J1 Praktis Topikal KSSM: Matematik Tingkatan 3 – Jawapan

(b) x3 = 8134 × 272 × x0

x3 = 33 × 272 × 1 x3 = 27 × 272

x3 = 273

x = 27

(c) 20

+ 3–2 × 2713

= 1 + 19

× 3 = 113

BAB 2 Bentuk Piawai

1 B 2 A 3 C 4 C 5 D6 B 7 D 8 B 9 D

10 (a) Palsu/False (b) Benar/True

(b) Palsu/False (d) Benar/True

11 (a) ✗ (b) ✓ (c) ✗ (d) ✗12

106.235 × 10–2

10 623.5

106.235 × 102

10.6235106.235 ÷ 10–2

1.062351 062 350 ÷ 105

•

•

•

•

•

•

•

13 (a) 73 600, 0.00851, 22 000

(b) (i) 370.9 × 10–5

10–8 = 3.709 × 10–3 ÷ 10–8

= 3.709 × 10–3–(–8) = 3.709 × 105 = 370 900 (ii) 29 × 24 × 60 = 41 760 = 4.176 × 104

(c) Jumlah bilang mentol/Total number of bulbs: 9 560 × 7 = 66 920 Bilangan mentol yang tidak pecah

Number of unbroken bulbs

66 920 – 420 = 66 500 66 500 × RM8.60 = RM571 900

14 (a) (i) (a) 4.26 × 10–4 (b) 2.64 × 108

(ii) 0.003564 (b) (i) 150.892 × 10–2 = (1.50892 × 102) × 10–2

= 1.50892 × 102+(–2) = 1.50892 × 100

Bandingkan 1.50892 × 100 dengan 1.50892 × 10s,Compare 1.50892 × 100 to 1.50892 × 10s,

∴ s = 0 (ii) Isi padu sfera/Volume of sphere

= 43

(3.142)(103)

= 4 189.333 = 4 190 m3

(c) Tinggi/Height: 3.65 × 10–4 m = 3.65 × 10–4 × 102 = 3.65 × 10–2 cm

Luas segi tiga = 12

× tapak × tinggi

Area of triangle = 12

× base × height

BAB 1 Indeks

1 D 2 B 3 D 4 B 5 B6 C 7 A 8 D 9 A 10 D

11 n3 × n4

n8

(n2)4

n3

n9 ÷ n6

n7nn–2

•

•

•

•

•

•

•

12 (a) Palsu/False

(b) (i) 32 (ii) 0 (iii) 8

13 x4

x–4 , (x4)2 , x0

x–8

14 (a) (i) (m4)12 = m2 (ii) s7 ÷ s2 = s5 (iii) k6 × k3 = k9

(b) 52y ÷ 5y–2 = 625 52y–(y–2) = 54

5y+2 = 54

y + 2 = 4 y = 2(c) 34x+2 × (3x)2 = 81 34x+2+2x = 34

36x+2 = 34

6x + 2 = 4 6x = 2

x = 13

15 (a) (72)–3 =

1(7

2)3 = 1

(3438 )

= 8343

(b) (8k4)13 × k

53

3√k = 2k

43 × k

53

k13

= 2k

43

+53

k13

= 2k83

(c) 1ab

= 8–3

1ab

= 183

Bandingkan/Compare, a = 8, b = 3 a + b = 8 + 3 = 11

16 (a) (i) s8t7 (ii) 12y2

(iii) a3b4

(b) (313 × 81

23

2512 )–2

= (313 × (34)

23

√25 )–2 = ( 3

13

+83

5 )–2

= (275 )–2

= 25

729

(c) 3t × 27 × 9 = 34t–1

3t × 33 × 32 = 34t–1

3t+3+2 = 34t–1

t + 5 = 4t – 1 6 = 3t

t = 63

= 2

17 (a) (i) h–3 × h8

h4 = h(–3)+8–4 = h

(ii) 4–

12 × 27 ÷ (82)

13

1

√4 × 27 ÷ [(23)2]

13

= 2–1 × 27 ÷ 22 = 2–1+7–2 = 16

Jawapan

J2 J3 Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2019

7.7745 × 102 = 12

× ST × (3.65 × 10–2)

(7.7745 × 102) = ST × 3.65 × 10–2

ST = 1.5549 × 103

2.65 ×10–2 = 4.26 × 104 cm

15 (a) 0.000007195 = 7.195 × 10–6 A x 10n, 1 ≤ A < 10

Bandingkan 7.195 × 10–6 dengan m × 10n,Compare 7.195 × 10–6 to m × 10n.

m = 7.195, n = –6

(b) (2.1 × 10–2)2

5 ×10–7 = 2.12 × (10–2)2

5 × 10–7

= 4.41 × 10–4

5 ×10–7 = (4.41 ÷ 5) × (10–4 ÷ 10–7)

= 0.882 × 10(–4)–(–7) = (8.82 × 10–1) × 103

= 8.82 × 102 atau/or 882(c) Jumlah bilang kertas (1 minit):

Total number of papers (1 minit):

5 × (6 × 104) = 300 000 = 3 × 105 kertas/papers

Tempoh masa/Duration

5 × (8 × 60 minit/minutes) = 2 400 = 2.4 × 103 minit/minutes

Jumlah keras dibalut/Total of wrapped papers: (3 × 105) × (2.4 × 103) = (3 × 2.4) × 105 × 103

= 7.2 × 105+3 = 7.2 × 108

16 (a) Kad 1/Card 1: 38 983 (3 a.b./s.f.) = 39 000 Kad 2/Card 2: 273.3 × 10–2 = 2.733 Beza/Difference: 39 000 – 2.733 = 38 997.267 (b) (i) 320 000 + 5.4 × 106 = 3.2 × 105 + 5.4 × 106

= 0.32 × 106 + 5.4 × 106

= 5.72 × 106

(ii) Isi padu silinder/Volume of cylinder: V = π × jejari2 × tinggi

V = π × radius2 × height

= 227

× (162 ) × 210

= 422.4 = 4.224 × 102 mm3

(c) Isi padu prisma/Volume of prism:

= 12

× 4 × 3 × 1 = 6 m3

Jisim prisma/Mass of prism

= 6 × 3 015 = 18 090 kg Bilangan kubus terbentuk/Number of cubes formed:

= 18 090 × 103

4.5 1 kg = 1 000 g = 103 g

= 4 020 000 = 4.02 × 106

BAB 3Matematik Pengguna: Simpanan dan Pelaburan, Kredit dan Hutang

1 D 2 B 3 C 4 A 5 C 6 (a) Akaun simpanan/Savings account

(b) Akaun semasa/Current account

(c) Akaun simpanan tetap/Fixed deposit account

(d) Akaun semasa/Current account

7 (a) Palsu/False

(b) Benar/True (c) Palsu/False

(d) Benar/True

8 Jenis

PelaburanType of

investment

PernyataanStatement

Bebas risiko, tahap pulangan rendah, tahap kecairan tinggiRisk-free, low return levels, high liquidity levels

Tahap risiko rendah, tahap pulangan sederhana, tahap kecairan tinggiLow risk level, simple rate of return, high liquidity level

Tahap risiko rendah, tahap pulangan tinggi, tahap kecairan rendahLow risk level, high rate of return, low liquidity level

SimpananSavings

Simpanan tetapFixed deposit

Unit amanahUnit trust

HartanahReal estate

(a)

(b)

(c)

(d)

•

•

•

•

•

•

•

9 (a) P = 15 000

r = 5100

= 0.05

t = 3 n = 1

MV = P( rn)nt

= RM15 000(1 + 0.051 )3

= RM17 364.375

Faedah/Interest = RM17 364.375 – RM15 000 = RM2 364.375

(b) Tahun/Year Faedah/Interest

Tahun 1Year 1

P1 = RM5 000

I1 = RM5 000 × 2.9%

= RM145

Tahun 2Year 2

P2 = RM5 000 + RM145 = RM5 145

I2 = RM5 145 × 2.9%

= RM149.205

Tahun 3Year 3

P3 = RM5 145 + RM149.205

= RM5 294.205 I

3 = RM5 294.205 × 2.9%

= RM153.532

Tahun 4Year 4

P4 = RM5 294.205 + RM153.532

= RM 5 447.737 I

4 = RM5 447.737 × 2.9%

= RM157.984

Jumlah faedahTotal interest

RM(145 + 149.205 + 153.532 + 157.984) = RM605.721

(c) Bulan pertama/First month

Faedah/Interest = RM20 000 × 3100

× 112

= RM50.00 Jumlah pinjaman pada akhir bulan pertama

Total loan at the end of 1st month

= RM20 000 + RM50 = RM20 050 Baki pinjaman/Loan balance = RM20 050 – RM400 = RM19 650 Bulan kedua/Second month Pinjaman/Loan = RM19 650

Faedah/Interest = RM19 650 × 3100

× 112

= RM49.125

J2 J3 Praktis Topikal KSSM: Matematik Tingkatan 3 – Jawapan

Jumlah pinjaman pada akhir bulan keduaTotal loan at the end of 2nd month

= RM19 650 + RM49.125 = RM19 699.125 Baki pinjaman/Loan balance = RM19 699.125 – RM400 = RM19 299.125

10 (a) Faedah mudah/Simple interest,

I = Prt = RMx × 4 × 10100

= RM0.4x

Jumlah amaun/Total amount = RM0.4x + RMx = 49 000 RM1.4x = 49 000

x = 49 0001.4

= RM35 000

(b) Pelan 1/Plan 1

A = 10 000(1 + 0.062 )2×3

= 10 000(1 + 0.03)6

= 10 000(1.03)6 = RM11 940.52 Pelan 2/Plan 2

A = 10 000(1 + 0.121 )1×3

= 10 000(1 + 0.12)3

= 10 000(1.12)3 = RM14 049.28 Pelan 2 menawarkan lebih pulangan

Plan 2 offers more returns

(c) I = Prt

121 = P × 2.2100

× 1

∴ P = RM5 500

MV = P(1 + rn)nt

= 5 500(1 + 0.0221 )2

= 5 744.662 Faedah/Interest

RM5 744.662 – 5 500 = RM244.66211 (a) A = P + PRt

Jumlah bayaran balik/Total payment

= RM30 000 + (RM30 000 × 3.6% × 10) = RM30 000 + RM10 800 = RM40 800

(b) (i) Bayaran minimum/Minimum payment

= 5% × RM3 400 = RM170 (ii) Baki belum dijelaskan/Outstanding balance

= RM3 400 – RM170 = RM3 230 Tempoh dikenakan caj kewangan

Period subjected to financial charges

= 15 hari/days = (15 ÷ 365) tahun/years

Faedah yang dikenakan/Interest charged

= RM3 230 × [(18 ÷ 100) × (15 ÷ 365)] = RM23.89 Jumlah terkini (baki tertunggak) dalam bulan

Jun/Current amount (balance outstanding in June

= RM3 230 + RM23.89 = RM3 253.89(c) Pinjaman/Loan = RM72 000 – (10% × RM72 000) = RM72 000 – RM7 200 = RM64 800 Kadar faedah selama 7 tahun/Interest rate for 7 years

= 3.5% × RM64 800 × 10 = RM22 680 Jumlah pinjaman = pinjaman + faedah

Total loan = loan + interest

= RM64 800 + RM22 680 = RM87 480 Bayaran ansuran bulanan/Monthly payment

= RM87 480 ÷ (12 × 7) = RM1 041.4312 (a) P = 15 000

r = 4100

= 0.04

t = 1

n = 123

= 4

MV = P(1 + rn)nt

MV = 15 000(1 + 0.044 )4(1)

= RM15 609.06

(b) Modal/Capital = RM1.80 × 2 000 = RM3 600 Dividen/Interest = 6% × RM3 600 = RM216 Peningkatan harga saham/Increase of share price

= RM2 – RM1.80 = RM0.20 Keuntungan modal/Capital gains

= RM0.20 × 2 000 = RM400 Jumlah keuntungan/Total profit

= RM216 + RM400 = RM616 Nilai pulangan pelaburan/Return on investment

= 6163 600

× 100 = 17.11%

(c) Jumlah pinjaman bank/Total bank loan

= RM150 000 – RM15 000 = RM135 000 Keuntungan/Profit

= RM250 000 – (70% × 135 000) – RM5 000 – RM5 000 = RM250 000 – RM94 500 – RM10 000 = RM145 500 Nilai pulangan pelaburan/Return on investment

= 145 500150 000

× 100 = 97%

BAB 4 Lukisan Berskala

1 A 2 D 3 D 4 D5 C 6 D

7

•

B

A

A B•

•AB

• 1 : 2

• 1 : 1

• 1 : 12

• AB

•

•

•

J4 J5 Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2019

8 (a) ✗ (b) ✓ (c) ✓ (d) ✓

9 (a) Panjang/Length: 12

× 20 cm = 10 cm

Lebar/Width: 12

× 10 cm = 5 cm

Tinggi/Height: 12

× 6 cm = 3 cm

(b) Isi padu/Volume: 10 × 5 × 3 = 150 cm3

(c) (5 × 100) × (4 × 100) × 5(10 × 3 × 5)

= 1 000 000150

= 6 666 23

≈ 6 667

10 (a) (i) (a) 1212

: 60 × 10012

= 1 : 500

(b) 7 × 500 = 3 500 cm = 35 m

(ii) 12

× [(12 + 6) × 500] × (7 × 500)

= 15 750 000 cm2 = 1 575 m2

(b) (i) Lukisan berskala/Scale drawing

Lukisan sebenar/Actual drawing =

1(2

3)

Lukisan berskala/Scale drawing

72 =

1(2

3)

Lukisan berskala/Scale drawing = 72(2

3)

72 × 32

= 108 cm

(ii) Peratus pengurangan perimeter lukisan berskala ialah 40% kerana peratus pengurangan lukisan sebenar adalah sama dengan lukisan berskala.The percentage of decrease of the scale drawing's perimeter is 40% because the percentage of decrease of actual drawing is equal to the scale drawing.

Panjang sisi/Length of sides

= (72 ÷ 8) × 60% = 9 cm × 0.6 = 5.4 cm Perimeter: 5.4 cm × 8 = 43.2 cm Perimeter (lukisan/drawing):

43.2 cm × 32

= 64.8 cm Peratus/Percentage:

64.8108

× 100% = 60%

100% – 60% = 40%

11 (a) 66

: 84 × 1 000 × 1006

= 1 : 1 400 000

(b) 21 × 100 cm100

= 21 m

12 × 100 cm100

= 12 m

Luas tapak/Base area: 21 × 12 = 252 m2

(c) (i) (5.4 × 330) × (4.8 × 330) = 2 822 688 cm2

= 282.2688 m2

(ii) 20 cm = 0.2 m Luas 1 jubin/The area of 1 tile: 0.2 m × 0.2 m = 0.04 m2

Bilangan jubin/The number of tiles: 282.2688 m2 ÷ 0.04 m2 = 7 056.72 ≈ 7 057 keping jubin/tiles

Anggaran kos/Estimated cost: 7 057 × RM12 = RM84 68412 (a)

(b) (i) Jejari/Radius = 3.98192

× 320 000 000

= 637 104 000 cm ÷ 100 = 6 371 040 m = 6 371.04 km (ii) (2.3 + 1.6) × 400 000 = 1 560 000 cm

1 560 000100 × 1 000

= 15.6 km

(c) Tinggi segi tiga/Height of triangle:

√392 – 152 = √1 521 – 225 = √1 296 = 36 cm Luas lukisan berskala/Area of scale drawing:

12

× 303

× 363

= 60 cm2

BAB 5 Nisbah Trigonometri

1 C 2 C 3 C 4 C5 A 6 B

7 (a) ✗ (b) ✓ (c) ✗ (d) ✓

8 (a) (i) kos/cos y = 1213

(ii) tan y = 512

(b) AC2 = 172 – 15 2 = 64

AC = 8 cm

kos/cos θ = 817

9 (a) (i) 3 sin 30o – 5 kos/cos 60o = 3 × ( 12

) – 5 × ( 12

) = –1

(ii) 5 – 3 kos/cos 0o

4 sin 90o – tan 45o = 5 – 3 × 1

4 × 1 – 1 = 2

3(b) Tinggi LK/The height of LK:

12

× 50 × LK = 510

LK = 51025

= 20.4 cm

J4 J5 Praktis Topikal KSSM: Matematik Tingkatan 3 – Jawapan

tan ∠LJK = 20.450

∠LJK = tan–1 ( 20.450 )

∠LJK = 22.195o

kos/cos 22.19o = JKJL

0.9259 = 50JL

JL = 500.9259

= 54.00 cm

(c) tan 60o = 13VX

VX = 13tan 60o

= 7.5055 m

VW = 7.5055 ÷ 2 = 3.7528 m

tan θ = 133.7528

θ = 73.9o

10 (a) (i) 3 tan 45o = 3 × 1 = 3

(ii) 8

√3 sin 60o – 3 kos/cos 90o

= 8

√3 × √3

2 – 3 × 0 = 8

2 = 4

(b) (i) QRTQ

= tan 63o

QR4

= tan 63o

QR = 4 tan 63o = 7.85 m Lebar sungai/Width of river = 7.85 m

(ii)

12

TQ

TV = kos/cos 63o

12

(4)

TV = kos/cos 63o

TV = 2kos/cos 63

= 4.405 m

(c) tan 65o = PQ72

PQ = 72 tan 65o = 154.4045 km

tan θ = 154.404534

θ = tan–1 4.5413 θ = 77.58o

11 (a) (i) PR = √102 – 62 = 8 cm

tan β = 86

= 43

(ii) sin α + kos/cos β = 610

+ 610

= 1210

= 65

(b) (i) ∠MNP = 180o – 15o – 15o = 150o

∠PNO = 180o – 150o = 30o

sin 30o = OP10

OP = 10(0.5) = 5 cm Tiang/Pole = 5 cm (ii) ON = 10 kos/cos 30o

tan θ = OPPN

tan θ = 8.66035

tan θ = 1.7321 (c) (i) OM2 = 242 + 102

OM = √676 = 26 cm

sin y = 1026

= 513

(ii) PM = 26 – 13 = 13 cm

BAB 6 Sudut dan Tangen bagi Bulatan

1 D 2 A 3 C 4 C5 C 6 C

7

Q

R

S

Dicangkum oleh lengkok CDSubtended by arc CD

Sudut pada pusatAngle at the centre

Sudut pada lilitanAngle at the circumference

•

•

•

•

•

•

Separuh daripada ∠PHalf of ∠P

P • •

8 (a) Palsu/False (b) Benar/True (c) Benar/True (d) Palsu/False

9 (a) ✗ (b) ✗ (c) ✓ (d) ✗10 (a) RS, TU

(b) ✓ ✓

11 (a) Benar/True (b) Palsu/False (c) Palsu/False (d) Benar/True

12 (a) x = 27o

y = 71o

(b) r = 90o

s = 50o

13 (a) ∠ACB = 55o

∠AOB = 55o × 2 = 110o

∠OAB = (180o – 110o) ÷ 2 = 35o

x = 22o + 35o

= 57o

(b) (i) x = 180o – 85o = 95o

∠RST = 85o

x = 180o – 85o – 30o = 65o

(ii) x = 180o – 90o – 66o = 24o

180o – 96o = 84o

y = (180o – 84o) ÷ 2 = 48o

(c) 5a = 75o

a = 75o

5 = 15o

b = 180o – 90o – 75o = 15o

180o – 60o = 120o

c = 120o – 75o = 45o

14 (a) ∠IKL = 50o

∠KIL = (180o – 50o) ÷ 2 = 65o

∠JIK = 180o – 50o – 65o = 65o

x = 180o – 35o – 65o = 80o

(b) (i) x = 70o – 30o = 40o

(ii) 180o – (180o – 70o – 30o) = 180o – (180o – 100o) = 180o – 80o = 100o

y = 100o

x + y = 40o + 100o = 140o

(c) (i) ∠QPS = 180o – 94o = 86o

(ii) ∠OUW = 180o – 73o – 86o = 21o

(iii) ∠OTV = ∠OVT = 23o

∠UVT = 90o

∠WUV = 180o – 90o – 23o – 21o = 46o

15 (a) a = 42o

b = 35o

c = 180o – 42o – 73o = 65o

a + b + c = 42o + 35o + 65o = 142o

J6 J7 Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2019

(b) (i) x = 180o – 80o = 100o

∠QPU = ∠QSR ∠URS = ∠PSR y = 80o – 48o = 32o

(ii) ∆EDF = 31o (Lengkok sama panjang)∆EDF = 31o (Same length curve)

Dalam/In ∆DEF, y + 31o + 47o = 180o

y + 78o = 180o

y = 102o

(c) x = 360o – 90o – 110o = 70o

∠EDC = 360o – 90o – 90o – 70o = 110o

y = 110o ÷ 2 = 55o

x + y = 70o + 55o = 125o

BAB 7 Pelan dan Dongakan

1 B 2 A 3 B 4 D 5 (a) ✗ (b) ✗ (c) ✓ (d) ✗

6 Pelan/Plan Objek/Object

•

•

•

•

•

•

• •

7 (a)

PepejalSolid

Dongakan sisiSide elevation

(✓) atau/or (✗) ✗ ✓ ✓

(b) (i) M/G L K J/H

C/BDE

F/A

4 cm 1 cm 1 cm

8 cm

(ii) E Q/P L/M

SR

C/D

B/A H/G

J/K

5 cm 3 cm

2 cm

3 cm

2 cm

8 (a) (i) Silinder/Cylinder

(ii) Kon/Cone

(iii) Prisma segi tiga/Triangular prism

(b) (i) A/D P/S

R/UB/C

5 cm

6 cm Q/T3 cm

(ii) R/P Q

TU/S/KC/D/E

B/A/F L/J6 cm

2 cm

5 cm 3 cm 9 (a) ✓, ✗, ✗ (b) (i) B/F C/K/G

M

LD/J/H

A/E

6 cm

4 cm 4 cm

(ii)

A/B P/Q T/U D/C

J/K

L/MH/GE/F

S/R W/V2 cm

7 cm

4 cm 4 cm

1 cm 1 cm

AP = TD = 1 cm10 (a)

J6 J7 Praktis Topikal KSSM: Matematik Tingkatan 3 – Jawapan

(b) (i)

>>M/Q

L/P

D/F/G

A/K B/J3 cm

N/R

C/E/H

1 cm

3 cm

2 cm

(ii) Q M

L/NP/R

G/H

V/Z

U/W

K/J S/A/B

T/D/CF/E

4 cm

1 cm

3 cm

2 cm

>>

>>1 cm

> >

2 cm

BAB 8 Lokus dalam Dua Dimensi

1 D 2 B 3 B 4 B 5 (a) Garis lurus Straight line

(b) Bulatan Circle

(c) Lengkok Arc

(d) Bulatan Circle

6 (a)

Lokus M/Locus M

(b)

Lokus M/Locus M

7 (a) BGHD (b) AGJC (c) EHJF (d) G 8 (a) (i) (a) Lokus V ialah bulatan berjejari 3 cm dan

berpusat di Q. Locus V is a circle with radius 3 cm and centre Q.

(b) Lokus W ialah garis pembahagi dua sama sudut KLM.

Locus W is an angle bisector of KLM.

(ii) Satu garis lurus dengan keadaan jaraknya sentiasa 2 m dari pagar itu.A straight line such that its distance is always 2 m from the fence.

(b) (i) P Q

R

S

U

T

Lokus XLokus X

(ii)

Q

R

SP

=

=

Lokus LLocus L

(c)

Q • • R

• P

= =

•

•

• •

S

S

2 cm

9 (a) (i) (a) Bulatan/Circle

(b) Pembahagi dua sama serenjang Perpendicular bisector

(c) Elips/Bulatan bujur/Elipse

(b) A B

C

=

=

=

=

Lokus MLocus M

Lokus NLocus N

(c) (i) BD (ii), (iii)

A B

CD

Lokus YLocus Y Lokus Z

Locus Z

10 (a) Lokus X/Locus of X = Lengkung PLN/Curve PLN

Lokus Y/Locus of Y = Garis PON/Line PON

∴ P dan N/P and N

(b) (i) Garis lurus EC/Straight line EC

J8 J9 Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2019

(ii)

A B

C

D

E

Lokus YLocus Y

Lokus ZLocus Z

(c)

•

••

•••

DR

C

Q

BP

A

ST

•

Lokus YLocus Y

Lokus XLocus X

Lokus ZLocus Z

BAB 9 Garis Lurus

1 B 2 C 3 B 4 D5 A 6 D

7 (a) Jarak mencancang/Vertical distance = 7 (b) Jarak mengufuk/Horizontal distance = 5

(c) Kecerunan/Gradient = 75

(d) Rumus kecerunan/Gradient’s formula

= Jarak mencancang/Vertical distance

Jarak mengufuk/Horizontal distance 8 (a) (i) ✗ (ii) ✓

(b)

x

4 – y

8 = 1 y = x + 5

• •

••

––––––

––

6

4

2

0–6 –4 –2

y

x

–––––––––

2 4

–2

–4

–6

y

x–2 0

9 (a) (i) Pintasan-y = 3 y-intercept = 3

(ii) m = 4 – 12 – (–4)

m = 12

(b) x 0 1 2 3 4

y 9 6 3 0 –3

y

x

10 –

–

8 –

–

6 –

–

4 –

2 –

0

–2 –

–4 –

–4 –2 2 4

(c) (i) JKL // IH ∴ m

JKL = m

IH

mJKL

= 10 –41 – (–5)

= 66

= 1

Gantikan m = 1 dan (3, 4) ke dalam y = mx + cSubstitute m = 1 and (3, 4) into y = mx + c

4 = 1(3) + c c = 1 y = x + 1 (ii) Pintasan-x, y = 0/x-intercept, y = 0

y = x + 1 0 = x + 1 x = –1 Pintasan–x/x-intercept = –110 (a) (i) (1, –2) (ii) (–2, 1)

(b) (i) m = 1 – (–2)2 – 0

= 32

Pintasan-y, c = –2/y-intercept, c = –2

y = 32

x – 2

(ii) y = 32

x – 2, (x = –8, y = –16)

–16 = 32

(–8) – 2

Disebelah kiri/On the left-hand side: –16 Di sebelah kanan/On the right-hand side:

32

(–8) – 2 = 3(–4) – 2 = –12 – 2 = – 14

–16 ≠ –14 Maka, P(–8, –16) tidak terletak pada garis lurus

itu.Hence, P(–8, –16) is not on the straight line.

(c) (i) A(–3, 2), B(0, 4)

m = 4 – 20 – (–3)

= 23

(ii) 4 = 23

(9) + c

c = –2

y = 23

x – 2

(iii) Pintasan-y/y-intercept = –211 (a) 6x – 5 = 0

x = 56

2y = –5

y = – 52

J8 J9 Praktis Topikal KSSM: Matematik Tingkatan 3 – Jawapan

Pintasan-x/x-intercept = 56

Pintasan-y/y-intercept = – 52

(b) y = 2x3

+ 53

m1 = 2

3 9y = kx + 2

m2 = k

9 m

1 = m

2

23

= k9

k = 6 (c) (i) x = –2 (ii) CD (–2, 0), pintasan-y/y-intercept (0, y):

y – 00 + 2

= 32

y = 3 Pintasan-y/y-intercept = 3 (iii) AB (–2, 5):

5 = 32

(–2) + c

c = 8

y = 32

x + 8

12 (a) Pintasan-y/y-intercept = 3 y = –2x + 3

(b) (i) 2(6) = 3x + 6 x = 2 P = (2, 6) (ii) 2y = 3x + 6

y = 32

x + 3

∴ m = 32

PS//QR

y = 32

x + c

6 = 32

(5) + c

c = – 32

y = 32

x – 32

atau/or 2y = 3x – 3

(c) (i) Garis lurus EH adalah selari dengan paksi-x, ∴ y = pintasan-yThe straight line EH parallel to x-axis, ∴ y = y-intercept

y = 6 (ii) OP = 6 ∴ G (0, 2)

mEF

= mGH

= 6 – 23 – 0

= 43

y = 43

x + c ; (6, 15))

15 = 43

(6) + c

15 = 8 + c c = 7

y = 43

x + 7 atau/or 3y = 4x + 21

Pintasan-y/y-intercept = 7

Fokus PISA/TIMSS

1 p5q–3 × (8p6)23

4q = p5q–3 × 4p4

4q = p5q–3 × p4q–1 = p9q–4

2 4n–4 = 232n–3

(22)n–4 = 2(25)n–3

22n–8 = 2 × 25(3–n)

22n–8 = 21+15–5n

22n–8 = 216–5n

22n+5n = 216+8

27n = 224

7n = 24 Bandingkan/Compare

n = 247

3 D 5 000 pasangan menghasilkan 5 000 anak burung di awal

tahun pertama. Maka, pertambahan populasi burung pada tahun itu adalah 50% daripada 10 000 burung.5 000 couples produce 5 000 chicks at the beginning of the first year. Therefore, the bird's population is increased by 50% of 10 000 birds.

Tahun 1/1st year: B = 10 000 × [ 150100

× (100% – 20%)] = 10 000 × (1.5 × 0.8)Tahun 2/2nd year: B = 10 000 × (1.5 × 0.8)2

Tahun 3/3rd year: B = 10 000 × (1.5 × 0.8)3

Tahun 4/4th year: B = 10 000 × (1.5 × 0.8)4

Tahun 5/5th year: B = 10 000 × (1.5 × 0.8)5

4 (a) Encik Iqbal: Jumlah unit sahak/Total unit of shares

= RM50 0001.60

= 31 250 unit saham/unit of shares

Kos purata seunit saham/Average cost per share = RM1.60 Puan Melinda: Harga saham yang dibeli sebulan:

Price of shares purchased per month:

RM50 0005

= RM10 000

SahamShare

BulanMonth

Feb Apr Mei/May Jul Okt/Oct

1.65 1.55 1.60 1.75 1.40

Unit sahamUnit of share

RM10 000RM1.65= 6 060

RM10 000RM1.55= 6 451

RM10 000RM1.60= 6 250

RM10 000RM1.75= 5 714

RM10 000RM1.40= 7 142

Jumlah unit saham/Total unit of share: 6 060 + 6 451 + 6 250 + 5 714 + 7 142 = 31 617 Kos purata seunit saham/Average cost per share

RM50 00031 617

= RM1.58

(b) Puan Melinda(c) Puan Melinda mengamalkan strategi pemurataan

yang boleh membantu beliau memiliki lebih banyak saham dengan kos purata seunit yang lebih rendah.Puan Melinda practises an averaging strategy that can help her gains more shares with a lower average cost per share.

5 94 × 7 = 658 cm2

6 300 cm ÷ 50 = 6 cm (Diameter) Jejari/Radius = 6 cm ÷ 2 = 3 cm

2πr = 2 × 227

× 3 = 1867

cm

J10 J11 Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2019

7 tan ∠XZY = x + 25x + 10

= x + 25(x + 2)

= 15

∠XZY = tan–1 (15)

∠XZY = 11.31o atau/or 11o19'

8 CD = √122 – 52

CD = 13 m BC = 55 – 13 = 42 m

tan ∠DCE = 512

∠DCE = 22o37'

kos/cos ∠ACB = 2042

∠ACB = 61o34' ∠BCD = 180o – 22o37' – 61o34' = 95o49'

9 (a) 100% + 20% = 120% 120% → 36 km/j/km/h

100% → 1 × 36 ÷ 1.2 = 30 km/j/km/h

Kelajuan angin/Wind speed = 30 km/j/km/h

(b) sin 55o = 70x

x = 70sin 55o

= 85.454 m ∴ Panjang tali Length of rope = 85.45 m

Payung terjunParachute

70 mx m

KapalBoat

55o (

10 Andaikan r = jejari Assume r = radius

O

rr – 10

15 15

(r – 10)2 + 152 = r2

r2 – 20r + 100 + 225 = r2

–20r = –325 r = 16.25Maka diameter balak/Hence diameter of the log:2 × 16.25 cm = 32.5 cm

11 10 kubus/cubes

12 C 1 pintu di hadapan, maka pintu tidak kelihatan dari

belakang.1 door in front, therefore the door is not visible as viewed from the back.

1 tingkap di tepi dinding kanan di bahagian hadapan.1 window on the edge of the right wall at the front.

13

•

•

•

=

=Tiang lampuLamp post

Balai polisPolice station

RestoranRestaurant

KlinikClinic

14 Tak linear/Non-linear

15 (a) A M lebih laju daripada N. Maka, jarak yang dilalui oleh

M lebih jauh berbanding N dalam tempoh 4 jam.M is faster than N. Therefore, the distance travelled by M is further than N in 4 hours.

(b) Jarak/Distance

Masa/Time

M

N

P

Penilaian Akhir Tahun

Bahagian A

1 B 2 C 3 B 4 B 5 A6 C 7 C 8 B 9 B 10 D

11 B 12 C 13 D 14 C 15 D16 B 17 A 18 D 19 C 20 B

Bahagian B 1 (a) ✗ (b) ✓ (c) ✗ (d) ✓ 2 (a) Benar/True

(b) Benar/True

(c) Palsu/False

(d) Benar/True

3

ON

TR

NOMN

RSTR

Tembereng minorMinor segment

JejariRadius

PerentasChord

Sektor minorMinor sector

•

•

•

•

•

•

•

•

4 (a) 11 × 2 = 22 cm

(b) 3 × 115

= 0.2 cm

(c) 12 × 3 = 36 cm

(d) 13

× 15 = 5 cm

5 (a) Nonagon(b) 9(c) 9(d) 27

Bahagian C 1 (a) (i) (S, L), (S, 6), (S, 9), (2, L), (2, 6), (2, 9), (A, L), (A, 6),

(A, 9), (1, L), (1, 6), (1, 9), (7, L), (7, 6), (7, 9) (ii) (2, 6), (2, 9), (A, 6), (A, 9), (1, 6), (1, 9), (7, 6), (7, 9)

Kebarangkalian/Probability = 815

(b) (i) 6423 = (3√64)2 = (4)2 = 16

(ii) 32k + 4 ÷ 2713 × 9–2 = 81

32k + 4 ÷ 3 × (32)–2 = 34

32k + 4 – 1 + (–4) = 34

2k – 1 = 4

k = 52

atau/or 212

J10 J11 Praktis Topikal KSSM: Matematik Tingkatan 3 – Jawapan

(c)

Bank M N

Bayaran balik pinjamanLoan repayment

A = P + Prt= RM50 000 + (50 000)( 3.2

100)(7)

= RM50 000 + RM11 200= RM61 200

A = P + Prt= RM50 000 + (50 000)( 3

100)(9)

= RM50 000 + RM13 500= RM63 500

Ansuran bulananMonthly instalment

RM61 200 ÷ (7 × 12)= RM61 200 ÷ 84= RM728.57

RM63 500 ÷ (9 × 12)= RM63 500 ÷ 108= RM587.96

Jawapan dengan sebab yang munasabah diterima seperti:Answers with reasonable reasons are accepted such as:

1 Bank N kerana Bank N mengenakan faedah dan ansuran bulanan yang lebih rendah berbanding dengan Bank M.

Bank N because Bank N imposes lower interest and

monthly instalments than Bank M.

2 Bank M kerana tempoh bayaran yang dikenakan lebih singkat. Tempoh bayaran yang panjang menyebabkan amaun faedah yang dibayar adalah lebih tinggi.

Bank M due to the shortened charge period. The long term payment causes the amount of interest paid higher.

2 (a) (i) ✓

✗

(ii) Integer terbesar/Largest integer: 254

Integer terkecil/Smallest integer: –4

254

× (–4) = –25

(b) 9qr + 2sp – 6pq – 3rs = 9qr – 6pq + 2sp – 3rs = 3q(3r – 2p) + s(2p – 3r) = 3q(3r – 2p) – s(–2p + 3r) = 3q(3r – 2p) – s(3r – 2p) = (3q – s)(3r – 2p)

(c) (i) Koordinat P ialah (0, 24), maka koordinat Q ialah (10, 24)Coordinate P is (0, 24), therefore coordinate Q is (10, 24)

M titik tengah OQ, maka koordinate M ialahM is the midpoint of OQ, there coordinate Q is

(10 + 02

, 24 + 02

) = (5, 12)

(ii) QM = OM = √52 – 122 = √25 + 144 = √169 = 13 unit/units

PN = ON

= (12)(24)

= 12 unit/units

M = (5, 12) NM = 5 unit/units

Perimeter PQMN = 12 + 10 + 13 + 5 = 40 unit/units

3 (a) (i)

SebutanTerm

1 2 3 4 5 n

JujukanSequence

12 = 1 22 = 4 32 = 9 42 = 16 52 = 25 n2

Tn = n2

T7 = 72 = 49

(ii) • • • • •• • • • •• • • • •• • • • •• • • • •• • • • •• • • • •

••

••

••

••

••

••

••

(b) (i) x 0 1 2 3 4 5

y 7 11 17 25 35 47

(ii)

y

x

50 —

45 —

40 —

35 —

30 —

25 —

20 —

15 —

10 —

5 —

0 1 2 3 4 5

x

x

x

x

x

x

(c) (i) m = 3 – 13 – 2

= 2

c = –3 y = 2x – 3 (ii) (3x – y = 8) × 2 6x – 2y = 16 ➀ 5x – 2y = 12 ➁, ➀– ➁ x = 4 3(4) – y = 8 y = 4 x = 4, y = 4 Titik persilangan/Intersection point = (4, 4)

4 (a) (i) Translasi/Translation ( 4–3)

J12 Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2019

(ii) (a)

E

F

GH

E'F'

H' G'

P

(b) F'G'FG

= 36

= 12

(b) (i) Garis lurus DB/A straight line DB

(ii), (iii)

A B

CD

Q •

Lokus NLocus of N

Lokus MLocus of M

(c) Andaikan b = buku, p = penAssume that b = book, p = pen

3b + 3p = 18 --------① 4b + 2p = 19 --------② ① ÷ 3, b + p = 6 p = 6 – b --------③ Gantikan p = 6 – b ke dalam persamaan ②

Substitute p = 6 – b into equation ②

4b + 2(6 – b) = 19 4b + 12 – 2b = 19 2b = 7 b = 3.5 , Buku/Book = RM3.50 Gantikan b = 3.5 ke dalam persamaan ③

Substitute b = 3.5 into equation ③

p = 6 – 3.5 = 2.5 , Pen = RM2.50 RM3.50 + RM2.50 = RM6 ∴ Muna mempunyai wang yang mencukupi untuk

membeli barangan itu.∴ Muna has enough money to buy the items.

5 (a) 7 ≥ 5 – 8y3

21 ≥ 5 – 8y 16 ≥ – 8y

16–8

≤ y

y ≥ –2

(b) (i)

4 cm

6 cm (ii)

2 cm

4 cm

4 cm

2 cm

1 cm

H/C

J/N

E/D/MK/L

F/A

G/B

(c) ∠PQT = 60o, α = 30o

sin α = 0.5

kos/cos α = √32

sin2α + kos2/cos2α = 0.52 + (√32 )2

= 1

6 (a) ∠CDB = 180o – 65o = 115o

∠CBD = 180o – 115o – 30o = 35o

x = 90o – 35o = 55o

∠EOB = 2 × 65o = 130o

y = (180o – 130o) ÷ 2 = 25o

(b) (i) 384 400 km = 380 000 km

(ii) = 380 00058 000

= 3.8 × 105

5.8 × 104

= 0.655172 × 10 = 6.552 jam/hours

(c) Isi padu silinder/The volume of the cylinder

(9 × 9 × 9) – 27637

= 729 – 27637

= 45247

πj2t = 45247

( 227 )j2(9) = 4524

7 198j2 = 3 168 j2 = √16 j = 16 = 4 cm Diameter = 4 × 2 = 8 cm