C
First, prove OAB ≡ OAC (AAS), which are isosceles.
So AB = AC, corresponding sides in congruent triangles.
14 a AD = BD (equal radii), AC = BC (equal radii), CD is
common. Therefore, ACD ≡ BCD (SSS).
b AC = BC, ∠ACE = ∠BCE (corresponding angles), CE is
common. Therefore, ACE ≡ BCE (SAS).
c ∠BEC = ∠CEA = 90° (equal angles on a straight line)
∴ CD ⊥ AB and BE = EA (matching sides in congruent
triangles).
15 a It is the angle between a tangent and a radius.
b yes
c 180
d ∠OCP is a straight angle.
e yes
f collinear
Exercise 6G
∠ADC
b ∠ADC
c ∠ADC
∠AFC
e ∠AEC
f ∠AEC
∠AOB
b ∠ACB
c 80°
d 61°
180°
b 90°
c 60°
d 7°
50°
b 40°
c 80°
d 60°
250°
f 112.5°
g 38°
h 120°
18°
70°
b 25°
c 10°
∠ABC = 72°, ∠ABD = 22°
∠ABC = 70°, ∠ABD = 45°
∠ABC = 72°, ∠ABD = 35°
∠ADC = 75°, ∠ABC = 75°
∠ABC = 57.5°, ∠ADC = 57.5°
∠AOD = 170°, ∠ABD = 85°
100°
b 94.5°
c 100°
119°
e 70°
f 66°
58°
b 53°
c 51°
45°
e 19°
f 21°
70°
b 90°
The angle in a semicircle is 90°.
The second circle is the specific case of the first circle when
the angle at the centre is 180°.
11 a i false
ii true
iii true
iv false
b i false
ii true
iii true
iv false
12 a 2x
b 360 − 2x
1 a
d
2 a
3 a
4 a
e
i
5 a
6 a
b
c
7 a
b
c
8 a
d
9 a
d
10 a
c
d
∠AOC = 180° − 2x° (AOC is isosceles)
∠BOC = 180° − 2y° (BOC is isosceles)
∠AOB = 360° − ∠AOC − ∠BOC = 2x° + 2y°
∠AOB = 2(x° + y °) = 2∠ACB
∠BOC = 180 − 2x (BOC is isosceles),
∠AOB = 180 − ∠BOC = 180 − (180 − 2 x) = 2x
b ∠AOC = 180 − 2x (AOC is isosceles),
∠BOC = 180 − 2y (BOC is isosceles).
Reflex ∠AOB = 360 − ∠AOC − ∠BOC
= 360 − (180 − 2 x) − (180 − 2y ) = 2x + 2y = 2(x + y)
= 2∠ACB.
c ∠OBC = x + y, ∠COB = 180 − 2(x + y),
∠AOB = 180 − 2x − (180 − 2(x + y)) = 2y
15 ∠AOB = 180 − 2x (AOB is isosceles), ∠BOC = 180 − 2y
(BOC is isosceles)
∠AOB + ∠BOC = 180 (supplementary angles), therefore
(180 − 2x) + (180 − 2y ) = 180, 360 − 2x − 2y = 180,
2x + 2y = 180, 2(x + y) = 180, x + y = 90.
13 a
b
c
d
14 a
Exercise 6H
1 a ∠ACD
b ∠ACD
c ∠ACD
2 a ∠ABD and ∠ACD
b 85°
c ∠BAC and ∠BDC
d 17°
3 a Supplementary angles sum to 180°.
b 117°
c 109°
d Yes, 117° + 109° + 63° + 71° = 360°
4 a x = 37
b x = 20
c x = 110
d x = 40
e x = 22.5 f x = 55
5 a x = 60
b x = 90
c x = 30°
d x = 88
e x = 72, y = 108
f x = 123
6 a 72°
b 43°
c 69°
d 57°
e 52°
f 48°
g 30°
h 47°
i 108°
7 a a = 30, b = 100
b a = 54, b = 90
c a = 105, b = 105, c = 75
d a = 55, b = 70
e a = 118, b = 21
f a = 45, b = 35
8 a 80°
b 71°
c ∠CBE + ∠ABE = 180° (angles on a straight line)
∠CBE + ∠CDE = 180° (opposite angles of cyclic
quadrilateral BCDE )
∴ ∠CBE + ∠ABE = ∠CBE + ∠CDE
∴ ∠ABE = ∠CDE
9 a ∠ACD = ∠ABD = x and ∠DAC = ∠DBC = y (angles on
the same arc)
b Using angle sum of ACD, ∠ADC = 180° − (x ° + y °).
c ∠ABC and ∠ADC are supplementary.
10 a i 80°
ii 100°
iii 80°
b ∠BAF + ∠DCB = 180°, therefore AF || CD (cointerior
angles are supplementary).
11 a ∠PCB = 90° (angle in a semicircle)
808
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
b ∠A = ∠P (angles on the same arc)
a
c sin P =
2r
a
a
d As ∠A = ∠P, sin A =
; therefore, 2r =
.
2r
sin A
Exercise 6I
1
2
3
4
5
6
7
a once
b 90°
c 5 cm
a ∠BAP
b ∠BPX
c ∠ABP
d ∠APY
a 180°
b 360°
a a = 19
b a = 62
c a = 70
a a = 30
b a = 18
c a = 25
d a = 63
a 50°
b 59°
a a = 73, b = 42, c = 65
b a = 26, b = 83, c = 71
c a = 69, b = 65, c = 46
8 a 5 cm
b 11.2 cm
9 a a = 115
b a = 163
c a = 33
d a = 28
e a = 26
f a = 26
g a = 36
h a = 26
i a = 30
10 a a = 70
b a = 50
c a = 73
d a = 40
e a = 19
f a = 54
11 4 cm
12 a OA and OB are radii of the circle.
b ∠OAP = ∠OBP = 90°
c ∠OAP = ∠OBP = 90°
OP is common
OA = OB
∴ OAP ≡ OBP (RHS)
d AP and BP are corresponding sides in congruent triangles.
13 a ∠OPB = 90° − x ° (tangent meets radii at right angles)
b ∠BOP = 2x ° (using angle sum in an isosceles triangle)
c ∠BAP = x ° (angle at centre is twice the angle at the
circumference)
14 ∠BAP = ∠BPY (angle in alternate segment)
∠BPY = ∠DPX (vertically opposite angles)
∠DPX = ∠DCP (angle in alternate segment)
∴ ∠BAP = ∠DCP, so AB || DC (alternate angles are
equal).
15 AP = TP and TP = BP; hence, AP = BP.
16 a Let ∠ACB = x, therefore ∠ABC = 90 − x. Construct OP.
OP ⊥ PM (tangent). ∠OPC = x (OPC is isosceles).
Construct OM. OAM ≡ OPM (RHS), therefore, AM =
PM. ∠BPM = 180 − 90 − x = 90 − x. Therefore, BPM
is isosceles with PM = BM. Therefore, AM = BM.
Exercise 6J
1 a 3
21
2 a
2
b 6
5
b
2
c 7
33
c
7
d 8
27
d
7
3 a AP × CP = BP × DP
b AP × BP = DP × CP
c AP × BP = CP 2
4 a 5
b 10
c
112
15
5 a
143
8
b
178
9
c
161
9
6 a
32
3
b
16
3
c
35
2
c
81
7
7 a
65
64
8 a
7
d
74
7
b
77
209
b
10
e
Answers
Number and Algebra
153
20
f ( 65) − 1
9 a x(x + 5) = 7 × 8, x 2 + 5x = 56, x 2 + 5x − 56 = 0
b x(x + 11) = 10 × 22, x 2 + 11x = 220, x 2 + 11x − 220 = 0
c x(x + 23) = 41 2, x 2 + 23x = 1681, x 2 + 23x − 1681 = 0
10 The third secant rule states that, for this diagram,
AP 2 = DP × CP and BP 2 = DP × CP, so BP = AP.
11 AP × BP = DP × CP
AP × BP = AP × CP
BP = CP
12 a ∠A = ∠D and ∠B = ∠C (angles on the same arc)
b ∠P is the same for both triangles (vertically opposite),
so ABP ||| DCP (matching angles are equal).
c
AP BP
=
DP CP
d
AP BP cross multiplying gives AP × CP = BP × DP
=
,
DP CP
13 a ∠B = ∠C (angles on the same arc)
b PBD ||| PCA (matching angles are equal)
AP CP
=
, so AP × BP = DP × CP.
c
DP BP
14 a yes
b alternate segment theorem
c BPC ||| CPA (matching angles are equal)
d
BP CP
=
, so CP 2 = AP × BP.
CP AP
15 d = (4r1r2 ) = 2 r1r2
Challenges
1 21 units2
2 BD = 5 cm, CE = 19 cm
3 ∠ADE = ∠ABE, ∠EFD = ∠BFA, ∠DEB = ∠DAB,
∠DFB = ∠EFA
4 42.5%
809
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
810
Chapter 1
5 In ABC, AB = AC (given), CE = EB (given), AE is common.
∴ ABE ≡ ACE (SSS)
A
∴ ∠AEB = ∠AEC = 90°
(equal angles on a straight line)
Using Pythagoras’ theorem in AEB:
AB 2 = AE 2 + BE 2
E
AB 2 − AE 2 = BE 2
B
C
2
2
AB − AE = BE × BE
D
AB 2 − AE 2 = BE × CE (BE = CE )
6 a ∠FDE = ∠DFC = ∠ABC (alternate and corresponding
angles in parallel lines)
∠FED = ∠EFB = ∠ACB (alternate and corresponding
angles in parallel lines)
∠DFE = ∠BAC (angle sum of a triangle)
ABC ||| FDE (matching angles are equal)
b i 4 :1
ii 16 : 1
c 4n−1 : 1
Multiple-choice questions
1 C
5 B
9 D
2 B
6 A
10 B
b 1.2 km
c i
AC 3
=
DE 2
∴
AB 3
= (ratio of sides in similar triangles)
DB 2
∴
x +1 3
=
2
x
∴ 2( x + 1) = 3 x
ii 2
d 44.4%
2 a i 132.84° ii 47.16°
b 12 cm
c 25 cm
d (2 525 + 20 ) cm
Chapter 7
Pre-test
3 B
7 E
4 C
8 C
Short-answer questions
1 a 65
b 120
c x = 62, y = 118 d 46
2 a 148
b 112
3 a ABC ≡ DEF (AAS)
b ABC ≡ ADC (SAS)
c ABD ≡ CDB (SSS)
4 a AB = CD (given), ∠BAC = ∠DCA (alternate angles), AC is
common. Therefore, ABC ≡ CDA (SAS).
b ∠BAC = ∠DCA (alternate angles), therefore AB || DC
(alternate angles are equal).
5 a ABC ||| DEF (sides about equal angles are in
proportion), x = 19.5
b ABE ||| ACD (matching angles are equal), x = 6.25
c ABC ||| DEC (matching angles are equal), x = 8.82
100
d ABD ||| DBC (matching angles are equal), x =
7
6 a 65°
b 7
c 6
7 a 25°
b a = 50, b = 40
c 70°
d a = 30, b = 120
e 115°
f 54°
8 a x = 26, y = 58, z = 64
b a = 65, b = 130, c = 50, d = 8
c t = 63
40
9 a 5
b 6
c
3
Extended-response questions
1 a ∠BAC = ∠BDE = 90°
∠B is common
ABC ||| DBE (matching angles are equal)
1 a
d
2 a
d
3 a
c
4 a
0.89
b 9.51
3.25
e 2.37
1.1
b 3.6
5.4
e 7.7
H = b, O = a, A = c
H = a, O = c, A = b
hypotenuse b adjacent
c
f
c
f
b
0.27
7.75
22.3
2.8
H = a, O = b, A = c
c tan θ
2
5 a sin θ =
3
6 a 44.4
7
3
b cos θ =
c tan θ =
8
4
b 7.1
c 53.1
d 75.5
e 61.9
7 a 180°
f 24.8
b 270°
d 225°
c 45°
Exercise 7A
1 a 0.799
b 0.951
c
e 0.274
f 11.664
g
2 a sin θ
b cos θ
c
3 a 1.80
b 2.94
c
e 22.33
f 12.47
4 a 1.15
b 3.86
c
e 2.25
f 2.79
g
i 37.02
j 9.30
k
5 a 8.55
b 4.26
c
e 5.55
f 1.52
g
i 0.06
j 12.12
k
6 a x = 2.5 cm, y = 4.33 cm
b x = 12.26 cm, y = 6.11 cm
c x = 0.20 m, y = 0.11 m
7 a 125 m
b 327 m
8 1.85 m
9 22.3 m
1.192
0.196
tan θ
3.42
d 0.931
h 0.999
13.74
1.97
10.17
13.06
22.38
9.81
d
h
l
d
h
l
d 2.38
5.07
13.52
13.15
10.04
6.28
15.20
810
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Number and Algebra
∴ 1 = (sin θ)2 + (cos θ)2
b b a
ii yes
c i , ,
c c c
iii cos θ = sin (90° − θ)
Exercise 7B
1 a 60°
b 30°
c 45°
2 a 23.58°
b 60°
c 11.31°
d 5.74°
e 25.84°
f 45°
g 14.48°
h 31.79°
3 a tangent
b cosine
c sine
4 a 60°
b 45°
c 48°35′
d 30°
e 52°7′
f 32°44′
5 a α = 60°, θ = 30°
b α = 45°, θ = 45°
c α = 53.1°, θ = 36.9°
d α = 22.6°, θ = 67.4°
e α = 28.1°, θ = 61.9°
f α = 53.1°, θ = 36.9°
6 a 44.4°, 45.6°
b 74.7°, 15.3°
c 58.3°, 31.7°
d 23.9°, 66.1°
e 82.9°, 7.1°
f 42.4°, 47.6°
7 70.02°
8 31.1°
9 47.1°
10 a 66.4°
b 114.1°
c 32.0°
11 a 1
b 30°
12 a i 45°
ii 33.7°
b 11.3°
13 a Once one angle is known, the other can be determined by
subtracting the known angle from 90°.
b α = 63.4°, β = 26.6°
x
c 2x
b tan 45 = = 1
14 a
x
45°
45°
d sin 45° =
15 a θ = 30°
d i
v
1
2
1
x
1
=
; cos 45° also equals
.
2x
2
2
b α = 60° c
3
ii
3
3
1
2
vi
e AB = 1 x + 3 x
2
2
iii
3
3
2
iv
3
2
Exercise 7C
1 1866.03 m
2 39 m
3 28.31 m
4 4°16′
5 320 m
6 1509.53 m
7 32°
8 a 1.17 m
b 1.50 m
9 8.69 cm
10 299 m
11 a 1.45°
b 3.44°
12 yes
13 89.12 m
14 a i 8.7 cm ii 5 cm
b i 17.3 cm ii 20 cm
c Answers may vary.
15 321.1 km/h
16 a i 18°
ii 72°
b i 0.77 m ii 2.38 m
c 3.85 m
d 4.05 m
Answers
10 7.54 m
11 28.5 m
12 26.4 cm
13 a 4.5 cm
b 8.5 mm
14 The student rounded tan 65° too early.
15 a 3.7
b 6.5
c 7.7
a
16 a i a = c sin θ ii b = c cos θ iii tan θ =
b
c sinθ sinθ
iv tan θ =
=
v Answers may vary.
c cosθ cosθ
b i a = c sin θ ii b = c cos θ iii c 2 = a 2 + b 2
iv c2 = (c sin θ)2 + (c cos θ)2
c2 = c2 (sin θ)2 + c2 (cos θ)2
c 1.99°
iii 36°
iii 2.02 m
e proof
iv 54°
iv 1.47 m
c 90°
g 270°
c 139°
d 135°
h 315°
d 162°
c 335°
iii 157.5°
d 164°
iv 247.5°
Exercise 7D
1 a 0°
b 45°
e 180°
f 225°
2 a 050°
b 060°
e 227°
f 289°
3 a 200°
b 082°
4 b i 022.5° ii 337.5°
5 a 1.7 km
b 3.6 km
6 a 121°
b 301°
7 a 3.83 km
b 6.21 km
8 a 14.77 cm b 2.6 cm
9 a 217°
b 37°
10 a 1.414 km b 1.414 km
11 a 1.62 km
b 5.92 km
12 10.032 km
13 a i 045°
ii 236.3°
b i 296.6° ii 116.6°
14 a i 2.5 km ii 2.82 km
b i 4.33 km ii 1.03 km
c i 45.2° ii 7.6 km
15 a 229.7°, 18.2 km
b 55.1°, 12.3 km
16 a 212.98 m
b i 99.32 m ii 69.20 m
c 30.11 m
17 a 38.30 km b 57.86 km
18 a 4.34 km
b 2.07 km
c 2.914 km
c 2.16 km
iii
iii
iii
iii
26.6°
101.3°
5.32 km
5.36 km
iv 315°
iv 246.8°
c 33.50°
c 4.81 km
811
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
812
Chapter 1
Exercise 7E
3
2
m –1
i
1 a, b
D
C
2√3
2√2
A
2
c 35.3°
d
2 61.4°
3 a 37.609 m b
4 a 57.409 m b
5 a i 26.57° ii
6 a 7.31 m
b
7 138.56 m
8 a i 2.25 m ii
9 a i 1.331 km ii
10 a camera C b
11 a 5.5 m
b
12 a 45°
b
13 a i 1.55
ii
b 34.34°
14 22°
2
2
B
A
C
2√2
45°
c 3.43 m
c 34.7°
c 35.26°
iii 2.82
d 0.2°
d 1.73 units
15 a
3
2
5 a 140°
i
e
i
6 a
e
i
7 a
e
8 a
138°
172°
30°
45°
37°
positive
negative
3
2
e −
3
2
1
3
e i 90° − θ
sin (135°) ≈ 0.7, cos (135°) ≈ –0.7
sin (160°) ≈ 0.34, cos (160°) ≈ –0.94
149°
b 128°
c 135°
d 93°
41°
f 56°
g 29°
h 69°
0.34
b 0.98
c 0.63
d 0.77, −0.77
0.91, −0.91
f 0.42, −0.42 g 1.19, −1.19
2.75, −2.75
i 0.21, −0.21
1
1
3
4 a
b
c 1
d
2
2
2
1
1
1
e
f
g
h
3
3
2
2
b 160°
f 99°
c 115°
g 143°
d 155°
h 124°
b 55°
f 9°
c 86°
g 21°
d 70°
h 78°
b positive
f positive
c positive
g negative
d negative
h negative
2
2
c
3
d
f – 3
g
2
2
h −
b
l
d 120°
d 14
c 60°
ii
12
13
iii
5
12
b
2 7
7
c
2 21
21
ii
b
c
iii
iv 0.59
viii 0.37
iv 38°
b
c
f
2 5
5
Exercise 7G
a
b
c
=
=
sin A sin B sin C
2 a 1.9
b 3.6
1
c 2.5
3 a 50.3°
b 39.5°
c 29.2°
4 a 7.9
b 16.5
c 19.1
d 9.2
e 8.4
f 22.7
5 a 38.0°
b 51.5°
c 28.8°
d 44.3°
e 47.5°
f 48.1°
6 a 1.367 km b 74°
c 2.089 km
7 27.0°
8 131.0 m
9 a ∠ABC = 80°, ∠ACB = 40°
b 122 km
10 a ∠ABC = 80°
b 61.3 km
c 53.9 km
11 a 147.5°
b 102.8°
c 126.1°
d 100.5°
e 123.9°
f 137.7°
12 Impossible to find θ as such a triangle does not exist.
13 37.6° or 142.4°
14 a 59.4° or 120.6°
b
B
B
2
2
2
2
1
2
p undefined
1
2
16 a − 5
b − 5
c −1
2
3
7
17 a i 0.17
ii 0.17
iii 0.59
v 0.99
vi 0.99
vii 0.37
b sin a = cos b when a + b = 90.
c i 90° − θ
ii 90° − θ
d i 70°
ii 5°
iii 19°
Exercise 7F
1 b
c
2 a
e
3 a
e
h
k −
b 30°
b
2
5
13
b i
b 40.98°
b 0.346 km
3
3
n 1
o 0
9 a 30°, 150° b 45°, 135° c 60°, 120°
10 a 120°
b 135°
c 150°
e 135°
f 150°
20 3
11 a 3 2
b 3 2
c
3
e 5 3
f 3
12 a 45°
13 a
3
14 a 13
45.47°
57.91°
11.18 cm b 10.14°
6.87 m
2.59 m
1.677 km
609.07 m
34.5°
1.41 units
1.27
j –
3
A
3
35°
59.4°
2
C
35°
120.6°
A
C
2
812
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
c 31.3°
e
d A triangle can have only one obtuse angle.
C
31.3°
10
120°
A
6
b They are equal because sin 60° and sin 120° are equal.
c For example, same side lengths with included angle 140°.
13 a 65.4°, 114.6°
b
B
B
11 m
B
A
Exercise 7H
65.4°
8m
11 m
C
114.6°
A
C
8m
2
1 a a 2 = 32 + 42 − 2 × 3 × 4 × cos 105°
52 + 92 − 7 2
b cos θ =
2×5×9
2 a 9.6
b 1.5
c 100.3°
d 36.2°
3 a 16.07 cm b 8.85 m
c 14.78 cm
d 4.56 m
e 2.86 km
f 8.14 m
4 a 81.79°
b 104.48° c 64.62°
d 61.20°
e 92.20°
f 48.83°
5 310 m
6 32.2°, 49.6°, 98.2°
7 a 145.9°
b 208.2°
8 383 km
9 7.76 m
10 a cosine rule b sine rule c sine rule
d cosine rule
e sine rule f cosine rule
11 Obtuse, as cos of an obtuse angle gives a negative result.
14 a i 540°
ii 108°
iii 11.89 cm
v 72°, 36° vi 19.24 cm2 vii 43.0 cm2
b 65.0 cm2 c Answers may vary.
b 2 + c2 − a 2
b 121.9°
2bc
13 a AP = b − x b c2 = x2 + h2
c a2 = h2 + (b − x)2
x
2
2
2
2
2
2
d a = c − x + (b − x) = c + b − 2bx
e cos =
c
f x = c cos C substitute into part d.
4 a
e
i
5 a
b
c
d
e
f
g
h
i
j
k
l
6 a
e
i
7 a
e
i
8 a
e
9 a
c
12 a cos A =
Exercise 7I
1
2
3
4
a
a
a
a
e
5 a
6 a
e
7 a
8 a
9 a
10 a
11 a
3.7
b 28.8
c 48.0
α
b θ
c β
56.44°
b 45.58°
c 58.05°
b 26.4 m2
c 0.9 km2
d 13.7 m2
4.4 cm2
2
2
318.4 m
f 76.2 cm
11.9 cm2 b 105.6 m2 c 1.6 km2
5.7
b 7.9
c 9.1
d 18.2
10.6
f 1.3
59.09 cm2 b 1.56 mm2 c 361.25 km2
35.03 cm2 b 51.68 m2 c 6.37 km2
965.88 m2 b 214.66 m2 c 0.72 km2
17.3 m2
b 48 cm2
c 124.8 km
Area = ab sin θ
1
3 2
b Area = a2 sin 60° =
a
2
4
1
1
c Area = a2 sin (180° − 2θ ) = a2 sin 2θ
2
2
12 a i 129.9 cm2
ii 129.9 cm2
Answers
Number and Algebra
iv 8.09 cm
Exercise 7J
1 a quadrant 1 b quadrant 3
2 a quadrants 1 and 2
c quadrants 2 and 3
e quadrants 1 and 3
3
c
b
d
f
quadrant 4 d quadrant 2
quadrants 2 and 4
quadrants 1 and 4
quadrants 3 and 4
θ
0°
90°
180°
270°
360°
sin θ
0
1
0
−1
0
cos θ
1
0
−1
0
1
tan θ
0
undefined
0
undefined
0
0.139
b 0.995
c −0.530
d −0.574
−0.799
f −0.259
g 0.777
h −0.087
0.900
j −1.036
k 0.900
l −0.424
quadrant 2, sin θ positive, cos θ negative, tan θ negative
quadrant 4, sin θ negative, cos θ positive, tan θ negative
quadrant 3, sin θ negative, cos θ negative, tan θ positive
quadrant 1, sin θ positive, cos θ positive, tan θ positive
quadrant 4, sin θ negative, cos θ positive, tan θ negative
quadrant 2, sin θ positive, cos θ negative, tan θ negative
quadrant 3, sin θ negative, cos θ negative, tan θ positive
quadrant 3, sin θ negative, cos θ negative, tan θ positive
quadrant 3, sin θ negative, cos θ negative, tan θ positive
quadrant 1, sin θ positive, cos θ positive, tan θ positive
quadrant 4, sin θ negative, cos θ positive, tan θ negative
quadrant 2, sin θ positive, cos θ negative, tan θ negative
−sin 80°
b cos 60°
c tan 40°
d sin 40°
−cos 55° f −tan 45° g −sin 15°
h −cos 58°
tan 47°
j sin 68°
k cos 66°
l −tan 57°
30°
b 60°
c 24°
d 40°
71°
f 76°
g 50°
h 25°
82°
42°
b 47°
c 34°
d 9°
33°
f 62°
g 14°
h 58°
0 < θ < 90°
b 90° < θ < 180°
270° < θ < 360°
d 180° < θ < 270°
813
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Chapter 1
814
10
150° 315° 350° 195° 235° 140° 100° 35° 55°
θ2
11 a quadrant 4
d quadrant 2
b quadrant 1 c quadrant 2
e quadrant 1 f quadrant 3
12 a 45°
b i −
c 30°
d i −
e 60°
f i
2
2
b 0
13 a
e −1
−
i
−
f
2
2
2
2
2
2
ii −
iii −
1
2
iii − 3
ii
3
2
ii −
3
2
c −
3
2
iii 1
3
2
1
2
g −
d −
1
2
b
3
3
1
2
p −1
1 a
θ
0°
30°
60°
90°
120°
150°
sin θ
0
0.5
0.87
1
0.87
0.5
θ
180°
210°
240°
270°
300°
330°
360°
sin θ
0
−0.5
−0.87
−1
−0.87
−0.5
0
sin θ
1
–1
90°
0.5
0
θ
180°
210°
240°
270°
300°
330°
360°
cos θ
−1
−0.87
−0.5
0
0.5
0.87
1
90°
180°
270°
360°
120°
150°
−0.5 − 0.87
cos θ
3 a
b
c
4 a
b
sinθ
and cos θ = 0 at 90° and 270°, the value of
cosθ
Exercise 7K
0
60°
0.87
90°
180°
270°
360°
θ
–1
sin θ
and, hence, tan θ is undefined at these values.
cos θ
16 a true
b true
c false
d true
e true
f false
g true
h false
i false
j true
k true
l false
17 a i Check with your teacher.
ii True for these values.
b i sin 60° = cos 30° = 0.866, sin 80° = cos 10° = 0.985,
sin 110° = cos (−20°) = 0.940, sin 195° = cos (−105°) = −0.259
ii Their values are the same. iii They add to 90°.
iv sin θ = cos (90° − θ)
v True for these values.
b
30°
1
0
sinθ
14 As tanθ =
and both sin θ and cos θ are negative over
cosθ
this range, tan θ is positive in the third quadrant.
15 As tanθ =
0°
3
2
l
n undefined o 1
θ
cos θ
1
h − 3
3
3
k −
j −1
m0
2 a
θ
5 a
b
6 a
e
i
7 a
e
8 a
e
9 a
e
i
10 a
i maximum = 1, minimum = −1 ii 0°, 180°, 360°
i maximum = 1, minimum = −1 ii 90°, 270°
i 90° < θ < 270°
ii 180° < θ < 360°
i 0.82
ii −0.98
iii 0.87
iv −0.77
v −0.17 vi 0.26
vii −0.42
viii 0.57
i 37°, 323°
ii 53°, 307°
iii 73°, 287°
iv 84°, 276°
v 114°, 246°
vi 102°, 258°
vii 143°, 217°
viii 127°, 233°
i 0.42
ii 0.91
iii −0.64
iv −0.77
v 0.34
vi −0.82
vii −0.64
viii 0.94
i 37°, 143°
ii 12°, 168°
iii 17°, 163°
iv 64°, 116°
v 204°, 336°
vi 233°, 307°
vii 224°, 316°
viii 186°, 354°
true
b false
c false
d true
false
f true
g true
h true
true
j false
k true
l true
110°
b 60°
c 350°
d 260°
27°
f 326°
g 233°
h 357°
280°
b 350°
c 195°
d 75°
136°
f 213°
g 24°
h 161°
30°
b 60°
c 15°
d 70°
55°
f 80°
g 55°
h 25°
36°
j 72°
k 63°
l 14°
θ
0°
30°
45°
60°
90°
sin θ
0
1
2
2
2
3
2
1
cos θ
1
3
2
2
2
1
2
0
b i
1
2
ii −
1
2
iii −
2
2
iv 0
814
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
v
1
2
3
ix −
2
vi − 3
2
1
x −
2
viii
2
xi −
2
1
xii −
2
3
3
2
3
xv −
xiv
xvi
2
2
2
2
45°, 315°
b 60°, 120°
c 30°, 150°
210°, 330°
e 120°, 240°
f 150°, 210°
17.5°, 162.5°
b 44.4°, 135.6°
53.1°, 306.9°
d 36.9°, 323.1°
191.5°, 348.5°
f 233.1°, 306.9°
113.6°, 246.4°
h 49.5°, 310.5°
28.7°, 151.3°
0, the maximum value of sin θ is 1.
0, the minimum value of cos θ is −1.
Graph is reflected in the x-axis.
Graph is reflected in the x-axis.
Graph is dilated and constricted from the x-axis.
Graph is dilated and constricted from the y-axis.
Graph is translated up and down from the x-axis.
Graph is translated left and right from the y-axis.
xiii −
11 a
d
12 a
c
e
g
i
13 a
b
14 a
b
c
d
e
f
2
2
vii 0
12 a i sin 60°
b i
c
13 a
b
c
i
i
i
i
ii −cos 30°
iii −tan 45°
iv −sin 45°
3
3
ii −
iii −1
iv − 2
2
2
2
negative ii positive
iii negative
iv positive
0.77
ii −0.97
53°, 127°
ii 197°, 343°
iii no value
true
ii true
iii false
Answers
Number and Algebra
Extended-response questions
1 a
Waterfall
N
3 km
5 km
Entrance
325°
Challenges
1
2
3
4
5
6
a 120°, 60°
225°
Use the cosine rule.
514 m
a 2 h 9 min
17.93°
b
d
2 a
c
e
b 8.7 cm
b 308°
Chapter 8
Multiple-choice questions
1 D
5 A
9 D
2 B
6 B
10 C
Pre-test
3 E
7 C
4 D
8 A
Short-answer questions
a 14.74
b 13.17
c x = 11.55, y = 5.42
a 45.6°
b 64.8°
a 4
b 5 3
6.1 m
A = 115°, B = 315°, C = 250°, D = 30°
a 98.3 km
b 228.8 km c 336.8°
a 15.43 m
b 52°
a i 15.5 cm ii 135.0 cm2
b i 14.9 cm ii 111.3 cm2
9 28.1 m
10 a 52.6°
b 105.4°
11 a 12.5
b 42.8°
1
2
3
4
5
6
7
8
2.9 km west
c 7.7 km
i 21.9 m
ii 38.0°
33.646°
b 3177.54 m2
41.00 m
d 61.60 m
i 65.66°, 114.34°
ii 80.2 m, 43.1 m
1 a 3
2 a 4x + 2y
3 a 2a
1
3
4 a 4m + 4n
b 2
b 2xy − x
b −2m
6x
f −
y
b −6x + 12
c 6x 2 + 2x
e 3x − 7
f −2x + 1
g 7x + 10
e −
2
c 5
c 3x2 − 3y2
c 18a2
d 4a − 8a2
h 3x − 21
2
x + 2x − 3
b x − 10x + 21
6x2 − x − 12
d 35x2 + 19x + 2
7(x + 1)
b −9x (1 + 3x)
(x + 3)(x − 3)
e (x + 5)(x + 4)
1
x+4
3
b
c
7 a
2
2
4
8 a 1
b −23
c 9
1
5
9 a −
b 3
c −
2
2
1
, −7
d 0, 5
e −2, 3
f
2
5 a
c
6 a
d
d −6x2y
c a(a + b + 3)
f (x − 2)(x + 7)
d 1
815
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
816
Chapter 1
Exercise 8A
1 a x2 + 2x
2 a a2 + ab
d (y + x)2
3 a 6x
x
e
2
i −18x
4 a 2x + 10
d −4x + 8
g −10x + 6
j 3x 2 − x
m −6x 2 − 4x
p −4x + 16x 2
1
3
14
v -2 x −
9
s -2 x −
5 a
d
6 a
d
g
j
m
p
s
7 a
d
g
j
m
8 a
e
9 a
c
e
g
i
k
m
o
q
s
u
10 a
d
g
j
2
b x2 + 4x + 3
b 2x − 5
a−b
2
b −20x
c 2x2
x
f
g −4x
3
j 7x
k 5x
b 3x − 12
e 6x − 3
h −20x − 15
c x2 + 8x + 16
c a2 − b2
e
k 2x − 2x 2
n −18x 2 + 6x
8
q 4x +
5
3
t −2 x +
2
9 2
w x + 6x
4
2
d −4x2
h −3x
l −13x
c −5x − 15
f 12x + 4
i 2x 2 + 5x
l 6x − 3x 2
o −10x + 10x 2
15
r 6x −
4
3
u -9 x +
8
14
6
x
x − x2
5
5
2
2x + 3x
b 6x − 3x
c 2x + 7x
8x 2 + 7x
e 2x 2 − 2x
f 25x − 12x 2
2
2
x + 10x + 16
b x + 7x + 12
c x2 + 12x + 35
2
2
x + 5x − 24
e x + x − 30
f x2 + x − 6
2
2
x − 4x − 21
h x − 10x + 24
i x2 − 13x + 40
2
2
x + 10x + 25
k x + 14x + 49
l x2 + 12x + 36
2
2
x − 6x + 9
n x − 16x + 64
o x2 − 20x + 100
2
2
x − 16
q x − 81
r 4x2 − 9
2
2
9x − 16
t 16x − 25
u 64x2 − 49
2
2
6x + 13x + 5
b 12x + 23x + 10 c 10x2 + 41x + 21
2
9x − 9x − 10
e 20x2 + 2x − 6
f 6x2 + 5x − 25
2
2
16x − 25
h 4x − 81
i 25x2 − 49
2
2
14x − 34x + 12 k 25x − 45x + 18 l 56x2 − 30x + 4
4x2 + 20x + 25
n 25x2 + 60x + 36 o 49x2 − 14x + 1
3
b 3
c 3
d 8
1
f 2
2x2 + 14x + 24
b 3x2 + 27x + 42
2
−2x − 20x − 32
d −4x2 − 44x − 72
2
5x + 5x − 60
f 3x2 + 6x − 45
2
−3a + 15a + 42
h −5a2 + 30a + 80
2
4a − 36a + 72
j 3y2 − 27y + 60
2
−2y + 22y − 48
l −6y2 + 42y − 72
2
12x + 48x + 45
n 18x2 + 12x − 48
2
−6x − 10x + 56
p 2x2 + 12x + 18
2
4m + 40m + 100
r 2a2 − 28a + 98
2
−3y + 30y − 75
t 12b2 − 12b + 3
2
−12y + 72y − 108
2x2 + 10x + 11
b 2x2 + 20x + 44 c 2y2 − 4y + 5
2
2y − y − 43
e −24a − 45
f b2 + 54b + 5
2
2
x + 10x + 18
h x − 14x + 40
i −4x2 + 36x − 78
2
−25x − 30x + 5
11 a
12 a
b
c
d
13 a
e
14 a
15 a
c
e
16 a
x2 − 12x + 36 cm2
b x2 + 10x − 200 cm2
2
(a + b) (a − b) = a − ab + ba − b2 = a2 − b2
(a + b)2 = (a + b)(a + b) = a2 + ab + ba + b2 = a2 + 2ab + b2
(a − b)2 = (a − b)(a − b) = a2 − ab − ba + b2 = a2 − 2ab + b2
(a + b)2 − (a − b)2 = a2 + ab + ba + b2 − (a2 − ab − ba + b2 )
= 2ab + 2ab = 4ab
618
b 220
c 567
d 1664
1386
f 891
g 3960
h 3480
−x2 + 7x b 10a − 28 c 4x2 + 12x + 9 d 4x + 8
x3 + 6x2 + 11x + 6
b x3 + 11x2 + 38x + 40
3
2
x + 2x − 15x − 36
d 2x3 − 13x2 + 17x + 12
3
2
2x − x − 63x + 90
f 6x3 − 35x2 + 47x − 12
2ab
b (a + b)2 − c2
c (a + b)2 − c2 = 2ab
c2 = a2 + 2ab + b2 − 2ab
c2 = a2 + b2
Exercise 8B
1 a 7
b 6
c 8
d −5
e 2a
f 3a
g −5a
h −3xy
2 a 3(x − 6)
b 4(x + 5)
c 7(a + b)
d 3(3a − 5)
e −5(x + 6) f −2(2y + 1) g −3(4a + 1)
h −b(2a + c)
i x(4x + 1) j x(5x − 2) k 6b(b − 3)
l 7a(2a − 3)
m 5a(2 − a) n 6x(2 − 5x) o −x(2 + x)
p −4y(1 + 2y)
q ab(b − a) r 2xy (xz − 2) s −12mn(m + n) t 3z2(2xy − 1)
3 a (x − 1)(5 − a)
b (x + 2)(b + 3)
c (x + 5)(a − 4)
d (x + 2)(x + 5)
e (x − 4)(x − 2)
f (x + 1)(3 − x)
g (x + 3)(a + 1)
h (x − 2)(x − 1)
i (x − 6)(1 − x)
4 a (x + 3)(x − 3)
b (x + 5)(x − 5)
c (y + 7)(y − 7)
d (y + 1)(y − 1)
e (2x − 3)(2x + 3)
f (6a − 5)(6a + 5)
g (1 + 9y)(1 − 9y)
h (10 − 3x)(10 + 3x)
i (5x − 2y)(5x + 2y)
j (8x − 5y)(8x + 5y)
k (3a + 7b)(3a − 7b)
l (12a − 7b)(12a + 7b)
5 a 2(x + 4)(x − 4)
b 5(x + 3)(x − 3)
c 6(y + 2)(y − 2)
d 3(y + 4)(y − 4)
e 3(x + 5y)(x − 5y)
f 3(a + 10b)(a − 10b)
h 7(3a + 4b)(3a − 4b)
g 3(2x + 3y)(2x − 3y )
i (x + 9)(x + 1)
j (x − 7)(x − 1)
k (a + 5)(a − 11)
l (a − 8)(a − 6)
m (4x + 5)(2x + 5)
n (y + 7)(3y + 7)
o (3x + 11)(7x + 11)
p 3x(3x − 10y)
6 a ( x + 7 )( x − 7 )
b ( x + 5 )( x − 5 )
c ( x + 19 )( x − 19 )
d ( x + 21)( x − 21)
e ( x + 14 )( x − 14 )
f
g ( x + 15 )( x − 15 )
h ( x + 11)( x − 11)
i
( x + 2 2 )( x − 2 2 )
j
( x + 3 2 )( x − 3 2 )
k ( x + 3 5 )( x − 3 5 )
l
( x + 2 5 )( x − 2 5 )
( x + 30 )( x − 30 )
816
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Number and Algebra
n ( x + 4 3 )( x − 4 3 )
o ( x + 5 2 )( x − 5 2 )
p ( x + 10 2 )( x − 10 2 )
q ( x + 2 + 6 )( x + 2 − 6 )
r
s ( x − 3 + 11)( x − 3 − 11) t
( x + 5 + 10 )( x + 5 − 10 )
( x − 1+ 7 )( x − 1− 7 )
u ( x − 6 + 15 )( x − 6 − 15 ) v ( x + 4 + 21)( x + 4 − 21)
w
7 a
d
g
( x + 1+ 19 )( x + 1− 19 ) x ( x − 7 + 26 )( x − 7 − 26 )
(x + 4)(x + a)
b (x + 7)(x + b)
c (x − 3)(x + a)
(x + 2)(x − a)
e (x + 5)(x − b)
f (x + 3)(x − 4a)
(x − a)(x − 4)
h (x − 2b)(x − 5)
i (x − 2a)(3x − 7)

8 a x +


c x +

2

3 
7

4 

x −


x −

2

3 
7

4 

b x +


d x +

3

2 
5

6 

x −


x −

3

2 
5

6 
e ( x − 2 + 2 5 )( x − 2 − 2 5 ) f ( x + 4 + 3 3 )( x + 4 − 3 3 )
g ( x + 1+ 5 3 )( x + 1− 5 3 ) h
( x − 7 + 2 10 )( x − 7 − 2 10 )
( 3 x + 2 )( 3 x − 2 )
j
( 5 x + 3 )( 5 x − 3 )
k ( 7 x + 5 )( 7 x − 5 )
l
( 6 x + 11)( 6 x − 11)
m ( 2 x + 3 )( 2 x − 3 )
n ( 5 x + 4 )( 5 x − 4 )
i
o ( 3 x + 10 )( 3 x − 10 ) p ( 13 x + 7 )( 13 x − 7 )
9 a (x + 2)(y − 3)
d (y − 4)(x − 3)
b (a − 4)(x + 3)
e (a − 3)(2x − 1)
c (a + 5)(x − 2)
f (2a − 5)(x + 4)
10 a 5( x + 2 6 )( x − 2 6 )
b 3( x + 3 6 )( x − 3 6 )
c 7( x + 3 2 )( x − 3 2 )
d 2( x + 4 3 )( x − 4 3 )
e 2( x + 3 + 5 )( x + 3 − 5 ) f
3( x − 1+ 7 )( x − 1− 7 )
g 4( x − 4 + 2 3 )( x − 4 − 2 3 )
h 5( x + 6 + 3 2 )( x + 6 − 3 2 )
11 a 60
b 35
c 69
d 104
e 64
f 40
g 153
h 1260
12 a 4 − ( x + 2 )2 = ( 2 − ( x + 2 ) ) ( 2 + ( x + 2 ) ) = − x ( x + 4 )
b i −x(x + 6)
iv (3 − x)(7 + x)
ii −x(x + 8)
v (8 − x)(6 + x)
iii x(10 − x)
vi (6 − x)(14 + x)
13 a ( x + a )2 = x 2 + 2ax + a2 ≠ x 2 + a2
b If x = 0, then (x + a)2 = x2 + a2; or if a = 0, then (x + a)2
= x2 + a2 is true for all real values of x.
4 1
1
14 x 2 − = (9 x 2 − 4 ) = (3 x + 2 )(3 x − 2 )
9 9
9
4 
2 
2
or x 2 − =  x +   x − 
9 
3 
3
1
1
= (3 x + 2) (3 x − 2)
3
3
1
= (3 x + 2)(3 x − 2)
9
−(2x + 5)
b −11(2y − 3)
c 16(a − 1)
20b
e −12s
f −28y
(5w + 7x)(−w − x)
h (4d + 3e)(−2d + 7e)
6f (2f + 6j )
j 0
x2 + 5y − y2 + 5x
= x2 − y2 + 5x + 5y
= (x − y) (x + y) + 5(x + y)
= (x + y )(x − y + 5)
b i (x + y)(x − y + 7)
ii (x + y)(x − y − 2)
iii (2x + 3y)(2x − 3y + 2) iv (5y + 2x)(5y − 2x + 3)
15 a
d
g
i
16 a
Answers
m ( x + 4 2 )( x − 4 2 )
Exercise 8C
1 a 9, 2
e −8, 3
b 10, 2
f −10, 3
c 5, −3
g −2, −5
d 4, −3
h −12, −3
x − 10
=1
x − 10
3( x − 7 )
b Possible answer:
=3
x −7
−5( x + 3 )
c Possible answer:
= −5
x +3
x+4
1
=
d Possible answer:
3( x + 4 ) 3
2 a Possible answer:
x
2
1
e
3
3 a
i x+1
b
f
x
3
1
4
c 3
g 5
x −3
2
b (x + 3)(x + 2)
j x−2
4 a (x + 6)(x + 1)
k
1
5
2
h
3
d
l 1 − 2x
c (x + 3)2
(x + 5)(x + 2)
e (x + 4)(x + 3)
f (x + 9)(x + 2)
(x − 1)(x + 6)
h (x + 3)(x − 2)
i (x + 4)(x − 2)
(x − 1)(x + 4)
k (x + 10)(x − 3)
l (x + 11)(x − 2)
(x − 2)(x − 5)
n (x − 4)(x − 2)
o (x − 4)(x − 3)
q (x − 6)(x − 3)
r (x − 2)(x − 9)
(x − 1)2
(x − 6)(x + 2)
t (x − 5)(x + 4)
u (x − 7)(x + 2)
(x − 4)(x + 3)
w (x + 8)(x − 4)
x (x − 5)(x + 2)
2(x + 5)(x + 2) b 3(x + 4)(x + 3)
c 2(x + 9)(x + 2)
5(x − 2)(x + 1) e 4(x − 5)(x + 1)
f 3(x − 5)(x + 2)
−2(x + 4)(x + 3) h −3(x − 2)(x − 1) i −2(x − 7)(x + 2)
−4(x − 2)(x + 1) k −5(x + 3)(x + 1) l −7(x − 6)(x − 1)
(x − 2)2
b (x + 3)2
c (x + 6)2
d (x − 7)2
2
2
2
(x − 9)
f (x − 10)
g 2(x + 11)
h 3(x − 4)2
2
2
2
5(x − 5)
j −3(x − 6) k −2(x − 7)
l −4(x + 9)2
1
7 a x+6
b x−3
c x−3
d
x+7
1
1
2
x+4
e
f
g
h
x −5
x −6
x −8
3
x −7
i
5
d
g
j
m
p
s
v
5 a
d
g
j
6 a
e
i
817
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Chapter 1
818
8 a
c
e
g
5( x + 2 )( x − 3 )
( x + 3 )( x + 6 )( x − 2 )
b
x −3
3
2( x − 1)
x+5
d
4
x+5
4
x+7
f
x+2
x −1
h
9 a x− 7
g x + 1− 2
10 a
d
11
6
x −2
a×c
Two numbers that multiply to
give a × c and add to give b
6x 2 + 13x + 6
36
9 and 4
x −4
x+6
8x + 18x + 4
32
16 and 2
12x 2 + x − 6
−72
−8 and 9
10x 2 − 11x − 6
−60
−15 and 4
21x 2 − 20x + 4
84
−6 and −14
15x 2 − 13x + 2
30
−3 and −10
1
3x + 4
e
1
ax 2 + bx + c
b x + 10
1
5x − 3
d
Exercise 8D
h x −3+ 5
2
c x −2 3
f
7x − 5
i x −6− 6
2( x + 3 )
3( x − 5 )
b
x −3
4
c
3
x −3
3
2
e
x −2
x+3
f
x+3
x −1
t 2 − 49 t 2 − 5t − 24 ( t − 7 )( t + 7 ) ( t − 8 )( t + 3 )
×
=
×
5t − 40 2t 2 − 8t − 42
5( t − 8 )
2( t − 7 )( t + 3 )
=
t+7
10
12 a x − 3
6
d
x −2
13 a
b x+1
4
e
x+5
a2 + 2ab + b2 a2 − ab
× 2 2
a2 + ab
a −b
=
( a + b )2
a (a + b)
×
c x−8
f x −7
5
b Answer will vary.
a( a − b )
(a − b) (a + b)
=1
2 a
d
g
j
3 a
d
g
j
m
p
s
v
4 a
(x + 2)(x + 5)
(x − 7)(x − 2)
(3x − 4)(2x + 1)
(x + 4)(5x − 2)
(3x + 1)(x + 3)
(3x − 2)(x − 1)
(3x + 1)(x − 4)
(2x − 7)(x − 1)
(2x + 1)(x − 5)
(4x − 5)(2x − 1)
(3x + 2)(2x + 3)
(2x − 5)(4x − 3)
(6x + 5)(3x + 2)
b (x + 4)(x + 6)
e (x − 3)(x − 4)
h (x − 4)(3x + 2)
k (5x + 6)(2x − 3)
b (2x + 1)(x + 1)
e (2x − 1)(x − 5)
h (3x + 1)(x − 1)
k (3x − 4)(x + 2)
n (13x + 6)(x − 1)
q (3x − 4)(2x + 3)
t (4x − 1)(x − 1)
w (3x − 2)(2x − 3)
b (4x + 3)(5x + 6)
d
g
5 a
d
g
6 a
(5x − 2)(6x + 5)
(6x − 5)(4x − 3)
2(3x + 4)(x + 5)
4(4x − 5)(2x − 3)
−5(5x + 4)(2x + 3)
2x − 5
e
h
b
e
h
b
d
2
3x + 2
e
2
7x − 2
f
4
2x − 3
g
x+4
3x + 1
h
3x − 1
2x + 3
i
5x + 4
7x − 2
j
3x − 2
5x − 2
k
2x + 3
7x + 1
l
2x − 3
4x − 5
a−b
a
b 1
( a + b )2
( a − b )2
d
( a + b )( a − b )
a2
15 a
3x − 8
( x + 3 )( x − 4 )
b
7 x − 36
( x + 2 )( x − 9 )
7 a 3(x − 5)(x − 2)
c x = 2 or x = 5
c
x − 12
( x + 4 )( x − 4 )
d
3 x − 23
( x + 3 )( x − 3 )( x − 5 )
8 a
x − 14
e
( x − 3 )( x + 2 )( x − 6 )
f
14 x + 9
( x + 3 )( x + 4 )( x − 8 )
9 − 3x
( x + 5 )( x − 5 )( x − 1)
h
14 a
c
g
4 x + 11
( x − 1)2 ( x + 4 )
3x + 4
x −3
1− x
3
5(2 x + 1)(2 x − 1)
e
( x + 5 )( x − 5 )
c
g 1
(8x + 3)(5x − 2)
(9x − 2)(5x − 4)
3(2x + 3)(x − 4)
8(2x − 1)(x − 1)
3(2x − 3)2
4x − 1
c (x + 3)(x + 7)
f (x − 5)(x + 3)
i (2x − 1)(4x + 3)
l (2x − 1)(6x − 5)
c (3x + 2)(x + 2)
f (5x − 3)(x + 1)
i (7x − 5)(x + 1)
l (2x − 3)(x + 4)
o (5x − 2)(x − 4)
r (5x − 2)(2x + 3)
u (4x − 5)(2x − 1)
x (3x − 2)(3x + 5)
c (7x − 2)(3x + 4)
f
i
c
f
i
c
(7x + 2)(4x − 3)
(5x − 2)(5x − 8)
3(8x + 1)(2x − 1)
10(3x − 2)(3x + 5)
5(4x − 1)(x − 1)
3x − 2
b −6 m. The cable is 6 m below the water.
b
3x + 2
4
d
4x − 3
5x + 1
f
x+2
5
h
(4 x − 5 )2
( x − 3 )2
818
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
9 −12x 2 − 5x + 3
= −(12x2 + 5x − 3)
= −(3x − 1)(4x + 3)
= (1 − 3x)(4x + 3)
a (3 − 2x)(4x + 5)
c (4 − 3x)(4x + 1)
e (2 − 7x)(2x − 5)
d ( x + 2 + 2 )( x + 2 − 2 )
e ( x + 5 + 22 )( x + 5 − 22 )
f
b (5 − 2x)(3x + 2)
d (3 − 4x)(2x − 3)
f (3 − 5x)(3x + 2)
( x + 2 + 10 )( x + 2 − 10 )
g not possible
h ( x − 3 + 3 )( x − 3 − 3 )
11 a
9x + 2
(2 x − 3 )(4 x + 1)
b
5 x + 15
(3 x − 1)(2 x + 5 )
i ( x − 6 + 34 )( x − 6 − 34 )
j not possible
c
16 x 2 + 5 x
(2 x − 5 )(4 x + 1)
d
7 x − 12 x 2
(3 x − 2 )(4 x − 1)
k ( x − 4 + 17 )( x − 4 − 17 )
e
8x − 5
(2 x + 1)(2 x − 1)(3 x − 2 )
f
11− 3 x
(3 x + 5 )(3 x − 5 )(3 x − 2 )
g
2
(2 x − 5 )(3 x − 2 )
h
12 x + 3
(5 x − 2 )(2 x − 3 )(2 x + 7 )
l not possible
1 a 9
e 16
b 36
c 1
d 4
f 25
25
g
4
9
h
4



b  x + 7 + 41   x + 7 − 41 
2
2



81
4
2 a (x + 2)2
 9 + 93   9 − 93 
d x +
 x +

2 
2 

 3+ 7   3− 7 
e x −
 x −

2 
2 

i
b (x + 4)2
e (x − 3)2
2
d (x − 6)
3 a ( x + 1+ 5 )( x + 1− 5 )
 5 + 23   5 − 23 
f x −
 x −

2 
2 

c (x + 5)2
f (x − 9)2
b ( x + 2 + 7 )( x + 2 − 7 )
c ( x + 4 + 10 )( x + 4 − 10 ) d ( x − 3 + 11)( x − 3 − 11)
e ( x − 6 + 22 )( x − 6 − 22 ) f ( x − 5 + 3 )( x − 5 − 3 )
4 a 9, (x + 3)2
2
g 16, (x − 4)2
j
 3+ 5   3− 5 
7 a x +
 x +

2 
2 

c  x + 5 + 33   x + 5 − 33 
2 
2 

Exercise 8E
d 16, (x + 4)
Answers
Number and Algebra
2
81 
9
, x + 
4 
2
9 
3
, x − 
4 
2
p
81  9  2
, x − 
4  2
 9 + 91   9 − 91 
h x −
 x −

2 
2 

b 36, (x + 6)2
e 25, (x − 5)2
c 4, (x + 2)2
f 1, (x − 1)2
h 36, (x − 6)2
i
25 
2
, x + 5 
4 
2
c 4( x − 1+ 5 )( x − 1− 5 )
d 3( x − 4 + 14 )( x − 4 − 14 )
e −2( x + 1+ 6 )( x + 1− 6 )
f −3( x + 5 + 2 6 )( x + 5 − 2 6 )
l
2
121 
11
, x + 
4 
2
g −4( x + 2 + 7 )( x + 2 − 7 ) h −2( x − 4 + 3 2 )( x − 4 − 3 2 )
k
2
m
 5 + 31   5 − 31 
g x −
 x −

2 
2 

n
2
49 
7
, x + 
4 
2
7
49 
, x − 
2
4 
2
o
1 
1
, x − 
4 
2
8 a 2( x + 3 + 5 )( x + 3 − 5 )
2
c ( x + 1+ 5 )( x + 1− 5 )
d ( x + 5 + 29 )( x + 5 − 29 )
e ( x − 4 + 3 )( x − 4 − 3 )
f
g ( x − 2 + 7 )( x − 2 − 7 )
h ( x − 4 + 21)( x − 4 − 21)
c ( x + 4 + 15 )( x + 4 − 15 )
−3( x − 4 + 11)( x − 4 − 11)



9 a 3 x + 3 + 5   x + 3 − 5 
2 
2 

 3 + 37   3 − 37 
b 5 x +
 x +

2 
2 

5 a ( x + 2 + 3 )( x + 2 − 3 ) b ( x + 3 + 7 )( x + 3 − 7 )
6 a not possible
i
b 3( x + 2 + 5 )( x + 2 − 5 )
( x − 6 + 26 )( x − 6 − 26 )
b not possible



c 2  x − 5 + 17   x − 5 − 17 
2 
2 

 7 + 37   7 − 37 
d 4x −
 x −

2 
2 

 7 + 57   7 − 57 
e −3  x +
 x +

2 
2 

819
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
820
Chapter 1
f
 7 + 65   7 − 65 
−2  x +
 x +

2 
2 




g −4  x − 3 + 29   x − 3 − 29 
2
2






h −3  x − 3 + 17   x − 3 − 17 
2 
2 

i
 5 + 41   5 − 41 
−2  x −
 x −

2 
2 

10 a x 2 − 2x − 24
= x2 − 2x + (−1)2 − (−1)2 − 24
= ( x − 1)2 − 25
= ( x − 1+ 5 )( x − 1− 5 )
= ( x + 4 )( x − 6 )
b Using a quadratic trinomial and finding two numbers that
multiply to −24 and add to −2.
11 a If the difference of two squares is taken, it involves the
square root of a negative number.
b i yes
ii yes
iii no
iv no
v no
vi yes
vii yes
viii no
c i m≤4
ii m ≤ 9
iii m ≤ 25


12 a 2( x + 4 )  x − 3 
 2



b 3  x + 2 + 13   x + 2 − 13 
3 
3 




c 4  x − 7 + 305   x − 7 − 305 
8
8



d Not able to be factorised.
 3 + 41   3 − 41 
e −2  x +
 x +

4 
4 

f
 7 + 13   7 − 13 
−3  x +
 x +

6 
6 

g Not able to be factorised.
 3 + 41   3 − 41 
 x −

h −2  x −
4 
4 

i
j
 7
2( x − 1)  x + 
 2
 2 + 19 
3 x +

3 

 2 − 19 
x +

3 

 5
k −2  x +  ( x − 1)
 2
l
Exercise 8F
1 a 0, −1
e −5, 4
i 2 2 , −2 2
2 a
c
e
g
i
k
3 a
f
4 a
d
b 0, 5
c 0, 4
f 1, −1
g
j
1 7
,−
2 3
k
2
x + 2x − 3 = 0
b
x 2 − 5x + 6 = 0
d
3x 2 − 14x + 8 = 0
f
x2 + x − 4 = 0
h
−x 2 − 4x + 12 = 0
j
3x 2 + 2x + 4 = 0
l
2
b 2
c
2
g 1
h
x = 0, 4
b x = 0, 3
x = 0, 4
e x = 0, 5
d 3, −2
5, − 5
h
3, − 3
5
2
3
3
, −
l − , −
4
5
8
4
x 2 − 3x − 10 = 0
5x 2 − 2x − 7 = 0
4x 2 + 4x − 3 = 0
2x 2 − 6x − 5 = 0
x 2 − 3x − 2 = 0
x 2 + 3x − 6 = 0
1
d 1
e 2
1
i 1
c x = 0, −2
f x = 0, −2
g x = 7, − 7
h x = 11, − 11
i x = 5, − 5
j x = 0, 2
k x = 0, −5
l x = 0, −
m
5 a
d
g
j
m
x = 2, −2
x = −2, −1
x = 5, 2
x = 5, −4
x = −2
x=7
n
b
e
h
k
n
3
6 a x = − , −4
2
7
c x = 5,
2
5
e x =− , 3
3
4 5
g x= ,−
3 2
7 a x = −2, −6
d x=2
8 a
d
g
j
x = 6, −4
x = −2, −5
x = 4, −4
x = −5
x = 3, −3
x = −3, −2
x = −6, 2
x = 8, −3
x = −5
x = 12
1 7
b x =− , −
2 2
1
d x = , 11
2
3
f x =− , 2
5
3
h x = , −4
7
b x = −1, 11
e
b
e
h
k
3
x = , −2
2
x = 8, −4
x = 5, 3
x = −1, −9
x=8
2
3
m x = 3, −1
n x = − , −4
9 a x = 12, −7
b x = −5, 14
5
d x = , -4
2
4
e x =- , 2
5
1
h x = 1,
2
g x = −3, 1
10 a x = ± 2, x = ± 3
o
c
f
i
l
o
1
7
x = 6, −6
x = 2, 4
x = −5, 3
x = 4, 8
x=4
x = −9
b x=± 6
c x=3
2 5
f x= ,
3 2
c x=3
f x = 3, −3
i x = 5, −1
l x = 8, −8
1 3
o x =− , −
4 2
c x = −9, 2
f x = 2, −
5
6
i x = 3, −2
c no solutions
 4
−3  x +  ( x + 1)
 3
820
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
11 a i x = 1, −2
ii x = 1, −2
b no difference
c 3x 2 − 15x − 18 = 3(x 2 − 5x − 6) and, as seen in part a, the
coefficient of 3 makes no difference when solving.
12 This is a perfect square (x + 8)2, which has only 1 solution; i.e.
x = −8.
13 The student has applied the null factor law incorrectly; i.e.
when the product does not equal zero. Correct solution is:
x2 − 2x − 8 = 7
x2 − 2x − 15 = 0
(x − 5)(x + 3) = 0
x = 5 or x = −3
1
14 a x = −2, −1
b x=1
c x= ,5
2
3
d x = 8, −6
e x = −6, −2
f x = , −4
2
g x = 8, −3
h x = 5, −3
i x=2
j x = 4, −3
k x = 5, −2
l x = −5, 3
Exercise 8H
1 a 1
e 4
b 9
c 100
d 625
f 25
25
g
4
h
2 a ( x + 3 )( x − 3 ) = 0
c ( x + 10 )( x − 10 ) = 0
9
4
b ( x + 7 )( x − 7 ) = 0
d ( x + 1+ 5 )( x + 1− 5 ) = 0
Answers
Number and Algebra
e ( x + 3 + 11)( x + 3 − 11) = 0
f
( x − 1+ 2 )( x − 1− 2 ) = 0
3 a x = 2, − 2
b x = 7, − 7
c x = 10 , − 10
d x = 3 − 5, 3 + 5
e x = 4 − 6, 4 + 6
f
4 a x = −3 − 6 , −3 + 6
x = −5 − 14 , −5 + 14
b x = −2 − 2 , −2 + 2
c x = −5 − 10 , −5 + 10
d x = −2 − 6 , −2 + 6
e x = −4 − 19 , −4 + 19
f
Exercise 8G
g x = 4 − 17 , 4 + 17
h x = 6 − 39 , 6 + 39
1 b x+5
c x(x + 5) = 24
d x2 + 5x − 24 = 0, x = −8, 3
e breadth = 3 m, length = 8 m
2 a breadth = 6 m, length = 10 m
b breadth = 9 m, length = 7 m
c breadth = 14 mm, length = 11 mm
3 height = 8 cm, base = 6 cm
4 height = 2 m, base = 7 m
5 8 and 9 or −9 and −8
6 12 and 14
7 15 m
8 a 6
b 13
c 14
9 1m
10 father 64, son 8
11 5 cm
12 a 55
b i 7
ii 13
iii 23
13 a 3.75 m
b t = 1 second, 3 seconds
c The ball will reach this height both on the way up and on
the way down.
d t = 0 seconds, 4 seconds e t = 2 seconds
f The ball reaches a maximum height of 4 m.
g No, 4 metres is the maximum height. When h = 5, there is
no solution.
14 a x = 0, 100
b The ball starts at the tee; i.e. at ground level, and hits the
ground again 100 metres from the tee.
c x = 2 m or 98 m
15 5 m × 45 m
16 150 m × 200 m
i
x = 1− 17 , 1+ 17
x = −3 − 14 , −3 + 14
j
x = 5 − 7, 5 + 7
k x = 3 − 5, 3 + 5
l
x = 4 − 7, 4 + 7
m x = −3 − 13 , −3 + 13
n x = −10 − 87 , −10 + 87
o x = 7 − 55 , 7 + 55
5 a x = −4 − 2 3 , −4 + 2 3
b x = −3 − 2 2 , −3 + 2 2
c x = 5 − 2 5, 5 + 2 5
d x = 2 − 3 2, 2 + 3 2
e x = 5 − 2 7, 5 + 2 7
f
g x = 1− 4 2 , 1 + 4 2
h x = −6 − 3 6 , −6 + 3 6
i
x = −4 − 2 6 , −4 + 2 6
x = −3 − 5 2 , −3 + 5 2
6 a 2
e 0
i 0
b 2
f 2
j 2
c 0
g 2
k 2
d 0
h 0
l 0
7 a x=
−3 + 5 −3 − 5
−5 + 17 −5 − 17
,
,
b x=
2
2
2
2
c x=
3 + 17 3 − 17
−7 + 29 −7 − 29
,
,
d x=
2
2
2
2
e x=
1+ 13 1− 13
,
2
2
g x = 7 + 41 , 7 − 41
2
2
i
x=
−1+ 17 −1− 17
,
2
2
3
3
k x = + 3, − 3
2
2
f
x=
h x=
j
l
−5 + 33 −5 − 33
,
2
2
9 + 61 9 − 61
,
2
2
−9 + 3 5 −9 − 3 5
,
2
2
5
5
x = − + 5, − − 5
2
2
x=
821
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
822
Chapter 1
8 a No real solution.
b x=
−5 ± 17
2
g x=
−9 ± 69
2
j
c x=
5 ± 17
2
d x=
e x=
−5 ± 21
2
f
−3 ± 29
2
c No real solutions.
e x = −5 ± 2 5
x = 3± 5
−3 + 89
3 + 89
cm, length =
cm
2
2
i 1.5 km
ii 1.5 km
i 0 km or 400 km
ii 200 km
200 ± 100 2 km
x2 + 4x + 5 = 0
(x + 2)2 + 1 = 0, no real solutions
2
b  x − 3  + 3 = 0 , no real solutions
 2 4
13 Factorise by quadratic trinomial; i.e. (x + 6)(x − 5) = 0,
6 × (−5) = −30, and 6 + (−5) = −1. Therefore, x = −6, 5.
b x = −4 ± 10
d x = 1± 6
14 a x = 3 ± 7
c x = −2 ± 11
e x =4±2 3
f x = −5 ± 2 6
15 a Use the dimensions of rectangle BCDE and ACDF and the
corresponding side lengths in similar rectangles.
b a = 1+ 5
2
6
2
d x=
−5 ± 17
2
f
a = 3, b = 2, c = 1
a = 5, b = 3, c = −2
a = 2, b = −1, c = −5
−8
b −31
−23
e 41
1
b 0
2
b 0
2
f 2
2
j 1
5 a x=
−3 ± 17
2
d x=4
l x = −3 ± 19
5
6 a x = −2 ± 3
b x = 3± 5
c x = −3 ± 11
x=
−1 ± 13
2
b x=
b
d
f
c
f
c
c
g
k
a = 2, b = 1, c = 4
a = 4, b = −3, c = 2
a = −3, b = 4, c = −5
49
−44
2
1
d 2
0
h 0
0
l 2
−7 ± 65
2
e x = −1, −4
c x = 7 ± 29
2
f x = −1, −7
x=
−5 ± 37
6
i x=
2 ± 22
3
d x=
−3 ± 3 5
2
e x = 2±2 2
f x=
4 ± 10
3
g x=
1± 7
2
h x=
3±2 3
3
i x=
4 ± 31
5
c x=
−5 ± 57
8
7 − 5 + 105
2
3±2 3
3
5 ± 17
d x=
4
8 a x=
g x=
1 ± 11
5
−2 ± 10
2
−2 ± 13
e x=
3
b x=
h x = 3 ± 41
4
f x =1± 6
i x=
5 ± 19
6
9 3 + 53 , −3 + 53
2
2
10 6 2 + 10 units
11 63 cm
−b
12 When b 2 − 4ac = 0, the solution reduces to x = ; i.e.
2a
a single solution.
13 Student’s answers will vary.
14 k = 6 or −6
15 a i k > 4
ii k = 4
9
b i k>
8
c i −2 < k < 2
d i no values
3± 5
2
Exercise 8I
1 a
c
e
2 a
d
3 a
4 a
e
i
4
k x =− , 1
3
b x = −1 ± 5
c x = 4 ± 11
e x=
5 ± 65
4
b x=
10 breadth =
16 a x = −1 ±
h x=
−5 ± 61
2
d x =4± 5
f No real solutions.
9 a x=
11 a
b
c
12 a
−7 ± 65
8
9
ii k =
8
ii ±2
ii no values
iii k < 4
9
8
iii k > 2, k < −2
iii All values of k.
iii k <
Challenges
1
2
3
4
5
6
7
8
b = −4, c = 1
47
a ± 2, ± 1
b ±3
a x = 0, 1
b x = 1, −2
144 cm2
25 km/h
1.6
x 2 − 2x + 2 = (x − 1)2 + 1, as (x − 1)2 ≥ 0, (x − 1)2 + 1 > 0
Multiple-choice questions
1 D
5 B
9 E
2 B
6 D
10 C
3 C
7 C
11 A
4 A
8 C
12 B
822
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Number and Algebra
2 a A = 63π m2
1 a
d
2 a
3 a
c
−2x + 26
b 3x2 + 11x − 20 c 25x2 − 4
2
x − 12x + 36 e 7x + 22
f 12x 2 − 23x + 10
2
2
x + 4x + 4
b 4x + 18x
c x 2 + 3x + 21
(x + 7)(x − 7)
b (3x + 4)(3x − 4)
(2x + 1)(2x − 1)
d 3(x + 5)(x − 5)
e 2(x + 3)(x − 3)
f
g −2( x + 2 5 )( x − 2 5 )
h (x + 5)(x − 3)
i
4 a
c
5 a
c
( x + 11)( x − 11)
( x − 3 + 10 )( x − 3 − 10 )
(x − 6)(x − 2)
b (x + 12)(x − 2)
−3(x − 6)(x − 1)
(3x + 2)(x + 5)
b (2x − 3)(2x + 5)
(6x + 1)(2x − 3)
d (3x − 2)(4x − 5)
2x
6 a
x+3
b
x −4
4
7 a ( x + 4 + 6 )( x + 4 − 6 )
Pre-test
1 a x 2 + 2x − 3



d  x + 3 + 17   x + 3 − 17 
2 
2 

12 a 1 solution
13 a x =
c x=
−2
−1
0
1
2
3
y
9
4
1
0
1
4
9
b (9x − 7)(9x + 7)
d (x − 5)(x − 4)
f 3(x − 3)(x − 4)
3
1
5
or 2 e x = or −
2
3
2
f
x =−
2
4
or
7
3
d x = −3
8 a 0, 2
9 a
2
1
–4 –3 –2 –1–1 O 1 2 3 4 5 6
–2
–3
–4
–5
–6
–7
–8
–9
b x = 3±2 2
d x=
−5 ± 3 5
2
b 2 solutions c 0 solutions d 2 solutions
−3 ± 33
2
b x = 1± 5
2 ± 14
2
d x=
b
x
y
3
2
1
1± 37
6
i 15 + 2x m
ii 12 + 2x m
Area = 4x2 + 54x + 180 m2
Trench = 4x2 + 54x m2
Minimum width is 1 m.
c (1, −1)
b 0
y
c x = 5, −5
f x = −9, 4
Extended-response questions
1 a
b
c
d
−3
c x = 5 or − 5
 7 − 31 
x +

2 

b x = 0, 3
e x=4
3 ± 17
2
x
3 a (x − 3)(x + 3)
c (x + 4)(x + 1)
e (x + 8)(x − 2)
2 5
1 3
h x= ,−
i x= ,−
g x = −2, 1
3 2
9 2
2
9 a x = 3, −3
b x = 5, −1
c x = 4, −7
d x = −3, 6
10 length = 8 m, width = 6 m
c x=
c 3x 2 + 2x − 8
6 a x = 2 or −1
b x = 3 or −5
c x = 3 or −4
d x = −1 or −4 e x = 1 or 2
f x = 4 or 3
7 a x = 0 or 4
b x=4



e  x + 5 + 13   x + 5 − 13 
2
2



11 a x = −2 ± 7
b 2x 2 − 9x − 5
2
d x =−
c ( x − 3 + 2 3 )( x − 3 − 2 3 )
8 a x = 0, −4
d x = 3, 7
Chapter 9
4 a ( x + 4 − 15 )( x + 4 + 15 ) b ( x − 1− 8 )( x − 1 + 8 )
5 a x = 1 or −2
b x = 2 or −3
c x = −5 or −7
b ( x + 5 + 29 )( x + 5 − 29 )
 7 + 31 
f x +

2 

b 0.46 m
c i 420 = 3πr 2 + 12πr
ii 3πr 2 + 12πr − 420 = 0
2
iii r = 4.97 m; i.e. πr + 4πr − 140 = 0
Answers
Short-answer questions
–3 –2 –1–1 O 1 2 3 4
–2
–3
–4
–5
x
823
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Chapter 1
824
Exercise 9A
2 a
b
c
d
e
3 a
1
x
−3
−2
−1
0
1
2
3
y
9
4
1
0
1
4
9
y
10
9
8
7
6
5
4
3
2
1
b
c
d
e
–4 –3 –2 –1 O
x
1 2 3 4
f
maximum
(−2, 4)
2
−5, 1
x = −2
i (2, −5)
ii x = 2
iii −1, 5
iv −3
i (2, 0)
iii 2
i (2, 5)
iii no x-intercept
i (−3, 0)
iii −3
i (2, −2)
iii 1, 3
i (0, 3)
iii −3, 3
ii
iv
ii
iv
ii
iv
ii
iv
ii
iv
x=2
−1
x=2
7
x = −3
4
x=2
6
x=0
3
4
Formula
Max or
min
Reflected in the
x-axis (yes/no)
Turning
point
y value when
x=1
Wider or narrower than
y = x2
narrower
a
y = 3x
2
min
no
(0, 0)
y=3
b
1
y = x2
2
min
no
(0, 0)
y=
c
y = 2x
2
1
2
wider
min
no
(0, 0)
y=2
narrower
d
y = −4x
2
max
yes
(0, 0)
y = −4
narrower
e
1
y = − x2
3
max
yes
(0, 0)
y=−
f
y = −2x
2
max
yes
(0, 0)
y = −2
1
3
wider
narrower
5
Formula
2
a
y = (x + 3)
b
y = (x − 1)2
c
y = (x − 2)
2
d
y = (x + 4)2
y-intercept (x = 0)
x-intercept
x = −3
9
−3
x=1
1
1
Turning point
Axis of symmetry
(−3, 0)
(1, 0)
(2, 0)
x=2
4
2
(−4, 0)
x = −4
16
−4
6
Turning point
y-intercept (x = 0)
y value when x = 1
2
(0, 3)
3
y=4
2
(0, −1)
−1
y=0
2
(0, 2)
2
y=3
2
(0, −4)
−4
y = −3
Formula
a
b
c
d
y=x +3
y=x −1
y=x +2
y=x −4
824
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Number and Algebra
x=0
x=0
x = −1
x=0
(0, 0)
(0, 0)
(−1, 0)
(0, 2)
0
0
1
2
H
A
y = x2 + 1
b
e
h
k
b
e
h
k
b
e
h
k
b
f
x=0
x=0
x = −3
x=0
(0, 7)
(0, −4)
(−3, 0)
(0, −16)
7
−4
−9
−16
C
E
c
f
i
l
c
f
i
l
c
f
i
l
c
g
x=0
x=2
x=0
x = −4
(0, 0)
(2, 0)
(0, −3)
(−4, 0)
0
4
−3
−16
G
B
b y = x 2 − 2 c y = 2x 2
12 a 0, ± 3, ± 5 b 5, 10, 12
13 a i
y
iii
22
20
18
16
14
12
10
8
6
4
2
d D
h F
–5 –4 –3 –2 –1–2O
d y = 9 − x2
iv
1 2 3 4 5
–5 –4 –3 –2 –1–2O
x
1 2 3 4 5
x
y
y
22
20
18
16
14
12
10
8
6
4
2
22
20
18
16
14
12
10
8
6
4
2
–5 –4 –3 –2 –1–2O
x
y
v
ii
1 2 3 4 5
22
20
18
16
14
12
10
8
6
4
2
22
20
18
16
14
12
10
8
6
4
2
–5 –4 –3 –2 –1–2O
y
Answers
7 a
d
g
j
8 a
d
g
j
9 a
d
g
j
10 a
e
11 a
1 2 3 4 5
x
–5 –4 –3 –2 –1–2O
1 2 3 4 5
x
825
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
826
Chapter 1
vi
15 a i
y
y
22
20
18
16
14
12
10
8
6
4
2
11
10
9
8
7
6
5
4
3
2
1
–5 –4 –3 –2 –1–2O
1 2 3 4 5
x
b The constant a determines the narrowness of the graph.
14 a i
y
–5 –4 –3 –2 –1–1O
ii
ii
1 2 3 4 5
8
7
6
5
4
3
2
1
1 2 3 4 5
x
b The constant k determines whether the graph moves up or
down from y = x 2.
16 Answers could be:
a y = x2 − 4
b y = (x − 5)2
2
c y=x +3
17 a y = x 2 + 2
b y = −x 2 + 2
2
c y = (x + 1)
d y = (x − 2)2
2
e y = 2x
f y = −3x 2
1
2
g y = (x + 1) + 2
h y = ( x − 4 )2 − 2
8
18 parabola on its side
20
18
16
14
12
10
8
6
4
2
–5 –4 –3 –2 –1–2O
–5 –4 –3 –2 –1–1O
–2
–3
–4
–5
x
y
x
y
20
18
16
14
12
10
8
6
4
2
–5 –4 –3 –2 –1–2O
1 2 3 4 5
1 2 3 4 5
x
b The constant h determines whether the graph moves left or
right from y = x 2.
y
4
3
2
1
O
–1
–2
–3
–4
1 2 3 4 5 6 7 8 9
x
826
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Exercise 9B
1 a
d
g
2 a
d
g
j
3 a
d
g
4 a
d
b
e
h
b
e
h
k
b
e
h
(0, 0)
(0, −3)
(−5, 0)
3
−1
−16
2
up
down
right
c
f
i
c
f
i
l
c
f
(0, 0)
(0, 7)
(0, 0)
−3
1
−25
6
right
down
up
1
O
(0, 3)
(2, 0)
(0, 0)
−3
4
−2
−63
left
left
–5 –4 –3 –2 –1
–1
–2
–3
–4
–5
–6
–7
–8
–9
–10
y
7
6
5
4
3
2
1
–4 –3 –2 –1–1O
–2
b
e
(1, 2)
x
1 2 3 4
–4 –3 –2 –1–1O
–4 –3 –2 –1–1O
–2
–3
–4
1 2 3 4
x
8
7
6
5
4
3
2
1
(1, 3)
(1, 12 )
1 2 3 4
y
f
y
10
9
8
7
6
5
4
3
2
1
x
x
–8
–9
c
1 2 3 4 5
(1, 3)
O
–4 –3 –2 –1–1
1 2 3 4
(1, – 13 )
y
10
9
8
7
6
5
4
3
2
1
y
1
O
–4 –3 –2 –1
–1
–2
–3
–4
–5
–6
–7
y
Answers
Number and Algebra
1 2 3 4
x
(1, –3)
x
827
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
828
Chapter 1
g
j
y
1
O
–4 –3 –2 –1
–1
–2
–3
–4
–5
–6
–7
–8
–9
–10
h
(1, 0)
10
9
8
7
6
5
4
3
2
1
x
1 2 3 4
O
–3 –2 –1–1
y
x
1 2 3 4 5
k
1
O
–4 –3 –2 –1
–1
–2
–3
–4
–5
–6
–7
–8
–9
–10
y
y
1
x
1 2 3 4
–6 –5 –4 –3 –2 –1–1O
–2
–3
–4
–5
–6
–7
–8
–9
–10
(1, –4)
i
y
l
10
9
8
7
6
5
4
3
2
1
O
–6 –5 –4 –3 –2 –1–1
1
x
y
1
1
x
–2 –1–1O
–2
–3
–4
–5
–6
–7
–8
–9
1 2 3 4 5 6
x
828
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
5 a
d
g
j
6 a
(−3, 1)
(4, −2)
(3, 3)
(2, −5)
b
e
h
k
(−2, −4)
(3, −5)
(2, 6)
(−1, −1)
c
f
i
l
d
(1, 3)
(2, 2)
(−1, 4)
(4, −10)
y
11
10
9
8
7
6
5
4
3
2
(1, 2)
1
y
10
9
8
7
6
5
4
3
2
1
(–1, 1)
–4 –3 –2 –1 O
–3 –2 –1–1O
1 2 3 4
b
x
e
9
8
7
6
5
4
3
2
1
–1–2O
c
1 2
(4, 1)
x
1 2 3 4 5 6 7 8
f
y
7
6
5
4
3
2
1
11
10
9
8
7
6
5
4
3
2
1
–6 –5 –4 –3 –2 –1–1O
x
x
y
(–3, 2)
1 2 3 4
y
18
16
14
12
10
8
6
4
2
y
–6 –5 –4 –3 –2 –1–1O
(–2, –1)
Answers
Number and Algebra
–3 –2 –1–1O
–2
–3
–4
1
1 2 3 4 5
x
(1, –4)
x
829
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
830
Chapter 1
g
j
y
3
2
1
–3 –2 –1–1O
–2
–3
–4
–5
–6
–7
h
y
(1, 3)
1
1 2 3 4 5
x
k
y
1
–3 –2 –1–1O
–2
–3
–4
–5
–6
–7
–8
–9
–3 –2 –1–1O
–2
–3
–4
–5
–6
–7
–8
–9
(2, 1)
1 2 3 4 5
x
y
(2, 1)
1 2 3 4 5
i
x
–1–2O
–4
–6
–8
–10
–12
–14
–16
–18
–20
x
1 2 3 4 5 6 7 8
(4, –2)
y
l
1
–7 –6 –5 –4 –3 –2 –1–1O
(–3, –2)
–2
–3
–4
–5
–6
–7
–8
–9
–10
–11
1
y
(–2, 2)
x
2
–6 –5 –4 –3 –2 –1–2 O 1 2
–4
–6
–8
–10
–12
–14
x
7 a y = −x 2
d y = x2 + 4
g y = −(x + 3)2
b y = (x + 2)2
e y = (x − 1)2
h y = (x + 5)2 − 3
c y = x2 − 5
f y = −x 2 + 2
i y = (x − 6)2 + 1
8 a y = 6x 2
b y = x2 + 4
1
e y = x2
2
h y = (x − 1)2
c y = (x − 3) 2
d y = −(x + 2)2
g y = x2 − 1
f y = −x 2 + 2
i y = −7x 2
830
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
9 a maximum
b (5, 25)
c 0
d 25 m
e i 21 m ii 21 m
iii 0 m
10 a y = x 2 − 9 b y = (x − 2)2
11 (−3, 6), (3, 6)
12 a (1, 0)
b (−2, 0)
c (−3, 0)
e (0, −2)
f (0, 5)
g (−4, −1)
i (5, 4)
j (−2, 3)
k (−3, −5)
13 a translate 3 units right
b translate 2 units left
c translate 3 units down
d translate 7 units up
e reflect in x-axis
f translate 2 units left and 4 units down
g translate 5 units right and 8 units up
h reflect in x-axis, translate 3 units left
i reflect in x-axis, translate 6 units up
14 a (h, k)
b (0, ah 2 + k)
15 a
y
24
22 22
20
18
16
14
12
10
8
6
4
2
–1–2O
c
y
(3, 4)
4
2
–1–2O 1 2 3 4 5 6 7
–4
–6
–8
–10
–12
–14
–14
–16
–18
d (0, −4)
h (−2, 3)
l (3, −3)
Answers
Number and Algebra
d
x
y
2
–7 –6 –5 –4 –3 –2 –1–2O 1 2
(–3, –4)
–4
–6
–8
–10
–12
–14
–16
–18
–20
–22 –22
x
(3, 4)
x
1 2 3 4 5 6
b
e
y
20
18
16
14
12
10
8 8.5
6
4
(3, 4)
2
y
24
22
20
18
17
16
14
12
10
8
6
(–2, 5) 4
2
–5 –4 –3 –2 –1–2O
1
–3 –2–2O
2 4 6 8
x
x
831
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
832
Chapter 1
f
j
y
4
2
(3, 4)
O 2 4 6 8 10
–4 –2–2
–4 –0.5
–6
–8
–10
–12
–14
g
y
18
16
14
12
10
8
6
4 2.5
2
(1, 2)
–5 –4 –3 –2 –1–2O 1 2 3 4 5 6 7
x
x
y
k
4 4
2
–4 –3 –2 –1–2O
–4
–6
–8
–10
–12
h
y
(–2, 1) 2
1 2 3 4
x
–5 –4 –3 –2 –1–2O 1 2 3
–4
–6
–7
–8
–10
–12
–14
x
y
O
–4 –3 –2 –1–2
1 2 3 4
–3
–4
–6
–8
–10
–12
–14
–16
–18
x
l
y
4
2
(2, 3)
–1–2O 1 2 3 4 5
–4
–6
–8
–10
–12
–13
–14
x
y
i
– 5
2
6 5
4
2
–4 –3 –2 –1–2O
–4
–6
–8
–10
–12
–14
5
2
1 2 3 4
Exercise 9C
x
1 a
d
g
2 a
d
g
3 a
d
g
x = −1, x = 2
x = 0, x = 3
x =± 5
x = −2, x = −1
x = 0, x = 4
x = ±3
2
0
−9
b
e
h
b
e
h
b
e
h
x = 3, x = 4
x = 0, x = 5
x =± 7
x = −4, x = 2
x = 0, x = 6
x = ±5
−8
0
−25
c
f
i
c
f
i
c
f
i
x = −1, x = −5
x = 0, x = −2
x = ±2 2
x=4
x = 0, x = −5
x = ± 10
16
0
−10
832
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
4 a (x + 6)(x − 8)
c x=1
5 a
y
b x = −6, x = 8
d (1, −49)
e
b
y
x
–4 –2–2 O 2 4 6
–4
–6
–8
–10 (1, –9)
x
1 2 3 4 5 6 7
y
10
8
6
4
2
12
10
8
6
4
2
–1–2O
Answers
Number and Algebra
f
14
12
10
8
6
4
2
y
15
10
5
–2–2O
–4
2 4 6 8 10
–6 –4 –2–5O 2 4 6 8 10
–10
–15
(0, –21)
–20
–25
x
g
c
y
10
8
6
4
2
y
18
16
14
12
10
8
6
4
2
–8 –6 –4 –2–2
d
(0, 15)
O
2
–10 –8 –6 –4 –2–2O
–4
–6
–8
–10
x
h
y
15
10
5
–4 –2–5 O 2 4 6 8 10
–10
–15
–20
–25
x
x
2
x
y
30
25
20
15
10
5
–2–5O
(0, 24)
2 4 6 8 10
(5, –1)
x
833
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
834
Chapter 1
i
y
25
20
15
10
5
–2–5 O 2 4 6 8 10 12
–10
–15
(6, –16)
–20
6 a
y
d
35
30
25
20
15
10
5
x
–12–10 –8 –6 –4 –2–5O
(–5.5, –0.25) –10
y
y
e
5
4
3
2
1
25
20
15
10
5
–1–5 O 1 2 3 4 5 6 7 8 9 10
(4.5, –0.25)
–10
x
–5 –4 –3 –2 –1–1O
–2
(–2.5, –2.25) –3
1
x
y
b
f
6
5
4
3
2
1
y
15
12
10
5
–1–1 O 1 2 3 4 5
(2.5, –0.25)
–2
c
x
2
–12–10 –8 –6 –4 –2–5O
–10
–15
–20
–25
–30
(–6.5, –30.25)
x
y
20
15
12
10
5
–2–5O
–10
–15
–20
–25
–30
g
2 4 6 8 10 12
(6.5, –30.25)
x
y
25
20
15
10
5
x
–4 –2–5O 2 4 6 8
–10 –12
–15
(2, –16)
x
834
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
h
l
y
40
35
30
25
20
15
10
5
5
O
–6 –4 –2–1
–5
i
y
x
2 4 6 8
(0.5, –2.25)
y
7 a
y
25
20
15
10
5
–4 –2–5O 2 4 6 8 10
–10
–15 –14
–20
(2.5, –20.25)
j
16
14
12
10
8
6
4
2
x
x
–6 –4 –2–2 O 2 4 6
y
x
–12–10 –8 –6 –4 –2–5O 2
–10
–15
–20 –22
–25
–30
–35
–40
(–4.5, –42.25)
Answers
Number and Algebra
b
35
30
25
20
15
10
5
y
10
8
6
4
2
–8 –6 –4 –2–5O –4 2 4
(–1.5, –6.25)
k
–8 –6 –4 –2–2O
–4
–6
–8
–10
x
y
2
x
5
O
–10 –8 –6 –4 –2–5
–10
–15
–20
–25
–30
–35
–40
(–3.5, –42.25)
2 4
x
c
y
12
10
6
4
2
–4 –2–2O
–4
2 4 6 8
x
835
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
836
Chapter 1
d
b
y
y
8
6
4
2
–4 –2–2O
–4
–6
18
16
14
12
10
8
6
4
2
x
2 4 6 8
(2.5, –6.25)
e
y
–10 –8 –6 –4 –2–2 O 2
14
12
10
6
4
2
c
–6 –4 –2–2 O 2 4
(–1.5, –2.25)
f
35
30
25
20
15
10
5
y
–2–5O
x
2
y
x
120
100
80
60
40
20
–22–20–18–16–14–12–10 –8 –6 –4 –2–2 O 2
y
9 a
12
10
6
4
2
–6 –4 –2–2O
2 4 6 8 10 12
d
–8 –6 –4 –2–2O
–4
–6
–8
–10
–12
(–3.5, –12.25)
8 a
y
x
6
4
2
x
x
y
8
6
4
2
2 4
x
O
–6 –4 –2–2
–4
–6
–8
–10
2 4 6
x
836
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
b
c
y
y
4
2
4
2
–6 –4 –2–2O
–4
–6
–8
–10
–12
–14
–16
c
x
2 4 6
O
–4 –2–2
–4
–6
–8
–10
–12
d
y
y
–3 –2 –1–1O
–2
–3
–4
O
–4 –2–2
–4
–6
–8
–10
–12
x
1 2 3
e
(3.5, −4.5) b (3.5, −6.75)
(−3, −4)
e (0, −196)
(1, 0)
h (1, 0)
a = −1, b = −3, TP (2, −1)
a = 5, b = −1, TP (−2, −9)
a = 2, b = −6, TP (2, −16)
–4 –2–2 O 2 4 6
–4
–6
–8
c x-intercepts: ± 5 2 , TP (0, −50)
10
8
6
4
2
–4 –2–2O
–4
–6
f
b
y
2
2 4
x
–4 –2–2O
–4
–6
–8
–10
–12
x
10
8
6
4
2
b x-intercepts: ± 11, TP (0, −11)
y
2 4
y
c (−3, −3)
f (0, 196)
12 a x-intercepts: ± 2 , TP (0, −2)
13 a
x
2 4 6
4 (1.5, 2.25)
2
3
2
1
10 a
d
g
11 a
b
c
Answers
Number and Algebra
2 4
x
x
y
25
20
15
10
5
–2–5O
–10
2 4 6 8 10
x
14 a = −2, TP (1, 18)
15 Coefficient does not change the x-intercept.
16 a y = x 2 − 2x + 1 = (x − 1)2
Only one x-intercept, which is the turning point.
b Graph has a minimum (0, 2), therefore its lowest point is
2 units above the x-axis.
837
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
838
Chapter 1
17 a x = 4, x = −2
b (1, −9), (1, 9)
c same x-coordinate; y-coordinate is reflected in the x-axis
18 a 0
19 a y = x(x − 4)
c y = x(x + 6)
7 a
y
 b b2 
c − , − 
 2
4
b 0, −b
b y = x(x − 2)
d y = (x + 3)(x − 3)
e y = (x + 2)(x − 2)
f
y = ( x + 5 )( x − 5 )
g y = (x + 4)(x − 2)
i y = (x + 1)(x − 3)
h y =(x − 1)(x − 5)
j y = −x(x − 4)
k y = −(x + 2)(x − 6)
l
O
y = −(x − 10 )( x + 10 )
(2, –4)
b
Exercise 9D
1 a y = x 2 + 2x − 5
= x 2 + 2x + 1 − 1 − 5
= (x + 1) 2 − 6
TP (−1, −6)
b y = x 2 + 4x − 1
= x 2 + 4x + 4 − 4 − 1
= (x + 2)2 − 5
TP (−2, −5)
c y = x 2 − 6x + 10
= x 2 − 6x + 9 − 9 +10
= (x − 3)2 + 1
TP (3, 1)
d y = x2 − 3x − 7
x
4
y
7
–7
–1
x
O
(–4, –9)
c
y
15
9 9
= x2 − 3x + − − 7
4 4
 3  2 37
= x −  −
 2
4
–5
–3
(–4, –1)
 3 −37 
TP =  ,
 2 4 
2 a x = ±3
d x = −5, x = 3
d
b
x =± 3
x = 6± 5
e
x =± 2 −4
f
b
e
h
b
e
h
k
b
max (1, 3)
min (−5, 10)
min (3, −7)
−2
−16
1
13
x = −7, x = −1
c max (−1, −2)
f max (7, 2)
c
f
i
l
c
7
−55
−5
−5
x = 9, x = −3
d x = −2 ± 5
e
x = 1± 10
f
x = 5± 3
g x=4
h x = −6
j no x-intercept
k
x = 2± 5
y
c x = 3, x = −1
min (3, 5)
min (−2, −5)
max (3, 8)
6
9
3
−8
x = 5, x = 1
3 a
d
g
4 a
d
g
j
5 a
x
O
5
O 1
x
5
(3, –4)
e
y
80
i no x-intercept
l
x = 3 ± 10
6 a x = −1, x = −5
b x =± 7 −3
c −4 ± 21
e no x-intercept
d x = −1± 7
f x = 6 ± 41
(–8, 16)
O
x
838
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
f
l
y
y
(2, 4)
51
O
(–7, 2)
y
g
8 a
5
x
4
O
x
Answers
Number and Algebra
y
3
(2, 1)
x
–3
b
y
h
O
–1
(–2, –1)
x
O
y
15
–1 O
(3, 6)
3
x
x
O
–3
(1, – 4)
y
i
c
O
(5, –4)
y
x
9
–29
–3 O
y
j
d
O
y
16
x
(–4, –9)
x
–25
O
k
e
(–9, 25) y
–14
–4 O
–56
x
x
4
y
–2 O
4
x
–8
(1, –9)
839
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
840
Chapter 1
f
c
y
–3 O
–15
y
6
x
5
(1, 5)
20
7
(4, 4)
x
O
–1
y
d
y
g
–7
x
O
(1, –16)
x
O
e
y
(–4, –9)
h
y
O
–2 – 2√2
x
–2 + 2√2
–4
5
(–2, –8)
–5
–1
x
O
f
y
(–3, –4)
y
i
O
–12
1
x
3 + √5
2
3 – √5
2
O
3,
2
– 54
g
(–6, –36)
x
y
y
9 a
–5 + √17
2
1
–2 – √3
b
y
y
O
–3 – √14
– 52 , – 17
4
h
(–2, –3)
x
O
–5 − √17
2
O
x
–2 + √3
2
–5
x
–3 + √14
–1 O
–2 1
2
2
x
, –9
4
(–3, –14)
840
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
y
i
– 32 ,
3
4
3 2 11
e y = −  x +  −

4
2
y
3
x
O
10 a 2
b 1
c 1
e 0
f 2
11 a (−1 + 6 , 0 ), (−1 − 6 , 0 )
d x = −2 ± 10
Answers
Number and Algebra
x
(– 32, – 114 ) O
–5
d 0
b x = 3, 1 c x = 7, x = −1
e −1 ± 6
12 a y = −(x − 2)2 + 7
f
x=
3± 5
2
f
5  2 33

y = − x +  +

4
2
y
y
(– 52 , 334 )
(–2, 7)
3
–2 − √7
O
–2 + √7
x
–5 − √33
2
2 –5 + √33
2
x
O
b y = −(x − 1)2 + 3
13 a k > 0
y
2
c y = −(x − 3)2 + 5
y
2
b 2 − 4c
b

=x+  −


4
2
15 a y = 4(x + 1)2 − 1
y
(3, 5)
O 3 − √5
2
2
b 2 4c
b

=x+  − +


4
4
2
x
1 + √3
O
c k<0
2
 b  b
14 x 2 + bx + c = x 2 + bx +   −   + c
2
2
(1, 3)
1 − √3
b k=0
3
x
3 + √5
–1
2
–3
2
–4
x
O
d y = −(x − 4)2 + 8
(–1, –1)
y
b y = 3(x − 2)2 − 2
y
(4, 8)
10
4 − 2√2
O
4 + 2√2
x
2 + √3
2
O
–8
2 –√3
2
x
(2, –2)
841
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
842
Chapter 1
c y = 2(x + 3)2 − 17
3  2 131

h y = 5  x −  +
20
10
5  2 191

g y = 6 x +  +

24
12 
y
–3 – √ 2
17
–3 +
√172
y
y
x
O
9
(– 125 , 191
24 )
7
x
O
(103 , 131
20 )
x
O
(–3, –17)
i
2

1  25
d y = 2 x +  −
 4
8
y
– 32
O
 6  2 36
5 x +  −
 5
5
j
y
y
– 12
5
x
1
(– 14 , – 258)
O
( – 65 , – 365 )
 7  2 25
e y = 2 x −  −
 4
8
x
– 10
7
(– 57 , – 257 )
– 32 ,
– 32 −
3
1
2
3
x
O
 3  2 35
k y = −3  x +  +
 2
4
y
O
 5  2 25
y = 7x +  −
 7
7
y
35
4
√35
2√3
2
– 32 +
√35
2√3
x
O
x
( 74 , – 258 )
f y = 4(x − 1)2 + 16
y
l
y
 5  2 21
y = −4  x −  +
 4 4
5
4
21
4
20
5 − √21
4
(1, 16)
O
–1
O
5 + √21
4
x
x
842
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Exercise 9E
e
1 a 2 intercepts
2 a −1 ± 2
b 0 intercepts
b 2.5, −1
3 a zero
4 a 1 intercept
d 2 intercepts
g 2 intercepts
j 0 intercepts
5 a 3
d −4
g 0
6 a (−1, 3)
b
b
e
h
k
b
e
h
b
−1 ± 17
c
4
positive
no intercepts
no intercepts
2 intercepts
2 intercepts
5
8
0
(−2, −5)
 3 1
− , 6 
 2 4
d (1, −5)
e
 3
1
g  − , −5 
 4
8
3
9
h  ,− 
 8 16 
j
1
3
 , −2 
4
4
 1 1
 − , 
3 3
k
7 a
y
c 1 intercept
−3 ± 15
3
negative
2 intercepts
1 intercept
2 intercepts
2 intercepts
−2
−10
−7
(2, −1)
 7 1
 ,5 
2 4
–1.08 O
d
c
c
f
i
l
c
f
i
c
f
i (0, −9)
–5.08
x
–11
Answers
Number and Algebra
(2, –19)
f
y
O
–3.86
x
0.86
–10
(–1.5, –16.8)
g
y
(1, 11)
l (0, 2)
8
y
O 0.55
–5
–4.55
x
–0.91 O
h
2.91
7
y
O
–2.29
0.29
x
–2
O
–3.12
i
(–1, –5)
c
y
(–1, 9)
(–2, –13)
b
x
y
y
x
1.12
(1, 7)
3
O
–0.65
–3
x
1.15
–0.32
(0.25, –3.25)
d
j
O
2.32
x
y
y
(–0.25, 12.1)
–1.35
O
3.35
12
x
–9
(1, –11)
–2.71
O
2.21
x
843
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
844
Chapter 1
k
e
y
– 13 ,
x
(2, –3)
O
x
O
–0.67
y
1
3
–11
l
f
y
y
(–1, 1)
O
–1.45
x
(2, –4)
O
x
–0.55
–16
–4
8 a
g
y
y
– 12 , 2
9
3
h
y
x
– 32 O
b
– 23 ,
y
9 a 1±
x
1
3
c 1±
c
2
3
2
O
x
2 3
3
b −1±
1
O
x
O
10
2
10
2
y
– 52
x
O
–25
d
y
O
–25
5
3
x
d
− 3 ± 15
2
e 2±
6
2
f 1±
30
5
10 y = (x + 1)2 − 6 = x 2 + 2x − 5
11 a anything with b 2 − 4ac > 0
b anything with b 2 − 4ac = 0
c anything with b 2 − 4ac < 0
−b
12 number under square root = 0, therefore x =
2a
(one solution)
13 x =
−b ± b 2 − 4 c
2
844
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
14 y = −
3 a
b2
+c
4a
d (cm)
3
b
c
15 x 2 +   x + = 0
a a
–12 –9 –6 –3–3O
–6
–9
–12
–15
–18
–21
–24
–27
b
c
x2 +   x = −
a
a
 b  b2
b2
c
x 2 +  + 2 − 2 = −
 a  4a 4a
a
2

b
b2 c
x +  = 2 −
 2a  4 a a
2

b  b2 − 4 ac
x +  =
 2a 
4 a2
−b ± b2 − 4 ac
as required
2a
O
–20
–5
Exercise 9F
20
18
16
14
12
10
8
6
4
2
–1–2O
(2, 20)
d
1 2 3 4
b h = 20 m
2 a
b 9m
60
100
140
180
220
x (m)
e 20 m
A (cm2)
t (s)
–2–5O
e 25 cm2
6 a 100 − 2x
d A (m2 )
h (m)
–2–2O
20
25
20
15
10
5
c 4s
10
8
6
4
2 2.75
(100, 20)
d 200 m
5 a 2 × length = 20 − 2x
length = 10 − x
b A = x(10 − x)
c 0 < x < 10
h (m)
x (cm)
c 27 cm
b (0, 0) (200, 0)
20
15
10
5
b
b2 − 4 ac
x+ =±
2a
2a
1 a
3 6 9 12
b 18 cm
4 a (100, 20)
c
h (m)
b
b2 − 4 ac
x+ =±
2a
4 a2
x=
Answers
Number and Algebra
(10, 9)
2 4 6 8 10
x (cm)
f 5 cm by 5 cm
b A = x(100 − 2x)
c 0 < x < 50
(25, 1250)
2 4 6 8 10 12 14 16 18 20 22
d (m)
0
c 22 m
e 1250 m2
50
x (m)
f breadth = 25 m, length = 50 m
845
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
846
Chapter 1
7 a 20 − x
c P
b P = x(20 − x )
13 a P = x(64 − x) so maximum occurs at x = 32.
Maximum product = 32 × (64 − 32) = 1024.
14 a A = (20 − 2x)(10 − 2x)
b min x = 0, max x = 5
d
Turning point occurs
c
A (cm2)
for an x value greater
200
than 5.
e 1 cm
(10, 100)
0
x
20
d i x = 0 or 20
e 100
8 a (20, 0)
b
h (m)
0
ii x = 10
0
15 a 6 m
20
40
b
0
(50, –250)
b i 2
ii none
c i (27.6, −200) and (72.4, −200)
ii (1.0, −10) and (99.0, −10)
d The highway meets the edge of the river (50 m along).
30
0
√6
t (s)
17 5
2 seconds
10 a
h (m)
1
m
24
Exercise 9G
1 a one
2 a (2, 12)
3 a
7
O1
7
b zero, one or two
b (−1, −3)
t (s)
3
y=x+2
2
1
(4, –9)
b i 1 second
ii 7 seconds
iii 4 seconds
c 9 m below sea level
d at 3 and 5 seconds
11 a
c
d
12 a
b
y
5
4
y = x2 – x – 1
P
0
1
2
3
4
5
H
0
0
1
3
6
10
b H=
x (m)
100
d 10 m
h (m)
c
b No, the maximum height reached is 4.5 m.
16 a y (m)
x (m)
–10
c 40 m
6 seconds
9 a
x (cm)
5
P × ( P − 1)
2
1225 handshakes
87 people
1m
No, 1 metre is the minimum height the kite falls to.
−5 −4 −3 −2 −1−1
−2
O
x
1
2
3 4
5
b (−1, 1) and (3, 5) −3
x 2 − x − 1= x + 2
c
x 2 − 2 x − 1= 2
x2 − 2x − 3 = 0
( x − 3 )( x + 1) = 0
x − 3 = 0 or x + 1= 0
x = 3 or x = −1
When x = 3, y = 3 + 2 = 5
When x = −1, y = −1 + 2 = 1
846
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
4 a x2 + 3x + 6 = 0
b x 2 − 5 x + 3 = 0 c x 2 + 3 x − 12 = 0
x2 + 2x + 3 = 0
d
+ x − 1= 0 e
f
5 a b2 – 4ac < 0
b b2 – 4ac > 0
c b2 – 4ac = 0
6 a (−3, 6) and (2, 6)
b (−2, 12) and (6, 12)
1
c no solutions
d (−3, −2) and ( − , −2)
2
3
f no solutions
e ( , 0)
2
7 a x = 0, y = 0 and x = 3, y = 9
b x = 0, y = 0 and x = −2, y = 4
c x = −3, y = 9 and x = 6, y = 36
d x = 0, y = 5 and x = 3, y = 8
e x = −6, y = 34 and x = −2, y = 22
f x = −2, y = −3 and x = 3, y = 17
g no solutions
h no solutions
65
9
and x = −1, y = 8
i x= − ,y=
2
2
2x2
x2 − x + 4 = 0
5
25
j x= − ,y= −
and x = 3, y = 1
3
3
k
l
8 a
b
c
d
e
f
9 a
x = −3, y = 6
x = −1, y = 2
x = −4, y = 16 and x = 2, y = 4
x = −1, y = 1 and x = 2, y = 4
1
1
x = −1, y = 1 and x = , y =
3
9
13
1
x = −2, y = 7 and x = − , y =
4
2
2
16
x = −2, y = 0 and x = , y =
3
9
x = −8, y = −55 and x = 2, y = 5
i no solutions
ii x = −0.73, y = 1.54 and x = 2.73, y = 8.46
iii x = −1.37, y = −2.10 and x = 0.37, y = 3.10
iv x = −2.62, y = 8.24 and x = −0.38, y = 3.76
b i x=
−1 ± 21
−1± 21
, y=
2
2
c
d
13 a
14 a
15 a 1 + 4k
b i k> −
1
4
ii k = −
1
4
iii k < −
iii c < −4
1
4
16 a
b
17 a
b
Discriminant from resulting equation is less than 0.
k≥2
m = 2 or m = −6
The tangents are on different sides of the parabola, where
one has a positive gradient and the other has a negative
gradient.
c m > 2 or m < −6
Exercise 9H
1 a f (x) = 8x
c f (x) = 2
x
e f (x) = 2x + 1
2 a
d
3 a
c
e
4 a
c
e
g
i
5 a
d
6 a
d
7 a
ii x =
3± 5
, y = 3± 5
2
iii x =
−1 ± 13
, y = −1± 13
2
b
iv x =
−1± 17
, y = ± 17
2
c
10 a 2
b 0
c 2
d 0
e 1
f 2
11 Yes, the ball will hit the roof. This can be explained in a
number of ways, using the discriminant we can see that the
path of the ball intersects the equation of the roof y = 10.6.
1
7
12 a x = −1, y = −2 and x = − , y = −
2
4
5
15
,y= −
and x = 2, y = −4
2
4
x = 1, y = 8 and x = 2, y = 7
x = −6, y = −14 and x = 2, y = 2
1
1
(−1, 4) and ( , 5 )
b 212 m
2
2
(3, −4)
b i c > −4 ii c = −4
b x=
Answers
Number and Algebra
b f (x) = 9 − x2
d f (x) = x(2x − 3 )
true
b true
c false
false
e true
y≥0
b y>0
y>9
d 0≤y≤1
y≥0
function
b function
function
d function
not a function
f function
not a function
h function
not a function
4
b 10
c 28
1
e −2
f 3a + 4
5
2
0
b 2
c −4
230
e 0.176
f 2k3 − k2 + k
f (0) = 0, f (2) = 8, f (−4) = −16, f (a) = 4a,
f (a + 1) = 4a + 4
f (0) = 1, f (2) = −3, f (−4) = −15, f (a) = 1 − a2,
f (a + 1) = −a2 − 2a
1
,
f (0) = 1, f (2) = 4, f (−4) =
16
a
a+1
f (a) = 2 , f (a + 1) = 2
d f (0) = undefined, f (2) = 1, f (−4) = − 1 ,
2
2
2
f (a) = , f (a + 1) =
a
( a + 1)
e f (0) = −12, f (2) = 0, f (−4) = −12, f (a) = a2+4a −12,
f (a + 1) = a2 + 6a − 7
847
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
848
Chapter 1
f f (0) = 9, f (2) = 25, f (−4) = 73, f (a) = 4a2 + 9,
f (a + 1) = 4a2 + 8a + 13
8 a all real x
b all real x
c all real x
d all real x
e all real x
f all real x
g all real x
h x≠0
9 a all real y
b y≥0
c y≥0
d all real y
e y>0
f y>0
g y≤2
h y≠0
10 a i 5
ii −2
iii 3
iv −15
v 5
5
represents the x value of the point where the line
b a=
3
graphs intersect.
11 a i false
ii false
b i false
ii true
c i false
ii false
12 4x + 2h − 3
13 a They all pass the vertical line test, as each x value has only
one y value.
b vertical lines in the form x = a
c The y value of the vertex is the maximum or minimum value
of the parabola and therefore is essential when finding the
range.
1
d i y ≥ − 4 ii y ≥ −12
iii y ≤ 1.125 iv y ≥ 1
4
1
14 a x ≠ 1
b x≠−
c x≠1
2
15 a x ≥ 0
b x≥2
c x ≥ −2
d x≤2
17 a i
y
(−1, 2)
(1, 2)
ii
y
4
2
−2
2
iii
y
6
(1, 6)
4
2
−2
−4
x
O
2
4
−2
−4
b
y
b iii
c i y≥0
d i 8, 8, 2
6
4
ii 0 ≤ y ≤ 4
ii 34, 18, −2
iii y ≥ −4
Exercise 9I
2
y
1
O
–2
x
O
−2
16 a f (a) = f (−a) = a2 + 1
a2
–4
x
O
2
4
x
2
–2
c The y-axis is the axis of symmetry for the function.
–2
O
2
x
–2
848
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Number and Algebra
b x = ±4
d y = ± 11
3 a (0, 0) or O b r
4 a (0, 0)
d x =±
c x =± 3
e y = ± 57
f y = ±2
b r=3
c y =± 5
11
3
(
27 ±3 3
=
2
2
e
y
Answers
2 a x =± 5
O
–3
3
√2
,
3
√2
)
x
3
(– , – )
3
√2
y
3
√2
–3
3
12
–3
O
y
x
3
√10
(1, 3)
–3
f −3 ≤ x ≤ 3
5 a (0, 0)
d x = ±3
c y =±
b r=5
e
19
2
5
13
O
5
y
√6
x
(
−2√30, √30
5
5
)
O
−√6
–5
√6
(
2√30, −√30
5
5
−√6
f
6 a
d
7 a
d
g
−5 ≤ x ≤ 5
r=6
r= 5
x2 + y2 = 4
x 2 + y 2 = 2601
x 2 + y 2 = 1.21
g
b
e
b
e
−5 ≤ y ≤ 5
r=9
r = 14
x 2 + y 2 = 49
x2 + y2 = 6
2
c
f
c
f
r = 12
r = 20 = 2 5
x 2 + y 2 = 10 000
x 2 + y 2 = 10
b ±4
c ± 3
10 a r = 2 2
b r=2
c r=3
e r =2 3
f
r =2 5
)
√5
(2, 1)
b (−1, 3 ), (−1, − 3 )
9 a ±1
x
y
2
1 
15
1
 1 15   1
15   15
c  ,
, −  , −
,− 
, ,−
 d 



 2 2   2
2
2
2
2
2
e (0, −2)
f (2, 0), (−2, 0)
d r = 10
14
h x + y = 0.25
8 a (1, 3 ), (1, − 3 )
x
√10
(–1, –3)
–√10
y
–5
O
–√10
g −3 ≤ y ≤ 3
d ± 11
O
−√5
(−1, −2)
√5
x
−√5
chord length = 3 2 units
15 a m = ± 3
b m > 3 or m < − 3
c − 3 <m< 3
16 a D
d C
b A
e F
c E
f B
17 a y = ± 16 − x 2 = ± 42 − x 2
b x = ± 3 − y2 = ± ( 3 )2 − y2
849
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
850
Chapter 1
18 a Radius of graph is 2 so points are 2 units from (0, 0); i.e. < 2.
b Radius of graph is 1 so points are 1unit from (0, 0); i.e. −1 is
the left-most point, which is not as far as −2.
19 a i
vii
y
√7
y
2
−2
O
−√7
x
2
x
O
−√7
viii
y
√5
ii
√5
y
x
O
5
−√5
−5
x
5
O
ix
y
2√3
iii
y
−1
x
1
O
−1
iv
x
O
2√3
−2√3
y
√10
−√10 O
v
√10
b i y = 25 − x 2
ii
y = − 16 − x 2 iii x = 4 − y2
iv x = − 1− y2
v
y = 3 − x2
x
vii x = 10 − y2 viii x = − 8 − y2
vi y = − 5 − x 2
ix y = − 18 − x 2
y
Exercise 9J
−4
O
4
x
1 a
−4
vi
x
−2
−1
0
1
2
y
1
4
1
2
1
2
4
y
b
6
−6
O
−6
y
4
3
2
1
x
−2 −1 O
y = 2x
1 2
x
850
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
2 a
6
x
−2
−1
0
1
2
y
1
9
1
3
1
3
9
b
y
−2 −1 O
x
O
−1
(1, −2)
(1, −3)
y
10
8
6
4
2
7
b
c
a
x
1 2
a
c
(1, −5)
b
y = 3x
Answers
Number and Algebra
y
(1, 12 )
(1, 13 )
1
x
O
3 a
x
−2
−1
0
1
2
y1 = 2x
1
4
1
2
1
2
4
1
2
−1
−2
−4
2
1
1
2
1
4
−
y2 = −2x
y3 = 2−x
1
4
−
4
b
y
y = 2−x
y = 2x
4
3
2
1
−1 −1O
−2
−3
−4
−2
4 a a −2 =
1
y = −2x
1
≠ −a 2
a2
b False since 3−2 =
1
.
32
2
x
(1, 16 )

1
ii −1, 

3
iii (0, 1)
iv (2, 9)
b i (4, −16)

1
ii −1, − 

2
iii (0, −1)
iv (2, −4)
 1
c i 1, 
 4
ii (−3, 64)
iii (0, 1)
 1
iv 1, 
 4
8 a i (0, 1)
9 a
10 a
b
c
11 a
b (2, 9)
(2, 4)
1000
i 2000
i 2 years
N = 2t
c (1, −4)
d (−3, 8)
ii 8000
ii 4 years
b N = 210 = 1024
c 14 seconds
12 x = 2.322
13 a C
b A
c D
d E
e F
f B
14 Substitute (2, 5) into the equation y = 22 = 4 ≠ 5.
15 y = 1
16 It is the asymptote.
17
y
c 5−3, 3−2, 2−1
3
1
d −9, − 125, −
4
5
y
c
(1, 5)
(1, 4)
a
b
1
O
x
Compared to y = 2x, y = x + 3 meets the y-axis at 3; cuts
the x-axis at −3; and maintains a constant gradient of 1.
Therefore, the two graphs intersect twice.
(1, 2)
1
O
x
851
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
852
Chapter 1
18 a
3 a
y
8
6
4
2
−3 −2 −1 O
x
1 2 3
 1 x
b i y = 
3
c i y = 4−x
They are the same graph.
 1 x
ii y =  
5
 1 x
iii y =  
 10 
ii y = 7−x
iii y = 11−x
1
4
1
2
1
2
4
−8
−4
−2
−1 −
−4
−2
−1 −
y
1
2
1
2
b
8
4
1
2
y
8
6
4
2
x
 1
  = ( a−1 ) x = a− x as required (or similar)
a
d 1 = a−1 , thus
a
1
1
−
2
4
x
−4 −3 −2 −1−2O
−4
−6
−8
Exercise 9K
1 2 3 4
x
1 a
x
−2
−1
−
1
2
1
2
1
2
y
1
−
2
−1
−2
2
1
1
2
b
4 a 1 ÷ 0.1, 1 ÷ 0.01, 1 ÷ 0.001, 1 ÷ 0.00001
b x=
1
100
c 0.099
5 a
d 998
y
(1, 1)
y
x
O
2
1
(−1, −1)
−2 −1−1O
−2
1 2
x
b
y
2 a
x
−3
−1
−
1
3
1
3
1
3
y
−1
−3
−9
9
3
1
b
(1, 2)
x
O
(−1, −2)
y
9
6
3
−3 −2 −1−3O
−6
−9
c
1 2 3
y
x
(1, 3)
O
x
(−1, −3)
852
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
d
y


1
7 a 10, −  b −5,


2
(1, 4)
x
O
(−1, −4)
e
y
(−1, 1)
(1, −1)
f

5
d 9, − 

9
b (3, 1)

 3
c  − , −2 

 2

 1
d  − , −6 

 2
9 a yes
b yes
c no
1 
10 a  , 2
2 
1 
b  , 6
6 
c (−1, −1)
d no
 1

d − , −10 
 10


 1
f  − , −2  ,

 2

 1
  1
, − 2
g  , 2  , −

 2
  2
 1
  1

h  , 5  , −
, − 5
 5
  5

1

3
 1 
b − , 4 
 2 

1
c 4, − 

2
  1 
1
f  , −4  ,  − , 4 
  2 
2
e (1, −2), (−1, 2)
g (2, −1), (−2, 1)
h ( 2 , − 2 ), (− 2 , 2 )
12 a E
b C
c D
d B
e A
f F
13 yes; x = 0 or y = 0
14 a zero
b zero
c infinity
d infinity
15 The greater the coefficient, the closer the graph is to the
asymptote.
(−1, 2)
x
(1, −2)
16 a i x =
1± 5
−1± 5
, y=
2
2
iii x = −1± 2 , y = 1± 2
b no intersection, ∆ < 0
g
1 
 , 2
2 
e (1, 1), (−1, −1)

d −6,

y
O

5
c −7, 

7
8 a (1, 3)
2

11 a  , −3 
3

x
O
5

4
Answers
Number and Algebra
ii x = 1± 2 , y = −1± 2
c y = −x + 2, y = −x − 2
y
Exercise 9L
(−1, 3)
1 a
e
i
2 a
x
O
(1, −3)
3 a
h
y
27
−216
−5
0
c (−1, −2)
b 54
c −2
d −128
−1
0
1
2
y
−8
−1
0
1
8
x
 1
b  4, 
 2
d −1000
h −3
−2
y
8
6
4
2
–2 –1–2 O 1 2
–4
–6
–8
(1, −4)
6 a (2, 1)
c −64
g 1
x
(−1, 4)
O
b 125
f 2
x

1
d −6, − 

3
853
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
854
Chapter 1
b
x
−2
−1
0
1
2
y
8
1
0
−1
−8
y
c
8
6
4
2
–2 –1–2 O 1 2
–4
–6
–8
4 a (0, 2)
5 a x=4
e x = 10
1
i x=
2
6 a x=1
e x=4
b D
b x=5
f x = −3
1
j x=
3
b x=3
f x = 6.1
7 a x=
b x=
8 a
3
21
3
6
y
x
(1, –2)
d
c x = −3
g x=6
1
k x=
5
c x = −1.4
g x = −2.9
d x = −2
h x = −3
1
l x=
2
d x=3
h x = −4.3
c x=
d x=
3
−2
3
x
O
y
(1, 12 )
O
18
y
e
(1, 3)
y
x
O
1
(1, 10
)
O
b
x
x
y
(1, 4)
O
f
y
x
O
(1, –1
4)
x
854
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
9 a
8
1
x
O
–1
y
f
y
b
10 a
y
O
2
x
3√−−
24
O
Answers
Number and Algebra
y
x
1
x
–1 O
–8
c
b
y
y
O
–3
x
2
x
O
–8
–27
c
d
y
y
O
32
3
x
–27
–4
x
O
d
e
y
y
–2 O
x
10
–8
3√−−
–5
O
x
855
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
856
Chapter 1
e
18 a
y
2
O
y
x
1
O
(1, –8)
–9
f
x
3
y
b
y
32
19
O
–4
(2, 27)
x
x
–1 O
11 384 cm2
12 a 54 mm
d
b 5 weeks
c after 8 weeks
h (mm)
c
(8, 1024)
y
56
O
t (weeks)
8
–4
13 a x = 1
b x=5
c x = −7
d x=3
e x = −2
f x = −14
14 6371 km
15 a (−x)2 = (−x) × (−x) = x2
b (−x)3 = (−x) × (−x) × (−x) = −x3
c If n is odd, then (−x)n = −xn for all values of x.
16 a V = πr3
b
3
1
17 a a = − , d = 0
2
c a = 1, d = −2
d a = 1, d = 2
y
–2 O
(–4, –4)
(1, π)
b a = 2, d = 3
x
28
8000 20
= 3 units
π
π
O
856
d
V
c Volume increases by a
factor of 8; i.e. (2)3.
(−3, 2)
O
r
x
e
2 a
y
y
3
x
O
−1
(–2, 1)
x
O
–1
Answers
Number and Algebra
(3, −1)
y
b
(−2, 3)
–7
3
y
f
x
−2 O
c
x
O
–3
y
2√6 − 3
−3 O
(–1, –8)
–9
−2√6 − 3
y
g
y
d
6
x
O
–6
x
5
(1, −3)
(−3, 2)
(–3, –9)
x
O
−2
–18
√21 − 3
–√21 − 3
h
y
y
e
18
1 + √5
(1, 16)
(−2, 1)
3
2√2 − 2
x
O
x
O
1 − √5
−2√2 − 2
f
y
10
Exercise 9M
1 a up
e right
i right
b down
f left
j left
c right
g up
k down
d left
h down
l up
4
−2√5
O
−2
2√5
x
857
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
858
Chapter 1
y
g
y
c
2√2
−4
−1 O
x
2
−2√2
h
x
O
−4
y = −5
y
d
5 + 2√15
(2, 5)
2 − √39
2 + √39
O
y
1
2
x
5 − 2√15
y=0
O
x
y
i
e
y
(−3, 1)
8
x
−1 O
−5
O
3 a
x
y=0
y
f
x
O
−1
−2
b
y
2
y = −2
O
y=0
x
y
2
1
O
g
y=1
x
y
3
2
1
O
y=1
x
858
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
h
1
x
O
O
2
− 12
y = –3
−3
i
y
d
y
y
y=0
x=2
e
y
x
O
−3 78
2
y = −4
4
y=1
1
−2−1 O
y
4 a
y=2
2
− 12
O
x = −1
f
y
4
3
y
b
x
O
y = −3
−4
1
O
−1
x=1
x
g
y = −1
2
y=2
5
3
y
O
1
3
O
x=3
y
x=0
x = −3
x
x
x=0
c
x
Answers
Number and Algebra
y=0
5
2
3
x
x
859
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
860
Chapter 1
h
d
y
x = −4
y
O
− 34
y = −1
e
i
y
y
x
O
y = −1
6
−3
y=6
29
5
f
O
29
6
5
y
5 a
y=5
5
x
1
x=5
x
O
y
y=1
6 a
x = −1 y
x
O
y=2
2
b
x
x
O
−4 −3
y=0
−1
4
y
1
x
1
−1 O − 2
x
O
−3
y = −3
−4
c
x = −2
b
y
y
y=0
O
x
−4
−2
O
−1
y = −1
x
−2
−8
860
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
c
y
− 43
Answers
Number and Algebra
 3 2
29  3 
29
vii  x +  + ( y − 3 )2 = , C − , 3  , r =
 2 
 2
4
2
O
2
 5 2
7
49  5 
viii  x +  + ( y − 2 )2 = , C − , 2  , r =
 2 
 2
2
4
x
3
2
2

3
1  3  3  1 3 
ix  x −  +  y +  = , C  , −  , r =
 2  2 2 2 2
2
y = −2
2


2
2 

x  x − 3  +  y − 5  = 25 , C  3 , 5  , r = 5
2 2
 2  2
2
2
x=3
7 a y=
1
−1
x −2
b y=
b (x + 2)2 + (y − 3)2 = −2, radius can’t be negative
1
+3
x +1
c y=
1 3
+
x −1 2
 −3 − 5 1− 5   −3 + 5 1+ 5 
,
,
, 

8 a 
 2
2   2
2 
b
(
5 , 3 + 5 ) , (− 5 , 3 − 5 )
 −1− 11
  −1+ 11

, 11 , 
, − 11
c 

  2

2
Exercise 9N
1 a
b
c
d
e
f
2 a
 3 6
d (1, 2 ), − , − 
 5 5
b
c
e (−6, 3), (−2, −1)
f (3, 0), (−3, −2)
d
9 a max x = 5, min x = 1
b max y = 0, min y = −4
10 a (x − 2)2 + (y − 1)2 = 8
c (x + 5)2 + (y + 3)2 = 18
e y=
1
−1
x +2
b (x + 2)2 + y2 = 25
1
d y=
+1
x −1
−1
f y=
x+3
1
11 a Solving = − x would require x2 = −1, which is not
x
possible.
b Circle has centre (1, −2) and radius 2, so maximum 3
value on the circle is 0, which is less than 1.
c Exponential graph rises more quickly than the straight line
and this line sits below the curve.
2
1
− 1=
gives a quadaratic with ∆ < 0, thus
d Solving
x+3
3x
no points of intersection.
12 a i (x + 2)2 + (y − 1)2 = 4 , C (−2, 1), r = 2
2
2
ii (x + 4) + (y + 5) = 36, C (−4, −5), r = 6
iii (x − 3)2 + (y − 2)2 = 16, C (3, 2), r = 4
iv (x − 1)2 + (y + 3)2 = 15, C (1, −3), r = 15
v (x + 5)2 + (y + 4)2 = 24, C (−5, −4), r = 2 6
vi (x + 3)2 + (y + 3)2 = 18, C (−3, −3), r = 3 2
3 a
b
c
d
e
f
g
4 a
b
5 a
b
6 a
b
direct proportion
indirect proportion
indirect proportion
direct proportion
neither
indirect proportion
Straight line with y-intercept; neither direct nor inverse
(indirect) proportion.
Straight line starting at (0, 0); direct proportion.
Upward sloping curve so as x increases, y increases;
neither direct nor inverse (indirect) proportion.
Hyperbola shape so as x increases, y decreases; inverse
(indirect) proportion.
Fixed distance from home, zero gradient, stationary.
Decreasing distance from home, negative constant
gradient, lower constant speed.
Increasing distance from home, positive constant gradient,
higher constant speed.
Increasing distance from home, positive varying gradient,
increasing speed, accelerating.
Increasing distance from home, positive varying gradient,
decreasing speed, decelerating.
Decreasing distance from home, negative varying gradient,
decreasing speed, decelerating.
Decreasing distance from home, negative varying gradient,
increasing speed, accelerating.
i p = 4q ii p = 60
iii q = 25
i p = 50q ii p = 750
iii q = 4
72
ii y = 2
iii x = 24
i k = 72, y =
x
50
i k = 50, y =
ii y = 0.5
iii x = 0.5
x
Positive variable rate of change, increasing speed,
accelerating.
Positive constant rate of change, constant speed.
861
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
862
Chapter 1
c Positive varying rate of change, decreasing speed,
decelerating.
d Zero rate of change, stationary.
e Negative varying rate of change, increasing speed,
accelerating.
f Negative constant rate of change, constant speed.
g Negative varying rate of change, decreasing speed,
decelerating.
7 a y is increasing at an increasing rate.
b y is increasing at a decreasing rate.
c y is decreasing at an increasing rate.
d y is decreasing at a decreasing rate.
8 a k = $244/tonne
b P = 244n
c $33 184
d 1175 tonnes
9 a C = 74
s
b $4.93
c
y (lbs)
(100, 220)
y = 2.2x
200
150
100
50
x (kg)
0
20
d
c $2.47
Time (min)
y=x
(100, 100)
80
60
y=
60
40
20
(80, 10)
0
20
0
20
40
60
80
100
x (km)
b
y=
Cost per person
200
320
x
11 a
b
c
d
e
f
100
800
x
60
40
80
(10, 80)
80
10 a
y (km/h)
100
40
20
60
40
Words/min
80
decreasing at a decreasing rate
increasing at an increasing rate
increasing at a decreasing rate
decreasing at an increasing rate
decreasing at a constant rate
increasing at a constant rate
12 a P
(2, 160)
150
100
t
50
b P
(8, 40)
0
1
2
3 4 5 6
No. of people
7
8
t
862
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Number and Algebra
Answers
e
Depth
c P
t
Time
f
Depth
d P
t
13 a
Time
Distance
Depth
14 Corrected graphs are shown with a dashed line.
a
Time
Time
Vertical line incorrect. Can’t change
distance instantaneously.
Depth
b
Distance
b
Time
c
Time
Depth
Graph correct.
Distance
c
Time
d
Depth
Time
Can’t be in two places simultaneously. Curve must
increase in gradient, turn, decrease in gradient.
Time
Distance
d
Time
863
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
864
Chapter 1
No breaks in curve as continuous motion describes.
Final deceleration segment needs a curve becoming flatter,
showing a decreasing gradient.
15 A & d: School bus; distance increases at an increasing rate
(acceleration), then a constant rate (steady speed) and then a
decreasing rate (deceleration) becoming a zero rate (stopped).
B & a: Soccer player; distance increases at a constant rate
(steady speed), then a zero rate (stopped) and then at an
increasing rate (acceleration).
C & c: Motor bike; distance increases at a constant rate (steady
speed), then at an increasing rate (acceleration), and then at a
decreasing rate (deceleration) becoming a zero rate (stopped).
D & b: Rocket booster; distance increases at an increasing rate
(upward acceleration), then a decreasing rate (deceleration
when detached) becoming zero (fleetingly stopped). Distance
then decreases at an increasing rate (acceleration towards
Earth) and finally distance decreases at a constant rate (steady
fall to Earth with parachute).
16 Various solutions; check with your teacher.
5 a a =± b
b a = ± c2 − b 2
6 a no
b b=
d c =±
7 a
c
e
8 a
a
b
A
y
x ≠ 0, y ≠ 2
x ≠ 4, y ≠ 0
x ≠ 3, y ≠ 2
t≠0
2A
−y
h
d x = z 2 − y2
4A
π
f
x=
V
h
ii −10
9
c F = C + 32
5
e No; C and F can take any value.
10 a i 5
b 32
d 212
11 a x = ± 4 − y2
b y = ± 4 − x2
Exercise 9O
c −2 ≤ x ≤ 2, −2 ≤ y ≤ 2
1 a A > 0, b > 0, h > 0
d i −3 ≤ x ≤ 3, −3 ≤ y ≤ 3 ii −5 ≤ x ≤ 5, −5 ≤ y ≤ 5
e Semicircle below x-axis with centre (0, 0) and radius 5.
b b=
2A
h
c h=
2A
b
d
e
2 a
b
c
d
i 5
ii 5
No, can’t divide by 0.
i 0
ii 16
all real numbers
y≥0
x= ± y
e no
1
3 a i −
3
b i 4
c x ≠ 0, y ≠ 0
4 a a=
1
2
1
ii
4
ii
P−b
2
c a = 2M − b
e t=d
s
2( s − ut )
g a=
t2
2A
−b
i a=
h
Q2
k h=
2g
864
b
d
f
b
b x=
c x = 180 − y
e x=
c yes; c ≠ 0.
e yes; b ≠ 0.
x ≠ O, y ≠ 0
x ≠ O, y ≠ −3
x ≠ −2, y ≠ 0
V≠0
9 a x=
a
c2
c a =± b +1
1 1 1
12 a No; 2 + 3 = 5 but + ≠ .
2 3 5
xy
b z=
x+y
c x=
iii 1
iv 4
v 81
yz
y−z
ii −3
13 a i 2
b i no
c r=3
iii undefined
iii undefined
x = 9 − y2
l
h=
V
πr 2
y
ii no
3V
4π
b i AB = 2r 2 (1− cos 0° )
= 2r 2 (1 − 1)
=0
ii AB = 2r 2 (1− cos 180°)
= 2r 2 (1− ( −1))
= 4r 2
h b = ± c2 − a 2
j
3
14 Yes; x ≥ 0 in the first equation but x > 0 in the second equation.
15 a i 2.6
ii 3.1
P − 2
b b=
2
b2 − D
d c=
4a
A
f r=
π
iii
= 2r
c r=
AB2
2(1− cos θ )
d 9
e r > 0, AB > 0, 0° < θ < 360°.
Exercise 9P
c
1 a false
b true
c true
d false
2 C(2, 0), D(4, 1)
3 a A horizontal line cuts the function at only one point.
b A horizontal line cuts the function at only one point.
c A horizontal line cuts the function at only one point.
4 a
y
y
2
–4
O
–2
2
4
x
6
–2
–4
4
y=x
y = f –1(x)
2
O
–2
y=x
4
6
–4
y = f –1(x)
6
Answers
Number and Algebra
2
4
d
y
6
x
6
y=x
4
–2
2
–4
y = f –1(x)
b
–4
y
y=f
O
–4
6
x
–4
2
–2
4
y=x
4
–2
2
–2
6
–4
O
–2
2
4
6
–1(x)
x
5 Note: Graphs not shown for this question.
a f −1(x) = x + 4
x
b f −1(x) =
6
x −4
c f −1(x) =
2
d f −1(x) = 4x
e f −1(x) = 2 − x
6− x
f f −1(x) =
3
g f −1(x) = 2x − 1
h f −1(x) = 2(x − 1)
865
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
866
Chapter 1
6 a
d
y
y
8
8
y=
6
y=
1
x
6
x3
4
y=x
4
y=x
2
2
f –1(x) = 3√
x
–4
O
–2
f –1(x) =
x
2
6
4
8
–4
–2
2
4
6
x
8
–2
–4
–4
e
y
b
O
–2
1
x
y
8
8
y = x3 – 1
6
y = 2x3
6
y=x
4
4
y=x
2
2
x + 1
f –1(x) = 3√

f –1(x) = 3√ 2x
–4
O
–2
2
4
6
x
8
–4
x
8
8
1
y = (x + 2)
6
6
y=x
4
y=x
–x3
2
4
–8
2
–6
–4
O
–2
–2
O
–2
6
y
f
y
8
–4
4
–4
–4
y=
2
–2
–2
c
O
–2
2
4
6
8
x
2
4
6
8
x
1
f –1(x) = x – 2
–4
–6
–2
f –1(x) = – 3√
x
–4
866
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
g
b A horizontal line cuts the function at more than one point.
c
y
y
8
(3, 9)
6
f –1(x) = x2 for x ≥ 0
y=x
y = x2
4
2
–4
y=√
x
O
–2
2
4
x
8
–2
y
6
y=x
4
y = 1x + 1
2
–4
O
–2
–2
f −1(x) = x ; domain: 0 ≤ x ≤ 9; range: 0 ≤ y ≤ 3
x = 0 and x = 6
vertex (3, −9 )
A horizontal line cuts the function at more than one point.
x ≥ 3 or x ≤ 3
Either the increasing section of the parabola or the
decreasing section can be taken for an inverse.
11 As a parabola does not have an inverse unless the domain
has been restricted, the vertex is important because it is from
this point that the largest domain for an inverse can be found,
as an inverse exists for the increasing or the decreasing part
of the function.
1
12 a x ≥ 1
b x ≥ −1
c x≥−
2
1
1
e x≤
d x≤
2
2
d
10 a
b
c
d
e
–4
h
x
O
6
Answers
Number and Algebra
2
4
f –1(x) =
1
(x – 1)
6
8
x
( x − 6)
b f −1(x) =
( x + 1) + 1
c f −1(x) =
( x + 4) − 2
d f −1(x) =
9 1

 x + 4  + 2
e f −1(x) =
( x + 16 ) − 1
13 a f −1(x) =
–4
–6
7 ax≥1
b x≥2
c x≥0
8 a Domain of f −1(x) is 2 ≤ x ≤ 4; range of
f −1(x) is −1 ≤ y ≤ 1.
b Domain of f −1(x) is 1 ≤ x ≤ 5; range of
f −1(x) is −3 ≤ y ≤ −1.
c Domain of f −1(x) is 0 ≤ x ≤ 5; range of
f −1(x) is −2 ≤ y ≤ 0.
9 a
y
Challenges
2
1
1 a − ≤x≤
3
2
c x<
b x < − 3 or x > 1
4
3
7 − 41
7 + 41
or x >
2
2
2 (x − 2)2 − (y − 3)2 ≤ 16
y
5
4
3 + 2√3
3
(2, 3)
2 − √7
2
O
1
O 1
–3 –2 –1
–1
2
3
x
3 − 2√3
3 a b 2 − 4ac < 0
x
2 + √7
b b 2 − 4ac = 0
c b 2 − 4ac > 0
867
868
Chapter 1
c
4 a (−2 − 5 , −1− 5 ), (−2 + 5 , −1 + 5 )
y

 

b  −1− 17 , 1− 17  ,  −1+ 17 , 1+ 17 

2
2  
2
2 
c (−2, −1), (1, 2)
d (−4, −3), (−2, −1)
5 (x − 2)2 + (y + 3)2 = −15 + 9 + 4 = −2, which is impossible.
6 a k= 1
3
b k<
1
3
−2 O
c k> 1
3
7 a k = ± 20 = ±2 5
−8
b k > 2 5 or k < −2 5
c −2 5 < k < 2 5
(1, −9)
4 a i maximum at (1, −3)
iv
y
3
b y = ( x + 2 )2 − 3
4
8 a y = −(x + 1)(x − 3)
x
4
ii (0, −4)
iii no x-intercepts
c y = x2 − 2x − 3
x
O
9  3 , − 73  a = 2, b = −3, c = −8
4
8
(1, −3)
−4
10 20
11 a
D
M
1
2
3
4
5
6
1
3
7
15
31
63
b i minimum at (−3, −8)
iii −1 and −5
iv
b M=2 −1
D
ii 10
y
Multiple-choice questions
1 B
6 A
11 E
16 C
2
7
12
17
3
8
13
18
D
C
C
E
10
E
D
D
D
4 D
9 D
14 B
5 A
10 A
15 A
x
−1 O
−5
Short-answer questions
1 a
c
2 a
c
3 a
minimum at (1, −4)
−1, 3
minimum at (2, 0)
maximum at (−1, −2)
b
d
b
d
y
x=1
−3
maximum at (0, 5)
minimum at (3, 4)
b
(−3, −8)
5 a
y
y
16
1
O
2 – √3
−2 O
2
x
−4
O
x
x
(2, −3)
b
2 + √3
y
−4
−3 − √17
2
x
O
−2 −3 +2√17
(− 32 , − 174 )
868
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
6 a 1
7 a i 5
b 0
ii (2, −3)
iv
c 2
iii 0.8, 3.2
d 0
10
y
y
3
5
0.8
3.2
(
x
− 3 ,− 6
√5
√5
3
6
,
√5 √5
O
−3
O
(
)
3
)
x
Answers
Number and Algebra
−3
(2, −3)
 3 25 
ii  , 
2 4
b i 4
11 a
iii −1, 4
y
(1, 4)
iv
y
3 , 25
2 4
1
4
−1
8 a 100 − x
d A (m2)
O
x
4
b
b A = x (100 − x )
y
c 0 < x < 100
(1, −3)
100
x (m)
c
y
f 50 m by 50 m
9 a
y
O
5
O
x
12 a
(1, 2)
y
O
√7
−√7
x
y
−5
b
(1, 15 )
1
5
−5
x
O
−1
(50, 2500)
0
e 2500 m2
x
O
O
x
(−1, −2)
√7
x
−√7
869
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
870
Chapter 1
b
15 a x = −6, y = 10 and x = 2, y = 10
b no solutions
c x = 0, y = 0 and x = 4, y = 8
1
10
d x = −1, y = 2 and x = , y =
3
9
y
(−1, 3)
x
O
(1, −3)
4 
13 a  , 3 
3 
b ( 2 , 2 2 ) and (− 2 , −2 2 )
14 a
y
2 + √3
16 a show b2 − 4ac < 0
b
c
17 a
b
c
d
e
f
g
h
18 a
show b2 − 4ac = 0
k = −2
x=3
x = −2
x = 1.8
x = −1.7
x=4
x=3
x = −2
x = −2.5
y
(1, 3)
(−1, 2)
−1 O 2 − √3
x
x
O
b
y
3.5
y=3
b
x
O
c
y
y
O
2
x
− 53
x
O
−2
−2.5
−3
–8
y = −3
x = −2
870
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
c
y
10
O
3√
5
x
24 a A: y is increasing at an increasing rate.
B: y is increasing at a decreasing rate.
C: y is decreasing at an increasing rate.
D: y is decreasing at a decreasing rate.
E: y is decreasing at a constant rate.
b A: i , v and viii
B: v, ix
C: ii and v
D: iii and ix
E: i, iv and vii
F: vii and ix
G: ii, vi and vii
p−a
d
25 a b =
b t=
s
2
d b=
d
y
2A
−a
h
e a = c2 − b 2
b a=
26 a no
y
x2
Answers
Number and Algebra
A
π
c r=
f x−±
y
k
c yes, x ≠ o
y
d x=±
a
e a ≠ o; when y ≠ o then a must have same sign as y.
O
2
x
Extended-response questions
−3
−7
−2.875
33
k 3 − 2k 2 + k − 3
domain: all real x values; range: all real y values
domain: all real x values; range: y = 4 only
domain: all real x values; range: y ≤ 4
domain: all real x, x ≠ 0; range: all real y, y ≠ 0
domain: x ≥ 0; range: y ≥ 0
x +5
x
b f −1(x) =
21 a f −1(x) =
4
5
3
x
d f −1(x) =
x
4
e f −1(x) = 1 − x
22 x ≥ 0 or x ≤ 0
72
ii y = 18
iii x = 102.86
23 a i y =
x
b CNY = 6.47 × AUD; AUD $154 559.50
1200
or xy = 1200, y = 15
c i k = 1200, y =
x
240
ii k = 240, y =
or xy = 240, x = 20
x
c f −1(x) =
0
e
2 a
c
e
c 30 ≤ h ≤ 80
(400, 80)
80
–4
19 a
b
c
d
e
20 a
b
c
d
e
b 0 ≤ x ≤ 400
1 a (200, 30)
d
h (m)
(200, 30)
400
x (m)
f 30 m
b i 150 g
d
A (g)
400 m
450 g
after 2 years
9.8 years
g 80 m
ii 5.6 g
450
t (s)
0
Chapter 10
3
Pre-test
1 a 8
1
e
27
2 a x7
d 2
g
8x3
y3
b 2
c 1
f 32
g 64
8
b 12x
e 12x3
h
y
2
d 9000
1
h
6
c 2x
f x9y3
i
y
24 x
871
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Chapter 1
872
3 a 3
e −3
b 4
f 3
2
4 a A = 10 000 × 1.05n
d −2
c 4
b $16 289
x 2 + 4x + 3
b 6x 2 + x − 2
15x 2 + 2x − 1
e 0
5(1 − 2x)
b (x − 8)(x + 6)
(x + 1)(2x − 3)
e 3(x − 2)(5x + 4)
7
b 5
c 41
16 rem. 3 b 39 rem. 3 c 55 rem. 19
5 a
d
6 a
d
7 a
8 a
c
f
c
f
10 a
4x 2 − 25
4x + 8
(x − 2)(x + 5)
−(x − 2)(x + 3)
Time (min)
0
1
2
3
4
5
Population
1
2
4
8
16
32
b P = 2t
Exercise 10A
c 256
d 14 min (round up)
e log210 000
11 Given that a is a positive number and y = loga b, then ay = b,
1
where y can be any negative number. Example: log2 = −2.
4
12 No. A positive number to any power is always > 0.
13 a 16
b 26
c 6
14 loga1 = 0 and dividing by 0 is not possible.
1
15 a
1
4
b
1
5
c
1
2
d
1
3
e
1
2
f
1
3
g
2
3
h
4
3
i
1
2
j
1
2
k
6
5
l
4
7
x
0
1
2
3
4
5
x
1
2
4
8
16
32
3x
1
3
9
27
81
243
x
1
4
16
64
256
1024
5x
1
5
25
125
625
3125
x
1
10
100
1000
2
4
10
2 a 4
3 a
1
10000
1
27
4 a 24 = 16
e
b 4
b
f
b log381 = 4
e log10
6 a 4
b 2
e
i
m
7 a
e
i
m
8 a
d
9 a
e
f
j
n
b
f
j
n
b
e
b
f
m3
q 3
2
2
0
−2
−4
−3
−1
1.672
−0.770
5
2
1
j
9
n 2
r 10
1
= −1
10
c 6
g
k
o
c
g
k
o
c
f
c
g
d 4
1
4
e 10−1 = 0.1
d log416 = 2
i 1000
c
1
1
g
25
64
b 102 = 100
1
4
5 a log28 = 3
1
2
0
−3
−2
−1
−3
0.699
−0.097
3
3
c 3
1
2
d 2−2 =
10 000 100 000
3
5
0
−2
−4
−5
−2
2.210
−1.431
6
16
1
k
4
o 4
s 2
d
1
32
1
36
c 33 = 27
h
1
9
c log232 = 5
f
3− 2 =
f
log5
1
= −3
125
d 3
h
l
p
d
h
l
p
3
3
1
−3
−1
−1
−2
d 4
h 81
1
l
343
p 8
t −1
Exercise 10B
1 a
b
c
d
e
2 a
b
3 a
b
c
d
4 a
x>0
y can be any real number.
y→ −∞
x=0
x-intercept is 1.
y = log2 x is the reflection of y = 2 x in the line y = x.
The coordinates of each point on y = log2 x are the
coordinates in reverse order of a point on y = 2 x ;
e.g. (1, 0) and (0, 1).
x>4
x=4
y = log2 x is translated 4 units to the right.
y = log2 x is translated 5 units up.
Physical
values
1
10
100
1000
10 000
100 000
log scale
log10
(physical
value)
0
1
2
3
4
5
b
d
f
h
1 unit added
4.2
4
0.301 added
c
e
g
i
factor of 100
2
0.903
3.01
872
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Number and Algebra
x
y=2
x
–3
–2
–1
1
8
1
4
1
2
0
1
2
3
y
10
1
2
4
8
Answers
b
5 a
y = 10 x
9
8
1
8
1
4
1
2
1
2
4
8
y = log2x
–3
–2
–1
0
1
2
3
b
6
5
4
y
4
3
y = 2x
2
y=x
3
–2 –1 O
–1
y = log2x
1
x
1
2
3
2
3
4
5
6
7
8
9 10
c y = 10 x , y-intercept = 1; y = log10 x , x-intercept = 1
d (10, 1) y = log10 10 = 1;, (1, 0 ) log10 1= 0
e y = 0 is the asymptote for y = 10 x ; x = 0 is the asymptote
for y = log10 x.
–3
–4
c (1) y = log2 x is the mirror image (i.e. the reflection) of
y = 2 x in the line y = x.
(2) The coordinates of each point on y = 2 x are reversed
to give the coordinates of a point on y = log2 x.
d y = 2 x has y-intercept of 1; y = log2 x has x-intercept
of 1.
e y = 0 is the asymptote for y = 2 x ; x = 0 is the asymptote
for y = log2 x.
f y = 2 x has no limitations on the values of x; x > 0 for
y = log2 x.
g y > 0 for y = 2 x ; y = log2 x has no limitations on the
values of y.
6 a
–3
–2
–1
0
1
2
3
y = 10
0.001
0.01
0.1
1
10
100
1000
x
0.001
0.01
0.1
1
10
100
1000
y = log2x
–3
–2
–1
0
1
2
3
x
x
1
–2
4
–2
x
y = log10x
1
2
–4 –3 –2 –1 O
–1
y=x
7
x
y = 10 x has no limitations on the values of x; x > 0 for
y = log10 x.
f
g y > 0 for y = 10 x ; y = log10 x has no limitations on the
values of y.
7
Graph
a
b
c
d
i
x>4
x=4
4 units right,
3 units up
(4.125, 0), no
y-intercept
ii
x>3
x=3
3 units right,
2 units up
(3.25, 0), no
y-intercept
iii
x > –8 x = –8
8 units left,
1.2 units down
(–5.7, 0), (0, 1.8)
iv
x > –4
4 units left,
3.5 units down
(7.31, 0),
(0, –1.5)
x = –4
873
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
874
Chapter 1
8 a y = log2 ( x − 3 )
d y = log2 x + 4
y
y
7
3
asymptote
x=3
6
(4, 6)
(7, 2)
2
5
(5, 1)
1
x
O
1
2
3
4
5
6
7
8
asymptote 4
x=0
3
–1
(1, 4)
2
–2
1
–3
–1
b y = log2 x − 3
O
–1
asymptote
x=0
1
2
6
x
x
2
3
4
5
3
4
5 6
(4, –1)
7
8
asymptote
x=4
5
9 10
(8, 5)
4
3
(1, –3)
–4
(5, 3)
2
–5
1
–6
(4.125, 0)
O
–1
c y = log2 ( x + 4 )
(–2, 1)
1
2
3
4
5
x
6
7
8
9
10
–2
y
asymptote
x = –4
1
y
–2
–3
–2 O
–1
e y = log2 ( x − 4 ) + 3
y
1
1 ,0
16
4
3
2
1
–5 –4 –3 –2 –1–1O
–2
–3
f
(4, 3)
3
x
1 2 3 4 5
y = log5 ( x − 2 ) + 1
y
asymptote
x=2
(7, 2)
2
1
O
(3, 1)
(2.2, 0)
1
2
3
4
x
5
6
7
8
–1
–2
874
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
11 a
b
c
d
e
f
g y = log4 ( x + 3 ) − 2
y
2
asymptote
x = –3
1
x
–3 O
–1
–6
3
6
9
12
15
18
21
(0, –1.2)
(–2, –2)
–2
–3
b H+  = 10−pH
h y = log3 ( x + 4 ) − 1
c 100 times more H+ 
y
d
3
asymptote
x = –4
2
(–1, 0)
–5 –4
0 dB
60 dB
100 million times louder
1000 times louder
noise intensity doubles in loudness
Band practice: 2 hours; belt sander: 1 hour; hand drill:
30 minutes; chainsaw: 15 min; rock concert: 1 min; iPod
at peak volume: 30 s; jet engine at take-off: pain and
immediate, permanent hearing damage.
12 a i pH = 2.4 acidic
ii pH = 4 acidic
iii pH = 7.4 alkaline
iv pH = 10 alkaline
Answers
Number and Algebra
–3 –2 –1 O
–1
(–3, –1)
–2
e H+  = 1.49 × 10−3 moles/litre, pH = 2.83
(5, 1)
1 (0, 0.3)
f
x
1
2
3
4
5
1
th of H+  for pH increase of 1.
10
6
 10−7.9 − 10−8.1 

 × 100 = 58% increase in acidity
10−8.1
13 a
–3
–4
9 loga a x = x loga a
= x ×1
=x
∴ alog x = x
a
loga alog
a
x
= loga x
loga x × loga a = loga x
loga x × 1 = loga x
loga x = loga x
LHS = RHS
∴ alog x = x
b Similarities: Both have an asymptote; each family
of graphs intersect an axis at a common point;
x → ∞, y → ∞ for both (but at different rates); and both
curves have roughly a concave shape.
a
10 a
b
c
d
e
M=4
107 = 10 000 000 times more intense
I = I 0 × 10 M
I × 107
= 103 = 1000
Ratio of intensities = 0
I 0 × 104
I 0 × 10 M 1
=2
I 0 × 10 M 2
10 M 1− M 2 = 2
M 1− M 2 = log10 2
M 1− M 2 = 0.301
f 102.7 = 501.2 ≈ 500 times stronger
g 39.8 ≈ 40 times greater. Meeberrie is mostly an
unpopulated area.
c Differences: Logarithmic graphs are the inverse of
exponential graphs.
Exponential graphs are ‘concave up’ and logarithmic
graphs are ‘concave down’.
A logarithmic graph is the reflection of the exponential
graph in the line y = x.
The coordinates of a point on an exponential graph are in
the reverse order to the coordinates of the reflected point
that is on the logarithmic graph.
Logarithmic graphs are valid for x > 0 and all real values
of y, whereas exponential graphs are valid for all real
values of x and y > 0.
875
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
876
Chapter 1
14 a y = log2 ( x + 3 )
b y = log2 ( x − 2 ) + 3
c y = log5 ( x + 2 ) + 4
d y = log3 ( x + 3 ) − 1
15 For 0 < x < 1 and a given y value, the x-coordinate will decrease
as the base a increases; hence, the log curve with the larger
base turns more quickly towards the y-axis. For example, when
y = –1, x = 0.1 on y = log10 x and x = 0.5 on y = log2 x, so
points y = log10 x will be closer to the y-axis than y = log2 x.
For x = 1, y = loga x will pass through (1, 0) for all base values
(a > 1).
For x > 1 and fixed, the values of loga x become smaller as the
base a increases; hence, the log curve with the larger base is
closer to the the x-axis. For example, when x = 10, log10 10 = 1
and log2 10 = 3.3, so y = log10 x will be nearer to the x-axis
than y = log2 x.
16 a y = log5 ( x + 4 ) + 2
b y = log4 ( x − 1) + 3
1
1
d y = log3 ( x − 7 ) − 2 or y = (log3 ( x − 7 ) − 4 )
2
2
e y = 3x − 4 + 1
y = 2x − 3 − 5
h y = 4x − 5 − 6
j
1
1
1
y = 2 x − 3 − or y = (2 x − 3 − 2 )
4
2
4
1
1
 x −8
 x −8
+ 7)
y = log3 
+ 3.5 or y = (log3 
 5 
 5 
2
2
Exercsie 10C
1 a logbxy = logbx + logby
b logb
x
= logb x − logb y
y
c logabm = m × logab
d logaa = 1
e logc1 = 0
f
logb
c
f
i
c
f
2
3
−1
3
0
2 a
d
g
3 a
d
4 a
2
4
4
2
−4
loga6
d logb18
5 a loga2
d logb2
6 a loga9
d loga16
b
e
h
b
e
8 a
d
9 a
d
1
3
−2
−1
1
2
b
e
b
e
10 a log320
d log72
3
2
3
d
2
d 1
h 3
k 1
l
c
f
c
f
b log1048
e log38
 3
g log2  
 4
11 a
2
3
−3
−2
1
2
j
c 0
g 0
1
2
−3
−5
3
2
c log102
f 0
h log56
5
2
1
e
3
b
c
f
4
3
4
5
1
= loga 1− loga x = 0 − loga x = − loga x , as required.
x
1
b loga = loga x −1 = − loga x , as required.
x
1
1
log x
13 loga n x = loga x n = loga x = a , as required.
n
n
14 a Recall the index law am × an = am + n.
g y = 5x − 2 + 8
i
i
b 0
f 1
12 a loga
c y = log2 ( x − 5 ) − 3
f
7 a 0
e 1
1
12
−1
5
12
1
= − loga b
b
b loga15
e logb15
b loga3
3
e logb  
2
c loga28
f logb17
c loga10
7
f logb  
5
b loga25
e loga32
c loga27
f loga1000
Now let x = am and y = an
so m = logax and n = logay
From (1): xy = am × an = am + n
So m + n = logaxy
From (2): m + n = logax + logay
So logaxy = logax + logay, as required.
b Recall the index law am÷ an = am − n.
Now let x = am and y = an
so m = logax and n = logay
From (1): x ÷ y =
(1)
(2)
(1)
(2)
x
= a m ÷ a n = a m −n
y
x
y
From (2): m − n = logax − logay
So m − n = loga
So loga
x
= loga x − loga y , as required.
y
n
c Recall the index law (a m ) = amn.
Let x = am
So m = logax
(1)
xn = amn
So mn = logax n
From (1): n logax = logax n, as required.
876
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Exercise 10D
b log525 = 2 c log4 2 = 1
2
log310 = x e log72 = x f log1.17 = x
3
b 4
c 2
d 8
e 2
f 0.1
0.845
b −0.222
c −0.125
1.277
e 0.780
f 0.897
1.465
b 3.459
c 1.594
6.871
e 1.177
f 2
1
b 1
c 3.969
1.727
e 6.579
f 1.528
2.585
b 1.893
c 1.209
d 1.129
1.559
f 6.579
g 3.322
h 1.262
0.356
j 3.969
k 3.106
l 1.137
2 days
b 2.548 days
c 3.322 days
14.21 years
b 23.84 years
c 47.19 years
10.48 years
b 22.20 years
c 91.17 years
b 8 years (rounded up)
A = 2000 × 1.1n
F = 300 000 × 0.92n
b 8.4 years
69 years b 1386 years
log10 7
log10 16
log10 1.3
ii
iii
i
log10 2
log10 3
log10 5
1 a log28 = 3
d
2 a
3 a
d
4 a
d
5 a
d
6 a
e
i
7 a
8 a
9 a
10 a
11 a
12 a
13 a
b i
1
log10 5
c i 1.631
ii
3
log10 2
ii 1.167
iii
−1
log10 3
iii −0.196
Exercise 10E
1 a 5
b i 3
ii −2
c 2
2 a linear
b quadratic
d quadratic e constant
g constant h quartic
3 a 4
b 3
d −2
e −9
4 a, b, f are polynomials.
5 a 14
b 92
6 a −5
b 11
7 a 0
b 92
8 a −2
b 25
d −17
e 17
9 a −
10 a
11 a
b
12 a
1
2
b −1
0
b 4
i 30 m
ii 24 m
Yes, when 5 < x < 7.
8
b n+1
9
13 a −
8
b −
20
27
iii 1
c
f
i
c
f
quartic
linear
cubic
−2
2
c
c
c
c
f
8
1
−4
−22
−351
iv −1
d 4
d −45
d 42
1
2
c −108
iii 0 m
c
c 1
c
5
8
d 1
d
27
64
e
14 a
c
e
g
15 a
b
c
16
216
f −
27
125
2k 3 − k 2 − 5k − 1
16a3 − 4a2 − 10a − 1
−16a3 − 4a2 + 10a − 1
2a3b3 − a2b2 − 5ab − 1
i 10
ii 2
iv −13
v −9
i 3
ii −11
a = 2 and b = −1
9
1
h −
8
2
2b3 − b2 − 5b − 1
−2a3 − a2 + 5a − 1
−54k3 − 9k2 + 15k − 1
−2a3b3 − a2b2 + 5ab − 1
1
−18
−22
g −
b
d
f
h
iii
vi
iii
Answers
Number and Algebra
Exercise 10F
1 a x 2 + 2x
b x − 3x 2
c x2 − 1
d x 2 + 6x − 55
e 6x 2 − 13x − 5
f 8x 2 − 26x + 15
2 a x4 − 5x 3 + 4x 2 − 3
b −x 6 − 3x 4 + x 3 − x 2 + 13
c −3x 8 − x 6 − 6x + 3
3 a, b, c are true.
4 a x 3 − 3x 2
b x4 − x2
c 2x 2 + 6x 3
d x3 − x4
e x 5 + 3x 4
f −3x 6 + 3x 3
g −2x 5 − 2x 4
h −x 7 + x 4
i −4x 7 + 8x10
5 a x 5 + x 3 + 2x 2 + 2
b x5 − x
c x 5 − x 4 − 3x 3 + 3x 2
d x 5 − x 3 − 2x 2 − 2x + 4
e x 5 + 2x 4 + 2x 3 − 2x 2 − 3x f x 5 − 2x 4 + 5x 3 − 4x 2
g x 6 − x 5 + x 4 − 4x 3 + 2x 2 − x + 2
h x 6 − 5x 5 − x 4 + 8x 3 − 5x 2 − 2x + 2
i x 8 − x 6 + x 5 − 2x 4 − x 3 + 3x 2 + x − 3
6 a x 5 − 2x 4 + 2x 3 − 3x 2 + 3x − 1
b x 6 + 2x 4 − 2x 3 + x 2 − 2x + 1
c x 4 − 4x 3 + 6x 2 − 4x + 1
7 a x 5 + 3x 4 − x 3 − 9x 2 − 2x + 8
b x 4 + 2x 3 − 3x 2 − 4x + 4
c x 6 + 4x 5 + 2x 4 − 12x 3 − 15x 2 + 8x + 16
8 a x 3 + x 2 − 4x + 1
b x 3 − x 2 + 6x − 1
c 2x 3 + 5x 2 − 23x + 5
d −x 5 + 5x 4 − 2x 3 + 5x 2 − x + 1
e −x 6 − 2x 4 − x 2 + 4
f −x 6 − x 4 − 10x 3 + 26x 2 − 10x + 1
9 (x 2 + x − 1)4 = x 8 + 4x 7 + 2x 6 − 8x 5 − 5x 4 + 8x 3 + 2x 2 − 4x + 1
10 (x 2 − x − 1)2 − (x 2 − x + 1)2 = x 4 − 2x 3 − x2 + 2x + 1 −
(x 4 − 2x3 + 3x 2 − 2x + 1) = 4x − 4x 2 as required (or could
use difference of two squares)
11 Yes. Multiplicative axiom ab = ba.
12 a 3
b 5
c 7
d 12
13 a m
b m
c m+n
d 2m
e 2m
f 3n
14 a x 4 − x 3 + x 2 − x
b x 5 + 2x 4 − 3x 3
c x 3 + 4x 2 + x − 6
d 6x 3 + 23x 2 − 5x − 4
e 15x 3 − 11x 2 − 48x + 20 f x 5 + 3x 4 − x 3 − 3x 2 − 2x − 6
877
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Chapter 1
878
Exercise 10G
1 a 1
b 3
c 0
2 a If 182 ÷ 3 = 60 rem. 2, then 182 = 3 × 60 + 2.
b If 2184 ÷ 5 = 436 rem. 4, then 2184 = 5 × 436 + 4.
c If 617 ÷ 7 = 88 rem. 1, then 617 = 7 × 88 + 1.
3 P(x) = (x − 1)(x 2 + 2x) + 3
4 P(x) = (x + 1)(3x 2 − 4x + 5) − 3
5 a 2x 3 − x 2 + 3x − 2 = (x − 2)(2x 2 + 3x + 9) + 16
b 2x 3 + 2x 2 − x − 3 = (x + 2)(2x 2 − 2x + 3) − 9
c 5x 3 − 2x2 + 7x − 1 = (x + 3)(5x2 − 17x + 58) − 175
d −x 3 + x2 − 10x + 4 = (x − 4)(−x2 − 3x − 22) − 84
e −2x 3 − 2x2 − 5x + 7 = (x + 4)(−2x2 + 6x − 29) + 123
f −5x 3 + 11x2 − 2x − 20 = (x − 3)(−5x2 − 4x − 14) − 62
6 a 6x 4 − x 3 + 2x2 − x + 2
= (x − 3)(6x 3 + 17x2 + 53x + 158) + 476
b 8x 5 − 2x 4 + 3x 3 − x 2 − 4x − 6
= (x + 1)(8x 4 − 10x 3 + 13x 2 − 14x + 10) − 16
7 a x2 − 2x + 3 −
5
x +2
c x 3 − 3 x 2 + 9 x − 27 +
b x2 + 2x + 2 −
79
x +3
1
x −1
d x 3 + 4 x 2 + 15 x + 60 +
8 −1, 1, 2
9
6x2 − 7 x − 3
x − 5 6 x 3 − 37 x 2 + 32 x + 15
)
6 x 3 − 30 x 2
−7 x 2 + 32 x
−7 x 2 + 35 x
−3 x + 15
−3 x + 15
0
Remainder of 0, as required.
13
8
253
−8
b −
c − 41
16
27
x3 − x 2 + 3x + 2 = (x 2 − 1)(x − 1) + 4x + 1
2x3 + x 2 − 5x − 1 = (x 2 +3)(2x + 1) − 11x − 4
5x4 − x 2 + 2 = 5x (x 3 − 2) − x 2 + 10x + 2
10 a 4
11 a
12 a
b
c
b
a
a
0
a
e
5 a
Exercise 10I
1 P (−1) = 0
2 a x = −3 or 1
b x = −2 or 2
d x = −3 or 4
3 a −3, 1, 2
e x = −3
b −7, −2, 1
1
or 4
2
f x = −4 or −3
c −4, 3, 4
2 1
e − ,− ,3
3 2
2 1 2
f − , ,
7 4 5
1 1
d − ,− ,3
2 3
12 1 1
g − ,− ,−
11 2 3
c x= −
3
2 1
h − ,− ,
5 19 2
4 a (x − 3)(x − 2)(x + 1); −1, 2, 3
b (x + 1)(x + 2)(x + 3); −3, −2, −1
c (x − 3)(x − 2)(x − 1); 1, 2, 3
d (x − 4)(x − 3)(x − 1); 1, 3, 4
e (x − 6)(x + 1)(x + 2); −2, −1, 6
f (x − 2)(x + 3)(x + 5); −5, −3, 2
5 a x = 1 or 1 + 5 or 1 − 5
b x = −2
6 a x = −1, 3 or 5
b x = −3, −2 or 1
7 a x = −4, 1 or 3
b −2, −1 or 3
8 a 3
b 4
c n
9 a x 2(x − 1); 0, 1
b x 2(x + 1); −1, 0
c x(x − 4)(x + 3); −3, 0, 4
d 2x3(x + 1) 2; −1, 0
4
2
2 2
10 0 = x + x = x (x + 1)
No solution to x2 + 1 = 0, thus x = 0 is the only solution.
11 The discriminant of the quadratic is negative, implying solutions
from the quadratic factor are not real; x = 2 is the only solution.
12 a x = −4, −3, −2 or 1
b x = −2 or 3
1
c x = −3, −2, 1 or 3
d x = −2, , 1 or 2
2
Exercise 10J
Exercise 10H
1
2
3
4
240
x −4
6 b, c and e are factors of P(x).
7 b, d, f, g
8 a x+1
b x − 1, x + 1 or x + 2
c x+2
d x−2
9 a x − 2, x − 1 and x + 1
b x − 3, x − 1 and x + 2
c x − 3, x − 2 and x + 1
d x − 5, x − 1 and x + 4
10 a −4
b −2
c −14
d 96
11 −38
12 a 5
b 1
c 5
d −3
13 a −2
b 23
14 a a = −1 and b = 2
b a = 3 and b = −4
1 a
−1
3
b 41
b −2
c −19
d −141
3
−127
3
b 11
f −33
b 20
c 27
g −13
c 36
d 57
h −69
d 5
y
3
−3
−1 O
2
x
878
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Number and Algebra
b y-intercept: 12
x-intercepts: −1, 3, 4
y
3
−5
Answers
b
y
O
2
x
5
12
2 a y-intercept = 12
x-intercepts = −1, 3, 4
b y-intercept = −21
x-intercepts = −3, −1, 7
c y-intercept = 0
x-intercepts = −2, 0, 4
d y-intercept = 0
x-intercepts = −7, 0, 5
3 a−c
x
−2
−1
y = x2
4
1
−1
3
x
4
c y-intercept: 10
x-intercepts: −2, 1, 5
y
10
−2 O
1
2
0
1
2
1
2
1
4
0
1
4
1
4
−
O
y = x3
−8
−1
−
1
8
0
1
8
1
8
y = x4
16
1
1
16
0
1
16
1
16
1
x
5
d y-intercept: 3
x-intercepts: −3, 1, 2
y
3
O
−3
1
x
2
y
10
5
−2
−1 −5O
−10
1
2
e y-intercept: 0
x-intercepts: −3, 0, 2
x
y
4 a y-intercept: 6
x-intercepts: −2, 1, 3
y
−3
O
2
x
6
−2
O
1
3
x
f y-intercept: 0
x-intercepts: −1, 0, 5
y
−1 O
5
x
879
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
880
Chapter 1
g y-intercept: 0
x-intercepts: −3, 0, 1
7 a
y
60
y
O
−3
40
30
20
x
1
−2 −1O
1 2 3
x
5
4
−20
h y-intercept: 0
x-intercepts: −1, 0, 3
b
y
y
80
−1 O
x
3
60
40
i y-intercept: 8
x-intercepts: −4, −2, 1
20
y
−4
−2
O
−2
x
1
8 a
j y-intercept:
b
y
−1
−1−2O
−4
−6
−8
−9
−10
y
3
2
x
O 0.5
c
5 a
y
8
6
4
2
2
3
2
1
x-intercepts: −3, −1,
2
−3
x
O1 2 3 4
−4 −3 −2 −1
8
b
y
−4
−2 −2
−4
d
y
−2
y
x
2
O
2
x
y
x
2
−2
1
−1
O
−1
1
x
O
−2
−2
−3
1
c y = x ( x − 2 )( x + 3 )
2
x
O1 2
−2 −1
−4
−1
−1
6 a y = (x − 2)(x + 1)(x + 4)
1
x
−2
e
f
y
y
4
15
13
2
10
b y = (x + 3)(x − 1)(x − 3)
1
d y = − ( x + 3 )( x + 1)( x − 2 )
2
−4 −3 −2
O
x
5
O
5
−5 (2, −3)
x
880
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
9 a
b
y
iv
y
Answers
Number and Algebra
y
15
4
2
10
2
O
2 3 4
O
−2
x
x
5
−2
−2
−1
c
d
y
6
4
2
12
5
−5 −2O
x
−2 −1−2O
−4
−4
c i y-intercept = 8
ii y = (x − 2)(x − 1)(x + 1)(x + 4)
iii x-intercepts: −4, −1, 1, 2
iv
y
10 8
−4 −3 −2 −1 O
−10
−20
−30
−40
x
3 5
−5
e
f
y
4
2
2
O
x
2
−2
−2
O
2
x
d i y-intercept = 225
ii y = (x − 5)(x − 3)(x + 3)(x + 5)
iii x-intercepts: −5, −3, 3, 5
iv
y
250 225
200
150
100
50
−2
10 a i y-intercept = −6
ii y = (x − 1)(x + 2)(x + 3)
iii x-intercepts: −3, −2, 1
iv
y
−6 −5 −4 −3 −2 −1 O
−50
−100
O
−4
−2
2
−2
−4
−6
−8
b i y-intercept = 15
ii y = (x − 5)(x − 3)(x + 1)
iii x-intercepts: −1, 3, 5
x
1 2
y
4
−2
x
−5
y
15
O 3 5
x
x
1 2 3 4 5 6
Challenges
1 a 2
b 8
b −1.43
1
3 a = 2 × 3 4 , b = 1 log2 3
4
log
2
10
4
= log1.12
log10 1.1
2 a 1.43
1
2
c −2.71
c
d 3
d x ≥ −2.81
5 −2
6 a = 5, b = −2
7 Proof using long division required.
a (x3 − a3) ÷ (x − a) = x2 + ax + a2
b (x3 + a3) ÷ (x + a) = x2 − ax + a2
881
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Chapter 1
882
8 a 2 ≤ x ≤ 5 or x ≤ −1
b −4 < x < 1 or x > 4
v
y
1
9 y = ( x − 3 )2 ( x + 2 )
9
1
10 y = − x 2 ( x − 3 )( x + 3 )
10
Multiple-choice questions
1 C
5 D
9 E
2 B
6 D
10 E
(–3, 1)
3 E
7 A
11 D
4 A
8 B
2
3
4
5
b log101000 = 3
7 a pH = log10
c
d
8 a
e
increase of 216%
794 times stronger intensity
loga8
b logb21
c logb144
loga4
f loga1000 g 2
3
j 4 loga2 = loga16
i
2
9 a x = log36
b x = log1.22
10 a log10 13
log10 2
x=2
(10, 2)
1
–1
x
1 2 3 4 5 6 7 8 9 10 11 12 13
(3, −1)
–2
log10 2
log10 0.8
1 3
b x = − , or 5
3 2
17 a (x − 1)(x + 2)(x + 3) = 0 x = −3, −2 or 1
b (x + 2)(x − 5)(x − 6) = 0 x = −2, 5 or 6
18 a
b
y
y
16 a x = −2, 1 or 3
5 4
1
–3
–4
b
d loga10
h 1
11 a −1
b 1
c −2
d −34
12 a x 4 + 3x 2 + 2
b x5 − x4 − 3x3
c x 5 + x 4 − 3x 3 − x 2 − x + 3
d x 6 + 2x 4 − 4x 3 + x 2 − 4x + 3
13 a x 3 + x 2 + 2x + 3 = (x − 1)(x 2 + 2x + 4) + 7
b x 3 − 3x 2 − x + 1 = (x + 1)(x 2 − 4x + 3) − 2
c 2x 3 − x 2 + 4x − 7 = (x + 2)(2x 2 − 5x + 14) − 35
d −2x 3 − x 2 − 3x − 4 = −(x − 3)(2x 2 + 7x + 24) − 76
14 a −3
b −39
c −91
d 41
15 b, c and d are factors.
3
–3 –2 –1 O
1
 H + 
b 6.31 × 10−5 moles/L
y
2
1 2 3 4 5
6 y = log5 ( x + 2 ) + 3
1
c log3 = −2
9
1
a 34 = 81
b 4 −2 =
c 10−1 = 0.1
16
a 3
b 4
c 4
d 0
e −3
f −3
g −1
h −4
i −2
(1) y = log2 x is the reflection of y = 2x in the line y = x.
(2) The coordinates of each point on y = log2 x are the
coordinates in reverse order of a point on y = 2x.
a y = log2 ( x − 2 ) − 1
i x>2
ii x = 2
iii moved right 2 units and down 1 unit
iv 4
v
4
x
–5 –4 –3 –2 –1–1O
–2
–3
Short-answer questions
1 a log216 = 4
6
5
4
3
2
1
x = –4
−1
b y = log3 ( x + 4 ) + 1
i x > –4
ii x = –4
iii moved left 4 units and up 1 unit
iv –3.67, 2.26
O
1
x
−1O
−1
−5
−2
−10
4 5
x
882
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
c
d
y
2
y
A
D
10
10
5
−5 −2 O
3
5
B
5
x
−5
x
O
−10
Extended-response questions
1 a
b
c
2 a
b
c
d
e
f
g
B
i $121 000
ii $115 369
i 7.27
ii 6.17
i 32
ii 0
There is no remainder; i.e. P (1) = 0.
x2 − 4x − 21
(x − 7)(x − 1)(x + 3)
x = 7, 1 or −3
P(0) = 21
iii $272 034
iii 16.89
∠DBC = ∠BDA (alternate angles in parallel lines)
∠BDC = ∠DBA (alternate angles in parallel lines)
BD is common
∴ BAD ≡ DCB (AAS)
Using congruence BC = AD and AB = DC, corresponding
sides in congruent triangles.
3 a x = 6.75 b x = 2
4 a x=8
b x=5
c a = 32, b = 65
d x = 40
e a = 55
f a = 90, b = 60, c = 70
5 a x = 20
b x=8
c a = 63, b = 55
6 a x=
47
5
b x=
29
3
c x=
39
5
Extended-response question
1 a CD = 6 cm (chords of equal length are equidistant from
the centre)
b OA = OD (radii of circle)
OB = OC (radii of circle)
AB = DC (chords of equal length are equidistant from the
centre)
∴ OAB ≡ OCD (SSS)
c OM = 4 cm, Area = 12 cm2
d 30.6%
e ∠BOD = 106.2°
y
21
20
−3
7
−4 −2 O 1 2 4 6 8
−20
−40
−60
C
Answers
Number and Algebra
x
Semester review 2
Chapter 7: Trigonometry
Multiple-choice questions
Chapter 6: Geometrical figures and circle geometry
Multiple-choice questions
1 D
2 B
3 C
4 E
Short-answer questions
1 a AB = DE (given)
AC = DF (given)
∠BAC = 60° = ∠EDF (given)
∴ ABC ≡ DEF (SAS)
a = 35 (corresponding angles in congruent triangles)
b BC = DC (given)
AC is common
∠ABC = 90° = ∠ADC
∴ ABC ≡ ADC (RHS)
x = 3 (corresponding sides in congruent triangles)
5C
1 E
4 D
2 B
5 C
3 A
Short-answer questions
1
2
3
4
5
6
a x = 19.5
a i 150°
a 32.174 m
a x = 9.8
95.1°
a tan θ
1
c i
2
7 a ≈ 0.34
b
ii
b
b
θ = 43.8, y = 9.4
330°
b i 310°
52.2°
θ = 125.3
b i θ = 155 ii θ = 35
ii 130°
iii θ = 42
2
− 3
ii
iii
2
3
b θ ≈ 233, 307 c yes
883
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
884
Chapter 1
Extended-response question
1 a 104.3 m
Short-answer questions
b
1 a dilated by a factor of 3
8 km
y
20°
108°
3
13 km
52°
(1, 3)
Start
x
O
c 17.242 km d 206°
b reflected in x-axis and translated 2 units left
y
Chapter 8: Quadratic expressions and quadratic
equations
Multiple-choice questions
1 C
2 B
3 D
4 B
x
O
−2
5D
−4
Short-answer questions
1 a 9x 2 − 1
2 a (2x − y)(2x + y)
c 3(x − 4)(x + 4)
e (x − 5)2
3 a (3x + 4)(x − 2)
4 a x = 0, 3
4x 2 − 20x + 25
c −x 2 + 30x − 5
(x + 2 + 7 )(x + 2 − 7 )
(x − 2)(x + 7)
2(x − 6)(x − 2)
(3x − 1)(2x + 3)
c (5x − 4)(2x − 3)
1
c x = 0, −5
x = −4,
2
x = 7, − 7
f x=2
1
x = −2,
3
x = 3, 7
c x = −4, 5
b
b
d
f
b
b
d x = 4, −4
e
g x = 8, −3
h
5 a x = −8, 5
b
6 a i (x − 3 + 5 )(x − 3 − 5 )
ii (x + 2)2 + 3, does not factorise further
 3
5  3
5
 x + +

iii  x + −
 2 2  2 2 
b i x = 3± 5
7 a x=
ii no solutions
−3 ± 57
4
iii x =
1 D
4 C
2 A
5 B
x
O
2 a −5
d
b −5, 1
c (−2, −9)
y
x
O 1
−5
(−2, −9) −9
c x=3
3 a maximum at (3, 8)
d
y (3, 8)
Chapter 9: Non-linear relationships, functions and
their graphs
Multiple-choice questions
(1, 6)
5
−3 ± 5
2
b x = 2 ± 10
b 44 m2
y
−5
Extended-response question
1 a 4x 2 + 40x
d x = 2.2
c translated 5 units up
O 1
5
b −10
c 1, 5
x
−10
3 D
6 C
884
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
4 a y = (x + 3)2 − 7
c
Answers
Number and Algebra
y
y
(1, 3)
2
O
−3 − √7
x
1
−3 + √7
x
O
−7
(−3, −7)
d
y
(−1, 5)
 5 2 7
b y = x −  +
 2 4
y
1
1)
5
(1,
x
O
8
e
( 52 , 74 )
y
(1, 2)
x
O
x
O
5 a Discriminant = 72, thus two x-intercepts.
b
y
(−1, −2)
f
y
(−1, 6)
−1.1 O
x
3.1
x
O
−9
(1, −9)
(1, −6)
6 a
y
7 a −4
b −2
d 30
e 2k2 + k − 7
8 a A > 0, a > 0, b > 0, h > 0
2
−2
O
2
x
b
9 a
b
c
d
e
y
√10
−√10
O
−√10
√10
x
10 a
b h=
2A
a+b
2A
−b
e 3
h
domain: all real x; range: all real y
domain: all real x; range: all real y
domain: all real x; range: y ≤ 4
domain: all real x, x ≠ 0; range: all real y, y ≠ 0
domain: x ≥ –2 range: y ≥ 0
x
b f −1(x) = 5 − x
f −1(x) =
4
c 4
−2
c 6
d a=
c f −1(x) = 3 4 − x
e f −1(x) =
1
1− x
d f −1(x) =
3
2x
885
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
886
Chapter 1
11 x ≥ 1 or x ≤ 1
12 a ( 3 , 2 3 ), (− 3 , −2 3 )
b
y
b (4, 16)

1  1
c  , 4   − , −4 

2   2
13 a
8
y
O
−1 + 2√3
x
2 + √15
O
2 − √15
(2, −1)
−1 − 2√3
b
x
2
c
y
y
(1, 17)
O
9
−2
y=1
1
O
c
x
Extended-response questions
y
1 a
x = −2
−2 −
x
1
h
(200, 427.5)
5
3
−
x
O
27.5
5
2
−3
O 10
110
(60, −62.5)
y = −3
14 x = 3, y = 22 and x = −5, y = −2
b no solutions
c x = −1, y = 1
15 a
y
(1, 2)
O
x
x
b 27.5 m
c 10 m and 110 m from start
d 427.5 m
e 62.5 m
2 a i Cost increases as the amount of hire time increases. If hire
time doubles, then the cost will also double.
ii C = 10n
iii Cost = $37
b i y=2
ii x = 24
c i Speed is inversely proportional to the time taken to travel
a certain distance. If the speed increases, then the time
decreases.
ii 97.2 min = 1 h 37.2 min
886
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press
Number and Algebra
ii
y
4 a i 6
ii 0
iii −49
b i 2x 6 + 6x 5 − 11x 4 − 25x 3 + 34x + 24
ii 4x 6 − 12x 4 − 16x 3 + 9x 2 + 24x + 16
5 P(x) = (x − 3)(x 2 − x − 1) + 4
6 a −24, not a factor
b 0, a factor
c −40, not a factor
x
x
iii
iv
y
x
x
v (either answer correct)
x = −4, −2, 1
x > –4
x = –4
y = log3 ( x + 4 ) + 2
d −3.89, 3.26
9 a P (−3) = 0
b P(x ) = −(x + 3)(2x − 1)(x − 4)
P(x)
(−5, 196)
Distance
Distance
c
Time
b x = 0, 5 , − 2
2
3
1
d x = −1, , 2
2
7 a x = −1, 3, −6
c
8 a
b
c
y
iv −5
Answers
d i
y
O
Time
−5
−3
−12
1
4
6
x
Chapter 10: Logarithms and polynomials
Multiple-choice questions
1 E
2 A
3 D
4C
(6, −198)
5D
Short-answer questions
1 a 3
e 2
b −2
f 2
c −1
g 1
2 a x=3
b x=3
c x = 81
3 a i x = log330
b i x = 2.460
ii x = log2.44
ii x = 9.955
d 0
h 3
Extended-response question
1 a 100 000 times louder
b 4.8 dB
c 355 times louder
887
© David Greenwood et al. 2014
ISBN 978-1-107-67670-1
Photocopying is restricted under law and this material must not be transferred to another party.
Cambridge University Press