BUKU TEKS ADDMATHS FORM 4 (ANSWERS)

Consider triangle BCD,
h
a = sin C

Thus, h = a sin C…1

Consider triangle ABD,
h
c = sin A

Thus, h = c sin A…2
1 = 2, a sin C = c sin A
a c
sin A = sin C

sin A = sin C
a c

Self Practice 9.1 (Page 244)

1. (a) p = q = r
sin P sin Q sin R

(b) k = l = m
sin K sin L sin M

(c) 6 = 8
sin 40° sin 120°

Mind Challenge (Page 245)
The non-included angle is the angle that is opposite with one of the given two sides.

Example:

7 cm 4 cm

40°

Self Practice 9.2 (Page 246)

1. (a) m = 6.7
sin 55° sin 78°

m = 6.7 × sin 55°
sin 78°

= 5.611 cm

(b) sin m = sin 40°
8 6.5
sin 40°
sin m = 6.5 × 8

m = 52.29°
m 12.4
(c) sin 43° = sin 115°

m= 12.4 × sin 43°
sin 115°
= 9.331 cm

2

2. (a) XY = 100
sin 30.5° sin 66.5°
100
XY = sin 66.5° × sin 30.5°

= 55.344 m

Inquiry 2 (Page 247)

Condition Number of triangles Ambiguous
case
(i) a , h Zero No
(ii) a = h Yes
(iii) a . h One Yes
Two if h , a , 10 No
(iv) a , c One if a . 10 Yes
(v) a = c Two if h , a , 10 No
(vi) a . c Zero if a , h No
One No
One

Self Practice 9.3 (Page 249)

1. (a) The ambiguous case exist where the non-included angle ˙B = 62.5° and side AC is

shorter than side BC but is longer than the height.

(b) Ambiguous case does not exist because the side of PQ is longer than the side of QR.

2. (a) sin ˙QRP = sin 35°
15.5 10.5

sin ˙QRP = sin 35° × 15.5
10.5

˙QRP = 57.86° or 180° − 57.86°

= 57.86° or 122.14°
(b) When ˙R = 57.86°

˙Q = 180° − 35° −57.86°

PR = 87.14°
sin 87.14° 10.5
= sin 35°

PR = 10.5 × sin 87.14°
sin 35°

= 18.283 cm

When ˙R = 122.14°
˙Q = 180° − 35° −122.14°

PR = 22.86°
sin 22.86° 10.5
= sin 35°

PR = 10.5 × sin 22.86°
sin 35°

= 7.112 cm

3

Self Practice 9.4 (Page 250)

1. Let the water sprinklers as A and B, the water tap as P, the angle between the water sprinklers
and water tap as q and the distance between the water tap and the furthest water sprinkler be j.

sin q = sin 25° j = 6
5 6 sin 134.38° sin 25°

sin q = sin 25° ×5 j= 6 × sin 134.38°
6 sin 25°
q = 20.62° = 10.147 m
25° P
˙ABP = 180° − 25° – 20.62° 5m
= 134.38°
jm
Thus, the distance between the water tap and the
␪
furthest water sprinkler is 10.147 m A 6m B

2. sin Q = sin 50°
100 80

sin Q = sin 50° × 100
80
Q = 180° – 73.25°

= 106.75° Q 80 m
xm 100 m
˙R = 180° − 50° – 106.75° R
50°
= 23.25° P
x 80
sin 23.25° = sin 50°

x= 80 × sin 23.25°
sin 50°

= 41.224 m

Intensive Practice 9.1 (Page 250-251)

1. ˙A = 180° – 77° – 39°
= 64°
a 40.5
sin 64° = sin 77°

a= 40.5 × sin 64°
sin 77°
= 37.359 cm
c 40.5
sin 39° = sin 77°

c= 40.5 × sin 39°
sin 77°
= 26.158 cm

2. (a) BE = !w102 – 62

= 8 cm

CE = !w102 – 82

= 6 cm

DE = !w172 – 82

= 15 cm

4

(b) kcooss ˙EAB = 6
10
˙EAB = 53.13°

˙BCE = 53.13°

˙BCD = 180° – 53.13°

= 126.87°

sin ˙ABD = sin 53.13°
21 17

˙ABD = 81.20°

sin ˙CBD = sin 126.87°
9 17

˙CBD = 25.06°
(c) Triangle BDC and triangle BDA have the same angle and two sides with the same length.

3. (a) ˙PQR R

sin Q = sin 40°
14 6! 3

sin Q = 14 sin 40° 14 cm 6Ίෆ3 m
6! 3
Q = 60° or 120°
˙PQR is obtuse, thus ˙PQR = 120° P 40°
Q
(b) ˙PRQ = 180° – 120° – 40°

PQ = 20°
sin 20° 14
= sin 120°

PQ = 14 sin 20°
sin 120°
= 5.529 cm

4. sin A = sin 48° 15 cm C
20 15 A
sin 48° 20 cm
sin A = 15 × 20 48°
cB
A = 82.25°

C = 180° – 48° – 82.25°

= 49.75°

c = 15
sin 49.75° sin 48°
c = 15.405 cm
Perimeter of the frame = 4(15.405) cm

= 61.62 cm

5. sin A = sin 120° A
150 250
sin 120° 250 m
sin A = 250 × 150 120° 60° 60° D

A = 31.31° B 150 m C

B = 180° – 120° – 31.31°

= 28.69°

5

b = 250
sin 28.69° sin 120°

b = 138.58 m
Since the distance from Aida's house and Anita's house to Puan Azizah's house are the

same, the triangle ACD is the same. Thus, the distance between Anita's house and Aida's

house is 138.58 m.

Mind Challenge (Page 252) B
a
The cosine rule can be used for right-angled triangles
a2 = b2 + c2 – 2bc cos 90° c
a2 = b2 + c2 (Pythagoras theorem)

Self Practice 9.5 (Page 253 – 254) Ab C

1. (a) x2 = 32 + 52 – 2(3)(5) cos 45°

x = 3.576 cm

(b) 180° – 55° 13' = 124° 47'

x2 = 92 + 122 –2(9)(12) cos 124° 47'
x = 18.661 cm

(c) x2 = 752 + 1002 –2(75)(100) cos 32°
x = 53.891 m

2. (a) kcos q = 202 + 142 – 15.72
2(20)(14)

q = 51.38°

(b) kcos q = 10.82 + 122 – 72
2(10.8)(12)

q = 35.26° 92 + 62 – 102
2(9)(6)
(c) kcos (180° – q) =

(180° – q) = 80.94°

q = 99.06°

3. PR2 = 92 + 12.52 – 2(9)(12.5) cos 42.3°
PR = 8.4162 cm

kcooss ˙PQR = 52 + 8.72 – 8.41622
2(5)(8.7)

˙PQR = 69.93°

Self Practice 9.6 (Page 255)

1. ALentdtahiekadnisjtaarnackedbieatwnteaerna tphaepbanoairadlsahbex.x.
x2 = 252 + 452 – 2(25)(45) kcos 38°

x = 29.614 m

MThauksa,,tjhaeradkisdtainacnetabreatwpaepeannthiaelbahoa2r9d.s6i1s42m9.6. 14 m.
102 + 52 – 82
2. kcos D = 2(10)(5)

D = 52.41°

AD2 = 102 + 102 – 2(10)(10) cos 52.41°
AD = 8.8317 m

6

The length of wire = 8.8317 + 10 + 10 + 5 + 8
= 41.832 m

3. x2 = 92 + 9.52 – 2(9)(9.5) cos 120°
x = 16.023 m
Total distance travelled by Amin = 2(8 + 16.023)
= 48.046 km

Intensive Practice 9.2 (Page 255-256)

1. x2 = 32 + 4.52 – 2(3)(4.5) cos 62° x 3 cm
x = 4.071 cm 62°

y2 = 32 + 4.52 – 2(3)(4.5) cos (180 – 62)° 4.5 cm
y = 6.475 cm
y

Thus the lengths of sides of the card are 4.071 cm and 6.475 cm.

2. KL2 = 152 + 202 – 2(15)(20) kos 35°

KL = 11.555 km

The distance between town K and town L is 11.555 km.
282 + 492 – 362
3. kcooss q = 2(28)(49)

q = 46.50°
TMhauksa,,thsuedauntgalentbaertawleaelunatnhekarpoaulteBsuonfgBauRnagyaaRdaayna kshapipalanBdunBguangOarkOidrkiiadlashhi4p6i.s5406°..50

4. MQ = !w52 – 42

=3m

kcooss ˙MQP = 32 + 72 – 82
2(3)(7)

˙MQP = 98.21°

˙PQN = 360° – 90° – 98.21°

= 171.79°

PN2 = 72 + 42 – 2(7)(4) kcos 171.79°

PN = 10.974 km

PTahnejalennggbthatouf=st8on+es5=+81+0.597+410.974

= 23.97=42m3.974 m

Inquiry 3 (Page 257)

Triangle Base Height Area
I b a sin C ab sin C
II p q sin R pq sin R
III z x sin Y xz sin Y

Self Practice 9.7 (Page 258)

1. (a) LAureaas = 1 (16.2)(18.4) sin 49°
2
= 112. 482 cm2

(b) LAuraesa = 1 (7)(10) sin 125°
2
= 28.670 cm2

7

(c) ˙Z = 180° – 60° – 35°

= 85°

XY = 10
sin 85° sin 35°
XY = 17.3681 cm

LAuraesa = 1 (10)(17.3681) sin 60°
2
= 75.206 cm2

2. 1 (17)(LM) sin 20° = 78.72
2
78.72 × 2
LM = 17 sin 20°

= 27.078 cm

3. ˙DBC = 180° − 2(55)° ˙ABC = 180° − 24.18° – 55°
= 70° = 100.82°
ALureaas osef gtriiatinggaleBBDDCC ALureaas osefgtriitainggaleABACBC
1 1
= 2 (10)(10) sin 70° = 2 (20)(10) sin 100.82°

= 46.985 cm2 = 98.222 cm2

Area of triangle ABD = Area of triangle ABC – Area of triangle BCD
= 98.222 – 46.985

= 51.237 cm2

4. ALureaas XYZ = 1 (5.5)(7) sin 70°30'
2

= 18.146 m2

Self Practice 9.8 (Page 260-261)

1. s= 5.4 + 6.1 + 7.3
2
= 9.4

ALureaas = !9.4(9.4 – 7.3)(9.4 – 5.4)(9.4 – 6.1)

= 16.142 cm2

2. s= 11 + 12 + 5 s= 5+5+9
2 2
= 14 = 9.5

LAureaas osefgtriitainggaleEGEGHH ALureaas osefgtriiatinggaleEEFFJ J

= !14(14 – 11)(14 – 12)(14 – 5) = !9.5(9.5 – 5)(9.5 – 9)(9.5 – 9)
= 27.496 cm2 = 9.808 cm2

Area of shaded region = 27.496 – 9.808
= 17.69 cm2

8

3. s= 3x + 4x + 2x
2
9
= 2 x

! ( )( )( )9x9x–4x 9x – 3x 9x – 2x = !w135
2 2 2 2
135
16 x4 = 135

x4 = 16

x=2

Self Practice 9.9 (Page 262) 18 m

1. x = !w162 + 20.52 16 m x 11.5 m
= 26 m

s= 18 + 11.5 + 26 20.5 m
2
= 27.75

ALureaas opfercmarapiedtani

= 1 (16)(20.5) + ! 27.75(27.75 – 18)(27.75 – 11.5) (27.75 – 26)
2
= 164 + 87.716

= 716 m2

2. VQ = ! 102 + 42

= ! 116

sin R = sin 80°
! 116 15

R = 45°

˙QVR = 180° – 80° – 45°

= 55°

ALureaas opferthmeuiknacalninecdonsduorfnagce

= 1 (! 116 )(15) sin 55°
2
= 66.169 cm2

Intensive Practice 9.3 (Page 262)

1. (a) 1 (9)(AC) sin q = 18
2
( )1 2
3 = 18
2
(9)(AC)

1 AC = 6 cm
2
(b) × base × height = 18

1 × 9 × height = 18 A
2 height = 4 cm B 3 cm D 6 cm ␪ C

Area of triangle ABD 9

= 1 (3)(4)
2
= 6 cm2

2. LAuraesa I = 1 (5)(5) sin 108°
2
= 11.8882 cm2

x2 = 52 + 52 – 2(5)(5) cos 108° 5 cm
x = 8.0902 cm

LAuraesa II = 1 (8.0902)(8.0902) sin 36° 108° x III
2 I
= 19.2357 cm2 5 cm
II

LAureaas opfenretagguolanrspeeknattaagon

= 2(11.8882) + 19.2357

= 43.012 cm2

3. 21 (8)(11) ssiinn q = 30
q = 0.6818

q = 42.99° aotaru 137.01°

TPahnejapnogsssibislie klentiggtah yoafntghemthuinrgdksiinde

x2 = 82 + 112 – 2(8)(11) kcos 42.99°

x = 7.501 cm

aotrau
x2 = 82 + 112 – 2(8)(11) kcos 137.01°

x = 17.713 cm

4. 3x + (x – 1) + (3x + 1) = 63

7x = 63

x=9
TMhauksa,,thseisilebnaggthi seogfistiidgeasioaflathe2t7ricamng,l8e acrme 2d7ancm28, 8cmcm. and 28 cm.

s = 27 + 8 + 28
= 31.5 2

LAureaas osefgtiritainggale

= ! 31.5(31.5 – 27)(31.5 – 8)(31.5 – 28)

= 107.977 cm2

5. kcos q = 52 + 82 – 72
2(5)(8)
q = 60°

˙DAE = 180° – 60° – 40°

x = 80°
sin 80° 12
= sin 60°

x = 13.6459 m A

ALureaas osefgtriiatinggaleBBDDCC 12 m
1
= 2 (5)(8) sin 60°

= 17.3205 m2 B 5 m ␪ 40° E
7m ␪ D
ALureaas osefgtriiatinggaleAABBEE x

= 1 (5 + 13.6459)(12) sin 40° C
2
= 71.9121 m2

LAureaas otafnlahnd==1177.3.3220055++7711.9.9112211

= 89.233 m2 10

6. s = 17 + 15 + 16
= 24 2

Area of the rack = ! 24(24 – 17)(24 – 15)(24 – 16)

1 = 109.9818 cm2
2
(15)(Height of rack) = 109.9818

Height of rack = 14.664 cm

Self Practice 9.10 (Page 265)

1. kcos A = 102 + 4.0272 – 6.5752
2(10)(4.027)

A = 25.01°

ckos B= 4.0272 + 6.5752 – 102
2(4.027)(6.575)

B = 139.98°

kcos C = 6.5752 + 102 – 4.0272
2(6.575)(10)

C = 15.01°
JumSlauhmsuodf uinttpeeridoarlamngalne = 25.01° + 139.98° + 15.01°

= 180° (Proved)

2. (a) A M 5 cm P
10 cm
A 5 cmO
C
16 cm 16 cm
B 10 cm M
B MP C M x cm C 10 cm
10 cm
AC = ! 162 + 125
x = ! 52 + 102 = 19.5192 cm
= ! 125 cm

(b) A
A y

y cm 19.519 cm 16 cm

B C B 5 cm M C

y = ! 162 + 52 BC = ! 102 + 102
= 16.7631 cm = 14.1421 cm

s= 19.519 + 16.7631 + 14.1421
2
= 25.2121

LAuraesa soaftaphlanAeBACBC

= ! 25.2121(25.2121 – 19.5192)(25.2121 – 16.7631)(25.2121 – 14.1421)
= 115.865 cm2

11

3. OA =( )˜ –4   OB( )˜ 1= 1
3 = ! 12
  ˜ = ! (–4)2 + 32 ˜ + 12
OB  
OA   =!2

= 5

˜˜˜
AB = AO + OB
4 1

( ) ( ) = – 3 + 1
( )   = – 25

˜ = ! 52 + (–2)2
AB  

= ! 29

kcos ˙AOB = 52 + 2 – 29
2(5)(!w2 )

˙AOB = 98.13°

LAureaas osefgtiritainggaleAOAOBB
1
= 2 (5)(!w2 ) sin 98.13°

= 3.5 units22

Intensive Practice 9.4 (Page 265)

1. (a) cos P = 4
5
P = 36.87°

ALureaas osefgtriiatinggalePPRRSS

= 1 (11)(14) sin 36.87°
2
= 46.20 cm2

LAureaas osefgtiritainggalePQPQT T
1
= 2 (4)(3)

= 6 cm2

ALureaas oQfRQSRTS=T4=6.4260.2–06– 6

= 40.20 cm2

(b) SR2 = 112 + 142 – 2(11)(14) cos 36.87°
SR = 8.4024 cm

sin S = sin 36.87°
14 8.402
S = 88.76°

R = 180° – 88.76° – 36.87°

= 54.37°

˙SUP = 180° – 54.37°

= 125.63°

12

2. sin P = sin 50° H
13 10 10 km
P = 84.78°
13 km P

H = 180° – 84.78° – 50°

= 45.22° K 50°
KP2 = 132 + 102 – 2(13)(10) kos 45.22°

KP = 9.266

The distance between the oil rig and the tanker is 9.266 km.

3. (a) AC = ! 82 + 42 = 8.9443
CQ = ! 82 + 62 = 10
AQ = ! 42 + 62 = 7.2111

s= 8.9443 + 10 + 7.2111
2
= 13.0777

ALureaas osaftpalhanAeCAQCQ

= ! 13.0777(13.0777 – 8.9443)(13.0777 – 10)(13.0777 – 7.2111)
= 31.241 cm2

(b) The plane that has the same area as plane ACQ is plane DBR.

4. ˙ABC = 360° – 60° – 225° N

= 75° 120°

AC2 = 202 + 302 – 2(20)(30) kos 75° A 20 km
N
AC = 31.455 km

sin ˙BAC = sin 75° B
30 31.455
˙BAC = 67.11°
30 km

The bearing of Port Cindai from Port Astaka C

= 120° + 67.11°

= 187.11°

5. x = 800 ␪x t
sin 20° sin 25° 45°
x = 647.4309 m 20° a
800 m
t = sin 45°
647.4309
t = 457.803 m

Thus, the height of the mountain from the level of Arman’s position is 457.803 m.

Mastery Practice (Page 267 – 269)

1. (a) ˙C = 180° – 72° – 50° B

= 58°
sina72° 5.8
= sin 58° 5.8 cm 50° a
C
a = 6.504 cm 72°
b
sinb50° = 5.8 A
sin 58°
b = 5.239 cm

13

(b) kcos P = 3.632 + 6.562 – 8.282 Q
2(3.63)(6.56)
˙P = 105.03°

kcos Q = 3.632 + 8.282 – 6.562 3.63 cm 8.28 cm
2(3.63)(8.28)
˙Q = 49.92°
P 6.56 cm R
˙R = 180° − 105.03° − 49.92°

= 25.05°

2. (a) Y = 180° − 55° 13'

= 124° 47'

X = 180° − 124° 47' − 31° 52'

= 23° 21'

x = sin 14 47'
sin 23° 21' 124°

x = 6.756 cm
(b) ˙P = 180° – 77°

= 103°

x2 = 32 + 62 – 2(3)(6) cos 103°
x = 7.287 cm

3. (a) cos ˙ADC = –0.3

AC2 = 72 + 102 – 2(7)(10) (–0.3)

AC = 13.82 cm

(b) cos ˙ADC = – 0.3

˙ADC = 107.46°

Area of triangle ADC = 1 (7)(10) sin 107.46°
2
= 33.387 cm2

4. (a)

13.4 cm Y 13.4 cm Y
10 cm 10 cm

X 42.2˚Z1 42.2˚ Z1
X

(b) sin Z = sin 42.2°
13.4 10
Z1 = 64.17°
Z2 = 180° – 64.17°

= 115.83°

(c) ˙Y = 180° – 42.2° – 115.83°

= 21.97°

Area of triangle XYZ
1
= 2 (13.4)(10) sin 21.97°

= 25.066 cm2

14

5. (a) sin ˙ACB = sin 30°
9 5
˙ACD = 64.16°

AD2 = 52 + 62 – 2(5)(6) cos 64.16°
AD = 5.903 cm
1
(b) 2 (5.903)(10) sin ˙DAE = 20
sin ˙DAE = 20 × 2

(5.903)(10)

˙DAE = 42.66°

6. (a) RM = !w62 – 32 PMQ T
M
= 5.1962 cm R
4 T
tan ˙TRM = 5.1962
5 cm
˙TRM = 37.59° P 6 cm R
(b) TR = !w42 + 5.1962 R

= 6.5575 cm

ckooss P = 52 + 62 – 6.55752
2(5)(6)
P = 72.54°

LAureaas osaftpahlanTePTRPR
1
= 2 (5)(6) sin 72.54°

= 14.309 cm2

7. (a) sin C = sin 50°
8 7
C = 61.1°

The obtuse angle of ACB = 180° – 61.1°
= 118.9°

(b)

B

A B
C1 8m

(c) BC = sin 61.1° 7m
7 A 50° 61.1° C
BC = 6.1283 m
A 20°
AC2 = 82 – 6.12832
15
AC = 5.142 m C
M
8. (a) 1 (5.2)(5.2) sin ˙BAC = 8.69 B
2 sin ˙BAC = 8.69 ×
2
5.22
˙BAC = 40°

(b) AM = ckooss 20°
5.2
AM = 4.8864 m

˙AVM = 180° – 25° – 50°

AV = 105° V
sin 50° 4.8864
= sin 105°

AV = 3.875 cm 25°
4.8864 cm
(c) s= 3.875 + 3 + 5.2 A M
2
= 6.0375

ALureaas opferthmeuskuarafnacVeAVBAB

= ! 6.0375(6.0375 – 3.875)(6.0375 – 3)(6.0375 – 5.2)
= 5.763 cm2

9. (a)

N

B 235°
35°

25

R

(b) sin B = sin 35°
25 16
B = 180° – 63.66°

= 116.34°

R = 180° – 116.34° – 35°

= 28.66°
x 16
sin 28.66° = sin 35°

x = 13.38 km

The boat travelled as far as 13.38 km.
y 16
(c) (i) sin 52.68° = sin 63.66° B
63.66°
y = 14.20 km 16 16
52.68°
(ii) Bearing of the light house from the second position
= 90o + 63.36o = 153.36o R

10. (a) (i) JL2 = 402 + 802 – 2(40)(80) cos 44°
JL = 58.277 km

(ii) sin M = sin 44°
80 65
M = 58.76°

(iii) ˙L = 180° – 44° – 58.76°

= 77.24° 1
2
AreLauoafs KLM = (80)(65) sin 77.24°

= 2535.79 km2

16

(b) The furthest petrol station from petrol station K is petrol station M because the distance

between petrol station K and M is opposite to the largest angle.

(c) 1 (80)(Shortest distance) = 2535.79
2
ShoJartreaskt tdeirsdtaenkcaet = 2535.79
40
= 63.395 km

11. (a) (i) sin E = sin 50.05°
7 6.5
= 55.65°

˙CED = 180° – 55.65°

= 124.35°
(ii) AB2 = 52 + 92 – 2(5)(9) kcos 50.05°

AB = 6.943 cm

(iii) ˙CDE = 180° – 50.05° – 124.35°

= 5.6° 1
2
ALreuaasofsetrgiai ntigglae CED = (7)(6.5) sin 5.6°

= 2.22 cm2

˙ACD = 180° – 50.05°

= 129.95° 1
2
ALreuaasofsetrgiai ntigglae ACD = (9)(7) sin 129.95°

ALreuaasofsetrgiai ntigglae AAEEDD = 24.148 cm2

= 2.22 + 24.148

= 26.368 cm2

(b) BЈ E

BC 6.5 cm
D
7 cm
9 cm

A

12. (a) sin ˙YXZ = sin ˙XYZ
4 12
( )10
sin ˙YXZ = 11 ×4
12
10
= 33

(b) ˙YXZ = 17.64°

˙XYW = 65.38°

˙XYZ = 180° – 65.38°

= 114.62°

˙XZY = 180° – 114.62° – 17.64°

= 47.74°

17

Area of triangle XYZ

= 1 (4)(12) sin 47.74°
2
= 17.762 cm2

Thus, 1 (4)(XW) = 17.762
2
XW = 8.881 cm

(c) Triangle ZXY where XZ is constant, XY = XY, ˙XZY = ˙XZY (Diagram I)
Triangle ZXY where XZ is constant, ZY= ZY, ˙ZXY = ˙ZXY (Diagram II)

XX

YЈ Y YZ
Diagram I Z

YЈ
Diagram II

18

CHAPTER 10 INDEX NUMBERS

Inquiry 1 (Page 273)

4. (a) Construction, I = 388 × 100 = 89.40
434

Manufacturing, I = 253 × 100 = 33.64
752
194
Agriculture, I = 174 × 100 = 111.49

Farming, I = 268 × 100 = 62.04
432

Service, I = 649 × 100 = 241.26
174

(b) The factors that cause such a change::

• Government policies on hiring foreign workers change according to sector

• Focus on the development of the country changes from year to year

• Job vacancies in a sector always change according to year

• Accept suitable answers

(c) Implication of the entry of foreign workers to the country:

• Positive effects
° Increases the productivity of manufacturing
° Fasten the process of economic development of the country
° Fills the chronic vacancy of labour manpower

• Negative effects

° Badly affects the local citizens' sustenance because the foreign workers' wages are
far lower

° The entry of illegal immigrants brings social problems
° Local citizens have to share facilities that are provided by the government

with foreign workers

• Accept suitable answers

(d) Steps to overcome negative effects of the entry of foreign workers in this country:

• Laws of employment of foreign workers must take into consideration of the wages of
local citizens so that there is no competition for job opportunities

• The government needs to attract local workers to get involved with 3-D work (Dirty,
Dangerous, Demanding)

• Improvements in education and training for the 3-D job sector in vocational institutes
to the youth

• Separate the needs of foreign workers with local citizens so that there is no
competition in using facilities by organising facilities according to needs and

suitability by prioritising local citizens

• Accept suitable answers

1

Mind Challenge (Page 274)

Index numbers can give a value of 100. This situation happens when the value that is taken in
the current year is the same as the value that was taken last year. This means that there is no
increase or decrease in the current year compared to the base year.

Self Practice 10.1 (Page 275)

1. I= 61 956 × 100
75 376
= 82.20

The decrease in number of registered commercial vehicles is 17.80% in 2017 as compared

to 2015.

2. I= RM4 033 × 100
RM3 578
= 112.72
The increase in a family's monthly expenditure is 12.72% in 2017 as compared to 2014.

3. 90.23 = Q2016 × 100
720 440 105
Q2016 = 650 053 107 metric tonnes

4. p= 225 × 100
150
= 150
95.8
5. I =2012/2011 101.4 × 100

= 94.48

Inquiry 2 (Page 277)

2. (a) Conjecture: The cases of work accidents increase from year to year regardless of whether
the accident happened during the commute to work or at the workplace.

(b) The effect if the rate of work accident in our country increases:
• The safety and health standards of workers in Malaysia are not observed well by the
authorities
• The country loses skilled labour if work accidents are not controlled
• The implication is serious to workers such as loss of income, loss of life and
impairment of the body
• Accept suitable answers

(c) Causes of increase of work accidents in our country:
• The attitudes of employers of which they do not take the safety and health standards
seriously at the work place
• Lack of awareness amongst workers regarding safe commute to work
• Lack of specific training programs for accident reduction and safety
• Accept suitable answers

(d) Methods to reduce the rate of work accidents in our country:
• The government needs to make the accident reduction and safety programs
compulsory in the work place to all employers.
• Employers need to take responsibility to educate and implement awareness
programmes to workers

2

• Implementation of safety and health awareness programs as well as safe driving to and
from work

• Workers need to cooperate and comply with safety procedures and apply them at all
times

• Accept suitable answers

Self Practice 10.1 (Page 277)

1. I= 140 × 100
125
= 112
110
2. I= 105 × 100

= 104.76

Intensive Practice 10.1 (Page 278)

1. I= 27.4 × 100
25.3
= 108.3
The increase of average temperature in town P is 8.3% on February 2017 as compared to
January 2017.

2. I =2015/2012 120 × 100
130
= 92.31
The decrease in price for an item is 7.69% in 2015 as compared to 2012.

3. 80 = 0.40 × 100 y z= 1.00 × 100
x 2.00 0.80
140 = × 100

x= 0.40 × 100 y = 140 × 2.00 = 125
80 100

= 0.5 = 2.80

4. p= RM5.80 × 100 q= RM7.65 × 100
RM5.80 RM5.80

= 100 = 131.90

r= RM7.80 × 100 s= RM7.30 × 100
RM5.80 RM5.80

= 134.48 = 125.86

5. I= 118 × 100
110
= 107.27

Inquiry 3 (Page 279) 50)
(120 × 30) + (127 × 15) + (108 × + (107 × 5)
2. Average price index = 114.4 100
=

3. The role of percentage represents the value of importance for each material and is known as
the weightage. If this value of percentage is the same for all materials, then the composite index is
the composite index without weightage.

3

Mind Challenge (Page 280)

The composite index with weightages has a different value of weightage for each item while the
composite index without weightages has the same weightage for each item. The importance of
weightages in the calculation of composite index is to get the correct value of indices according to
their respective importance.

Self Practice 10.3 (Page 281)

1. I– = 105 + 112 + 98
3
315
= 3

= 105

2. 120 = 136(3) + m(2) + 108(3)
3+2+3
2m = 228

m = 114

Brainstorming (Page 282)

The decrease in nickel prices in 2018 only gives a small effect towards the total cost of the
manufacturing of steel because the percentage of nickel in the manufacturing of steel is very low and
the increase of steel and chromium prices in 2018 cannot be covered by the decrease in nickel prices.

Self Practice 10.4 (Page 282)

1. (a) IA = 2.10 × 100 IB = 1.56 × 100
1.40 1.50

= 150 = 104

IC = 1.92 × 100 ID = 5.58 × 100
1.60 4.50

= 120 = 124

(b) I– = 150(10) + 104(20) + 120(40) + 124(30)
10 + 20 + 40 + 30
12 100
= 100

= 121

There is an increase of 21% for the prices of all items in 2016 compared to 2010.

(c) 121 = RM2.65 × 100
Q2010

Q2010 = RM2.19

Thus, the price of the roof tiles in 2010 is RM2.19.

2. (a) a= 23.00 × 100 b= 12.00 × 100
20.00 8.00

= 115 = 150

c= 18.00 × 100 d = 100 – 8 – 12 – 20 – 27
16.00 = 33

= 112.5

4

(b) I– = 120(8) + 115(12) + 150(20) + 112.5(27) + 130(33)
8 + 12 + 20 + 27 + 33
12 667.5
= 100

= 126.68

(c) 126.68 = Q2019 × 100
RM35
= RM44.34

Thus, the price of the souvenir in 2019 is RM44.34.

(d) 110

Intensive Practice 10.2 (Page 283-284)

1. I– = 120(60) + 130(40)
100
12 400
= 100

= 124

2. I– = 110(3) + 100(2) + 80(5)
3+2+5
930
= 10

= 93

3. 85(40) + 72(40) + 68(20) = 76.4
100

4. (a) 133 = 150(10) + 140(40) + m(30) + 115(20)
10 + 40 + 30 + 20
30m = 3900

m = 130

(b) I– = 150(10) + 126(40) + 130(30) + 138(20)
= 13 200 10 + 40 + 30 + 20

100

= 132

(c) 132 = Q2023 × 100
RM19.50
Q2023 = RM25.74

Thus, the cost of production of facial wash is RM25.74.

Mastery Practice (Page 285-287)

1. (a) 120 = 1.20 × 100 125 = y × 100 z= 2.20 × 100
x 0.80 2.00

x = 1.00 y = 1.00 = 110

(b) I– = 110(4) + 125(1) + 100(2) + 120(3)
4+1+2+3
1 125
= 10

= 112.5

5

2. n = 122

m= RM6.05 × 100
RM5.00

= 121

3. (a) P2018 × 100 = 120
7.60
P2018 = RM9.12

(b) I– = 90(50) + 160(30) + 120(10) + 140(10)
50 + 30 + 10 + 10

= 119

41 650 × 100 = 119
Q2015 Q2015 = 41 650

119 × 100

= 35 000
The cost of production in 2015 is RM35 000.

(c) 160 × 119 = 190.4
100

An increase of 90.4% from the year 2015 to the year 2020

4. (a) I =2015/2005 0.722 × 100
1.126

= 64.12
x
(b) 83 = 1.126 × 100

x = 0.935 million tonnes

(c) I =2020/2005 105 × 83
100

= 87.15

5. (a) RM10 × 100 = 125
RM8

The rate of increase from 2000 to 2015 is 25%. Thus the rate of price

increase from 2015 to 2020 is 50%.
x
RM10 × 100 = 150
x = RM15

The price of the item in 2020 is RM15.

(b) RM15 × 100 = 187.5 aotar u 125 × 150 = 187.5
RM8 100

6. (a) 108 = 107(2) + 118(x) + 94(1) + 105(2x)
2 + x + 1 + 2x

108 = 308 + 328x
3 + 3x

324 + 324x = 308 +328x

4x = 16

I– = x=4
128.4(2) + 118(4) + 94(1) + 94.5(8)
(b) 2+4+1+8

= 1 578.8
15

= 105.25 6

7. I =2017/2015 145 × 100
109

= 133.03

8. p= 231 × 100 q= 156 × 100 r= 170 × 150
165 120 100

= 140 = 130 = 255

9. (a) Q2020 = 3.68 + 2(3.68 – 2.45)
= 6.14 million

(b) I = 6.14 × 100
3.68

= 166.85

The number of visitors in the year 2020 increases by 66.85% compared to the year of

2017

10. 80x + 130y = 120 80x + 140z = 125
x+y x+z

80x + 130y = 120x + 120y 80x + 140z = 125x + 125z

x = 1 x = 1
y 4 z 3

x:y=1:4 x:z=1:3

Thus, the ratio of x : y : z = 1 : 4 : 3.

11. (a) P2018 × 100 = 110
P2014

166 × 100 = 110
P2014
P2014 = 150.91

P2014 × 100 = 80
P2010

1P5020.1901 × 100 = 80
P2010 = 188.64

Thus, the price of the safety helmets is RM188.64 in 2010 and RM150.91 in 2014.

(b) 166 × 100 = 88
188.64

The percentage of decrease in price in 2010 compared to 2018 is 12%

12. (a) I = RM172.5 × 100
RM150

= 115 115 × 115
100
(b) I =2020/2018

= 132.25

132.25 = Q2020 × 100
RM150

= RM198.38
Thus, the price for the service charge in 2020 is RM198.38.

7