ACE YR IGCSE A TOP APPR' TO MATHS

[ (1.5 + 9.5)50 ] = 134.058 cm 3
8 Volume = 2 × 25 Volume of the cylinder
2
= 6875 m = π × 4 × 3.5
3
= 6875000 l [3] = 175.952 cm 3
2
9 (a) Slanted height of the cone, l = √h + 9 Total volume
3
Surface area of the cone = π × 3 + π × 3 × = 309.97 cm
2
2
√h + 9 = 310 ml [5]
∼
= 9π + 3π √h + 9 (b) (i) 1 l ÷ 310 ml 3 cups [1]
2
Surface area of the hemisphere (ii) 1000 ml – (310 ml × 3) [1]
= 70 ml

= πr + 4πr 2 (c) Volume of the cylinder
2
= 3πr 2 = 250 – 134.058
2
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
2
9π + 3π√h + 9 = 3πr = 115.942 cm 3
2
√h + 9 = r – 3 115.942 = π × 4 × h
2
2
2
h + 9 = (r – 3) h = 2.306
2
2
2
h = √(r – 3) – 9 Height to the top of the cup
2
2
h = √r – 6r 2 = 5 – 2.306 [4]
4
= 2.69 cm
h = r √r – 6 12 (a) CX = BC – BX
2
2
2
2
a = 6 [7] = 5 – 2
2
2
1
(b) × 4 × π × r = 18π CX = 4.583 cm [2]
3
2 3 r = 3 [2] (b) tan ∠XCD = XD
(c) h = √3 – 6(3) CX
2
4
= 5.196 ∠XCD = tan ( 5 )
–1
Volume = 1 × π × 3 × 5.196 4.583
2
3 = 47.49° [2]
2
2
= 48.98 cm 3 [3] (c) AX = AD – XD 2
2
10 (a) ∠OBA = 90° = 7 – 5 2
The angle between a tangent and a radius is AX = 4.899
90°. [1] tan ∠ACX = AX
(b) ∠FOB = 2 × (180° – 90° – 25°) CX
= 130° ∠ACX = tan ( 4.899 )
–1
Reflex angle of ∠FOB = 360°−130° 4.583
= 230° [2] = 46.91° [4]
(c) AO = OB (d) Volume
sin ∠OAB = 1 × BD × CX × AX
AO = 8 3 2
sin 25° = 1 × 7 × 4.583 × 4.899
= 18.93 cm 3 2
AD = 18.93 + 8 = 26.93 cm [2] = 26.19 cm 3 [2]
(d) (i) Volume of the cylinder = π × 8 × 30 13 (a) (OB + 3) = 12 + OB
2
2
2
2
= 6032.6 cm [2] OB + 6OB + 9 – 144 – OB = 0
3
2
2
(ii) ED = AD × tan EAD OB = radius = 22.5 cm (shown) [3]
= 26.93 tan 25° (b) tan ∠AOB = 12
= 12.558 cm 22.5
Volume of the triangular prism ∠AOB = 28.072°
= ( EC × DA ) × 30 Perimeter of the shaded region
(
)
2
2 × 28.072
( 12.558 × 2 × 26.93 ) = 360 × 2π × 22.5 + (3 × 2) + (12 × 2)
= 2 × 30 = 52.05 cm [5]
= 10146 cm 2 [4] (c) Area of the shaded region
(iii) 10146 – 6032 ×100% = Area of ΔAOC – Area of sector AOC
)
6032 = ( 24 × 22.5 ) ( 2 × 28.072 × π × 22.5
–
2
= 68.2% [2] 2 360
11 (a) Volume of the hemisphere = 270 – 248.068
= 2 × π × 4 = 21.93 cm 2 [4]
3
3
Answers 171
Answers.indd 171 15/03/2022 11:08 AM

14 (a) (i) Cross-section area = 331.765 + 932.803
1 360 = 1264.6 cm 2 [4]
= 5 × × 5 × 5 × sin
2 5 (d) Bearing of S from P = 90°+ ∠SPQ
= 59.441 cm 2 [3] = 90° + 112.558°
(ii) Volume = 59.441 × 20 = 202.6° [1]
= 1188.82 cm 3 [1] 2 (a) sin ∠QRP = sin ∠QPR
(b) Mass =1188.82 × 13.5 g QP QR
= 16049.07 g sin ∠QRP = 100 sin 55
= 16.05 kg [2] 110
(c) 1 m × 0.9 = 0.9 m ∠QRP = 48.132°
3
3
= 900000 cm 3 ∠PQR = 180° – 55° – 48.132°
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
Number of prisms = 76.868° [4]
900000 (b) (i) Bearing of R from Q
=
1188.82 = 180° – (360° – 345°)
= 757.05 = 165° [1]
∼ 757 prisms [3] (ii) Bearing of Q from P
15 Volume of the box = 180° – (360° – 165° – 76.868°)
= (1.4 cm × 2 × 4) × (1.4 cm × 2 × 4) × (1.4 cm × 2) = 61.9° [2]
= 351.232 cm 3 (c) PR = QR
Volume of 16 balls sin PQR sin QPR
4 110 sin 76.868
= 16 × × π × 1.4 3 PR =
3 sin 55
21952 = 130.77 m
= π cm 3
375 Distance
Percentage of the balls that occupied the box = 130.77 × cos (345° – 270° – 48.132°)
21952 = 116.65 m [5]
= π ÷351.232
375 (d) Area = 1 (PQ)(QR) sin ∠PQR
1 2
= π% [6]
6 = 1 (100)(110) sin 76.868
2
6 6 Trigonometry = 5356.17 m 2 [2]
3 (a) ∠PQR = 360° – 90° – 240°
Trigonometry = 30° (shown) [1]
(b) (i) PR = PQ + QR – 2(PQ)(QR) cos ∠PQR
2
2
2
2
2
2
1 (a) SR = SQ + QR – 2(SQ)(QR) cos ∠SQR PR = 11.5 + 20 – 2(11.5)(20) cos 30°
2
2
2
75 = 45 + 45 – 2(45)(45) cos ∠SQR PR = 11.57 km [4]
2
2
2
∠SQR = 112.885° [4] (ii) sin ∠PRQ = sin 30°
sin ∠SPQ sin ∠PQS 11.5 11.57
(b) =
SQ PS ∠PRQ = 29.8°
sin ∠SPQ = ( 45 sin 38° ) Bearing of P from R
= 180° – 90° – 30° – 29.8°
30
∠SPQ = 180° – 67.442° = 30.2° [5]
= 112.558° (c) Total time used
PQ PS = 5.5 h – 20 min – 20 min
= 29
sin ∠PSQ sin ∠PQS = h
30 sin 29.442 6
PQ = Total distance travelled
sin 38°
= 23.95 cm [7] = 11.5 km + 20 km + 11.57 km
= 43.07 km
(c) Area of PQRS Speed of running
= Area of ΔPQS + Area of ΔSQR
1 1 = 43.07 km
= (PQ)(QS) sin ∠PQS + (SQ)(QR) 29
2 2 h
sin ∠SQR 6
1 1 = 8.91 km/h [3]
= (23.95)(45) sin 38° + (45)(45) 4 TS = TR + RS 2
2
2
2 2 2 2
sin 112.885° = 15 + 20
TS = 25 cm
Cambridge IGCSE
TM
172 Ace Your Mathematics

Answers.indd 172 15/03/2022 11:08 AM

PT Volume = Area of PSTW × TU
tan ∠PST = = 205.456 × 10
TS = 2054.56 m 3 [4]
tan ∠PST = 6 (d) WU = TU + TW
2
2
2
25
= 10 + 20
2
∠PST = 13.5° [4] WU = 22.361 2
5 (a) sin ∠PQS = sin ∠SPQ Let the angle of elevation = x
PS SQ 2.25
sin ∠PQS = 4.1 sin 32° tan x = 22.361
6.2
x = 5.75°
∠PQS = 20.51° [3] 7 (a) ∠DAC = 230° – 180° [4]
(b) Bearing of R from Q = 50°
= 180° + (148° – 20.51° – 72°) ∠ADC = 180° – 50° × 2
= 235.49° [2] = 80°
(c) PQ = SQ Bearing of C from D
sin ∠PSQ sin ∠SPQ = 50° + 80°
PQ = 6.2 sin 127.49° = 130° [3]
sin 32° (b) AC = AD + DC – 2(AD)(DC) cos ∠ADC
2
2
2
= 9.28 km [3] 1Rights Reserved.
= 58 + 58 – 2(58)(58) cos 80°
2
2
AC = 74.56 m
(d) QR = SR (c) AC = AB + BC – 2(AB)(BC) cos ∠ABC [4]
2
2
2
sin ∠QSR sin ∠SQR 74.56 = 52 + 75 – 2(52)(75) cos ∠ABC
2
2
2
QR = 7.5 sin (180° – 127.49°) ∠ABC = 69.2°
sin 72°
= 6.257 km Area of ABC = 1 (AB)(BC) sin ∠ABC
Total distance 2
= PQ + QR + RP = 2 (52)(75) sin 69.2°
= 9.28 + 6.257 + 7.5 + 4.1 = 1822.91 m 2 [6]
= 27.137 km [5] (d) (i) Shortest distance
(e) ∠SRQ = 180° – 32° – 72° – 20.51° = AD × sin ∠DAC
= 55.49° = 58 × sin 50°
Shortest distance = 44.43 m [2]
= QR × sin ∠SRQ (ii) Area of the whole garden
= 6.257 × sin 55.49 1
= 5.156 km [4] = 2 (58)(58) sin 80° + 1822.91
(f) Area = 3500 m 2 [3]
1
Penerbitan Pelangi Sdn Bhd. All 7π = ∠POR × 2 × π × 6
=
(PQ)(QR) sin ∠PQR
(a)
8
2 3 360
= 1 (9.28)(6.257) sin (72° + 20.51°) ∠POR = 70° [2]
2 (b) (i) Area of sector OPQR
= 29 km 2 [2] = 70 × π × 6
2
6 (a) Area ΔWST = 1 (TS)(TW) sin ∠STW 360
2 = 7π [2]
= 1 (18)(20) sin 40° (ii) Area of ΔOPR
2 1
= 115.7 m 2 [2] = 2 (6)(6) sin 70°
(b) SW = ST + TW – 2(ST)(TW) cos ∠STW = 16.91 cm 2 [2]
2
2
2
SW = 18 + 20 – 2(18)(20) cos 40° (c) Area of shaded segment PQR
2
2
2
SW = 13.13 m = 7π – 16.91
SW = PS + PW – 2(PS)(PW) cos ∠SPW = 5.081
2
2
2
13.13 = 14 + 17 – 2(14)(17) cos ∠SPW Percentage = 5.077 × 100%
2
2
2
∠SPW = 48.96° [8] π × 6 2
(c) Area of trapezium PSTW = 4.49% [3]
= Area of ΔWPS + Area of ΔWST 1
1 (d) Area of ΔABC = 2 (AB)(AC) sin ∠BAC
= (14)(17) sin 48.96° + 115.7
2 1
= 205.456 m 2 10 = 2 (5)(5) sin ∠BAC
∠BAC = 53.13°
Answers 173

Answers.indd 173 15/03/2022 11:08 AM

Area of shaded region (b) y = cos x
53.13 y
= × π × 5 – 10
2
360
= 1.59 cm 2 [4] 1
PS
9 sin ∠PRS =
RS
9
∠PRS = sin –1 ( ) 0 90° 180° 270° 360° x
18
= 30° –1
PS
cos ∠PSQ =
QS [2]
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
9 (c) y = tan x
QS =
cos 30° y
= 10.392
∠QSR = 180° – 90° – 30° – 30°
= 30°
Area of ΔQSR x
1 0 90° 180° 270° 360°
= (SQ)(SR) sin ∠QSR
2
1
= (10.392)(18) sin 30°
2 [2]
= 46.76 cm 2 [7] 2 (a) 360° – (210° – 180°)
sin ∠QPR sin ∠PRQ = 330° [2]
10 (a) =
QR PQ (b) sin x = 2
3
∠QPR = sin –1 ( 53.7 sin 47° ) x = 41.81°, 180° – 41.81°
40.2
= 77.68° x = 41.81°, 138.19° [3]
∠PQR = 180° – 47° – 77.68° 3 (a) cos x = 0.5
= 55.32° [4] x = 60°, 360° – 60° [3]
= 60°, 300°
(b) Area = 1 (PQ)(QR)sin ∠PQR (b) cos x = –0.5
2 x = 180° – 60°, 180° + 60°
= 1 (40.2)(53.7) sin 55.32° x = 120°, 240° [3]
2
= 887.61 cm (c) cos x = –0.25
2
= 0.0888 m 2 [2] x = 180° – 75.522°, 180° + 75.522°
AB x = 104.48°, 255.52° [3]
11 sin ∠ACB = (d)
AC y
AB = 10 sin ( 180 – 90 ) 2
2
= 7.071 1

A = Area of sector AEC – (Area of sector ABC – 0 90° 180° 270° 360° x
Area of ΔABC) –1
) (
45
= ( 360 × π × 10 – 1 × π × 7.071 – 7.071 2 ) –2 [2]
2
2
4
2
= 25 cm 2 [7] 4 (a) x = 360° – 50°
= 310° [2]
(b) –0.75 ≤ y < – 1
Trigonometry graph y
1 (a) y = sin x 1
y

1
0 90° 180° 270° 360° x

x
0 90° 180° 270° 360° –1

–1 [2]
[2]
Cambridge IGCSE
TM
174 Ace Your Mathematics






Answers.indd 174 15/03/2022 11:08 AM

(c) y y
1 6

Y
4
x
0 90° 180° 270° 360° 2 X


x
–1 0 2 4 6
[3]
→ → →
7 7 Vectors and Transformations (b) YZ = YO + OZ
–4
3
= ( ) ( )
+
Vectors –4 0
–1
1 (a) Coordinates of B = ( )
–4
2 + 1
→
= ( –5 – 20 ) |YZ| = √1 + 16
( ) = 4.12 [3]
3
= –25 [1] 4 (a) XP = XO + OP
→ →
→
→ → → ( ) ( )
–3
8
(b) BC = BA + AC = 5 + 7
–8
–1
= ( ) ( ) = ( )
+
5
20
–15
( ) 12
–9
= 5 |XP| = √25 + 144
→
→ = 13 [3]
|BC| = √81 + 25 → 1 →
= 10.3 [3] (b) XF = 2 FP
( ) ( )
1 –20 –5
=

2 Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
[1]
(a) (i)
→
→
→
→
4 4 1 XO + OF = 1 (FO + OP)
→ 20 2
(ii) YX = ( ) [1] OF – 1 FO = 1 OP – XO
→ →
→
→
–4
(b) 5 – AB = –1 2 2
–1 + 3A = 4B 3 → 1 → →
Solve the simultaneous equations. 2 OF = 2 OP + OX
AB = 6 → → →
6 OF = 1 OP + 2 OX
A = 3 3
B
( ) ( )
6
–1 + 3 ( ) = 4B = 1 –3 + 2 –8
3 –5
3 7
B
( )
–B + 18 = 4B 2 19
= – [3]
4B + B – 18 = 0 3
2
9 5. (a) –1
B = 2 or B = – y
4 8 P
A = 3 or A = – 8 [5]
3 6
→ → →
3 (a) OY = OX + XY 4
0
8
2
+
– 2
= ( ) ( ) ( ) 2 R Q (6, 2)
2
4
10
( )
4
= 4 0 2 4 6 8 10 x [1]
Answers 175
Answers.indd 175 15/03/2022 11:08 AM

→ –4 → → →
(b) QR = ( ) (c) OM = OA + AM
1
→
→ → = a + 1 AC
PS = QR 2
→ → → 1 → →
OS = OP + PS = a + 2 (AO + OC)
9
)
–4
= ( ) ( ) = a + 1 ( –a + b + 2 b
+
8
1
3
2
( ) 1 5
5
= 9 [2] = 2 a + 6 b [2]
→ ( ) → → →
2
6 (a) OQ = 1 [1] (d) MB = MO + OB
)
→ 4 = – ( 1 a + 5 b + b
(b) TR = ( ) [1] 1 2 1 6
0
→ → = – 2 a + 6 b [2]
(c) UR = 2 × UO → → →
→ → 9 (a) (i) UR = UP + PR
= 2 × (UT + TO) = a + 4b [1]
[( ) ( )]
2
0
→ → →
= 2 × –2 + 1 (ii) PS = PU + US
( ) 3 →
2
= 2 × –1 = –a + 2 PR
( ) 3
4
= –2 [3] = –a + 2 (4b)
→ → → = –a + 6b [2]
7 (a) AC = AD + DC → → →
= –x + y [1] (iii) RS = RU + US
→ → → = –a – 4b + 6b
(b) AB = AE + EB = –a + 2b [2]
1 → → → → →
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
EB + EB
y =

(b) TS = TR + RS
3
4 → 1 → →
EB = y = UR + RS
3 4
→ 3y = 1 (a + 4b) – a + 2b
EB = 4
4 3
→ → → = – a + 3b [2]
CE = CB + BE 4
3y → → →
= x – [3] 10 (a) (i) SP = SO + OP
4 = –s + p [1]
→ → → → → →
(c) DE = DA + AE (ii) TP = TO + OP
1 → 1
= x + EB = − s + p [2]
3 2
3y
→ → →
= x + 1 ( ) (b) (i) RT = RS + ST
3 4
→ →
= x + y [2] = 1 PS + ST
4 3
→ → → 1 ( )
1
8 (a) BA = BO + OA = 3 (s – p) + – 2 s
= –b + a [1] 1 1
→ → → = – 3 p – 6 s [2]
(b) AC = AO + OC → → →
→ → → (ii) QO = QP + PO
= AO + OB + BC 1 → →
= –a + b + 2 b = 3 SP + PO
5 3 1
= –a + b [2] = (–s + p) + (–p)
3 3
= – 1 s – 2 p [2]
3 3
Cambridge IGCSE
TM
176 Ace Your Mathematics





Answers.indd 176 15/03/2022 11:08 AM

→ → 11 11
(c) RT = k QO = – p + r [2]
1 5 5
1( )
– 1 p = k – 2 p → → →
3 3 (c) RS = PS – PR
k = 1 11 11
1 2 = – 5 p + 5 r – (–p + r)
→ → 6 6
RT = k QO = – 5 p + 5 r
2
2( )
– 1 s = k – 1 s PR:RS
6 3 –p: – 6 p
5
k = 1 PR:RS = 5:6 [3]
2
2
k = 1 → → →
2 (d) TR = TO + OR
)
(
Scalar multiples of one another. = – 1 11 r – 6 p + r
→ → 3 5 5
Vectors RT and QO are parallel. [2] = 4 r + 6 p
→ → → 15 15
11 (a) (i) OS = OQ + QS → →
= q + 3t [1] OQ = mTR
6
→ 2 3 p = m ( )
p
(ii) OP = q [1] 5 15
5 m = 1.5
→ → →
→
→
(b) PR = PO + OR OQ = nTR
→ 2 →
4
= PO + OS 2 r = n ( )
r
5 5 15
= – 2 q + 2 (q + 3t) n = 1.5
5 5 Scalar multiples of one another.
6
= t → →
5 Vectors TR and OR are parallel. [4]
6
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
k =
[2]
5 Transformations
(c) PR :QS
–11
6 1 ( ) [2]
t :3t 11
5
6 :15 2 p(2, 3) [2]
Area of ΔOPR:Area of ΔOQS 3 (a), (b) and (c)
36:225 [2] y
(d) Area of ΔOPR:Area of ΔOQS 8
20 cm :125 cm 2 7
2
Area of PRSQ 6
= 125 – 20 5
= 105 cm 2 [2] 4 Q
3
→ → → 2
12 (a) (i) PR = PO + OR S 1 P
= –p + r [1] x
→ → → –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8
(ii) OQ = OP + PQ R –1
–2
→ 2 → –3
= OP + PR
5 –4
2 –5
= p + (– p + r)
5 –6
–7
= 3 p + 2 r [2] –8
5 5 [7]
→ → →
(b) PS = PO + OS
= – p + 11 r – 6 p
5 5




Answers 177







Answers.indd 177 15/03/2022 11:08 AM

4 (a), (b) and (c) Clockwise, 90°
y Centre (0, 0) [3]
6
7 (ii) Translation ( ) [2]
6 –10
5 (b) (–4, 3), (–5, 3), (–4, 6) and (–5, 6) [2]
4 (c) (–2, –5), (–2, –7), (4, –5) and (4, –7) [3]
3
2 y
1 P 7
R x 6
–8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8
–1 Q 5
–2 P 4 Q
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
–3 S 3
–4 2
–5 1
–6 0 x
–7 –7 –6 –5 –4 –3 –2 –1 –1 1 2 3 4 5 6 7 8
[8] –2
5 (a)(i) and (ii) –3
y –4
7 –5 R
6 –6 R′
A’’ 5 –7
4 –8
3 –9
2 –10
1
x 8 (a)(i), (ii) and (iii) y
–8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8
–1
–2 7
–3 A’ 6
–4 5
–5 C 4
[4] 3
(b) Enlargement 2
Centre (8, –4) 1
Scale factor of 2 [3] x
6 (a)(i), (ii) and (iii) –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8
–1
y
A –2 B
7 –3
6 –4
5 –5
4 –6
A 3 (a)(iii) –7
2 –8
1 –9
x –10
–7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8
–1 [8]
–2 (b) 1:4 [2]
(a)(i) –3 (a)(ii)
–4
–5
–6 B 8 8 Probability
–7
–8 Probability
–9
–10 4
[7] 1 (a) 11 [1]
(b) Rotation 4
Anticlockwise, 90° (b) 11 × 165
Centre (6, 2) [3] = 60 [1]
7 (a) (i) Rotation 4 4 4
2 (a) × = [2]
6 6 9

Cambridge IGCSE
TM
178 Ace Your Mathematics






Answers.indd 178 15/03/2022 11:08 AM

3 3 1 2 1 2
(b) × = [2] (b) (i) × = [2]
6 6 4 5 3 15
+
(c) 1 + 4, 4 + 1, 2 + 3, 3 + 2 (ii) ( 3 × 1 ) ( 2 × 2 )
1 1 1 1 2 1 1 2 5 4 5 3
× + × + × + × 5
6 6 6 6 6 6 6 6 = 12 [3]
= 1 + 1 + 2 + 2 12 11 66
36 36 36 36 7 (a) (i) 26 × 25 = 325 [2]
= 1 [3] 14 12 12 14
6 (ii) ( × ) ( × )
+
3 (a) 0.3 × 0.3 + 0.5 × 0.5 + 0.2 × 0.2 26 25 26 25
= 0.38 [2] = 168 [2]
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
(b) 1 – 0.38 325
= 0.62 [1] (b) (i) ( 14 × 13 × 12 ) ( 12 × 11 × 10 )
+
(c) 0.3 × 0.2 + 0.2 × 0.3 26 25 24 26 25 24
= 0.12 [2] = 2184 + 1320
4 (a) 0.15 × 0.85 = 0.1275 [2] 15600 15600
(b) 0.15 × 0.15 × 0.85 = 0.019 [2] 73
= [3]
5 (a) First ball Second ball 12 325 11 14 14 12 11
+
3 (ii) ( 26 × 25 × 24 ) ( 26 × 25 × 24 ) +
9 White 12 14 11
( 26 × 25 × 24 )

4 6 Black = 231 [4]
10 White 9 650
8 (a) 2 × 2 × 2 = 8 [2]
15 15 15 3375
6 2 2 2 8
10 Black 4 White (b) (i) 3 × 3 × 3 = 27 [2]
9 2 1 2 1 1 1
(ii) ( 3 × 5 × 15 ) × 6 + ( 5 × 5 × 5 )
43
5 Black = 375 [3]
9 [3] 9 (a) 2 [1]
( 4 3 ) ( 6 5 ) 7
(b) (i) 10 × 9 + 10 × 9 (b) (i) 1 [1]


= 2 + 1 6 1
2
15 3 (ii) 6 = 3 [1]
= 7 [2] (c)
15 1
(ii) 1 – 7 6 Daniel stands
15 next to Harry
8
= [2] Harry
6. (a) 15 stands at 5 Daniel does not
1 2 an end 6 stand next to
4 Buy her breakfast 7 Harry
circle


3 Bus 3 Do not buy her 5 Harry
does not
5 4 breakfast 7 stand at 2 Daniel stands
an end 6 next to Harry
2 circle
5 Walk 2 4 Daniel does
3 Buy her breakfast 6 not stand next
to Harry
1 Do not buy her [3]
3 breakfast
[3]


Answers 179







Answers.indd 179 15/03/2022 11:08 AM

6 1 3
2 1 5 2 (b) (i) × = [1]
(d) × + × 14 10 70
7 6 7 6 (ii) 1 – ( 6 × 2 + 6 × 7 + 6 × 1 )
= 1 + 5 14 10 14 10 14 10
21 21 4
= 2 [2] = 7 [2]
7 3 7 5 2 6 2 6 7
2 1 (iii) × + × + × + ×
(e) = [2] 14 10 14 10 14 10 14 10
6 3 = 17 [3]
(f) 2 = 1 28
n – 1 6 1
n – 1 = 12 13 (a) (i) 6 [1]
n = 13 [2] (ii) 4 = 2
( 14 13 12 ) 6 3 [1]
10 (a) 1 – 26 × 25 × 24 (iii) 0 [1]
) (
)
= 43 [2] (b) (i) ( 3 × 2 × 2 + 3 × 1 × 2 +
50
( 14 13 12 ) 21 6 6 6 6
(b) 26 × 25 × 24 × 3 = 50 [2] ( 2 1 )
( 14 13 12 ) ( 12 11 14 ) 6 × 6 × 2
(c) 26 × 25 × 24 × 3 + 26 × 25 × 24 × 3 = 12 + 6 + 4

= 21 + 231 36 36 36
50 650 = 11 [3]
)
) (
= 252 [3] 3 18 2 3 1
325 (ii) ( × × 2 + × × 2 +
11 (a) (i) 2 [1] 6 5 6 5
8 ( 2 × 1 × 2 )
(ii) 7 [1] 6 12 5 6 4
8
( 1 1 ) = 30 + 30 + 30
(b) (i) 8 × 8 × 5 11

=
5
15
=
[2]
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. [3]
4
2
3
64
( 1 1 ) (c) 6 × 5 × 4
(ii) 8 × 8 × 12 = 1 [2]
= 3 [2] 5
16 130 13
12 (a) 14 (a) (i) 180 = 18 [1]
Box A Box B 35 7
3 (ii) 180 = 36 [1]
10 Red (b) 35 × 34
130 129
= 119 [2]
7 Black 1677
3 Red 10 (c) (i) 45 × 44 = 99 [2]
14 5 2 140 139 973
95
45
45
95
14 Black 10 Red (ii) 140 × 139 + 140 × 139
855
6 = 1946 [3]
14 White 8 Black 7 6 6 7 2
10 15 × x – 1 + x × x – 1 = 5
x
42 42 2
2 x(x – 1) + x(x – 1) = 5
10 Red 84 2
=
x(x – 1) 5
Black 2
7 x – x – 210 = 0
1 10 x = 15 or x = −14 (invalid) [4]
10
White
[5]
Cambridge IGCSE
TM
180 Ace Your Mathematics






Answers.indd 180 15/03/2022 11:08 AM

(c)
8 9 Statistics Points scored by 24 players

Statistics

1 (a) (i) 1 [1]
(ii) 1 [1]
0 × 10 + 1 × 20 + 2 × 12 + 3 × 5 + 4 × 3 60 61 70 76 80 86 90 95 100104 110
(iii) Points
50 [4]
= 1.42 [2] 5 (a) (i) 90 [1]
3k + 71
(b) = 2 (ii) 60 [1]
k + 50 (b) (i) 40 [1]
3k + 71 = 2k + 100 (ii) 30 [1]
k = 29 [2] (c) Class 2 [1]
8
(c) × 100% = 16% [1] (d) Class 1, their median is higher and the
50 interquartile range is also higher. [2]
2 (a) Number of words Frequency 6 x + 14 + 16 + 16 = 14 × 4
0 – 4 4 10, 14, 16, 16 x = 10 [3]
5 – 9 5 7 (a) 0 ≤ k ≤ 16 [1]
10 – 14 8 (b) 12 [1]
15 – 19 2 (c) 10 [1]
0 × 8 + 1 × 11 + 2k + 3 × 2 + 4 × 0 + 5 × 1
20 – 24 3 (d) 8 + 11 + k + 2 + 0 + 1
25 – 29 3 = 1.8
[3] 22 + 2k =1.8
9 22 + k
(b) × 100% = 36% [1]
25 k = 88 [2]
8
2
(a) 150 × 0.26 = 39
[1]
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. [1]
(c)
× 100% = 8%
25 (b) 6 + 18 + 34 + 26 × 150 = 126 [2]
(d) (i) 14 [1] 100
(ii) 13 [1] (c) Grade A* = 0.06 × 360° = 21.6°
(e) 0 0 1 2 3 5 7 8 8 9 Grade A = 0.18 × 360° = 64.8°
1 0 1 2 3 3 4 4 4 5 9 Grade B = 0.34 × 360° = 122.4°
Grade C = 0.26 × 360° = 93.6°
2 0 0 1 5 6 7 150 – 126
Fail = × 360° = 57.6° [5]
Key: 1 | 2 means 12 words [4] 150
3 (a) (i) 2 [1] 9 (a) (i) 78 km/h [1]
(ii) 2 [1] (ii) 83 – 71 = 12 [2]
(b) 100 km/h
(iii) 2 [1] 10 (a) [2]
(iv) 3 × 8 + 2 × 15 + 1 × 7 = 2.03 [2] Frequency
8 + 15 + 7
8 density
(b) × 360° = 96° [2]
30 5 4.8
(c) Mean 4.5
175 × 7 + 182.5 × 13 + 187.5 × 6 + 192.5 × 2
= 4
28
= 182.41 cm [4]
4 (a) (i) 43 [1] 3
(ii) 86 [1] 2.2
(iii) 86 [1] 2
(iv) Mean = 2045 = 85.208 [2] 1.4
24
5 1 0.8
(b) × 100% = 20.8% [2]
24
Age (a)
0 20 30 40 50 60
[6]


Answers 181







Answers.indd 181 15/03/2022 11:08 AM

24 + 22 (e)
(b) × 100% Cumulative
7 + 45 + 24 + 22 + 8 + 4 frequency
= 46 × 100% 80
110
= 41.8% [2] 70
(c) Mean
15 × 7 + 25 × 45 + 32.5 × 24 + 40 × 60
22 + 50 × 8 + 57.5 × 4 50
=
110 40
= 3520
110 30
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
= 32 [4] 20
11
10
0
150 155 160 165 170 175 180 185 190
Height (cm)
142 145 151.5 162 175
Height (cm) [3]
[5] (f) (i) 170 – 171 [1]
(ii) 174.5 – 163.5 = 11

12 (a) Height (cm) Frequency 13 (a) City A: 24 [2]
150 < h ≤ 155 5 City B: 44 [2]
(b) 20 [1]
155 < h ≤ 165 15 (c) (i) 42 – 44 [1]
165 < h ≤ 170 18 (ii) 30 [1]
(d) (i) 58 [1]
170 < h ≤ 175 24 (ii) 62 [1]
175 < h ≤ 185 6 (e) 11, 9, 9, 8, 6, 4, 2, 1 [3]
1832.5
185 < h ≤ 190 2 14 (a) 40 = 45.8125 minutes [4]

[2] (b) (i) 5, 11, 33, 40 [2]
11772.5 (ii) Correct curve or ruled polygon
(b) = 168.18 [4]
70 Cumulative
(c) 170 < h ≤ 175 [1] frequency
(d)
Height (cm) Cumulative frequency
40
h ≤ 155 5
h ≤ 165 20
30
h ≤ 170 38
h ≤ 175 62
h ≤ 185 68 20

h ≤ 190 70
[2] 10




0
30 35 40 45 50 55 60
Time (t mins)
[3]
(iii) Median: 47 – 48
Interquartile range: 49 – 43.5 = 5.5 [3]




Cambridge IGCSE
TM
182 Ace Your Mathematics






Answers.indd 182 15/03/2022 11:08 AM

(c) (i) 5, 6, 29 [2] (c) (i) 20, 27, 23, 5, 5 [2]
(ii) Frequency 20 × 10 + 27 × 25 + 23 × 40 + 5 ×
density 55 + 5 × 70
(ii) = 30.25
80 [4]
3.0
(d)
2.5 Frequency density

2.0 60
1.5 50
1.0 40

0.5 30
0 20
30 35 40 45 50 55 60
Time (t mins) 10
[2]
15 (a) (i) 26 – 28 [1] 0 10 20 30 40 50 60 70 80
(ii) 39 – 20 = 19 [2] m, marks
(iii) 23 [1] [4]
(iv) 5 [1]
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
(b) 62
[1]
























































Answers 183







Answers.indd 183 15/03/2022 11:08 AM

Penerbitan Pelangi Sdn Bhd. All Rights Reserved.








































































Cambridge IGCSE
TM
184 Ace Your Mathematics






Answers.indd 184 15/03/2022 11:08 AM

Format 210mm X 297mm Extent= 192 pgs (10.14 mm) (70gsm paper) Status: CRC Date: 14/3


Cambridge IGCSE TM redue70% DA1301



ACE YOUR
MATHEMATICS






Workbook


Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
Cambridge IGCSE Ace Your Mathematics is written to improve students’
TM
approach to mathematics by providing them with high-quality educational
materials that uphold top academic standards, while allowing them to acquire
valuable techniques to excel in their examinations.

This workbook is written distinctively based on the Cambridge IGCSE 2020– Cambridge IGCSE TM ACE YOUR MATHEMATICS
2022 and 2023–2024 syllabuses for Mathematics (0580/0980) course at
Extended Level. The scope, sequence and level of the workbook has been
constructed to match the Cambridge IGCSE Year 10 and Year 11 syllabuses.

The workbook provides ample practice exercises for each topic. Answers
and fully-worked solutions are provided for all questions to enhance students’
understanding in how to master the approach to certain questions. It is
highly recommended as it serves as a good aid in evaluating students’
proficiency in mathematical skills, concepts and processes, helping them
achieve excellent results in examinations.


About the Author
Kung Girly received her Bachelor of Laws (Honours) from the University of London and practiced law
for a few months. She then decided to venture into her true passion in education and teaching. TM

Throughout her teaching pathway, she has educated many students who sat for the IGCSE exams Cambridge IGCSE
with excellent results. She has also prepared and conducted seminars for exam-going students,
providing them the foundation and tips to excel in the examinations. She also provides coaching
for students participating in international mathematics competitions. ACE YOUR
Ms Kung has seven years of experience in teaching IGCSE Additional Mathematics, and currently
MATHEMATICS
serves as a Director at Teras Murni Education Group. She is actively involved in educating students
to ensure they meet their academic potential.







Workbook







www.dickenspublishing.co.uk
DA1301
ISBN: 978-1-78187-259-8

Suite G7-G8, Davina House, 137-149 Goswell Road,
London, EC1V 7ET, United Kingdom.
E-mail: [email protected] Kung Girly