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g y = –x + 4 y l i y=1x−2
4
5 8 y 2x – 3y = – 9
5 a y = 4x − 5
4 7
6
3 5 b y = −3x + 17

2 4 c y= 9x−6 d y = 17 x − 71
3 55 44
1 x 2
1234 5 1 6 Any line with the same gradient, e.g.
–3 –2 –1 0
–1 x

–2 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 a y = −3x − 5 b y = 2x + 13

–3 –2 c y=x −3 d y = −x − 4
x-intercept = –1 –3 2
–4 e x = −8 f y=6
y-intercept = 4 –5
–6
Review h x + 2y = 4 y –7 7 a, c
–8
3 3–32 8 a y = 2x − 2 b y = 2x

2 x-intercept = d y = 2x + 1
y-intercept = 2
1 c y = 2x − 4

–3 –2 –1 0 x 2 y = mx + c Gradient y-intercept 9 a Any line with gradient 2,
–1 1234 3

–2 a y = 1 x − 2 1 −2 e.g. y = 2 x − 5
x-intercept = 2–1 2 2 3
– b Any line with same y-intercept,

y-intercept = 2 b y = −2x + 1 −2 1 e.g. y = 2x + 3
2 4
i y x + –2y = 3 c y = 2x + 4 −5 c y=3
2 5
d y = 2x − 5 2 Exercise 10.5
−1 2
6 e y = 2x + 5 3

4 −1 1 y = −5x + 8
3
2 f y = x + 2 2 a Gradient AB = −2; Gradient
x PQ = 1; −2 × 1 = −1, so AB is
g y = 3x − 2 3 −2 22
0 1234 −4 2 perpendicular to PQ
4
Review x-intercept = –2 h y = −4x + 2 2
y-intercept = 6 6 −12
i y = 2x + 4 1 −3
8 b Gradient MN = 1; 1 × −2 = −1, so
j x = 4y – 2 2 y j y = 6x − 12 −12 6 22

1 x k y = 1 x − 3 MN is perpendicular to AB
123 8
–3 –2 –1 0 −1
–1 l y = −12x + 6 3 y = 3 x + 5

–2 –14 3 a y = 2x + 3 b y = −3x − 2 4 a y = –1x + ½ or x + 2y − 1 = 0
x-intercept = 2

y-intercept = 2–1 d y = − 3 x − 0.5 b x+y+1=0
2
c y = 3x − 1 5 Gradient A = −2, Gradient B = ½ −2 ×
½ = −1, so A is perpendicular to B
k x = 4–y + 2 e y = −3 x + 2 f y = 1x − 3
42
y 6 y = 5x − 18
8

7

6 g y = 0.75x − 0.75 7 Gradient AB = 10 ; Gradient AC = −1

5 h y = −2 i y=4 9
perpendicular
4 b y=1x+1 so AB is not to AC and
3 3
Review 2 4 a y = −4x − 1 figure cannot be a rectangle.
1x d y = 5x + 2
c y = −3x + 2 f y = −x + 2
–8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 e y = 3x + 1 h y=2x−1

–2 g y = 2x − 3 3
–3
–4
–5
–6
–7
–8

x-intercept = 4

y-intercept = –8

Answers 641

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Cambridge IGCSE Mathematicse-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity

Exercise 10.6 d 5 y y = 4x + 2 h 2 y y = –23 x – 1

1a y y = –5x + 10 4 1
12 –2 –1 0
x
11 3 –1 123

10 2

9 1 –2

8 –2 –1 0 x –3
–1 12 x-intercept = 1.5
y-intercept = –1
7

6 –2 i

Review 5 –3 1 y y = –4x – 2
4 x-intercept = –0.5
x
y-intercept = 2 –2 –1 0 1 2 3 4 5 6 7 8 9 10

3 e 3 y y = 3x + 1 –1

2 –2

1 2 –3
–1 0 x-intercept = 8 y-intercept = –2

–1 x 1 x j y = –25x + 1 y
12345 –3 –2 –1 0 123
3
–1
–2 2
x-intercept = 2
–2 1
y-intercept = 10
–4 –3 –2 –1 0 x
b y y = –3x –1 –3 –1 123
4 x-intercept = – 1–3
y-intercept = 1

3 –2
x-intercept = –2.5
2 f y y = –x + 2 y-intercept = 1
3

1 2 k 8 y y = –4x + 2
–1 0
Review x 1 x 7
–1 12345 –1 0 123 6
5
–2 –1
x-intercept = 3
–2 4
y-intercept = –1 x-intercept = 2 3
2
c 7 y y = –3x + 6 y-intercept = 2 1x

6 g y y = 2x – 3 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6

5 2 –2
1 –3
4 –4
–5
3 x –6
3 –7
2 –2 –1 0 12 –8
–1 y-intercept = 2

1 x –2
123
–1 0 –3 x-intercept = –8
–1
–4

Review –2 –5
x-intercept = 2
–6
y-intercept = 6 x-intercept = 1.5

y-intercept = –3

642 Answers

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e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity Answers

l 6 y y = –12x + 6 g x2− 3x − 28 h x2 + 5x − 24 d 9x2 −12xy + 4y2
i x2 − 1 j x2 − x − 72
5 k x2 − 13x + 42 l x2 − 9x − 52 e x2 + 4xy + 4y2
4 m y2 − 11y − 42 n z2 − 64
3 o t2 + 13t − 68 p h2 − 6h + 9 f y2 − 8x2 y + 16x4

2 r d2 + 3 d − 9 g x4 − 2x2y2 + y4

1 48 h 4 + 4y3 + y6
x
q g2 + 3 1 g−2 i 4x2 + 16xy2 + 16y4
0 11 2
–1 2 1 − 1 + 1
2 a 12 − 7x + x2 j 4x2 4xy 16 y 2
b 3 + 7x − 6x2
Review x-intercept = 0.5 c 6m2 − 17m + 7 k 9x2 − 3xy + y2
y-intercept = 6 d −8x2 + 2x + 3 16 4 4
e 8a2 − 2b2
f −8m2 − 2mn + 3n2 l a2 + ab + b2
4
2 a c=2 b c = −4 g x2 + 3 x + 1
c c = −9 d c = −8 m a2b2 + 2abc4 + c8
e c=4 f c=3 48 n 9x4y2 − 6x2y + 1
g c = −2 h c=2
h 2x2 − 2 x − 1 o 4x2 + 16xy + 16 y2
93
36
p x2 − 6x + 9
Exercise 10.7 i −2x4 + 6x2y − 4y2
j −36b2 − 26b + 42 2 a 4x − 12
1 a Length = 8.49 midpoint = (6, 9) k −4x4 + 2xy2 − 4x3 y + 2y3 b 2x2 + 2x − 19
b Length = 4.47 midpoint = (3, 8) l 6x2 + 9x − 15 c 2y2 + 8x2
c Length = 5.66 midpoint = (6, 5) d x2 + 8x − 2
Review d Length = 3.16 3 a 2x2 + 9x + 9 23
midpoint = (4.5, 9.5) b 3y2 + 10y + 7
e Length = 5 midpoint = (2.5, 5) c 7z2 + 15z + 2 e 6x2 + 13.8x + 3.6
f Length = 1.41 d 4t2 + 17t − 15 f −16x2 + 8xy + 2x − 2y2
midpoint = (11.5, 3.5) e 2w2 − 23w + 56 g −x2 + 3x − 22
g Length = 5 midpoint = (1, 3.5) f 16g2 − 1 h 4x2 − 12xy − 19y2
h Length = 6.08 midpoint = (4.5, 2) g 72x2 + 23x − 4 i −2x3 − x2 − 17x
i Length = 11.05 h 360c2 − 134c + 12 j 4x2 − 13x − 1
midpoint = (−2.5, 1.5) i −2m2 + 10m − 12

2 AB = 5.39 midpoint = (3, 4.5) 4 a 6x3 + 9x2 + 2x + 3 3 a −49 b9
CD = 4.47 midpoint = (−4, 6) b 15x4 − 18x2 + 3 c 66 d 36
EF = 8.60 midpoint = (−2.5, 2.5) c 6x3 + 9x2y − 2xy − 3y2 e0 f 321
GH = 7.07 midpoint = (3.5, 0.5)
IJ = 5.10 midpoint = (2.5, −3.5) 5 a 15x3 + 21x2 − 24x − 12 Exercise 10.10
KL = 12.6 midpoint = (1, −3) b x3 − 5x2 − 25x + 125
MN = 5.39 midpoint = (−3.5, −2) c 12x3 + x2 − 9x + 2 1 a (x + 12)(x + 2)
OP = 7.81 midpoint = (−4.5, −4) d 4x3 + 32x2 + 80x + 64 b (x + 2)(x + 1)
e 12x3 − 32x2 + 25x − 6 c (x + 4)(x + 3)
Review3 5.83 f 18x3 − 33x2 + 20x − 4 d (x + 7)(x + 5)
4B g x3 + 6x2 + 12x + 8 e (x + 9)(x + 3)
5B h 8x3 − 24x2 + 24x − 8 f (x + 6)(x + 1)
6 AB = 6.40 i x4y4 − x4 g (x + 6)(x + 5)
h (x + 8)(x + 2)
AC = 4.24 j 1 − x2 + x4 i (x + 10)(x + 1)
BC = 7.28 81 18 16 j (x + 7)(x + 1)
7 a=7 k (x + 20)(x + 4)
8 E = (−6, −2) 6 a V = (2x + 1)(x − 2)2 cm3 l (x + 7)(x + 6)
2
2 a (x − 6)(x − 2)
b 2x3 − 7.5x2 + 6x + 2 b (x − 4)(x − 5)
c (x − 4)(x − 3)
c 0.196 cm3 d (x − 4)(x − 2)
e (x − 8)(x − 4)
Exercise 10.8 Exercise 10.9

1 a x2 + 4x + 3 b x2 + 10x + 24 1 a x2 − 2xy + y2
c x2 + 19x + 90 d x2 + 15x + 36 b a2 + 2ab + b2
e x2 + 2x + 1 f x2 + 9x + 20 c 4x2 + 12xy + 9y2

Answers 643

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Cambridge IGCSE Mathematics e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity

f (x − 7)(x − 7) Examination practice 4 a Right-angled
g (x − 10)(x + 2) b Not right-angled
h (x − 9)(x + 2) Exam-style questions c Not right-angled
i (x − 8)(x + 4) d Right-angled
k (x + 3)(x − 2) 1 a x2 + 20x + 36 b 4x2 − 9 e Right-angled
l (x + 11)(x − 3) c 12y4 − 5y2 − 3
m (x + 12)(x − 2) Exercise 11.2
2 a i 6x(2x − 1)
Review3 a (y + 17)(y − 10) ii (y − 6)(y − 7) 1 53.2 inches b 3.16
b (p − 6)(p + 14) iii (d + 14)(d − 14) 2 3.03 m d 10.30
c (x − 12)(x − 12) 3 277 m
d (t + 18)(t − 2) b i x = 0 or x = 1 4 3.6 m
e (v + 15)(v + 5) 2 5 0.841 m
f (x − 10)(x + 10) 6 a 5.39
ii y = 6 or y = 7
Exercise 10.11 iii d = 14 or d = −14 c 9.90
7 P = 42.4 cm
1 a (x + 6)(x − 6) Past paper questions*
b (p + 9)(p − 9)
c (w + 4)(w − 4) 1a Exercise 11.3
d (q + 3)(q − 3)
e (k + 20)(k − 20) y 1 a Similar; all angles equal.
f (t + 11)(t − 11) b Similar; sides in proportion.
g (x + y)(x − y) 5 c Not similar; angles not equal.
h (9h + 4g)(9h − 4g) L4 d Not similar; sides not in
i 4(2p + 3q)(2p − 3q) proportion.
j (12s + c)(12s − c) 3 e Similar; angles equal.
k (8h + 7g)(8h − 7g) f Similar; sides in proportion.
l 3(3x + 4y)(3x − 4y) 2 A g Not similar; sides not in
m 2(10q + 7p)(10q − 7p) 1 x proportion.
n 5(2d + 5e)(2d − 5e) B h Similar; sides in proportion.
o (x2 + y2)(x2 − y2) –6 –5 –4 –3 –2–1 0 123456 i Similar; angles equal.
p x(y − x)(y + x) j Similar; all angles equal.
–2
2 71 –3
36 –4

ReviewExercise 10.12 b (−1, 0) 2 a x = 12 b y=5
c2 c p = 12 d a = 12
1 a x = 0 or x = 9 e b = 5.25 f c = 5.14
b x = 0 or x = −7 2 a (0, 5) b −1
c x = 0 or x = 21
d x = 4 or x = 5 3 (3w − 10)(3w +10)
e x = −7 or x = −1
f x = −3 or x = 2 4 (p − 6q)(m + n) 3 AC = 8.75 cm
g x = −2 or x = −1 4 CE = 4.51 cm
h x = −10 or x = −1 Chapter 11 5 BC = 2.97 m
i x = 3 or x = 4 6 lighthouse = 192 m
j x = 6 or x = 2 Exercise 11.1 7 r=8
k x = 10 or x = −10 8 x = 60
l t = −18 or t = 2 1 a x = 10 cm
m y = −17 or y = 10 c h = 2.59 cm b y = 13.4 cm Exercise 11.4
n p = −14 or p = 6 e t = 7.21 m d p = 1.62 cm
Review o w = 12 1 a 4 = 2 6 = 1.2
2 a x = 7.42 m b y = 3.63 cm
c t = 8.66 cm d p = 12 m 25
e a = 6 cm
b y = 4.47 cm e ratio of corresponding sides
3 a x = 2.80 cm d p = 8.54 km are not the same so the shapes
c h = 4.28 cm f h = 8.06 cm are not similar.
e k = 10.4 cm h f = 13 m b All sides of shape 1 have length x
g d = 6.08 m and all sides of shape 2 have length
y so the ratio of corresponding
sides will be equal and the shapes
are similar.

c 5 = 1.25 4 = 1.3ɺ

43

Ratios not equal, so not similar.

644 Answers * Cambridge International Examinations bears no responsibility for the example answers to
questions taken from its past question papers which are contained in this publication.

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e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity Answers

d 80 = 1.3ɺ 60 = 1.3ɺ Exercise 11.7 It follows that AC − BC = EC − DC,
so BC = CD.
60 45 BF = DF (corr sides of congruent
triangles)
Ratios of corresponding sides 1 a i SM ii PQ iii BC
equal, therefore they are similar. erefore BFCD is a kite (two pairs
b i MSR ii EFG iii OPQ of adjacent equal sides)

c ABCDEFG is congruent to Examination practice

e 12 = 1.5 9 = 1.5 SMNOPQR Exam-style questions

8 6 2 a A, C, I b D, F 1 215 m further
c B, G b E, H, L 2 4.21 m
Ratios of corresponding sides 3 a 35 cm b 37 cm
equal, therefore they are similar. 3 a DEF similar GHI 4 a a2 + b2 = c2
b ABCD similar EFGH
f ey are not similar because not c MNOP congruent STQR (7x)2 + (24x)2 = 1502
all corresponding angles are equal. d ABCDEFGH congruent 49x2 + 576x2 = 22500
PIJKLMNO and both similar to
2 a x=9 b y = 14 WXQRSTUV 625x2 = 22500
e ABC similar MON x2 = 36
Review c p = 3.30 d y = 7.46
b 336 cm
e x = 50, y = 16
Past paper questions*
f x = 22.4, y = 16.8
1 6.24
g x = 7.5, y = 12.5 4 2 432 cm2
3 12
h x = 178
Chapter 12
Exercise 11.5
Exercise 12.1
1 a 421.88 cm2 b 78.1 m2 a bc
c 1562.5 m2 d 375 cm2 1 a i Mode = 12
Exercise 11.8 ii Median = 9
2 a x = 24 cm b x = 30 m iii Mean = 8
c x = 2.5 cmd d x = 15 cm 1 Triangles ACB and PQR are
congruent because SSS is satisfied. b i Mode = 8
Review3 a Area will be 4 times larger. ii Median = 6
b Area will be 9 times larger. 2 Triangles ACB and PQR are iii Mean = 5.7
c Area will be smaller by a factor congruent because ASA is satisfied.
of 4. c i Mode = 2.1 and 8.2
3 Triangles ABC and PQR are ii Median = 4.15
4 8:3 congruent because RSH is satisfied. iii Mean = 4.79

Exercise 11.6 4 Triangles ABC and QPR are d i Mode = 12
congruent because SSS is satisfied. ii Median = 9
1 k2 : k3 iii mean = 11.7
5 e triangles are congruent because
2 a 4 b 16 : 1 c 64 : 1 SAS is satisfied. 2 Mean increased from 8 to 11.7
because of the extreme value of
3 216 cm2 6 Triangles ABC and QPR are 43 in(d). No change to mode or
congruent because RSH is satisfied. median.
4 172 cm2 OR
Triangles ABC and QPR are 3 a Andrew’s median = 54
5 a 16 mm b 157.9 cm2 congruent because ASA is satisfied. Barbara’s median = 48.5
c 83.2 cm3
b 21.3ɺ mm3 7 Triangles BAC and PQR are b Andrew’s mean = 84.25
6 a 20.83ɺ cm3 d 56.64 m3 congruent because SAS is satisfied. Barbara’s mean = 98.875
c 0.75 m3
8 Triangles ABC and QPR are 4 For example, 1, 2, 3, 4, 15
7 a 525 cm2 b 6860 cm3 congruent because RHS is satisfied. 5 Mode = none; mean = 96.4;
c 36 cm d 14.15 cm
11 Construct PM and NQ median = 103
Review8 Height 13 cm 11 cm 9 cm POM = QON (vertically opp) He will choose the median because
MO = NO (given) it’s the highest.
Surface x cm2 121x cm2 81x cm2 PO = QO (given) 6 4 451.6 cm
area 169 169  PMO congruent to QNO (S, A, S)
So, PM = QN (corresponding sides
Volume y cm3 1331y cm3 729 y cm3 of congruent triangles)
2197 2197
12 Since triangle FAB and FED are
9 x = 3.72 congruent:
Angle FAB = angle FED and that
makes Triangle CAE a right angled
isosceles triangle.

* Cambridge International Examinations bears no responsibility for the example answers to questions Answers 645
taken from its past question papers which are contained in this publication.

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Cambridge IGCSE Mathematicse-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity

7 2.38 kg 6a b Median = 17, Q1 = 12, Q3 = 21,
IQR = 9
8 91.26ɺ °C Stem Leaf
9 For example, 3, 4, 4, 6, 8 c Median = 14, Q1 = 5, Q3 = 18,
10 For example, 2, 3, 4, 7, 9 46 IQR = 13
5 004
11 mX + nY 5 5789 d Median = 3.4, Q1 = 2.45, Q3 = 4.95,
m+n 6 011233 IQR = 2.5
6 6689
Exercise 12.2 7 04 e Median = 15.65, Q1 = 13.9,
Q3 = 18.42, IQR = 4.53
1 a Ricky i mean = 0.152 Key
ii range = 0.089 2 Median = 6, Q1 = 4, Q3 = 8, IQR = 4
4 | 6 = 46 kilograms
Oliver i mean = 0.139 3 a Summer: median = 18.5,
Review ii range = 0.059 b 12 Q1 = 15.5, Q3 = 23.5
c Data has many modes. Winter: median = 11.5,
b Ricky d 74 − 46 = 28 Q1 = 9.25, Q3 = 12.75
c Oliver e 60.5 kg
b Summer: IQR = 8
2 a Archimedes median = 13 7a Winter: IQR = 3.5
Bernoulli median = 15
Stem Leaf c e lower IQR in winter shows
b Archimedes range = 16 that car numbers are more
Bernoulli range = 17 12 1 5 consistent. In poor weather
12 6 6 8 8 8 9 9 people either use their own
c Archimedes 13 0 1 2 3 3 4 transport or take transport more
d Archimedes 13 6 8 consistently.
14 0 0 2 2 3
3 Backlights. Footlights has the best 14 6 4 a Julian: median = 23, Q1 = 13,
mean but the range is large, whereas 15 0 Q3 = 24
Backlights and Brightlights have Aneesh: median = 18,
the same range but Backlights has a Q1 = 14 Q3 = 20
higher mean.
b Julian: IQR = 11
Exercise 12.3 Key Aneesh: IQR = 6
12 | 1 = 121 Components per hour
1 a Mean = 4.5 b Median = 4 c e IQR for the Algebraist is
Review c Mode = 4 and 5 d Range = 8 b 29 more consistent than that for the
c 132.5 Statistician and is therefore more
2 a Price Frequency Total likely to have a particular audience
while the variation is greater for
$6.50 180 $1170 Exercise 12.4 the Statistician and therefore could
$8 215 $1720 appeal to a varying audience.

1 Mean height = 141.7 cm 5 a i 6.5 ii 5.9

$10 124 $1240 2 a 5.28 min b 5 min 17 s b i 10.85 ii 14.05
$4130
c i 3.275 ii 3.65
3 Mean temperature = 57.36 °C
b $7.96 d At first glance it seems like country

3 a Mode = no letters 4 a Hawks mean mass = 76.7 kg Eagles driving gets much better fuel
b Median = 1 letter mean mass = 78.4 kg
c Mean = 0.85 letters consumption as it appears that the
d Range = 5 b 45 kg for both (this is group range
not actual data range) data is distributed more towards the

c e range of masses of the higher end of the stems. However,
players within each team is the same
for both teams. So, one can say that the smaller interval and the decimal
on average, the Eagles have a larger
4 a Mode = 1 mass than the Hawks. nature of the data mean that when
b Median = 2
c Mean = 2.12 you look at IQR, there is not such a

Review massive difference in consumption

5 a Mode = 8 given that the difference between
b Median = 6.5
c Mean = 6 03ɺ 5 Mean = 39.2 cm the two IQRs is only 0.375.
d If she wants to suggest the class 6 Mean age = 42.23 years
is doing better than it really is, Exercise 12.6
she would use the mode and say Exercise 12.5
something like: most students got 1
8 of 10.
1 a Median = 6, Q1 = 4, Q3 = 9, 145 147 149 150 152
IQR = 5

646 Answers

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University Answers

2 Examination practice Copy Copy Copy Copy3 125 ml < 1 litre < 0.65 litres < 780 ml
2
UniversitCyoPprye-ssC-aRmebrviidegwe2841.5 46.553.576Past paper questions*
4 60
3a
1 a 35 < t  40 5 a 14230 mm, 0.01423 km
Test 1 b 37.3 b 19 060 mg, 0.00001906 t
Test 2 c 2750 ml, 275 cl
Test 3 2 78, 78, 76, 68 d 4 000 000 mm2, 0.0004 ha
e 1300 mm2, 0.000 000 13 ha
3 a 137 g (3sf) f 10 000 mm3, 0.000 01 m3
bi
10 20 30 40 50 60 70 80 90 6 a 27 m3 b 27 000 000 cm3
Mass (m grams) Frequency
b Interpretations will vary, but
generally the students performed 0 < m  80 16 c 2.7 × 1010mm3
worst on Test 3. 80 < m  200 126
Review 200 < m  240 16 7 a 1.09 × 1012km3
b 1.09 × 1021m3
4 a 25 km b 47.5 km c 1.09 × 1030mm3

c 75% d 50% ii 8 a 1.13 × 102cm3
b 1.13 × 105cm3
e 10 km 1.2 c 1.13 × 10−13km3
1.0
f e data is evenly distributed 0.8
UniversitCyoPprye-ssC-a FrdeRemqnuseietbnycrvyiidegwe0.6
about the mean as its in the 0.4 9a6 b 20 g

middle of the box part of the

diagram. 10 a No b No c Yes

5 a 34 b 30 Exercise 13.2

c Team B d Team B 1 3 h 39 min

e Team A’s median is higher and 0.2
0m
their IQR is overall higher. 40 80 120 160 200 240 2 a 22.30 to 23.30
Mass (grams) b 09.15 to 10.45
6 Shamila spent 30 minutes or more c 19.45 to 21.10
studying every day. For 75 percent
of the days, she studied for more c 135 g 3 a 300 km b 120 km/h
than 45 minutes and on half the
Review days she studied for 50 minutes or Examination practice 4 9 min 47 s
more. Malika studied for less than
30 minutes on half the days. She Structured questions for 5 Monday 10 February 02.30
only studied for 45 minutes or more Units 1–3
on 25% of the days, suggesting she 6a
studied for a shorter time over the Answers for these questions are available
period. is could be because she in the Teacher’s Resource. Day Mon Tues Wed Thurs Fri
found the work easy and didn’t need Total time 7h 7h 7h 7h 8h
to study so much, or that she just Unit 4 worked 55min 55min 25min 53min 24min
doesn’t like to study.
UniversitCyoPprye-ssC-aRmebrviidegwe b 39 h 32 min c $223.36

Exercise 13.3

7 Reports will vary, but if you draw Chapter 13 1 a 20.02 b 45 min c 23 min
vertical lines on the graphs to show
the tolerances (at 16.95 and 16.75) Exercise 13.1 2 a 1 h 7 min 11:10
you can see that machines B and C b Aville
produce bars outside the tolerances.
Machine C produces the smallest 1 a 4000 g b 5000 m Beeston 11:45
rods, 75% of them are below the given
diameter. Machine A is the most c 3.5 cm d 8.1 cm Crossway 11:59
consistent with all rods within the
Review given limits. e 7300 mg f 5.760 t Darby 12:17

g 210 cm h 2000 kg c 14:25

i 1.40 m j 2.024 kg 3 a 00:17 b 12 h 40 min

k 0.121 g l 23 000 mm c 5 h 46 min

m 35 mm n 8036 m d i 01:29 or 13:34

o 9.077 g ii Unlikely to be 01:29 because it
- Cambridge
ess - Review
2 32.4 cm < 3.22 m < 3 2 m is in the middle of the night –
9
in the dark.

* Cambridge International Examinations bears no responsibility for the example answers to questions Answers 647
taken from its past question papers which are contained in this publication.

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Cambridge IGCSE Mathematics e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity

e i 1–6 February (Wed–Mon)  a 4 a 165 min b 4.8 kg
ii 1–4 February (Wed–Sat)  b 
j 47.9 < de − c (40m) + 30 = 25

Exercise 13.4 < 49.7 ⇒ m = − 0.125 kg

You cannot have a negative mass of

1 a 11.5  12 < 12.5 meat. As the graph assumes it will
b 7.5  8 < 8.5
c 99.5  100 < 100.5 2 a 78.5 cm b 79.5 cm always take at least 30 minutes to
d 8.5  9 < 9.5
e 71.5  72 < 72.5 3 37 kg  mass le < 39 kg cook any piece of meat, you cannot
f 126.5  127 < 127.5
4 a 3.605 cm  Length < 3.615 cm; use this graph for meat with a very
2 a 2.65  2.7 < 2.75 2.565 cm  Width < 2.575 cm
b 34.35  34.4 < 34.45 small mass that will take less than
c 4.95  5.0 < 5.05 b 9.246825 cm2  area <
d 1.05  1.1 < 1.15 9.308625 cm2 30 minutes to cook.
e −2.35  −2.3 < −2.25
f −7.25  −7.2 < −7.15 c 9.25 cm2  area < 9.31 cm2 5 a 10

3 a 131.5  132 < 132.5 Metres (in thousands) 8
b 250  300 < 350
Review c 402.5  405 < 407.5 5 a 511105787 km2  6
d 14.5 million  15 million < Surface area 511266084 km2
15.5 million 4
e 32.25  32.3 < 32.35 b 1.08652572 × 1012 km3  Volume
f 26.65  26.7 < 26.75 of Earth < 1.087036906 × 2
g 0.45  0.5 < 0.55 1012 km3
h 12.335  12.34 < 12.345
i 131.5  132 < 132.5 6 e smallest number of cupfuls is 0 5 15 20 25 30
j 0.1335  0.134 < 0.1345 426.4, and the largest is 433.6. Feet (in thousands)

4 250 kg  300 kg < 350 kg 7 maximum gradient = 0.0739(3sf) b 3600 (answer may vary +/− 100
minimum gradient = 0.06 foot.)
5 a 99.5 m  100 m < 100.5 m
b 15.25 seconds  15.3 seconds < 8 a 8.1 cm2  area of ∆ < 8.5 cm2 c 1050 m (answer may vary slightly
15.35 seconds b 5.76 cm  hypotenuse < 5.90 cm if answer to (b) varies from that
shown.)
6 4.45 m  L < 4.55 m 9 63.4°  x° < 63.6°

10 45.2%   45 × 100 < 46.7% 6a y
 98 50

11 332 kg  mean mass < 335 kg (3sf) Splooges in hundreds 40

ReviewExercise 13.5 12 117.36  number of 5s, < 117.84 30

1 a 30.8  a2 < 31.9 13 a Max = 232.875 20
Min = 128.625
b 13900  b3 < 14100 10
b i Max 5.32 and min 4.86
c 5.43  cd3 < 5.97 ii Only 1 can be used. e x
value of a is 5 to 1 sf. If we 0 5 10 15 20 25
d 609  (a2 + b2) < 615 find the maximum and
minimum values to 2 sf Squidges in hundreds
e 0.248 < c < 0.251 we get 5.3 and 4.9. is
b2 doesn’t tell us any more b 625 Squidges (answer may vary)
than the answer is 5 to 1 sf. c 224 000 Ploggs (answer may vary:

220, 000 – 228, 000)

f 2.66 < ab < 2.82 Exercise 13.7
cd
Exercise 13.6 1 $18.50
c b 2 $4163.00
g −43.5 < a − d 1 a 140 °F b 60 °F 3 £8520
c −16 °C d 38 °C 4 $384.52
< − 46.5 5 $2505.80

Review <  a ÷ c 2 a 4 lb b 4 kg
 d b 
h 2.66 c 36 kg d 126 lbs

< 2 82 e i correct

 a ii 18 lb = 8 kg Examination practice
 b 
i 48.9 < dc + iii 60 lb = 27 kg Exam-style questions

iv correct 1 a 4.116 × 103cm3  Volume of cube
< 4.038 × 103cm3
< 50.7 3 a $40 b $84
c £50 d £40

648 Answers

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e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity Answers

b 4.116 × 106mm3  Volume of cube c 10 2x + y = 10 e x = 7, y = 9 f x = 5, y = 3
< 4.038 × 106mm3 4 4 h x = 7, y = −6
8
g x = 6, y = 9 3 13
2 a 104 km/h 6 5 10
b 69 mph

3 a 129  (a + b) < 130 4 (3, 4) i x = −118, y = −5 j x = 29,
2 5x – 4y = –1 34 55 11 4
b 801  ab < 808

c 0.0529   a < 0.0534 y = 35
 b 12
–1 0 x
 1  12 5 k x = 1, y = −4 l x = −1, y = −4
 a 
d 122  b − < 123 m x = 5, y = −7 n x = −7, y = 3
32
ReviewPast paper questions* Solution is x = 3 and y = 4
o x = 3 , y = 29
1 249.5  j < 250.5 2 a x = −2, y = −2 b x = 3, y = 3 55
2 $2.20
3 6.1 cm c x = 3, y = −2 d x = −1, y = 6 4 a x = 3, y = 4 b x = 2, y = 4
4 95.5  l < 96.5 e x = 1 , y = −2 f x = 4 , y = 4 c x = −3, y = 5
5 10 e x = 3, y = 5 d x = 6, y = 3
7 33 g x = 5, y = 3
Chapter 14 i x = 2, y = 3
3 a x = 9 , y = −1 b x = 5 , y = −3 f x = 3, y = −4
Exercise 14.1 k x = −3, y = −2
h x = 2, y = 4
1a y m x = −1, y = 3
11 11 44 2 j x = −2, y = 1
10
c x = 7, y = 1 d x = 25, y = 22 o x = 5, y = 8 l x = 1 , y = 2
8 4 17 17 2

n x = −3, y = 4

4 a e scale can sometimes make it 5 a x = 209 y = −301
difficult to read off certain values, 12 , 80
such as fractions, accurately.
2x + y = 10 b x = −17.08, y = −65.05 (3dp)
b e equations must be solved
algebraically.

Exercise 14.2 c x = 0.015, y = −0.006 (3dp)

Review 6 1 a x = 2, y = 5 b x = 3, y = −2 d x = 112 , y = 504
(3, 4) c x = −10, y = 6 d x = 4 , y = −10 25 25
33 e x = 3, y = −2
4
2 x + 2y = 11 f x = −8, y = −2

x e x = −2, y = 4 f x = − 11, y = 17 g x = 6, y = −18
0 2 4 6 8 10 12 3
h x = −0.739, y = −8.217
Solution is x = 3 and y = 4
g x = 1, y = 1 h x = 19, y = 10 i x = 5.928, y = −15.985 (3dp)

by 22 17 17 6 a 90 and 30
b −14.5 and −19.5
4 2 a x = 4, y = 4 b x = 2, y = 6 c 31.5 and 20.5
c x = 1, y = 2 d x = 5, y = −1 d 14 and 20
3 x – y = –1 e x = 3, y = 4 f x = 1, y = 3
2 (1, 2)2x + y = 4 g x = 6, y = 3 h x = 5, y = 4 7 Pen drive $10 and hard drive $25
i x = 4, y = 3 j x = 4, y = 6
1 k x = 6, y = 6 l x = 4, y = 2 8 48 blocks (36 of 450 seats and 12 of
400 seats)

Review 3 a x = 2, y = 4 b x = 4, y = 3
c x = −5, y = −10 d x = 5, y = 5

–2 –1 01 2 x
3

Solution is x = 1 and y = 2

* Cambridge International Examinations bears no responsibility for the example answers to questions Answers 649
taken from its past question papers which are contained in this publication.

Copyright Material - Review Only - Not for Redistribution

Cambridge IGCSE Mathematicse-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity

Exercise 14.3 3 a t >91 b t > 109
4 4
1a x
–5 –4 –3 –2 –1 0 1 2 3 4 5 c t > 763 d r < 10
4 3

bx e d  − 305
–5 –4 –3 –2 –1 0 1 2 3 4 5 444

cp 4 Both give the answer 0.42
–3 –2 –1 0 1 2 3 4 5 6 7
Exercise 14.5

dy 1y

4

Review –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 3
2y – 3x ≥ 6
e q x
–8 –7 –6 –5 –4 –3 –2 –1 0 2 123 4
1
1

–3 –2 –1 0
–1

fx –2
–3

–6 –5 –4 –3 –2 –1 0 1

2y

4

gx 3

1.0 1.1 1.2 1.3 3.4 3.5 3.6 3.7 x + 2y < 4 2
1

h x –3 –2 –1 0 x
–3.4 –3.3 –3.2 –3.1 –1 123 4

2.8 2.9 3.0

–2

Review i k –3
–4.6 –4.5 –4.4
–3.2 –3.1 –3.0 3y

4

2 a {4, 5, …… 31, 32} b {8, 9, …… 18, 19} c {18, 19, …… −26, 27} 3
d {−3, −2, −1} e {−3, −2, − 1, 0} f {3, 4, …… 10, 11}
g {−6, −5, −4} h {4, 5, 6} i {3, 4} 2

Exercise 14.4 1 x–y≥0

–3 –2 –1 0 x
–1 123 4

c y  14 –2
15
1 a x<2 b x>3 d y > −2 –3
e c2 f x<4 g x<6 h p>3
i x > −15 j g4 l k< 7 4a y
k w<8 5
2 a y > 30 b q < 12 10
c g  11 d h < 19 4
e y  30 f x  −1 2 h y  −44 y > 3 – 3x
j v ≤ −13
i n < 48 g h−3 3 3
m e > 31 6 2
l k > 33 2
28 k z > 62
1

Review –5 –4 –3 –2 –1 0 x
–1 123 45

–2

–3

–4

–5

650 Answers

Copyright Material - Review Only - Not for Redistribution

University Answers

b y f Copy Copy Copy Copy 2
5
y 8y
5

4

3

2
1 –3 < x < 5
UniversitCyoPprye-ssC-aRmebrviidegwe4

3 7

2 6
5
1
–5 –4 –3 –2 –1 0 x 4 x+y=5
123 45
–1
x 3
–2 –10 –8 –6 –4 –2 0 246 8 10 2
–1
y=2
–2
–3 3x – 2y ≥ 6 1
–4 –3
–1 0
–5 –4 –5 –4 y=0 –1 x
–3 –2 123 45
cy –5
Review
6
5

4 g 3

3 x≤5 5y 8y
2
4 7
1 x=4
3
2 6
1
UniversitCyoPprye-ssC-aRmebr0vii≤ xde≤g2we–6 –5 –4 –3 –2 ––110 x 5
12345 6

–2

–3 x+y=5

–4 –10 –8 –6 –4 –2 0 x y=3 4
–1 2 4 6 8 10 3
–5 –2 2
–6 –3 R
–4
d –5 1

y –5 –4 –3 –2 –1 0 x
5 –1 123 45
4 y>3

3

Review 2 5 a above b below 4 y  − x + 4, y > 2x + 1, x  2

1 c above and below 5 (3, 0), (2, 0), (2, 1), (1, 1), (1, 2), (1, 0),
(0, 3), (0, 2)
–5 –4 –3 –2 –1 0 x 6 a y  4x + 5 b x+y<3
–1 123 45 c y ≥ 1x +1 d y ≤ −3 x
3
–2 2 6 y

UniversitCyoPprye-ssC-aRmebrviidegwe 5 +2
y=x
–3

–4 Exercise 14.6 y=4 4
3
–5 1

e y 2 3x + y = 4
7
y 1
7
6
x
6 5 –5 –4 –3 –2 –1 0 123 45
5 –1
4 x=4
–2
4 3 x + 2y = 6

3 –3
2
1 2 y=x –4
1
Review –10 –8 –6 –4 –2 0 –5
–1
–2 x –5 –4 –3 –2 –1 0 x (0, 4) (1, 4) (2, 4) (1, 3)
–3 2 4 6 8 10 –1 123 45

x + 3y ≤ 10

- Cambridge–4
ess - Review
–5

Answers 651

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Cambridge IGCSE Mathematics University

Exercise 14.7 5 10 y Copy Copy Copy Copy j x = 8 or 4
k x = 11 or −9
9 l x = 12 or −3
8 m x = 6 or 4
7 n x = 5 or 7
6 o x = −3 or 12
5
4 2 a x = 0.162 or −6.16
3 b x = −1.38 or −3.62
2 c x = −2.38 or −4.62
1 d x = −0.586 or −3.41
e x = 3.30 or −0.303
x f x = 3.41 or 0.586
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 g x = 7.16 or 0.84
h x = 2.73 or −0.732
Type A i x = 6.61 or −0.606
j x = 8.24 or −0.243
8 type A and 3 type B give 10m3 storage k x = 8.14 or −0.860
1 Greatest value: 3(6) + 2(6) = 30 UniversitCyoPprye-ssC-aRmTeybperviiBdegwe l x = −0.678 or −10.3
Least value: 3(−2) + 2(6) = 6
3 a x = 1.71 or 0.293
2a b x = 1.26 or −0.264
c x = 0.896 or −1.40
x+ 10 d x = −0.851 or 2.35
e x = −1.37 or 0.366
y = 9 y=x f x = 0.681 or −0.881
8
6 4 a x = 2.28 or 0.219
b x = 0.631 or 0.227
7 c x = 0.879 or −0.379
d x = 1.35 or −2.95
6 e x = −2.84 or −9.16
f x = 6.85 or 0.146
5
5 x = 1.61 cm (−5.61 is not a solution
4 because length cannot be negative)

3 6 a 4.53 metres b 248 months

Review 2 Exercise 14.8 Exercise 14.10

y=0 1 1 2 3 4 5 6 7 8 9 10 1 a (x + 3)2 + 5 b (x + 4)2 − 15 1 a (3x + 2)(x + 4)
c (x + 6)2 − 16 d (x + 3)2 − 4 b (2x + 3)(x − 1)
–4 –3 –2 –1 0 c (3x + 2)(2x − 1)
–1 e (x − 2)2 + 8 f (x − 1)2 − 18 d (3x + 8)(x + 2)
–2 e (2x − 5)(x + 2)
g ( + 5)2 − 21 h ( + 7)2 − 57 f (4x − 1)(4x + 9)
–3 24 24 g (3x + 1)(x + 5)
–4 h (4x − 1)(2x + 1)
UniversitCyoPprye-ssC-aRmebrviidegwe i (2x + 3)(x − 2)
b 2(6) + 0 = 12 i ( − 3)2 − 21 ( + 7)2 − 81 j (2x + 3)(x + 3)
24 24 k (3x + 8)(x − 2)
3 j l (5x − 3)(2x + 1)
m (5x + 1)(x + 1)
6 y  13 2 165 n (2x − 1)(x − 9)
 2 4 o (6x − 5)(2x + 3)
5 y=x+3 k x − −
4
l (x − 10)2 + 300

y=1 3 x≤5 2 a x = 0.74 or −6.74
2 b x = −0.54 or −7.46
c x = 3.41 or 0.59
1 d x = 1.14 or −6.14
e x = 2 or 1
x f x = 11.92 or 0.08
–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6

–1
3x + y = 6
Review –2

–3

–4 3 a x = 3.70 or −2.70
b x = 1.37 or −4.37
–5 c x = 0.16 or −6.16
d x = 1.77 or −2.27
–6 e x = 1.89 or 0.11
f x = 5.37 or −0.37
g x = 1.30 or −2.30
h x = 3 or −1
i x = 1.62 or –0.62
Greatest value = 5 UniversitCy NoumPpbreryoef f-lsagssC-aRmebrviidegwe
Least value = −1 Exercise 14.9
4
Review 1 a x = −3 or −4
200 y b x = −6 or −2
180 c x = −7 or −4
160 d x = −5 or 1
e x = −8 or 2
140 f x = 8 or −20
120 g x = 4 or 2
100 h x = 7 or −4
i x = 8 or −3
80
60
40
20

x
0 20 40 60 80 100 120 140 160 180 200

Number of T-shirts

120 T-shirts and 80 flags will
maximise income.
- Cambridge
652 Answers ess - Review

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e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity Answers

Exercise 14.11 e x y3 e 5(x y) 3x
x+4 f y +1 12 f2
1 As Exercise 14.10
x −6 x+5
2 a (3x − 7)(2x + 3) g x−4 h x −3 g 11y ha
b −(2x + 3)(x + 5) 8 40
c (2x + 3y)(2x + 3y)
d (3x + y)(2x − 7y) i8 j3 2 ia j 7x 18y
e (x2 − 9)(x2 − 4) = (x − 3)(x + 3) 32 2 63
(x − 2)(x + 2)
f 2(3x − 4y)(x − 5y) k x+3 l 23 2 a 19( 1)2 b 29pqr
g (3x + 2)(2x + 1) x+8 x +1 56 136
h (3x − 4)(x − 3)
i 3(x − 5)(x − 8) m 71 n 5y 4 c 93p d 71x
j (x − 1)(x − 2) x−4 y −7 70 84
k 4(x − 2)(x − 1) 33 − 5x
Review l (2x)(6x + 13) o3 7 62x2
54 e f 18

5a 3 4 b x2 + y2 63
7 1 d x2 + 1
f x3 y3 3a x+3 b 23
1 a 12a
Exercise 14.12 c x
19x 32
1a x by e1 c 6y d a2
2 4

c 5 d 10 Exercise 14.13 17 7
6x 5e
et u e f
6 f3
1a x2 b 3y2
gt hy c 4 14 4a 25 b 57
10 2 3z 2 (x + 1)(x + 4)
14 t2 (x 1)(x 2)
d
i 3z j 4t c 7 39 d 5
4 3 3 2x
(x + 2)(x 7)
e1 f1
2 a xy x 6
3 b 4y 7 f 2 + x2
3f gh2 e 6xy x
Review g 2e h
c1 dy
2 2 32

e 5x f 3b i2 j 1 g x2 + 2x + 5
2y2 2(x 1)

g 2x h 3b k 2d l r ( )x2 1 (27 y 14)
3y 7c 2 pq
h 63y2
2 1 3z2t 2 b 2xt
i 3de j 4b2 2a x3 3 2y x3
c 3 2x2 y
a 4xy d 64t 4 y4 i
5b 27
3a b ab

cb d ac e 3 y) f 1 4x2 y + 4xy − yz2 − z3
2 4 4(a b) j 12xyz2
4(x + y)5 (x
e abc 9b 1 2
2 f 4c ( )3 zt k x+3 l x+2
z2 t2
3y g 144 (x2 + y2 ) h z y
g (ab )2 h 4x Examination practice
Exam-style questions
Review 4x2z j9 Exercise 14.14
i 1 i 37
1 a 3y b 8t ii 5
3y 4 15

4 a 18 x3z c 12u dz
17z 3 b 35 14

2y

c 3v d x+3
7u2w 4 x+4

Answers 653

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Cambridge IGCSE Mathematics e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity

2a 10 a = 7 b = −69 c opp(65°) = q m or adj(25°)
24 opp(25°) = p m or adj(65°)
y hypotenuse = r m
11 x = 4 y = 0.5
556x + 6y
4 = y=x 1 Exercise 15.5
30 y = 1–2x +
Chapter 15
3 1 a 0.700 b 1.04
d1
2 Exercise 15.1 c 0.325 f 0.323
h0
e 0.279 b tan A = 3

1 1 6.8 m × 5.2 m g 0.00873 2
d tan x = 3
–1 0 1 2 3 4 5 6 x 2 a 3 cm b 2.4 cm 2 a tan A = 1
–1 b 15° 2 2
3 a 5.6 cm
C c tan A = 1 f tan C = a
Review b Greatest value for x + 2y = 8 2 Exercise 15.2 4
11 g tan D = p2
tan B = 4
(occurs at intersection of x = y and
5x + 6y = 30) 1a e tan x = n
m
Past paper questions* D

1 x > −9 tan y = m
n
2 (8, 2)
90 m 100 m 3 a 5.20 cm b 4.62 m
3 1, 2, 3, 4 c 35.7 m d 3.54 km
e 18 cm f 10.3 cm
4 23 −2x
12 A 75° 120 m 80° 4 a 20.8 cm b 16.1 cm
B c 9.17 cm d 7.85 cm
5 a i x  5, y  8, x + y  14, e 40.6 cm f 115 m
y  0.5x b BCD = 92°; ADC = 113° g 2.61 m h 95.8 km
c 80 m i 39.8 m
ii
2 a 20° b 3.4 m 5 a 1.0724 b 32.2 m
y 3 a 20 m b 34.8 m c 35°
6 32.3 m
Review 15
Exercise 15.3 7 a 1.73 b2

1 a 270° b 135° c 045° 8 0.45 m
2 a 262° b 135° c 230°
10 3 a 110° b 050° c 147 km 9 Adi is not correct, the pole is
5R e 280° 4.34 m tall.
d 025° b 288°
4 a 108° b 090° Exercise 15.6
5 a 9.6 km
1 a 40.4° b 60.0°
c 74.3° d 84.3°

Exercise 15.4 2 a 22° b 38°
c 38° d 70°
5 10 15 x 1

b i $480 Hypotenuse Opposite A Adjacent A 3 a a = 35.0° b b = 77.5°
ii 6 small boxes, 8 large boxes c c = 38.7° e f = 30°
ac ab d = 51.3°
h+4 by zx d e = 18.4°
6 h+5 cp qr
Review dl nm 4 71.8° (1dp)
7 (x −1) ec de 5 21.2° (1dp) b 26.7 (3sf)
3 fe fg 6 a 13.3 (3sf)

8 (x + 7) 7 AB = 6.32 (3sf)
(2x −1)(x + 2) ACB = 64.6° (1dp)
2 a opp (30°) = 5.7 cm
9 x = 1.58 or x = −0.380 b opp(40°) = x cm
adj(50°) = x cm

654 Answers * Cambridge International Examinations bears no responsibility for the example answers to
questions taken from its past question papers which are contained in this publication.

Copyright Material - Review Only - Not for Redistribution

e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity Answers

Exercise 15.7 13 a 1 b1 c1 2 AB = 13.856 cm (3dp)

1 d sin2x + cos2x = 1 3 a ABC = 59.0°
b AB = 1.749 (3dp)
abcde fg 14 a ACB = 45° b 2m c Capacity = 4.05 m3
c
C

sin A 4 7 12 20 8 4 13 45° 4 ABC = ACB = 38.9° and BAC = 102.1°
5 25 13 29 17 5 85

3 24 5 21 15 3 84 √2 m 1m 5 a 020° b 281.9 m
5 25 13 29 17 5 85 c 98 668 m2
cos A

tan A 4 7 12 20 8 4 13 6 a 3.5 m (1dp)
3 24 5 21 15 3 84 b DE = 6.1 m (1dp)
AB
2 a 0.0872 b 0.9962 1m 7 QT = 16 cm

Review c 0.5000 d 0.8660 d sin 45° = 1 8 a AOE = 72°
2 b AOM = 36°
e 0.8660 f 0.5000 c OM = 1.376 cm (3dp)
cos 45° = 1 d 0.688 cm2
g 0.9962 h 0.0872 2 e 6.882 cm2 (3dp)

3 a cos 42° = g b sin 60° = c tan 45° = 1
e a e y = 60°
g EG = 3 m
c cos 25° = RQ d sinθ = y hE f z = 30° 9 77.255 cm2
RP r
10 6.882a2 cm2
48° = q 30° = e
e cos r f sin f na2

HI x 30° 11  360° 
JI r 2m 2m  2n 
g cos 35° = h cos θ = tan
√3 m
4 a 0.845 m b 4.50 m Exercise 15.9

c 10.6 km d 4.54 cm 1 a −cos 60°
c −cos 44°
e 10.6 cm f 9.57 cm e −cos 92° b sin 145°
g sin 59° d sin 10°
g 14.1 cm h 106 cm D 60° F i cos 135° f cos 40°
1m h sin 81°
i 4.98 cm j 42.9 m G 1m j cos 30°

Review k 2.75 m l 137 m i sin 30° = 1
2
5 a 81.9° b 57.1°
c 22.0° d 30° cos 30° = 3 2 a 30, 150 b 90
b 44.9° 2 c 45, 315 d 78.7, 258.7
6 a 25.9° d 79.6° e 150, 210 f 191.5, 348,5
c 69.5° f 11.5° tan 30° = 1 g 109.5, 250.5 h 60, 240
e 26.9° 3 i 104, 284

7 1.93 m (2 d.p.) sin 60° = 3 3 a 45 b 120
8 a 10.1 km (3sf) 2
c 55 d 45
b 14.9 km (3sf) cos 60° = 1
9 a 14.1 m (3sf) 2 e 270 f 120

b 5.13 m (3sf) tan 60° = 3 g 270 h 90

10 552 m (3sf) i 696, 384

11 a x = 14.81 cm j Angle x sin x cos x tan x 4 30, 150, 210, 330
b y = 10.09 cm
31 5 41.4, 60, 300, 318.6
c AΔ = 44.99 m 23
d a = 29.52 cm 30° 1 Exercise 15.10
2
b = 52.80 cm
Review 60° 3 1 3 1 a 11.2 b 8.58
12 a i 0.577 ii 0.577 22 c 25.3 d 38.8°
ii 1.11
b i 1.11 ii −1.73 45° 1 1 1 2 a 10.6 cm b 5.73 cm
ii 0.249 22 c 4.42 cm d 5.32 cm
c i −1.73 e 6.46 cm f 155 mm

d i 0.249 sin x Exercise 15.8
cos x
∴ ta x = 3 a 54.7° b 66.8° or 113.2°

1 a ABC = 16.2° c 69.8° or 110.2° d 25.3° or 154.7°
b BC = 17.9 m
e 52.7° or 127.3° f 50.5°

Answers 655

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Cambridge IGCSE Mathematics University

4 C = 63° Exercise 15.13 Copy Copy Copy CopyPast paper questions*
AC = 15.9 cm
CB = 21.3 cm
UniversitCyoPprye-ssC-aRmebrviidegwe 1 a AC = 25 cm b EC = 13.0 cm 1 6.6 m
c 27.5° b AG = 75m
5 F = 25° 2 7.06
DE = 9.80 2 a EG = 50 m
EF = 14.9 cm c AGE = 35.3° 3 a 37.2°
b 11.7 cm2

6 R = 32.2° 3 a AC B = 53.1° 4 a 12.7 cm
P = 27.8° b 28.2°
QR = 7.0 cm b BC = 5 m
c CD = 4.2 m 5 a i 14.6 km
7 a Y is opposite a side shorter than X, d BM = 4.5 m ii North
so Y < X and therefore <40°. e BCD = 65°
North
Review b Y = 30.9° and Z = 109.1° 4 a 14.9 cm b 15.2 cm
c θ = 11.4°
c XY = 22.1 cm

8 a ACB = 51° 5 a AC = AB2 + BC2 C
b ABC = 52° A
c AC = 32.26 mm
b DA = DC2 − AC2 North

Exercise 15.11UniversitCyoPprye-ssC-aRmebrviidegwec DC = AD2 + AC2

1 AC = 8.62 cm d DAB = 90°  BD2 + DC2 − BC  DB
 2 × BD × DC 
−1 2 iii 260-264°
6 13.5
2 DE = 22.3 cm e BDC = cos
Chapter 16
3 P = 53.8° = cos−1  AD   AC 
 CD  sin−1  CD  Exercise 16.1
4 a 18.7 m f ADC or
b U = 32.1° 1 a Positive; weak
c T = 52.9° Examination practice b No correlation
c Negative; weak
5 a X = 60° b Y = 32.2° d Negative; strong
c Z = 87.8°
Exam-style questions 2 a+c
Relationship between width and
6 a Return = 14.4 km b 296° 1 AC = 9.8 m, BC = 6.9 m length of leaves
Review
7 51.2m on a bearing of 273 2 DAB = 47.9° Relationship between width
and length of leaves
Exercise 15.12 3 9.9 m
200
1 a 10.0 cm2 4 a X = 10.1m (to 3sf) b y = 20.6° 180
UniversitCyoPprye-ssC-aRmebrviidegwe 160
b 15.0 cm2 5 a i QX = 60 tan 4° = 50.3 m 140
c 52.0 cm2 ii 78.3 m 120
d 17.2 cm2 100
e 22.7 cm2 b i 250.3 m ii 257.4 m
f 24.2 cm2 iii 077° 80
60
6 a 5.16 m b 3.11 m2 40
20
2 108 cm2 7 a 7 cm b 51.1°
0 10 20 30 40 50 60 70 80
3 0.69 m2 8 a (90°, 1) b −1 Length (cm) Width (cm)
c
4 42.1 cm2 b Strong positive correlation.
d 40 cm
5 a 30.6 cm2 y
1

Review b 325.9 cm2 0.5
c 1.74 m2
0 x
30 60 90 120 150 180 210 240 270 300 330 360
–0.5
6 a 174 cm2 –1 y = – 1 y = sin x°
2

b 8.7 cm and 21.5 cm d 2 solutions
- Cambridge
ess - Review
7 a Q = 22.6° b P = 53.1° 9 a i AB = 107.3 km

ii PAB = 66.6° iii 143.4°

b i 5h ii 12 km/h

656 Answers * Cambridge International Examinations bears no responsibility for the example answers to
questions taken from its past question papers which are contained in this publication.

Copyright Material - Review Only - Not for Redistribution

University Answers

3 a Relationship between mass of dog c Copy Copy Copy Copy
and duration of morning walk
Scatter diagram showing the relationship
Mass (kg)/duration of morning walk (mins) between time watching TV and maths score
40 100
35 80
30 60
25 40
20 20
15
10 0
5 0 50 100 150 200 250
Time spent watching TV (min)
0 20 40 60 80UniversitCyoPMparsyseof-sMdaotsghCs(-ksgac)oreRm(%eb)rviidegwe
Review Duration of work (min) d 105 mins
e No way of knowing how accurate the estimate is as performance in
b No correlation
c e dogs are not a specific breed. test is affected by many factors.
4 a+c
Examination practice
Relationship between number of
assistants and queuing time Exam-style questions

300

250

200

150
UniPrivWcaeieti(rn)gstiitmeCy(seocoPpnrdsy)e-ssC-aRmebrviidegwe1 a+c

Relationship between price ( ) and area

12 000

10 000

8000

Review 100 6000

50

4000

2000

0
0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
Area (m2)

b Painting E because other paintings of a similar size are much cheaper.
d $6400
e Value is outside the range of the collected data.
UniversitCyoPprye-ssC-aRmebrviidegwe0 10 20 30 40
No. of sales assistants

b Strong negative correlation.
d Value is outside the range of the

collected data and waiting time
will be negative time!

5a

TV 122 34 215 54 56 78
watching (min)

Review Maths 64 92 30 83 76 78
score (%)

224 236 121 74 63 200
41 28 55 91 83 27

b Strong negative correlation.
- Cambridge
ess - Review Answers 657

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Cambridge IGCSE MathematicsUniversity

2 a+c Copy Copy Copy Copy4 $2085.75
5000 5 $474.30
Review Comparison of 1st and 2nd year maintenance 6 $8250
UnivReeprairssiitnCsyecooPnpdryyeea-rs (ys)C(-aminuRtmeseb)rviidegwe 7 Annie
4000 $319.20
Bonnie $315.00
3000 $300.30
Connie $403.20
$248.85
Donny

Elizabeth

8 All amounts in R million (3sf).

2000 per per 35 % tax per month

year month per year after tax

1000 ab c

(f ) 87.9 7.33 30.8 4.76

0 20 40 60 80 100 120 140 86.1 7.18 30.1 4.66
Maintenance hours (x)
85.1 7.09 29.8 4.61
b Strong negative correlation.
d 1 600 minutes 66.9 5.58 23.4 3.62
e Repair time is a negative number − value is outside the range of the

collected data.
f Approximately 130 hours – this is an extrapolated value so might not be

accurate.
URneivvieerwsitCyoPprye-ssC-aRmNheuobtmrvbmiieeralodsefgwHeieght (cURm)neivvieerwsitCyoPprye-ssC-aRmebrviidegwe 66.8 5.57 23.4 3.62

59.5 4.96 20.8 3.22

51.9 4.33 18.2 2.81

51.5 4.29 18.0 2.79

49.9 4.16 17.5 2.70

49.7 4.14 17.4 2.69

Past paper questions* 2 a and b d Bernard Fornas earned 4.48 × 10−2
million R (3sf) and Alan Clark
1 a and b 200 earned 2.54 × 10−2 million R (3sf)

40 192 Exercise 17.2
35
30 184 1 Employee a Net b%
25 income  net 
20 176  gross
15 ($)
10 168
5 B Willis 317.00 47
160
0 5 10 15 20 25 30 M Freeman 158.89 35
Temperature (°C) 152
J Malkovich 557.20 43
c strong negative 144
H Mirren 383.13 42
136
M Parker 363.64 43
128
120 2 a Mean weekly earnings: $ 836.63
b Median weekly earnings: $ 853.30
26 28 30 32 34 36 38 40 42 44 c Range of earnings: $ 832.50
Shoe size

c strong positive

Unit 5

Chapter 17 3 a Difference between gross and net
income:
Exercise 17.1 M Badru: 3954.52
B Singh: 724.79
1 $49.50
2 $332.50 b Percentage of gross income that
3 a $13.50 each takes home as net pay:
M Badru: 69.3%
c 9.35 B Singh: 57%
e $13.68
- Cambridge b $ 6.45
ess - Review d $12.15

658 Answers * Cambridge International Examinations bears no responsibility for the example answers to
questions taken from its past question papers which are contained in this publication.

Copyright Material - Review Only - Not for Redistribution

University Answers

Exercise 17.3 Exercise 17.4 Copy Copy Copy Copy3 4 years

UniversitCyoPprye-ssC-aRmebrviidegwe1 1 4 7% p.a.
Taxable income
a $98 000.00 Annual tax Monthly tax Principal Interest Time Interest 5 33 years 4 months b $96
b $120 000.00 $17 640.00 $1470.00 amount ($) rate (%) invested earned ($) ii $136.80
c $129 000.00 $21 600.00 $1800.00 6 a $32
d $135 000.00 $23 220.00 $1935.00 500 1 3 15.00 c i $40.80 b £3700
e $178 000.00 $24 510.00 $2042.50
$35 260.00 $2938.33 650 0.75 2.5 12.19 7 a $11 700
c 15.4% (1dp)
1000 1.25 5 62.50
Exercise 17.5
1200 4 6.75 324.00

2 a i Yes 875 5.5 3 144.38 1 a $100 b $60 c $460
ii No – he pays $6181.25 2 $2850
iii $6181.25 = $4681.25 + (40 000 900 6 2 108.00 3 a $141.83 b $2072
– 34 000) × 0.25 4 a ₤301 b 33.5% (1dp)
Review 699 7.25 3.75 190.04 5 a $3657.80 b 13.09% (2dp)
b $67 616.75
c i She owes additional tax. 1200 8 0.75 72.00

ii $238.25 150 000 9.5 1.5 21375.00

2 Exercise 17.6

UniversitCyoPprye-ssC-aRmebrviidegwe3 a Value-Added-Tax:Principal Interest Time Amount 1 a $10 035.20 b $9920.00
VAT is paid at each step in the amount ($) rate (%) invested repay ($) $5077.92
business chain. For the buyer it 2 a $4998.09 b
is the tax on the purchase price
but for the seller it is the tax on 500 4.5 2 545.00 3 $88 814.66
the ‘added value’ part of the price.
Rate/s at which charged vary from 650 5 2 715.00 4 $380 059.62 (2 dp)
country to country.
1000 6 2 1120.00 Exercise 17.7
b General sales tax:
Sales tax is paid only at the end 1200 12 1.5 1416.00 1 a 7.255 billion
of the consumer chain by the b 7.675 billion
consumer. Rate/s at which charged 875 15 1.5 1071.88 c 8.118 billion
vary from country to country.
900 15 3 1305.00 2 a 1724 pandas
c Customs and Excise duties: b 1484 pandas
Customs duties are taxes on 699 20 0.75 803.85
imported goods. Excise duties are
taxes on goods produced for sale, 1200 21.25 0.67 1370.85
or sold, within a country. Rate/s at
Review which charged vary from country 150 000 18 1.5 190500.00
to country.
3 a Time (days) 012345678
d Capital Gains Tax:
Capital gains tax is paid on the Total number of 1 2 4 8 16 32 64 128 256
profit made on the sale of assets.
Rate/s at which charged vary from microbes (millions)
country to country.

e Estate duties:
ese are taxes levied on people

who inherit money, property,
etc. Rate/s at which charged
vary from country to country.
UniNvumebrersofitmiCycroobePspr(ymiell-isonss)C-aRmebrviidegwe by

Review 60
50
40
30
20

10

- Cambridge 0 1 2 2.5 3 3.5 4 5 6 7 8 x
ess - Review Time (days)

Answers 659

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Cambridge IGCSE Mathematicse-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity

c i approximately 5.5 million ii approximately 12 million 8 11%

d just over 4 days Past paper questions*

4 a 6.5 minutes b 12 grams 1 $3826.38
2 $460
5 $27 085.85
Chapter 18
6 a $10 120 b $8565.57 c $5645.41 d $11 000(0.92)n
Exercise 18.1
7 $2903.70
1
8 a 7 137 564 b 10 years
x −3 −2 −1 0 1 2 3
9 15 hours a y = x2 + 1 10 5 2 1 2 5 10
b y = x2 + 3 12 7 4 3 4 7 12
Exercise 17.8 c y = x2 − 2 7 2 −1 −2 −1 2 7
d y = −x2 + 1 −8 −3 0 1 0 −3 −8
1 Cost price ($) Selling price ($) Profit ($) Profit (%) e y = 3 − x2 −6 −1 2 3 2 −1 −6
5.00 25.00
Review a 20.00 25.00 50.00 10.00
0.30 20.00
b 500.00 550.00 0.05 16.67

c 1.50 1.80

d 0.30 0.35

2 Cost price ($) Selling price ($) Loss ($) Loss % (b) 12 (a) y x2 1
400.00 300.00 100.00 25.00
a 0.75 0.65 13.33 (a) 10 (b) y x2 3
b 5.00 4.75 0.10 5.00 (c) y x2 2
c 6.50 5.85 0.25 10.00
d 0.65 (d) y x2 1
8 (e) y 3 x2
3 Percentage profit = 66.67%
(c)

6

Exercise 17.9 2 4

Review1 a $108.33 b $256.00 Original Sale price % discount 2
c $469.41 d $1125.00 price ($) ($)
11
2 $840 89.99 79.99 5 -3 -2 -1 0 123
125.99 120.00 25 -2
3 $3225 599.00 450.00 18
22.50 18.50 10
4 $360 65.80 58.99 5 -4
10 000.00 9500.00 (e) -6
5 $220.80 (d) -8

6 $433.55 for 10 and $43.36 each

7 28% Examination practice

8 a $67.38 b 60% f When the value of the constant
term changes the graph moves up
Exercise 17.10 Exam-style questions or down the y-axis.

1 1 a $366.56 b 9 hours 2aC bB
cA dD
2 a $12 b $14.40 eE
Original % Savings Sale
Review
price ($) discount ($) price ($) 3 7.5% Exercise 18.2

89.99 5 4.50 85.49 4 $33.60

125.99 10 12.60 113.39 5 $635 1x −1 0 1 2 3

599.00 12 71.88 527.12 6 a $30 000.00 b $2 977.53 y = x2 − 2x + 2 5 2 1 2 5
c $2 307.59
22.50 7.5 1.69 20.81

65.80 2.5 1.65 64.16 7 28.07%

10 000.00 23 2300.00 7700.00

660 Answers * Cambridge International Examinations bears no responsibility for the example answers to
questions taken from its past question papers which are contained in this publication.

Copyright Material - Review Only - Not for Redistribution

e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity Answers

y 3 5
y = x2 –2x + 2
x −3 −2 −1 0 1 2 x −6 −5 −4 −3 −2 −1 0
6 y = x2 + 2x − 3 0 −3 −4 −3 0 5 y = −x2 −5 0 3 4 3 0 −5
5 − 6x − 5
4 y
3 5 y = x2 – 6x – 5 y
2 y = x2 + 2x – 3 5
1 4 4
3
3

2

1 2

Review x 1
123 4
–4 –3 –2 –1 0
–1 0 1 2 3x –1 –8 –6 –4 –2 0 x
–1
–2

2 –3 –2

x −2 −1 0 1 2 3 4 5 6 –4 –3
x2 4 1 0 1 4 9 16 25 36
−5x 10 5 0 −5 −10 −15 −20 −25 −30 –5 –4
−4 −4 −4 −4 −4 −4 −4 −4 −4 −4
y = x2 − 10 2 −4 −8 −10 −10 −8 −4 2 4 –5
5x − 4
x 01234 6 a 6m
y = −x2 − 4x 0 −5 −12 −21 −32 b 2 seconds
c 3 seconds
y y y = –x2 – 4x x d 4.5 m
12 00 1 2 3 4 e e water surface is at h = 0.
–5
Review 10 –10 Exercise 18.3
y = x2 –5x – 4
–15 1a
8 –20
–25 y
6 –30 3
2
4 –35
1
2
x

–5 –4 –3 –2 –1–1 12345
–2
–4 –2 0 246 x –3
–2 8 –4
–4 –5
–6 –6
–8
Review
–10
–12

Answers 661

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Cambridge IGCSE Mathematics e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity

b e 3a
y
y y
6
10 6
8 5 4
6 4
4 3 2
2 2
1 –10 –8 –6 –4 –2 2 x
–10 –8 –6 –4 –2–2 –2 4
–4 –6 –5 –4 –3 –2 –1
–6 2 4 6 8 10 x –1
–8 –2
–10 –4
f
–6
y
Review x –8

–10

c –12
y
–14
5 4
4 3 b
3 2 y
2 1
1 2.5

–6 –5 –4 –3 –2 –1 2.0
–1
x 1.5
d
y –5 –4 –3 –2 –1–1 12345 1.0
–2
x –3 0.5
–4
Review –5 –2.5 –2.0 –1.5 –1.0 –0.5 x
–6 –0.5 0.5 1.0

g –1.0

6 y –1.5
5
4 8 –2.0
3
2 7 –2.5
1 c
6
–6 –5 –4 –3 –2 –1 y
5 4

4 3

x 3 2

Review 2 1

1 x –5 –4 –3 –2 –1 x
–1 12
–7 –6 –5 –4 –3 –2 –1
–1 –2

2 a y = −x2 − 4x + 5
b y = 4 − x2
c y = x2 − 3x − 4
d y = x2 − 2x − 3

662 Answers

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e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity Answers

d g y
y y 2

4 2.0 1 2
3
1
2 1.5 1 3

1

x 1.0 2
1 3
–6 –5 –4 –3 –2 –1 1
–2 0.5
3x

–4 x –6 –5 –4 –3 –2 ––1310 123456
0.5 1.0 1.5 2.0 2.5 –32
–6 –1.0 –0.5 –1
–0.5

–8 –1.0 –131
h –123
Review –10
y –2
e
y b

30 x

–5 –4 –3 –2 –1 x −5 −4 −3 −2 −1 1 2 3 4 5
–1
y = −1 0.2 0.25 1 0.5 1 −1 −0.5 −1 −0.25 −0.2
25 x 3 3

–2

20

–3 y

15

–4 1.00 y = –1
0.75 x
10 0.50
0.25
–5
–5 –4 –3 –2 –1 0
5 0.25
0.50
–6 0.75
–1.00
x 4 a (20, 0) x
2 4 6 8 10 b 0  ×  20
–4 –2 c −10  h  0 12345
–5
d
Review –10 y

–15 0x c
f
10 20 30 40 50 x −6 −4 −3 −2 −1 1 2 3 4 6
y –2
5

4 –4 y = −6 1 1.5 2 3 6 −6 −3 −2 −1.5 −1
x

3 –6

–8 y
6

2 5 y = –6
x
–10
1 4

–12 3
e width = 40 m
x f max height = 10 m 2
12345
–2 –1 Exercise 18.4 1
–1 –6 –5 –4 –3 –2 –1 0 x
1a 123456
Review –2 –1

–3 –2

–3

–4 x −6 −4 −3 −2 −1 1 2 3 4 6 –4

y= 2 −1 −0.5 −2 −1 −2 2 1 2 0.5 1 –5
x 3 3 33
–6

Answers 663

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Cambridge IGCSE Mathematicse-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity

d c 3a

x −6 −4 −3 −2 −1 1 2 3 4 6 y x 20 40 60 80 100 120
12 y 12 6 4 3 2.4 2

y = 4 −2 −1 −1 1 −2 −4 4 2 1 1 1 2 10
x 3 3 3 3 8

y b
4
6 y 240
14 x
4 y =

3 2

y= 4 –2.0 –1.5 –1.0 –0.5 12
x –2
x
2 0.5 1.0 1.5 2 10

1 8

Review–6 –4 –2 0 x –4 6
–1 246
d 4
y

6

2

–2 5 0 20 40 60 80 100 120 140x
4 0

–3 3 240
x
2 c y =
1
–4 4a
y –3 –2 –1
2a –1 x x −4 −3 −2 −1 − 1 1 1 2 3 4
8 1234 22

y 1 1 1 1 4 41111

6 –2 16 9 4 4 9 16

Review 4e b
y y

28 6

–8 –6 –4 –2 x 6 5
–2 2468 4

–4 4
2

–6 3
x

–8 –15 –10 –5 5 10 15 2
b f –2
1
y y
8
x
6 –4 –3 –2 –1 123 4
–1
6
4

4 c Graph is still disjoint but both
curves are above the x-axis on
Review 2 opposite sides of the y-axis.

2 d Division by 0 is meaningless
x e y = 0 (the x-axis) and x = 0

x –15 –10 –5 5 10 15 (the y-axis)
f x = 0 and y = 3
–8 –6 –4 –2 2468 –2
gi
–2
–4

–4

–6

–6

–8

–8

–10

664 Answers

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e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity Answers

y 3 a x = −1.3 and x = 2.3 b i x = −1.2 and x = 3.2
0.5 ii x = 0 or x = 2
9y

x 5a y
5
–6 –4 –2 246 8
–0.5 4 y = x2 – 2x – 4

–1.0 7 3

–1.5 6 2

–2.0 5 1

–2.5 4 –3 –2 –1 0 x
3 y = x2 – x – 3 –1 1 2345

–3.0 –2

Review ii 2 –3

y –4
9

8 1 –5
–6
7 –3 –2 –1 0 x
–1 1 2 34
(Students, graph should also include
6 the points (−3, 11) and (5, 11)
b i x = −1.2 and x = 3.2
5 –2 ii x = −1.8 and x = 3.8
–3 iii x = −1 and x = 3
4
Exercise 18.6
3 –4

2 b x = −2.6 and x = −0.4 1 a x = 2 and x = −1
b x = 2 and x = −2
x 5y c x = −2 and x = 1
–4 –3 –2 –1 0 1 2 3 4 y = x2 + 3x + 1 d x = 1.2 and x = −0.4

ReviewExercise 18.5 4 x 2 Students’ own graphs
3 1 a (0, 0) and (3, 9)
1 a x = −1 and x = 2 b (−1.4, −1.4) and (1.4, 1.4)
b x = −2.4 and x = 3.4 2 c (2, 0)
c x = −2 and x = 3 1
3 a x = 9.1 and x = 0.9
2a –4 –3 –2 –1 0 b x = −2 and x = 4
–1 c x = 3.8 and x = −1.8
x −3 −2 −1 0 1 2
y = −x2 − x + 1 −5 −1 1 1 −1 −5 –2 4 y
11
b –3
–4
y
4a 6y 10
y = 4 – x2 + 2x 9 y = x2 + 2x + 3

5 8

1 y = –x2 – x + 1 4 7

–3 –2 –1 0 123 x 3 6
–1 4 5
2
4

Review –2 1 3
2
–2 –1 0
–3 –1 x 1 x
–2 –5 –4 –3 –2 –1–10 1 2 3 4 5
1 234

–4 –2

–5 –3 –3
–4 y = –4
c x = −1.6 and x = 0.6
–4

–5

ere are no points of intersection.

Answers 665

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Cambridge IGCSE Mathematicse-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity

Exercise 18.7 ey

1 x −3 −2 −1 0 1 2 3 10 y = x3 – 2x2
y = 2x3 −54 −16 x
a y = −3x3 −2 0 2 16 54 –3 –2 –1 0
b y = x3 − 2 81 24 –10 123
c y = 3 + 2x3 −29 −10 3 0 −3 −24 −81
d y = x3 − 2x2 −51 −13
e y = 2x3 − 4x + 1 −45 −16 −3 −2 −1 6 25
f y = −x3 + x2 − 9 −41 −7
g y = x3 − 2x2 + 1 1 3 5 19 57 –20
h 27 3
−44 −15 −3 0 −1 0 9 –30

3 1 −1 9 43 –40
–50
−7 −9 −9 −13 −27

Review −2 1 0 1 10

fy

50

ay cy 40

60 40
30
40 y = 2x3 20 30
10 y = 2x3 – 4x + 1
20 y = x3 – 2
–3 –2 –1 0 20
–10 x
–3 –2 –1 0 x –20 123 10
–20 123 –30
–40 –3 –2 –1 0
–10 x
123

–40

–60 –20

by –30

100

Review 80 –40

60 dy –50
y = –3x3
60
40
50 gy

20 x 40 40
123 y = 3 + 2x3
–3 –2 –1 0 30
–20 30 y = –x3 + x2 – 9

20

20

–40 10

–60 10 x
123
–80 –3 –2 –1 0 x –3 –2 –1 0
–10 1 23 –10

–100

–20

–20

–30

Review –30 –40
–40

–50

–60

666 Answers

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e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity Answers

hy y (Please note, only part of the graph is
10 shown here).
10 y = x3 – 2x2 + 1
y = –x2 + 6x b x = 0 and x = 4.2
8
–3 –2 –1 0 x 6
–10 123
4

2

–20 –10 –8 –6 –4 –2 0 x
–2 2 4 6 8 10

–30 –4
–6
–40 y = x3 –8
10

–50

2a –10

Review x −1 −0.5 0 0.5 1 Exercise 18.8
y = x3 − 6x2 + 8x −15 −5.6 0 2.6 3
1x
−3 −2 −1 −0.5 −0.2 0 0.2 0.5 1 2 3

1.5 2 2.5 3 3.5 4 4.5 5 a y = 3 + x2 − 2 12.7 8 6 7.3 13.0 Ν/Α −7.0 −0.8 2 6 11.3
1.9 0 −1.9 −3 −2.6 0 5.6 15 x

b y = 3x − 1 −8.7 −5.5 −2 0.5 4.4 Ν/Α −4.4 −0.5 2 5.5 8.7
x 11.3 5 0 −3.3 −9.8 Ν/Α 9.8 3.8 2 3 6.7

b c y = −x + x2 + 2
x
y
20 d y = −x3 − 2x + 1 34 13 4 2.1 1.4 1 0.6 −0.1 −2 −11 −32

15 Note: e y-values are rounded to 1 decimal place.

10 a y = 3 + x2 − 2 b y = 3x − 1 c y = −x + x2 + 2 d y = −x3 − 2x + 1
x x x
y
5 y 12 y 40
y = x3 – 6x2 + 8x 25 10 10 30
–2 –1 0 x 20 8 y = 3x – 1
–5 x
12345 6
Review 15 8 20
6 6 10 y = –x3 – 2x + 1

10 x
4 –3 –2 –1 0 1 2 3
5
–10 2 4

–15 –10 –5 0 5 10 15 x 2 y = –x + x2 + 2 –10
3 x –20
–15 –5 + x2 – 2 –3 –2 –1 0 1 2 –1 0 1 2 3 x –30
y= x –2 –40
–3 –2
–10 –2

–20 –15 –4

–6 –4

c i x = 0, x = 2 and x = 4 –8 –6
ii x = 0.7, 1, and x = 4.3
–10 –8
3a
–10

x −4 −3 −2 −1 Exercise 18.9 2a2 b 0.8

y = x3 −6.4 −2.7 −0.8 −0.1 1ab c
10
10 y

y = 6x − x2 −40 −27 −16 −7 y 8 y = 10x
30 6

25

Review 0123456 y = 3–x 20 y = 3x 4
0 0.1 0.8 2.7 6.4 12.5 21.6 15
0589850
10 2

5 –0.2 0 0.2 0.4 0.6 0.8 x
1
10x = 8 − 5x when x = 0.67
–4 –3 –2 –1 0 x (0.66 − 0.68 also fine)
12 34

–5

c e two graphs are symmetrical

about the y-axis.

Answers 667

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Cambridge IGCSE Mathematics University

Copy
3 a Instructions will vary, but should 4a2 b 5.3 hours 6a
include determine whether the c 64 d 20 hours
graph is increasing or decreasing
using the value of a. If a is positive 5
the graph is decreasing, if a is 400
negative, the graph is increasing.
Use a + q to determine the
y-intercept. Work out the
asymptote by finding the line y = q.
If a < 0, the graph is below the
asymptote and if a > 0, the graph is
above the asymptote.
UniversitCyTeomPppry(e°-sC)sC-aRmebrviidegwe 18000

16000

14000

12000

Population 10000

300 8000

6000

4000

200 2000

0 12 3 4 5
0 Time (months)

Review bi y 100 b 3.25 months
c 64 000
2

–3 –2 –1 x Copy
–2 123

0 1234

Time (min)
UniversitCyoPprye-ssC-aRmebrviidegwe–4

–6

–8

–10 Exercise 18.10

ii 1a y
y 9

6 y = x2
8

7

4 6

Review 2 5

–3 –2 –1 x 4 (i) Copy
–2 123
(ii) 3
2

UnivePorpsuliattiCoynoPprye-ssC-aRmebrviidegwe–4 1

–6 –3 –2 –1 0 x
iii 123

y i 4 ii −1.75
1.0 b (−1.5, 2.25)

0.5 2a
y

300

–3 –2 –1 x 200
–0.5 123

Review –1.0 100

–1.5 Copy

–2.0 0x
1930 1940 1950 1960 1970 1980 1990 2000 2010

Year

e gradient at point (1950, 170) is −4.4 people per year.
b Rate of change of population in the village in 1950.
- Cambridge–2.5
ess - Review
–3.0

668 Answers

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e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity Answers

3a Exercise 18.13 h (−2, 15) max and (2, −17) min

y 1a5 b −4 i (0, 3) min and (4, 35) max
10 c0
e −3 d7 j (2, −4) min k (0, −25) min
g 21x2 + 2 f 8x − 4
8 im h x2 + x l  3, − 89  min
y = x3 + 1 4

6 m (3, 81) max and (4, 80) min

2 a 2x + 2 b 5x4 + 8x3 n (0, 0) min and (2, 4) max
c 2x − 1 d 2x − 9
4 e 16x3 + 24x2 f −10x + 20 2m
2A g 4x + 5 h 6x − 7
i 24x + 23 j 12x − 13 y
k 42x − 44 l 2x + 6
Review–3 –2 –1 0 x m 8x + 4 n 18x − 12 (3,81)
–2 123 o 3x2 + 6x p 14x6 + x5
55 (4,80)
3 0 x
q 10x − 20
–4 r 2x

3 67

–6  2 13 n
3 y
4 ,

–8

b3 5  1 , −3
3
Exercise 18.11 (2,4)
6 (1,5) and (2, −4) x
1 a 4x3 b 6x5
c 9x8 d 12x2 7 (2,11) and (−2,5) 03
e 24x f 49x6
g −16x3 h 84x11 8 a = 2, gradient at x = 4 is 92. dy at
i −80x4 dx
x = −3 is 50 dh
dt
Review Exercise 18.14 3 a = 7 − 10t

2a6 b3 1 a y = 6x − 9 b 2.45m
c 32 d −8 b y = −4x − 4
e −108 f 960 c y = 56x − 144 4 a 54 thousand
d y = 18.25x - 19.25
5 a Length = 2 − 2x and

3 (3, 27) width = 1 − 2x

Exercise 18.12 e y = 9x −1 V = length x width x
20 16
depth = x(2 − 2x)(1 − 2x)

1 a 4x3 + 5x4 b 9x2 − 20x3 b e width is only 1m and we are
c 42x5 + 18x d x2 − 28x6
2  34 , 0 subtracting two lots of x from this
19
e 30x4 − 32x3 f −14x + 18x5 length. So we can only subtract
11
34 something less than 0.5

g 36x2 + 16x7 h −120x11 − 80x9 c x = 0.211, V = 0.192
3
4 a  26 , 2 Examination practice
i 8x − 36x2 + 20x3 9
Exam-style questions
j − 32 x3 + 6 x2 − 3 x Exercise 18.15
11 7 2 1 a A: x = −2
Review
2 a 93 b 52 c 12 1 a (2, −3) min b (−3, −13) min B: y = −x
c (4, 14) max d (2, −8) min
3 (1, 5) and (−2, −4) e (1, −1) max C: y = x2 − 2
4 (0, 0) or ( 3, −9)
D: y = 2x + 1
4
f  − 3 , − 143 min b i (−2, 2)
2
ii (3, 7) and (−1, −1)

g  3 , 89  max c  − 1 , 1
10 20  3 3

dD eC

Answers 669

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Cambridge IGCSE Mathematicse-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity

2a 4 a vi b ii
x −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 d iv
ci p = 160, q = 10, r = 2.5
y 7 5.25 4 3.25 3 3.25 4 5.25 7
5a i
b y ii
9
y = x2 8 M
160

7 y = x2 + 3 140

6

5 120

4 100

3

Review 2 80

1 60

x 40
–3 –2 –1 0 1 2 3

c No, x2 will never equal x2 + 3 20
d i x = +2.4 or −2.4
0 t
ii x = +1.7 or −1.7 0123456 7

3 a i p = −10. iii Rate of change = 28.2
b t=1
ii q = 6.3

iii r = 9.2 Past paper questions
b
1 a i x −2 −1 0 1 2 3 4 5
x 0.6 1 1.5 2 2.5 3 3.5 4 4.5 5 y −5 1 5 7 7 5 1 −5
y −10 −5.9 −3.7 −2.3 −1.1 0.3 1.9 3.8 6.3 9.2

Review y y
10 9
8
8 7 (a)(ii) (c)(i)
6
6 5
4
4 3 (d)(i)
2
2 1 x
45
x
0 12345
–2

–4

Review –6 0 123
–8
–10 b x = −1.2, x = 4.2 c ii x = 1.5

c x = 2.9 d ii 1 iii y = x + 2

d Gradient = 6 2 a x −1.5 −1 −0.5 0 0.5 0.75 1 1.5 2
f(x) −24.9 −3 7.4 10 8.6 7.6 7 8.9 18

670 Answers

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e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity Answers

by c x = −0.68
d i a = −1 b = 2.5
20
ii x = −0.5
15
e 0.71

10 Chapter 19

5 Exercise 19.1

–1.5 –1 –0.5 0 x 1 a None
–5 0.5 1 1.5 2 b CD, HG
c CD, HG
Review –10 d AB
e AB, EF
–15 f AB, CD
g CD
–20 h AB, CD, GH Number of lines
of symmetry
–25 2
Shape 4
c i e.g. 5 or 0 or −5 ii e.g. 9 2
d x = 0.445 or x = −1.35 e 7.74 Square 3
Rectangle 1
3 a x −2 −1.5 −1 −0.75 −0.5 0.5 0.75 1 1.5 2 3 Equilateral triangle 0
y −1.75 −1.06 0 1.03 3.5 4.50 2.53 2 1.94 2.25 3.11 Isosceles triangle 1
Scalene triangle 0
Review by Kite 2
Parallelogram 5
5 Rhombus 6
Regular pentagon 8
4 Regular hexagon
Regular octagon

3

3

2

1 4

Review –2 –1 0 x
–1 123

–2

Answers 671

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Cambridge IGCSE Mathematicse-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity

5 Students’ own answers but might Exercise 19.3 4 a 17.3 cm b 4.25 m
include names such as Audi, Citroën, c 31.1 mm
Suzuki, Honda and Toyota. 1
5 13.5 cm

Exercise 19.2 6 AO = 9 cm
Area AOCB = 108 cm2
1a2
b5 7 x = 43°
c2
d6 Exercise 19.6
e2
f1 1 a x = 43°, y = 43°, z = 94°
g1
h1 b x = 124°, y = 34°

c x = 35° d x = 48°

Review2b 2 a x = 41.5° b x = 38°

Regular Lines of Order of 3 a Tangents subtended from the same
polygon symmetry rotational
symmetry point are equal in length.
Triangle 3 b i CAB = 70°
Quadrilateral 4 3
Pentagon 5 4 ii DAC = 20°
Hexagon 6 5 iii ADC = 70°
Octagon 8 6
Decagon 10 8 Exercise 19.7
10
1 a p = 50°, q = 65°, r = 65°
c Lines of symmetry = order of 2a4 b b = 40°
rotational symmetry in regular c c = 30°, d = 55°, e = 45°, f = 45°
polygons b infinite d p = 85°, q = 105°
e b = 60°
c infinite f x = 94°, y = 62°, z = 24°
g p = 85°, q = 65°

Review d Number of sides = lines of d 2 if base is a right-angled; 2 a AOB = 2x b OAB = 90° − x
symmetry = order of rotational c BAT = x
symmetry in regular polygons isosceles triangle

e2 f 2 3 a a = 70° b b = 125°

3 Audi = 1 Students’ own g infinite h7 c c = 60°, d = 60°, e = 80°, f = 40°
Citroën = 1 answers, this is an
Suzuki = 2 example. i2 4 a 90° − x b 180° − 2x
Honda = 1 c 2x − 90°
Toyota = 1 Exercise 19.4

1 Each has a rotational symmetry of 2 5 a Length of side = 30 mm;
area = 900 mm2
4 ABCDEFGHIJKLMNOPQRST 2 a Infinite b1
UVWXYZ c2 d8 b 193 mm2
a ABCDEMUVWY e Infinite f1
b HIOX 6 10 3 17.3 cm
c HIOSX
Exercise 19.5 7 a Draw the chords AD and BC.

1 a AB = 5 cm b AB = 30 cm ADX and BCX are angles in the
c AB = 2.4 m
5 Students’ answers will vary. same segment, so they are equal.

Similarly angle DAX is the same

2 Join OP and construct a line at right as angle CBX. AXD and BXC are
angles to OP that will be the chord.
Review vertically opposite angles, so they

3 O is the centre of both concentric are the same, too. is means that

circles. both triangles contain the same
Construct OX perpendicular to AD.
∴X is the mid-point of AD and BC three angle and so they are similar.
∴BX = XC and AX = XD
AB = AX – BX = XD – XC = CD b Using similarity DX = AX . You
CX BX

can then multiply through by CX

and BX

672 Answers

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University Answers

Exercise 19.8 b i 1:2 Copy
ii OQ; MQ = NQ; OM = ON; O 3 a Class interval Frequency

5 Parallelogram

Chapter 20

Exercise 20.1

1 a 2.5  m < 3.5 b 13
c

Mass of babies delivered by Maria in one month

30
UnFrieqvuenecyrsitCyoPprye-ssC-aRmebrviidegwe1 a 120°b 85° 28.5  l < 29.0 2
c 80° d 120°
e 90° f 90° 29.0  l < 29.5 1
g 30°
29.5  l < 30.0 6

2 AngleBTC = 180°−30°−(180°−60°) 30.0  l < 30.5 5
= 30° because angles in a triangle add
up to 180° 30.5  l < 31.0 2

31.0  l < 31.5 6

So angle TDC = 30° by the alternate 31.5  l < 32.0 3
segment theorem
5
CTD = 180°−60°−30° = 90° (angle 32.0  l < 32.5
sum in a triangle)
Review b Lengths of ribbon

So CD is diameter because the angle 20 8
in the segment is 90°
10 FrequencyCopy 6

3 CTD = 90° 4

UniversitCyFroeqPpurenycey-ssC-aRmebrviidegweSo TDC = 180°−90°−x = 90°−x0 2
0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5
So by the alternate segment theorem 0
CTB = 90°−x Mass of baby (Kg) 28.5 29.0 29.5 30.0 30.5 31.0 31.5 32.0 32.5
Ribbon length (cm)
But BCT = 180°−x 2 a 142 b 2  t < 4 min
So y + 180°−x + 90° − x = 180° c c Not very accurate; only 11 out of 30
So 2x − y = 90° Length of telephone calls were within 0.5cm of 30cm.
45
5 103° 40 4 6 8 10 12 14 16 4 Speed over the limit (km/h)
35 Time t (min)
ReviewExamination practice 30 FrequencyCopy 50
25 45
Exam-style questions 20 40
15 35
1 a and e 10 30
2 Order 3 5 25
3 a = 90°, b = 53°, c = 90°, d = 53° 20
02 15
Past paper questions* 10
d 5
1a2 0
b Class 0t 4t 8t< 12  t
interval <4 <8 12 < 16 10 20 30 40 50
Speed over limit (km/h)
Frequency 58 31 25 28
UniversitCy FroeqPpuernycey-ssC-aRmebrviidegwe
e Length of telephone calls 5 IQ test scores

28 70 30 Frequency 95 100 105 110 115 120 125 130 135 140
60 25 Test scores
3 52° 50 20
40 15
4 a i 43 30 10
20 5
ii w = 62 because YOZ is 10 0

Review an isosceles triangle and

YOZ + OYZ + YZO = 180° 0 4 8 12 16
Timt t (min)
so YOZ = 180 − 2 × 28 = 124. Copy

Angle at centre is twice angle f e smaller the class intervals the
more detailed the information
at circumference so w = 1 of represented.
2 e larger class intervals give a
good general picture of the data.
124 = 62- Cambridge
ess - Review
iii p = 30° because opposite sides

in a cyclic quadrilateral add

up to 180°

* Cambridge International Examinations bears no responsibility for the example answers to questions Answers 673
taken from its past question papers which are contained in this publication.

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Exercise 20.2 4 a 80 b 73 c7 Copy Copy Copy Copy

d Body fat is too low for intense physical activity.
Review1a
UniversitCyoPprye-ssC-aRmebrviidegwe e No – the expectation is that soldiers are physically active and therefore keep their

No. of Frequency Class Frequency body fat at a satisfactory level.
sweets (n) ( f ) width density
100 0.18 5 a Age (a) in years
100  n 18 Frequency
< 200 50 0.36
0 < a  15 12
50 0.64
200  n 18 15 < a  25 66
< 250 50 0.62
25 < a  35 90
250  n 32 50 0.42
< 300
100 0.2 35 < a  40 45

300  n 31 40 < a  70 60
< 350
b 156
21
350  n 6 a No – frequency density and not frequency given.
< 400
b Yes – one can see most of the bars are with the boundaries of the speed limits.
400  n 20
< 500 ci
Speed (km/h) Class Frequency
Frequency width density

b 0  s < 50 240 50 4.8

Number of sweets in jar

0.8
URneiFrveviqueeenrcywsidetnsCyityoPprye-ssC-aRmebrviideFregqwueencyURdennesiitvyvieerwsitCyoPprye-ssFrCeq-eancyRdmenesbitrvyiidegwe50  s < 653201521.3

65  s < 80 500 15 33.3

0.6 80  s < 95 780 15 52

0.4 95  s < 110 960 15 64

0.2 110  s < 125 819 15 54.6

0 200 300 400 500 125  s < 180 638 55 11.6
100 Number of sweets (n)

2 ii 240 below the minimum speed limit
d 15%
Mass of children
6 7a Speeds of cars

5 35

4 30

3 25

2 Frequency 20
density 15
1
10
0 12 15 18 21 24 27 30
69 Mass, m (kg) 5
Mass of actors
3 0
65 70 140 150 160 170 180 190 200 210
18 Mass, m (kg)
16 Height (h cm)
14
12 b Height (h cm) Frequency
10 140  h < 150 15
150  h < 160 35
8 160  h < 165 20
6 165  h < 170 18
4 170  h < 180 22
2 180  h < 190 12
0 190  h < 210 12
- Cambridge
60 ess - Review

c 150 − 160 d 75.7

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University Answers

Exercise 20.3 Exercise 20.4 Copy c Classification of fish caught

UniversCiutmuCylatiovePpfrryeqeu-esncsyC-aRmebrviidegwCeumulatUive fnrieCquuvenmcueylartivsei ftreCyquoenPcpyrye-ssC-aRmebrviidegwe1a1 a 30.0 cm b 27.5 cm c 33.5 cm
d 6 cm e 29.5 cm
Height in cm 6–15 16–20 21–25 26–40 m < 300

Number of 3 7 10 5 2 a i Paper 1: 48% Paper 2: 60% m ≥ 400
ii Paper 1: 28% Paper 2: 28%
plants iii Paper 1: 52% Paper 2: 66%

Cumulative 3 10 20 25 b Paper 1: >66% Paper 2: >79%
frequency

b 21–25 cm 3 a i 45 kg ii 330 girls 300 ≤ m < 400
c 30 b 10%

20 4a Speed Total number of fish caught = 18

Review 10 200 3 × 360° = 60°
180 18
0 10 20 30 40 160 Copy 9 × 360° = 180°
Height (cm) 140 18
120 6 × 360° = 120°
Median = 21 cm 100 18

2 a $36.25 80
b p = 12, q = 24, r = 35 60
c 40
20
Amount spent on books
10 20 30 40 50 60 70 80 90100110120 130140 2 Time taken by home owners to
40 Speed (km/h) complte a questionnaire
26
30 b Median = 102 km/h 24 2 4 6 8 10 12
Q1 = 92 Speed km/h 22 Time taken, t (min)
Q3 = 116 20
18
c IQR = 24 km/h d 14.5% 16
14
20 Examination practice 12 Frequency density
10
10 Exam-style questions
8
Review 0 10 20 30 40 50 60 1a Masses of fish caught Copy 6
Amount spent ($) 4
6 2

d Median amount spent $37 5 0

UniversitCyoPpryeC-usmsulaCti-vae freRqmueebncrvyiidegwe3 a 50Masses of children4

45 Frequency 3 Past paper questions
40
35 2
30
25 1i
20
15 1
10
5 0 Time 20 < t 35 < t 40 < t
200 250 300 350 400 450 500
(t seconds)  35  40  50
Mass (g)
Frequency 15 19 16

0 10 20 30 40 50 60 b
Mass in kg
Review b 19 kg Mass (m) in Number of Classification
c 24 grams fish
m < 300 3 Small
9 Medium Copy
300  m <
400 6 Large

m  400
- Cambridge
ess - Review Answers 675

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Cambridge IGCSE Mathematics University

ii 4 Copy Copy Copy Copy9 1200 men
10 a πr2 : 2πr
UniversitCyoPpryFere-qsuesncCy-daensiRtmyebrviidegwe3
= πr × r : πr × 2
2 =r:2
b 4 πr3 : 4πr2
1 3
= 4πr3 : 12πr2

= 4πr2 × r : 4πr2 × 3
=r:3

Exercise 21.4

Review 0t 1 (i) (ii)
20 25 30 35 40 45 50
a 1 : 200 0.005 : 1
Time (seconds)
b 1 : 250 0.004 : 1
2 a 56
b i 63 ii 24 c 1 : 25 000 0.00 004 : 1
UniversitCyoPprye-ssC-aRmebrviidegwe
d 1 : 200 000 0.000 005 : 1

Unit 6 3 a false b true c false e 1 : 28.6 0.035 : 1
d false e true
f 1 : 16 700 000 0.000 000 06 : 1

Chapter 21 4 a 1g b 1.33 g 2 a 4m b 6m
c 7:5 d 3:5 c 14 m d 48 m
Exercise 21.1
5 a 18 : 25 : 5 b 1.67 g c 4.17 g 3 a 0.0012 m = 0.12 cm = 1.2 mm
b 0.0003 km = 300 mm
1 a 1:1 b 1:5 c 25 : 3 6 a 20 ml b 2.5 ml c 0.0024 km = 2400 mm
d 3 : 10 e 3 : 20 f 1:5 d 0.00151 km = 1510 mm

2 a 12 : 5 b 5 : 12 7 15 750 kg

Review3 a 2:3 b 3:4 Exercise 21.3 4 a 100 mm × 250 mm
c 11 : 16 d 1:2 d 80 mm × 200 mm
1 a 40 : 160 b 1 200 : 300
5 a 1740 km b 1640 km
4 a 1 : 12 b 1:2 c 1:8 c 15 : 35 d 12 : 48 c 1520 km
d 7:6 e 10 : 3 f 5 : 12
e 150 : 450 f 22 : 16

5 a 1 : 10 g 220 : 80 h 230 : 460 : 1 610 6 a 1 cm = 150 cm = 1.5 m
c 100 : 1 Answers will vary due to
e 1 000 : 1 measuring variations.
UniversitCyoPprye-ssC-aRmebrviidegweb 1 : 1002 0.3 l = 300 ml
d 1 : 1 000 b i 8.4 m
f 1 : 60 3 Josh gets 27 Ahmed gets 18 ii 5.85 m
iii 2.7 m
4 Annie gets $50, Andrew gets $66.67 iv 3.15 m

6 a 1:2 b 1:8 and Amina gets $83.33 c i 27.92 m2
c 3:8 d 3 : 25 ii 20.88 m2
e 3 : 200 f 1 : 20 5 Students should draw a 16 cm iii 26.46 m2
g 8:5 h 2 : 15
line with 6 cm marked and 10 cm d 3.94 m2
e $162.49
marked.

Exercise 21.2 6 N (kg) P (kg) K (kg)

1 a x=9 b y = 24 a 0.25 0.375 0.375
c y=2 d x=6
e x = 176 f y = 65 b 1.25 1.875 1.875
g x = 35 h y = 180
Review i y = 1 400 j x = 105 c5 7.5 7.5 Exercise 21.5
k x = 1.25 l y=4
d 6.25 9.375 9.375 1 a 2.4 kg/$ b 0.12 l/km

7 1.8 m : 2.25 m : 1.35 m c $105/night d 0.25 km/min
8 37.5 cm
2 a x = 15 b x=8 e 27 students/teacher
c y = 20 d x = 2.4
e x = 0.6 f y = 3.25 f 3 hours/hole dug
g x = 5.6 h y = 7.2
- Cambridge 22.5 cm 2 a 9600 t b 48 000 t
ess - Review
3 a 120 l b 840 l

676 Answers

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University Answers

4 7.4 minutes train travelling from Valladolid Copy Copy Copy CopyExercise 21.10
5 12.75 km to Madrid, stopping at Segovia on
6 a 805 km the way.
7 a 312.5 km – e object travels slowly at first,
8 110 km/h then very quickly, then slowly
9 18.7 km/h again in the direction of y.
10 a 37.578 km/h Example: an Olympic runner
doing interval training.
– e object is moving at a constant
speed in the opposite direction to y
then it suddenly changes direction
and travels at a slightly faster speed
in the direction of y.
UniversitCyoPprye-ssC-aRmebrviidegwe 1 $6.75

b 76.67 km 2 60 min
b 3000 km
3 70 s
b 40.236 s
4 172.5 kg

5 10.5 km

Exercise 21.6 6 a 320 g flour, 64 g sultanas, 80 g
margarine, 99 ml milk, 32 g
1 a 700 m b 7 min sugar, 16 g salt

b 4:1

Review c 09:07 and 09:21 2 a 6 min b 10 km/h 7 250 g
c 3 min d 3.33 m/s
d Going to the supermarket 8 a 550 km b 17.31

2 a 45 min b 17:55 c 17:15 3 a For the first 50 minutes the taxi 9 a 13

3a travelled a distance of 10 km at b 13.12

Distance–time graph 12 km/h, then was stationary for c i 4 m ii 6.5 m
UniversitCyoPprye-ssC-DiastanRcme (ebmervtiries)degwe
30 50 minutes then took 20 minutes to d i 30 ii 6.59 m
25
20 return to starting point at 30 km/h. e 6.49 m
15
10 e taxi was then stationary for Exercise 21.11
5
40 minutes, then travelled 5 km in 1 Number 120 150 200 300 400
of people
40 minutes at a speed of 7.5 km/h
Days the 40 32 24 16 12
and was then stationary for water
will last
0 5 10 15 20 25 30 35 40 45 50 55 60 40 minutes.
Time (seconds)
b 130 minutes – the graph is
b 15 m c 5 m
horizontal.
Exercise 21.7
c 25 km

d i 12 km/h ii 10 km/h 2 a 8 days b 2 days

Review1 a and b iii 6 km/h iv 6.25 km/h 3 a 100 b 25
Answers will vary, examples: c8 d 250 cm
(from le to right) 4 a Other questions are possible, these
– e object is moving in the are just examples: What is the total 4 722.86 km/h
direction of y at a constant speed. time taken to attain a height of 16m?
Example: a helium-filled children’s When was the helicopter 5 3 h 36 min
balloon released in a large hall descending?
(with no breeze). When was the helicopter
– e object is stationary. Example: ascending?
a parked car. During what time period was the
– e object is moving in the vertical speed the greatest?
direction of y at a constant speed, At what speed was the helicopter
then suddenly changes direction, travelling between 2 and
moving at a much faster speed. 4 seconds?
Example: a squash ball travellingUniversitCyoPprye-ssC-aRmebrviidegwe Exercise 21.12
towards the court wall, hitting it
then bouncing back. 1 a y= 45 b y = 62.5
– e object is moving very quickly x x
in the direction of y at a constant
speed, then stops and is stationary c y= 2 d y = 0 28
for a while, then continues in the x x
same direction at the same speed
as before, then stops and is e y= 48
stationary again. Example: a x
Exercise 21.8
2 a k = 5120 b y = 10
1 a 1 500 m b 2 m/s c y = 23.70 d x = 5.98
c He was stationary. d 0.5 m/s
Review 3 x 0.1 0.25 0.5 0.0625
2 a 2 m/s2 b 35 m c 3.5 m/s y 25 4 1 64

3 a 1 m/s2 b 100 m c 15 m/s

Exercise 21.9 4 x 25 100 3.70 1

- Cambridge 1 a, c, d, e, f, h, i y 10 5 26 50
ess - Review
5 a 2.5 b 1000 c 0.125

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Cambridge IGCSE Mathematicse-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity

6 400 b 260 8 b x = 1.62 or x = −0.62
4 16 km c Negative solution can’t work as a
7 6.4 length must be positive
d perimeter = 5.24 cm
8 p and q are not inversely proportional Chapter 22
because p × q is not constant.
9 (0,1,2), (−7, −6, −5), (4, 5, 6)
Exercise 22.1
9 60 10 7 or −2

10 a false b false c true 1 a 4x = 32 b 12x = 96 11 3 cm by 8 cm
x=8
11 5 h x=8
c x + 12 = 55
x = 43 d x + 13 = 25 12 0.836 seconds

12 16 666. 7N (16.7 kN to 3sf) e x − 6 = 14 x = 12 13 1.96 seconds
x = 20
13 a 2 °C; f 9 − x = −5
b As temperature varies inversely g x =25
it will never reach −1 °C 7 x = 14 14 6 or −4
x = 17.5
Review h 28 = 4 15 2.75 cm
2 a y=3 x
c y = 46
14 m 3 5 n 28 x=7 16 7 and 8

P 24 40 P 24 6 b y = 12 Exercise 22.4

d y = 70

Exercise 21.13 3 a x = 13 b x=9 1 a x = m − bp b x = pr − n
c x=2 d x = 11
c x=m d x= cb
1 56 Exercise 22.2 4 a

2 24 1 Daughter = 15.5 years and e x = d − b − c
father = 46.5 years m
3 105
p
4 38 2 Silvia has 70 marbles; Jess has 350 f x = 3by g x = m
marbles.
5 40 cm long and 25 cm wide h x = np i x = mk
3 Kofi has $51.25 and Soumik has m 2
Examination practice $46.25
= 20
Exam-style questions 4 $250 and $500 j x p

Review1 Sandra receives 12 marshmallows 5 9 years 2 a x = m 3y b x = 4t c
34
2 Raja receives $40 6 Width = 15 cm and length = 22 cm
x = y +15 d x=5
3 300 cm = 3 m 7 48 km c 3 2

4 a 1.6 kg raisins b 1.2 kg dates 8 Pam = 12 years and Amira = e x = m + y f x = 2r − a
24 years 4c πr
5 960 males

6 9 cups 9 6.30 p.m. 3 m = E
c2
7 a 90 km/h b 18 km/h2 10 50 km
c 15 km R = 100I
e 18 km/h d 2 1 min Exercise 22.3 4 PT
2
f 17.5 km
1 −8 and −5 or 5 and 8 2k
Past paper questions* 5 m = v2

2 t = 2 seconds b = 2A − a
h
1 460 3 12 6

Review2 175 4 4 and 7 7 h = 3V
A
3a 5 6 cm
h = 3V
10 6 8 cm 8 πr 2
8
Speed 6 t 7 a 12 sides 9 B = 0.68
(m/s) 4 10 20 30 40 50 60 b n not an integer when the 10 h = 3.07
2 equation is solved
Time (seconds)
0

678 Answers * Cambridge International Examinations bears no responsibility for the example answers to
questions taken from its past question papers which are contained in this publication.

Copyright Material - Review Only - Not for Redistribution

e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity Answers

11 a 38 °C b 100 °C 10 a A = 1.13 m2 b A = 1.13 m2 e 726
c 0 °C b 6.18 25
c d = 4A
12 a 2.11 b x = m+ y π 4 a 26 b7
c 0.40 c 26 d 29
Exercise 22.6
Exercise 22.5 hg(4) = 4
1 5 gh(4) 5 5

1 a x= m i ii iii iv 6 a −56 + 16x2 − x4
a f (2) = f (−2) = f (0.5) = f (0) = b 56 − 16x2 + x4
c −56 + 16x2 − x4
c x = n − m d x ay a8 −4 3.5 2 d 56 − 16x2 + x4

b8 −12 0.5 −2

e x= ac f x = a + b2 c3 −5 0 −1 7 a −25 b3 c −7
b 2 34
Review d 11 11 3.5 3

g x= n h x = m2 e0 8 −0.75 0 d1 e −15
m y 3
f6 −10 −1.875 −2

i x = a2 j x = y2 + z 2 a −5 b −1 ( )8 a x2 + 36 2 b x8 + 36
5 c5 d −17

k x = (y + z)2  c  2 3 a0 b −4 c 0 d 76
  c5 d −3.9375
l x = a b

m x =  m a2 n x = y2 +1 4 a0 b −9 9 hgf (1) = 1 which is undefined.
 −b  3 c −2 d5
0

o x = y a2 a2 by 2 5 a 16 b 16 c1 10 a x +1 + 1
2 4 y2 x −1
p x = 6 x=4 ff (x) = +1
3 −1
b x L x − 1
1− x + +C x
2 a a = b a = x=1 x +1+ x −1
B 3 x +1−(x − 1)
7 =

Review c a = 5b d a = x(y + ) 8 x=6 = 2x
b −1 y −1 2

3− y 2 9 a x = −2 or 3 b x = −6 =x
y −1 mn
e a = f a= 10 a 2a b 2a + 4 b Same as f(x) as the function is
c 8a d 8a
E self inverse. So f −1(x) = x +− 11.
3 c= m 11 a 9 b x=2 x

4 a = c2 − b2 12 a 15 b3 c1 Exercise 22.8
2y
Exercise 22.7 1a x b 1 c 3x
5 1− y 7 3
1 a fg(x) = x + 3; gf(x) = x + 3
b fg(x) = 50x2 − 15x + 1; d x−3 7x
gf(x) = 10x2 − 15x + 5 4
6 s= A c fg(x) = 27x2 − 48x + 22; e 2(x − 5) f 2x − 2
gf(x) = 9x2 − 12x + 4
7 a y = 2x + 2 b y = 3x − c g x +2 29 4x 2
3 3 2 2+x
4x + z h i

Review c 3 2(b a) d fg (x) = 4x2 − 36 ; j 3 x −5 k x2 − 8
d = 3
3

y 3 gf (x) = 16x2 − 9 x +1
x −1
8 a E = 49 2E 9 l f 1(x) =
m
b v= 2 a −2x b −4

9 a V = 2 010 619 cm3 c 16 d −2 2 a f −1(x) = g(x)
b f −1(x) = g(x)
b r= V 3 a 9x + 4 b 18x2 + 1 c f −1(x) ≠ g(x)
πh c 3456 d 150 d f −1(x) = g(x)

Answers 679

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Cambridge IGCSE Mathematicse-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity

3 g −1 (x) = 3(x + 44) Past paper questions* 2a y
f −1 (x) = x 4
4ai
5 1 x=± y−4

ii ff −1(x) = x 2 y = ± πx2 A 2
π
iii f −1f(x) = x y=0 x
or 0 24 6
b i f −1(x) = x − 4 y = ± x2 − A –2

ii ff −1(x) = x π –4
iii f −1f(x) = x
c i f −1( ) = x + 7 3ai 8 b y x=2
C′
2 ii 4 D 4 B′
ii ff −1(x) = x A C
iii f −1f(x) = x b 4 or −4
c 1.176 or −4.68 2
Review d i f −1 (x) = 3 x − 2 d (x + 2) B D′

ii ff −1(x) = x 5 A′
e −2 x
iii f −1f(x) = x
f −1 ( ) = x2 + 1 6
ei 2 –2 0 24
–2
ii ff −1(x) = x
4 a i 11
iii f −1f(x) = x
ii 256
f −1 (x) = 9 b (x − 5) –4

fi x 2

ii ff −1(x) = x c 19 − 6x

iii f −1f(x) = x d −1, 0, 1, 2 c
y
g i f −1 (x) = 3 x +1 Chapter 23
4
ii ff −1(x) = x Exercise 23.1

iii f −1f(x) = x 1a

5a8 b 20 c 11 2
52
6 a −10 b –2 0 y=1
20 –2 x
Review c x = 1.54
ii 3 2 46

di −56 2
5

iii −7 4
5
–4

Examination practice 3a

Exam-style questions b y

1 $2 C′ B′3 A B C
2
2 16 5c coins and 34 10c coins D
D′ 1 x
3 a = 3.64 E′ E

4 a false b true

c true d false –3 –2 –1 0 1 2 3

5 a 14 b x = 1.26 or −0.26 b B′ = (−1, 3)
c A and A′ are invariant – they are
c x = 1.76 or −0.76
4−x the same point.
d x=1 e 3
Review
6a7 b 3−x c
7 −3 4
c4
4

680 Answers * Cambridge International Examinations bears no responsibility for the example answers to
questions taken from its past question papers which are contained in this publication.

Copyright Material - Review Only - Not for Redistribution

e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity Answers

4 a and b cy 4 a and b
2
y y
P 4 x=1
P′ 9

SQ 8
2
Q′ S′ x 7
R R′
y=2 –2 0 6 B"
–2 0 R" A (3, 5)
Q" S" x 5
6
P" –2 C 4)43
24 (–1,

2 a Centre of rotation A; angle of 2 A′ x
rotation 90° clockwise. 1 B (2, 1)

b Centre of rotation point on line C′
AC; angle of rotation 180°.
–6 –5 –4 –3 –2 ––110 1 2 3 4 5 6
c Centre of rotation point on
Review5a y line AC; angle of rotation 90° –2 B′
clockwise. –3

F 2 D D′ F′ 5 X′ (7, –1)
(a) Y′ (6, 4)
–3 –2 Z′ (3, –7)
F" 2 E E′ 3 a no b no c yes
6 a and b
y=1

1 Exercise 23.3

E" x 1a y

–1 0 1 23 9
(c) 8
M′ (–2, 8) 7
–1 D" N′ (2, 8)
Object Image

b F is at (−2, 3) b 6M N
F’ is at (2, 3) 5
4 O′ (2, 5)
P′ (–2, 5)
Exercise 23.2 Image Object

1a y  −6  3 3P O
 0  6 2
Review 2a A → B A C
4
B  0  −6 1 x
C  −7  1 –4 –3 –2 –1 0
B′ 2 b A → B A C 123 456

A x c A → B  0 A C  6 Exercise 23.4
02  5  −3
C′ –2 1 a Scale factor 2; centre of
enlargement = (8, 0)
b 3
b Scale factor 2; centre of
y enlargement = (3, −2)

4 (c) c Scale factor 2; centre of
C A enlargement (−3, 4)

2A B B C d Scale factor 1; centre of
A′ B′ (b) (a) 2
46
C′ (d) enlargement (0, 0)

–2 0 x

Review 2

–2

Answers 681

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Cambridge IGCSE Mathematicse-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity

2a 6 4 a  4  5  6
 2 b  −1 c  −1
y y
 −4  5
15  0 e  3  f  2
d  −3
14

13
12 A

11 Exercise 23.6

10

9

8 B  9  3  −6
7D  −21  −227   14
1 a b  c
6 

O x 5

by 4 AЈ C
3
Review  −9
2 DЈ BЈ  −3  2441   45 
 7    −10.5
1 CЈ x d e  f
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
0

7 a 9.6 cm wide 2 a DK = 2JK b JQ = 1 JF
b Length will be tripled. 4
c No; the image will not be in
proportion. c HP = 1 HF d 2GO = 1 GC
d 2.5 cm long and 1.5 cm wide 2 2

e 3DG 1CL f 6BE 2CL

8 a Scale factor is 0.75 3a  2 b  9
b 1.78 times smaller  8  21

O x Exercise 23.5
3y
 4  4  −4  45  0 75
7 1a  6 b  2 c  2  c 10.5 d  3 

6  −4  6  0 1 5  −36
 2  e  −4  4 e  6  f  −84
Review 5 d f

4 B  8  4 1 5  −−39355
3C  4 h  −2  6  
2 g g h 

1P A x 2

0 12 3456 789 P M S Exercise 23.7
B
4 Scale factor 1; centre of enlargement M
(0, −1) 2
A Q R 1 a  12 b  3
C D  −6  −5

5 Scale factor 1.5; centre of enlargement U TN J  12
(4, 2) P  −7

K 2

V 12  8
L  8   24

T WE F 3 a b

QL  −4   2
 −12  0
Review =  4 =  4 c d
 0  0
3 a AB DC

b BC =  1 AD =  1
 3  3

c ey are equal.

682 Answers

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 0 16 Exercise 23.9 4y
12  21
e f 1 a y = −x 3
b y=x−1 2
g 10 h  −2 c y=2−x 1 F x
 9   −7
2a –7 –6 –5 –4 –3 –2 –1–10 PD E
–2 12 3 4 5 6
–3
b a + 3b y –4
2 –5
4 a 2a + 3b 5 –6
4 –7
cb d a+b 3
2 2
5 a x+y 1
b 3 (x y)
−1x+ 3y 4 –5 –4 –3 –2 –1 0 5
44 –1
Review c –2 y
–3
6 a 2q − 2p b 2p + q c p − q –4 x 5
–5 12345 4

Exercise 23.8 3 A
B
1 a 4.12 b 3.61 c 4.24 2
d5 e 4.47 f 5 C
g 5.83
1

2 a 10.30 b 13.04 b –5 –4 –3 –2 –1 0 x
c5 d 10 –1 12345
y
BЈ C–Ј 2
3a 5 b 13 c 17 5
4 –3
4 a A(4, 2) B(–1, 3) C(6,–2) 3
2 –4
b AB =  −5 CB =  −7 AC =  2 1 AЈ
 1   5   −4
–5 –4 –3 –2 –1 0 –5
–1
5a a b −b –2 6
2 2 –3
–4 y
Review c ab d 3a 3b –5 x FЈ
2 4 –6 12345 10
–7 8 EЈ GЈ
6 a 10 b 8.60 6 DЈ
4 x
7 100 km/h 2
2 4 6 8 10
8 6.71 km/h (3sf) –10 –8 –6 –4 –2 0
–2
9 a b−a b3 –4
–6
c CD = CA + AD –8

So CD = −2a + 3b − a = 3b − 3a = 3AB 3a –10
–12
So CD is parallel to AB, so the y

triangles are similar. 5
4 BЈ
10 a −p + q b 2(−p + q)
c 2+1 3 3
3q 3p
d q+1p 2
2
B
Review 1

CЈ

–5 –4 –3 –2 –1 0 x
12345

–1

A

–2

AЈ
–3

C
–4

–5

Answers 683

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7 Examination practice 6a

y Exam-style questions

5 1 a NOT TO SCALE 10 y

4 9
3M
N 8
7
y
2 5 6
5 C(9,5)
1O 4C
4
PЈ Q P x 3
–5 –4 –3 –2 –1 0 1QЈ 2 34
5 B2 A 3
B(3,2)
–1
2
OЈ –2 1x
–4 –3 –2 –10 1 2 3 4 5 6 7
–3 1b A(3,1)
–1 D
a x
–2
Review –4 –2 –1 0 1 2 3 4 5 6 7 8 9 10
N–Ј5 MЈ –1 D(0,–1)

–3 –2

8 b a–b c |a| = 3.16

y 2 A: reflection about y = 0 (x-axis).  
 −3  3
6 7a i Translation
J5 B: translation  2  .
ii Enlargement scale factor 3
C: enlargement scale factor 2 centre
4 origin. centre origin
N K3
D: rotation + 90° about the origin. iii Rotation 90° centre (2, 1)

2  −3
and translation  1 
1 3   −6
ML a i  −3 ii  3

–6 –5 –4 –3 –2 –1 0 LЈ 1 2 MЈ3 4 5 x b Shapes B, D
–1
b 8 7y
KЈ NЈ
–2
6

–3 JЈ 5 A′(3, 4)
l (b)(i)
Review –4 m 4 B(1, 3)
3

–5 2 (a)(i)

9 13 y C(–1,12) A(5, 0) x

12 A′ A" –7 –6 –5 –4 –3 –2 –1–1 1 2 34 5 6
11 –2 (a)(ii) B′(3, 1)
–3
10 C′(1, –2)

9 A 4 a (–1, 2)
8 b Scale factor −2
7 –4

6 –5

C" 5 B B" C C′ –6
4 B′
3 (2, 5)
 –7
2  −2
5a b ii y = x
1x 2

–9 –8–7–6–5–4–3–2–10 1 2 3 4 5 6 7 8 9 b Rotation 180° about centre (6, 0).

10 13 y ci Past paper questions*

12 10 y
9
11 8 1 34
7
10 6
5
O′′ P′′ 9 4
8 3
M′ N′ 2 2 a i Rotation, 90°, anticlockwise
7 1x
6 MN
–2 0 1 2 3 4 5 6 7 8 9 10 about (0, 2)
Review N′′ M′′ 5 P′ –2
4 P OO′ –3
–4 ii Reflection in the line y = 1
3 –5
–6
2 –7 iii Enlargement, scale factor − 1 ,

1x

–9 –8–7–6–5–4–3–2–10 1 2 3 4 5 6 7 8 9 –10 –8 –6 –4 centre (0, 0) 2

ii 4 : 1

684 Answers * Cambridge International Examinations bears no responsibility for the example answers to
questions taken from its past question papers which are contained in this publication.

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b b 16 Exercise 24.3
1
y 1a 1
c 16 2
9
8 b2
3
7
c1
6 A Exercise 24.2 6
B5
d1
4 1 1 3
2
3 H HH e1
2 AЈ
1 H 2
2
B
1 1 T HT
0x 31
2
–7 –6 –5 –4 –3 –2 –1–1 1 2 3 4 5 6 7 8 9
4 outcomes

–2 1 H TH
D –3 2

–4 C 1 T
2

Review –5 1 T TT

2

–6 P(TT or HH) = 2 = 1 L
4 2 5
–7

3 a i −a + c 2a 8 B BB 8 × 8 11
ii –1a + 1c 10 10 10
33 8 a4
10 B 5
b AX = 1 (−a + c) = 1 AC
33 2 R BR 8 × 2 b1
10 10 10 4
so AX and AC are parallel and
pass through A therefore the 2 8 B RB 2 × 8 c 11
points lie on a straight line. 10 10 10 20

10 R 3a

2 R RR 2 × 2
10 10 10

Chapter 24 1 H T
b i P(RR) = 25
Exercise 24.1 1 12 3
8
1a ii P(RB) + P(BR) = 25

Review 1st 2nd 16 4
draw iii P(BB) = 25
draw
r 3a 5 R b3
r 12 5
b
g 5R

12 7 W
12
r 4a
bb
5 P C
g
7 12 R
12 W 57 8

r 7 W
gb 12

g 25
b i P(RR) = 144
b 9 possible outcomes c 3 8
d5 e 4 49
ii P(WW) = 144 bi 5
2a A AA 28
B AB 4a 4
A C AC ii 5
D AD b4 7
A BA 9
B BB iii 1
Review B C BC c1 4
D BD 9
A CA
B CB d He is equally likely either to buy
C C CC
D CD two birds, or to buy one of each.
A DA
B DB
D C DC
D DD

Answers 685

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5a 2a 1 C b1 c1
2 12 12
P V bi 1 B D
29 37 24 B 6a
1 A 1 C 1
iii 0 4 3 2 D
c1 B L
27 D 1
4 2 C M
3 Sn 1 R
C 5 B
C
37 6 A 1 D Sm
2 A
b i 32
65 1 B 1 C 1 D Sn
4 3 2 A
ii 93 M
130 C 1
D 1 1 L 5 R
iii 27 2 B 6
130 B
D
iv 37 A 1 A Sm
130 2
D 1 Sn
A
1 C 1 B 1 6L
4 3 2 B 1
B M 5 R
1 B
D 1 C 6
2 A
Review Sm
C
1 A 1 Sn
2
A B 6L

1 D 1 B 1 R 1 M
4 3 2 5
1 B
6 a 12 6
Sm
b3 C 1
2 Sn
c 21 1
ii 1 6 1 L
d 12 24 5
e7 B M

12 R
f 12
Sm
19
d1 Sm 1 Sn
24 5 L
M
= 25 R
T B
98
Exercise 24.4 b1
30
1a 1st card 2nd card Locker 2 Locker 3
7a

Locker 1

Review 12 Raju 1 Sam 1 Kerry
51
1 2 Kerry 1 Sam
13 3
51 a3 2 1 Raju 1 Kerry
5 1 2 Kerry 1 Raju
13 b9
51 4 17 3 Sam 1 Raju 1 Sam
2 Sam 1 Raju
13 C = 100 1
13 51
52 58 A 3 Kerry

13 22 20 b Conditional – once the first name
51

12
51
13 13
52 51

13
51

13

52 13 is chosen it cannot be chosen
51
13 again, so the second choice
51

13 12 depends on the first, and so on.
51 c 1 ways d 6 ways e 1
52 13
6
51
84
13 a i 0.58 b 11 or 0.275 15
51 5 a 12 outcomes 40

13 O S
51 K
S A
13
51 K O
A K
12 A
51 O 12 outcomes
S
A 9 a Friday Saturday
O
Review b i P(♥♥) = 13 × 12 = 3 S 0.21 Rain 0.83 Rain
52 51 51 K 0.17 No rain

ii P(♣♣) = 13 × 12 = 3 0.3 Rain
52 51 51 0.79 No rain

iii P(red, black) = 26 × 26 = 13 0.7 No rain
52 51 51
b i 0.1743 ii 0.4113

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10 a weather forecast over the next ci 1 ii 2 bi
2 days 6 3
G5
b P(rain both days) = 1 iii 1 iv 1 4
50 3
7 0
P(sun both days) = 96 F
125 Past paper questions* S
12
1a

P(1 fine day and 1 rain) = 53 First Pencil Second Pencil
250
5
11 a 13 red ii 28

3 5 Kite flies 6 Red Examination practice
4 8 14 8
13 Structured questions for
1 Good wind 8 Units 4–6
4 14 6
3 Kite does not fly. 13 blue Answers for these questions are available
8 b i 15 red in the Teacher’s Resource.
91
Review Not a good wind 1 blue

16 Kite flies

15 Kite does not fly. Blue
16
ii 9
b 15 c 33 d 31 13 7
32 64 128 13

Examination practice 2a2
3
Exam-style questions
b
1a 1 1
2 1 8 Late
3 3 Not late
1 1 4 Cycles 7 Late
6 5 2 8 Not late
3 Walks 3
6 8
ci 1 ii 17
1 21 24 24 5
6 3 8
2 1 4 3ai
6 5

16
61
Review 2
3
1 3 1 4
6 6 5

6

1 12
6 3
4
1 4 1 5
6 6

6

1 21
6 3
5 1 4
6 5 (25) (16)
F
6 S (18)

1
2
3
6 1 4 5 11 7
6 5

6

bi 5 ii 1 2
36 6

2a ii 9 iii 14
iv 11 v7
First card Second card Third card
25 100
1 2 13
Review 1 2 3 12
1 13
1 1 3 11
2 1 12
3 2 11
1
1 2 2
3

1
3

3

b6

* Cambridge International Examinations bears no responsibility for the example answers to questions Answers 687
taken from its past question papers which are contained in this publication.

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Glossary

A Chord A straight line joining one point on the circumference of
a circle to another point on the circumference.
Acute An angle greater than 0° but less than 90°.
Adjacent A “shorter” side of a right-angled triangle that is “next Circle A set of (joined) points that are the same distance (radius)
to” an angle other than the right-angle. from a given fixed point (the centre).

Algebra e use of letters and other symbols to write Class interval e difference between the upper and lower limits
mathematical information. of groups of data.

Alternate A pair of equal angles formed between two parallel Coefficient In a term which is a mixture of numbers and letters,
lines and on opposite sides of the transversal (a line that crosses the coefficient is the number that is multiplying the letters.
both parallel lines).
Review Column vector Number pair notation used to describe a vector
Alternate segment When an angle is drawn between a tangent
and a chord the alternate segment is the segment that the angle as the movement between two points: x units in the x-direction
does not cross.
(left or right) and y units in the in the y-direction (up or down)
Angle A measure of the amount of turning between two lines
that meet at a point. e.g.  x . See also vector.
 y
Apex In a pyramid, the apex is the point, above the base, at
which all the sloping sides meet. Combined events One event followed by another event.

Arc Part of the circumference of a circle. Commission Pay based on a percentage of sales made.

Area A measure of the amount of space contained within Common Denominator A common value that two or more
a plane shape. fractions need to be converted to in order to be able to add and/
or subtract fractions.
Asymptote A line that a graph approaches but never intersects.
Common Factor A term that can be divided exactly into two or
Average A single value used to represent a set of data. more other terms.
(A measure of the central tendency of the data.)
Complement e elements that are in the universal set but not
Axis of symmetry A line that divides a plane shape into two in a given set.
symmetrical halves or a “rod” about which a solid can rotate and
Reviewstill look the same in different positions. Composite function Applying a function to a value and then
another function to that result.
B
Composite number Integer with more than two factors i.e. it has
Bar graph A diagram used to display discrete data. more factors than just 1 and itself.
Base When working with indices, the base is the number that is
being raised to a power. Compound interest Interest paid on interest already earned and
Bearing An angle indicating the direction of travel between not just the original capital.
two points. e bearing begins from the “North” direction and
is measured clockwise round to the line joining start point Congruent Shapes that are identical in both shape and size.
and destination.
Bias Something that affects the chance of an event occurring in Constant term A term in an equation or expression that has a
favour of a desired outcome. fixed numerical value.
Bivariate data Two measurements, relating to an investigation,
taken at the same time. Continuous Data that can take any value in a range, such as
BODMAS/BIDMAS When more than one arithmetic operation height or weight.
appears in a sum, this mnemonic indicate the order in which
the operations should take place. Brackets, Of, Divide, Multiply, Conversion Changing one quantity or unit into its equivalent in
Add, Subtract. another unit.

C Correlation e relationship between bivariate data.

ReviewCategorical data Non-numerical data. Corresponding angles A pair of equal angles formed in the
“same manner”. ese occur with parallel lines and transversals,
Centre of rotation e point around which a plane shape can in similar and congruent triangles and other
rotate and show the same shape in different positions. similar shapes.

Corresponding sides Sides that occupy the same relative
position in similar or congruent shapes.

Cosine ratio For a given angle (other than the right-angle) in
a right-angled triangle, the cosine ratio is the length of the side
adjacent to the angle divided by the hypotenuse of the triangle.

Cosine rule A formula connecting the three sides of a triangle
and one of the angles.

688 Glossary

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Review Cost price e price that a trader pays for goods. Equidistant e same distance from.
Co-interior angles e two angles formed on the same side of a
Review transversal when it cuts two parallel lines. Co-interior angles add Equivalent Fraction e result of multiplying (or dividing) the
up to 180° (they are supplementary). top and bottom of a fraction by the same value.
Review Cube A cube is the result obtained when a number is multiplied
by itself and then multiplied by itself again. Estimate An approximate solution to a sum found by using
Cube root e number which, when multiplied by itself and rounded values.
then by itself again, gives the cube.
Cumulative frequency A “running total” of the frequencies. Estimated mean A calculated approximation for the mean
Cumulative frequency curve A curve formed using the of grouped data.
cumulative frequencies as the vertical axis value.
Cyclic quadrilateral A quadrilateral whose vertices all exactly Event e outcome that is being tested for in a
touch the circumference of a circle. probability “experiment”.

D Exchange rate e value to convert from one currency
to another.
Data A set of facts, numbers or other information.
Deductions Money that is subtracted from income before Expansion Multiplying out the terms inside a bracket by the
tax is calculated. term multiplying the bracket. ( is includes multiplying one
Denominator e number on the bottom of a fraction. bracket by another.)
Dependent variable A variable whose value depends on the
value of another variable. Experimental probability e chance of an event
Determinant In a two by two matrix, the product of the happening, calculated by running an “experiment” many
elements in the leading diagonal minus the product of the times.
elements in the other diagonal.
Difference between two squares A method of factorising (putting Exponent Another word for power or index, indicating how
into brackets) one squared term subtracted from another. many times a base number is multiplied by itself.
Differentiation e process of finding a derived function, which
tells you the gradient of the function at a point on the curve. Exponential A function formed when the variable is in
Directed numbers Numbers that have a (positive or negative) the index.
direction; once a direction is taken to be positive, the opposite
direction is negative e.g. −4°c is a directed number. Expression A group of terms linked by operation signs.
Direct proportion When two quantities increase or decrease
at the same rate. Extrapolation A value determined by continuing a line of best
Discount e amount by which an original selling price fit beyond the plotted data.
is reduced.
Discrete Data that can only take certain (usually integer) values. F

E Face A plane shape that forms part of a solid.
Factor A number that divides exactly into another number with
Earnings e amount earned for work done. (See commission, no remainder.
salary and wages.) Factorisation To re-write an expression using brackets.
Favourable combinations Combinations of outcomes that mean
Element A member of a set. a desired event has occurred.
Favourable outcomes Any outcomes that mean a desired event
Empty set A set that contains no elements. has occurred.
FOIL A mnemonic for the order of multiplying out terms in a
Enlargement A transformation of a shape that keeps the ratio double bracket expansion. First, Outside, Inside, Last.
of corresponding sides the same but increases or decreases the Formula A general “rule” expressed algebraically (such as how to
lengths of the sides. find the area of a shape).
Fraction Part of a whole.
Equation A mathematical “sentence” involving the use of Frequency e number of times a particular value occurs.
the “=” sign. Frequency density e frequency of a class divided by the width
of the class.
Equation of a line A formula that shows how the x coordinate is Frequency table A method of summarising data when values or
related to the y coordinate. classes occur more than once.
Function A set of rules or instructions for changing one number
(an input) into another (an output).
Function notation An alternative mathematical way of
writing equations.

Glossary 689

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ReviewG Line graph A chart where numerical values are plotted against
'number lines' on vertical and horizontal axes. Points are plotted
ReviewGradient e steepness of a line (or the steepness of a tangent and joined to form a line.
drawn at a point on a curve).
ReviewGross income Amount earned before deductions and tax. Line of best fit A trend line drawn onto a scatter graph that
Grouped e collection of individual data values into passes as close to as many data points as possible.
convenient groups. Used especially for continuous data.
Line segment A section of a line that is the shortest distance
H between two points.

Histogram A specialised graph used to illustrate grouped Line symmetry A line that divides a plane shape into two halves
continuous data. so that one half is the mirror image of the other. (See also axis
Hyperbola A graph where the variable is in the denominator of a of symmetry.)
fraction. (Also called reciprocal graphs.)
Hypotenuse e longest side of a right-angled triangle. Linear Equation A linear equation has no terms with a power in
x greater than one.
I
Linear inequalities Similar to linear equations but
Image e new position of a point a er a transformation. using <, >, ⩽ or ⩾.
Imperial A non-metric system of measurement.
Included angle In congruency, an angle that is formed by the Linear programming A method for finding region in a plane
meeting of two given sides. that satisfies a set of constraints defined as linear inequalities.
Included side In congruency, a side that connects two given 
angles. Loss When goods are sold for less than they were bought, the
Independent An event whose outcome is not affected/influenced loss is the cost price − selling price.
by what has occurred before.
Index Another word for power or exponent, indicating how Lower bound e exact smallest value that a number (given to a
many times a base number is multiplied by itself. specified accuracy) could be.
Integer Any of the negative and positive whole numbers,
including zero. Lower quartile e value of data at the 25th percentile.
y-intercept e point at which a line or curve crosses the y-axis.
x-intercept e point at which a line or curve crosses the Lowest Terms An equivalent fraction where the numerator and
x-axis. denominator are the smallest allowable whole numbers. Also
Interest e amount charged for borrowing, or earned for called simplest form.
investing, money.
Interest rate e percentage charged for borrowing, or earned M
for investing, money. (Usually an annual rate.)
Interquartile range e difference between the upper and Magnitude e size (of a vector) irrespective of direction.
lower quartiles.
Intersection In sets, the elements that are common to two Major segment e larger of two circle segments formed when a
or more sets. In algebra, the point where two lines or curves chord is drawn between two points on the circumference.
cross.
Inverse functions A function that does the opposite of the Maximum A turning point or vertex on a graph whose
original function. y-coordinate is greater than points immediately to its le
Inverse proportion When one quantity decreases in the same and right.
proportion as another quantity increases.
Irrational number A (decimal) number that does not terminate Mean An average that uses all the data.
or recur and cannot be written as a fraction.
Median An average, the middle value of data when it is arranged
L in increasing order.

Line A straight, one dimensional figure that extends to infinity Metric A measuring system based on multiples of ten.
in both directions. See “line segment”.
Midpoint Exactly half way between the ends of a line segment.

Minimum A turning point or vertex on a graph whose
y-coordinate is lower than points immediately to its le and
right.

Minor segment e smaller of two circle segments formed when
a chord is drawn between two points on the circumference.

Mixed Number A number with a whole number part and a
fraction part.

Modal class For grouped data, a class that has the highest
frequency.

Mode An average, the most frequently occurring value in a set
of data.

Multiple e result of multiplying a number by an integer.

690 Glossary

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