Input MT 3

CONTOH

Kandungan

1 Bab 1 Indeks 1–6
2Praktis Bab 2 Bentuk Piawai 7 – 13
14 – 20
Bab 3 Matematik Pengguna: Simpanan dan Pelaburan, Kredit 21 – 27
28 – 34
3 dan Hutang 35 – 41
CONTOH 42 – 47
4 Bab 4 Lukisan Berskala 48 – 54
55 – 60
Praktis Bab 5 Nisbah Trigonometri 61 – 78
J1 – J8
5

6 Bab 6 Sudut dan Tangen bagi Bulatan

7 Bab 7 Pelan dan Dongakan

Praktis Bab 8 Lokus dalam Dua Dimensi

8

9 Bab 9 Garis Lurus

Pentaksiran PT3

Jawapan

iii

Praktis Nama: Tarikh:

1 Bab 1 Indeks

AJaBwAabHsAeGmIuAaNsoalan. Hal. Buku Teks: 1 – 29

JaAwnaswbesreamll uqauessotiaolnasn. .
Answer all questions.
Ά CONTOH
1. (–6)3 = 7. Diberi x2m = x20 , cari nilai m.
A –216 x6
B –18
C 18 Given that x2m = x20 , find the value of m.
D 216 x6

2. Cari nilai x, jika 5x = 625. A6
Find the value of x, if 5x = 625. B7
A2 C 13
B4 D 14
C6
D8 8. Diberi 4x12 ÷ 12x–4 = 1 x2k, cari nilai k.
Given that 3 =
3. 85 × 812 = 4x12 ÷ 12x–4 1 x2k, find the
A 817 3
B 825 value of k.
C 830 A4
D 860 B5
C6
4. 6p4 × 3p2 × p = D8
A 9p7
B 9p9 9. (y4)2 × y0 =
C 18p7 A0
D 18p9 B1
C y6
5. 82 × 32 = D y8
A 114
B 112 10. Diberi (pr)4 = p20, cari nilai r.
C 244 Given that (pr)4 = p20, find the value of r.
D 242 A4 B5
C 10 D 80
6. 64y6 ÷ 8y2 =
A 8y3 11. (94 × 96)2 = B 948
B 8y4 A 920 D 8120
C 512y3 C 950
D 512y4
Nota / Note
an = a × a × a × .... × a

n faktor/n factors

1

1 2. Permudahkan: 1 7. Permudahkan:
Simplify: Simplify:
(m3n2)2 ÷ mn6
A mm44n2 √mn3 × 3√m6 × √m5n
B n2 A m2n4
B m4n2
C m5 C m5n
n2 D m5n2

D m5 18. A(3 6–  )3 2 =
n4
43

1
CONTOH 36
1 3. Permudahkan:
Simplify: B 1
6
( )a22× b2 × c2 C 6
b3 c a3
D 36
A ac
b2
64–  2 1
B ac 1 9. Diberi m2 = 3 × cari nilai m.
b4 81 2 ,
2 1
C ac2 Given that m2 = 64–  3 × find the value
b4 81 2 ,

of m.
3
D a2c A 4
b7
B 9
16
1 4. (p4)–3 =
A 4 C 4
p3 3
D
B 3 36
p4 9

C 1 20. Permudahkan:
12p
1 Simplify:
D p12 5n × 54 × 1252
A 54n + 2
13 B 5n + 8
C 5n + 10
1 5. 9 2 × 9 2 = D 58n
3
Nota / Note
A 9 4
4 Penyelesaian persamaan eksponen.

B 9 3 Solving exponential equation.
C 92 (a) Ungkapkan kedua-dua belah persamaan
D 94
kepada asas yang sama.
16. Cari nilai bagi: Express both sides of the equation to the
Find the value of:
1 2 same base.
–  4 × 27 3 (b) Samakan indeks mereka.
Equate their indices.
1 16 1 ax = ay,
2 2 ∴x=y
A 4 B 6

C 18 D 36

2

BAHAGIAN B

Jawab semua soalan.
Answer all questions.

1. (a) Isikan petak-petak di bawah dengan Berdasarkan jadual di sebelah,
jawapan yang betul. tulis operasi yang menghasilkan
jawapan seperti diberikan dalam
Fill in the boxes below with the correct peta bulatan.
answers. SP 1.2.5
Based on the table, write the operations
Jawapan/Answer: that produce the answer as given in the
circle map.

Jawapan/Answer:

(a5)2

a10


CONTOH 3 √ 5  )

(i) 5 4 = (

(ii) 9 = 1

(b) Rajah berikut menunjukkan dua
kumpulan indeks.

The following diagram shows two
groups of indices. SP 1.2.4

m8 5√m m–4 m2 × m3 (b) Padankan dengan jawapan yang
betul./Match the correct answers.
1 m6 m 1 (m4)2
m4 5 SP 1.2.6

AB Jawapan/Answer:

Berdasarkan contoh diberi, (i) 58 × 5–5 1
lengkapkan pasangan setara di 125
bawah.
Based on the example, complete the two (ii) 25–  3 1
equivalent pairs below. 2 5
125
Contoh/Example:

5√m = m 1 3. (a) Jadual berikut menunjukkan
5 beberapa indeks.

Jawapan/Answer: The following table shows some indices.
(i) m8 =
SP 1.2.2

(ii) 1 = x2 x 9 x4 x12
m4 2

2. (a) Jadual berikut menunjukkan operasi Lengkapkan operasi yang berikut
melibatkan indeks. dengan menggunakan indeks
dalam jadual di atas.
The following table shows the Complete the following operations by
operations involving indices. SP 1.2.2 using the indices in the table above.

a2 × a5 a4 ÷ a–6 a–8 × a–2 Jawapan/Answer:
(a5)2
a20 a14 (i) x16 ÷ x4 =
a2 a4
(ii) x4 × x 1 =
2

3

(b) Rajah berikut menunjukkan indeks. 4. (a) Isikan petak dengan nombor yang
The following diagram shows indices. betul.

SP 1.2.5 Fill in the box with the correct number.

p 5 1 (p2)5 p–4 SP 1.2.4
7 p4
Jawapan/Answer:
3√ p p10 p 1 7√ p5 (i) (–3) × (–3) × (–3) × (–3) × (–3)
3
= (–3)
AB

Lengkapkan dua pasangan setara (ii) 1 = m–4
berdasarkan contoh yang diberi.

Complete two equivalent pairs based on
the given example.
CONTOH m

(b) Padankan dengan jawapan yang
betul.
Contoh/Example: Match the correct answers. SP 1.2.3

3√ p = p 1 Jawapan/Answer:
3

Jawapan/Answer: 6–6
63
(i) 62 ÷ 6–3 65
(ii) (6–3 × 69) 12

BAHAGIAN C

Jawab semua soalan.
Answer all questions.

1. (a) Permudahkan/Simplify: SP 1.2.3

(i) m6 × m3 (ii) k8 ÷ k2 (iii) (p–3)2
(iii)
Jawapan/Answer: (ii)
(i)

(b) (i) Cari nilai bagi 10–2.
Find the value of 10–2. SP 1.2.4
(ii) Diberi bahawa 82x + 5 ÷ 83 = 4 096, cari nilai x.
Given that 82x + 5 ÷ 83 = 4 096, find the value of x. SP 1.2.7

Jawapan/Answer: (ii)
(i)

4

(c) Diberi y2 = 1 × 64– 23 , tentukan nilai y.

81 2 1 2
3
Given that y2 = 81 2 × 64– , determine the value of y. SP 1.2.7

Jawapan/Answer:

2. (a) (i) Permudahkan/Simplify: g–3 ÷ g4. SP 1.2.2
2 4 21 .
(ii) Cari nilai bagi ×
125 3
CONTOH 21
Find the value of 125 3 × 4 2 . SP 1.2.6

Jawapan/Answer: (ii)
(i)

(b) Diberi p ÷ 12 × 48 = 8 dan q–3 × 6 ÷ p2 = 3 , hitung nilai p dan q.
024
1 3
024
Given that p ÷ 12 × 48 = 8 and q–3 × 6 ÷ p2 = 1 , calculate the value of p and q. SP 1.2.7


Jawapan/Answer:

(c) Rajah berikut menunjukkan satu segi empat tepat, ABCD.
The following diagram shows a rectangle, ABCD. SP 1.2.7

AB

(x 2y 5) cm

DC

Diberi luas bagi ABCD ialah x 5 y7 cm2, hitung panjang, dalam cm, bagi CD.
2

Given that the area of ABCD is 5 cm2, calculate the length, in cm, of CD. SP 1.2.7

x 2 y7

Jawapan/Answer:

5

3. (a) Diberi 163 × 42n – 5 = 214, hitung nilai n.
Given that 163 × 42n – 5 = 214, calculate the value of n. SP 1.2.7
Jawapan/Answer:

(b) Diberi (√x )7 × y3 = 3 125 dan y√x = 5, hitung nilai x dan y.
Given that (√x )7 × y3 = 3 125 and y√x = 5, calculate the value of x and y. SP 1.2.7 Modul hebat 31
Jawapan/Answer:

(c) Diberi bahawa 4x × 162x – 3y = 32, ungkapkan x dalam sebutan y.
Given that 4x × 162x – 3y = 32, express x in terms of y. SP 1.2.7
Jawapan/Answer:


CONTOH
4. (a) Permudahkan/Simplify: SP 1.2.6 Modul hebat 31
(4p2q3)2 × (p3)–2 ÷ 2p5q2


Jawapan/Answer:

(b) Cari nilai bagi/Find the value of : 482n + 2 × 21 – 4n SP 1.2.6
122n
Jawapan/Answer:

(c) Diberi 25x × 1251 – x = 625, hitung nilai x.
Given that 25x × 1251 – x = 625, calculate the value of x. SP 1.2.7
Jawapan/Answer:

6

JawapanCONTOH

J11

CONTOH

PRAKTIS 1 Bahagian B

Bahagian A 1. (a) (i) 2
1. A 2. B 3. A 4. C 5. D (ii) 4
6. B 7. B 8. D 9. D 10. B
11. A 12. C 13. B 14. D 15. C (b) (i) –3
16. A 17. D 18. B 19. A 20. C (ii) 9

Bahagian B 2. (a) (i) 3
(ii) 5
1. (a) (i) (4√5 )3
(ii) 0 (b) (i) 8.04 × 105
(ii) 3.7 × 10–6
(b) (i) (m4)2
(ii) m–4 3. (a) (ii) 0.0004307 → 0.000431
(iv) 0.002004 → 0.00200

(b) (i) 0.000275
(ii) 670 000
CONTOH2.(a)a14 ; a4 ÷ a–6
a4
Bahagian C
(b) (i) 125
1. (a) (i) 28 500
(ii) 1 (ii) 1.02
125 (iii) 0.0700

3. (a) (i) x12 (b) (i) 1.512 × 106 cm
9 (ii) 3.266 × 1010 cm2
(ii) x2
(c) 6.696 × 106
(b) 5 = 7√p5 , p10 = (p2)5 atau/or 1 = p–4
p4 2. (a) (i) 9.25 × 10–3
p7 (ii) 3.47 × 100
(iii) 5.401 × 106
4. (a) (i) 5
(ii) 4 (b) 1.6294 × 10–25 kg
(c) 1.007 × 105
(b) (i) 65
(ii) 63 3. (a) (i) 0.000258
(ii) 734 000
Bahagian C (iii) 9 600 000

1. (a) (i) m9 (b) 3.854 × 10–26 kg
(ii) k6 (c) 1.74 × 10–3 m3
(iii) p–6 4. (a) 31.5
(b) (i) 1.26 × 10–2 m3
(b) (i) 0.01
(ii) x = 1 (ii) 7 135 kgm–3
(c) 3.24 × 104
(c) y= 3
4

2. (a) (i) g–7 PRAKTIS 3
(ii) 50
Bahagian A
(b) p = 2, q = 8 1. D 2. B 3. A 4. C 5. C
1 6. A 7. B 8. A 9. D 10. C
B 13. A 14. B 15. C
(c) x2 y2

3. (a) n = 3 11. C 12.
(b) x = 5, y = √5
12y + 5 16. D
(c) x= 10

4. (a) 8q4 Bahagian B
p7 1. (a) Saham/Shares

(b) 4 608 Amanah saham/Unit trusts
(c) x = –1 (b) (i) ✓

PRAKTIS 2 (ii) ✗

Bahagian A 2. 3. B 4. D 2. (a) (i) ✓
1. B C 8. B 9. B 5. C (ii) ✓
6. A 7. D 13. B 14. D 10. A (iii) ✓
11. C 12. A 18. B 15. C (iv) ✓
D
(b) (i) Saham syarikat/Company shares
(ii) Hartanah/Real estate

16. A 17.

3

Bahagian C 2. (a)
1. (a) (i) RM17 104.29
P
(ii) RM2 104.29
(b) x = 3.8 (b) (i) 1: 1
(c) 2.7% 4

2. (a) (i) RM1 080 (ii) 15.36 cm2
(ii) RM1 230 (c) (i) 1 : 400 000
∴ Perbezaan/Difference
RM1 230 – RM1 080
= RM150

(b) (i) RM290
(ii) RM5 550.76
(iii) RM5 901.33

3. (a) RM1.58
(b) RM971.58
(c) RM1.73

4. (a) RM45 000
(b) RM405 000
(c) 3.47%
CONTOH (ii) 212 km

3. (a) 2 880 cm2

(b) (i)

PRAKTIS 4

Bahagian A 3. D
1. A 2. C 8. D 4. A 5. B
6. D 7. C 13. B 9. B 10. A
11. C 12. D 14. D 15. A (ii)

Bahagian B

1. (a) (i) 1 : 3
(ii) 1 : 1
(b) (i) C

(ii) 1: 2
3

2. (a) (i) 1 : 3

(ii) 1: 3 PRAKTIS 5
5

(b) Bahagian A

1. C 2. B 3. D 4. A 5. B
6. A 7. A 8. B 9. C 10. D
11. C 12. C 13. B 14. A 15. B

Bahagian B

1. (a) (i) m
p

(ii) m
n

Bahagian C (b) (i) Ͻ
1. (a) (i) 18 m (ii) Ͼ

(ii) 9 cm 2. (a) (i) Sisi bersebelahan/Adjacent side
(b) (i) BC = 25 cm (ii) Sisi bertentangan/Opposite side

(ii) 760 cm (b) (i) ✗
(c) 625 cm2 (ii) ✗
(iii) ✓

Bahagian C

1. (a) (i) sin x = 12
13

4

(ii) kos/cos x = 5 3. (a) (i) x = 63°
13 (ii) y = 27°
(iii) z = 73°
(iii) tan x = 12
5 (b) (i) 12 cm
(ii) 20 cm
(b) (i) 6.401 m (iii) 24 cm
(ii) 13.614 m
(c) (i) 180° – 4x°
(c) (i) sin x = 24 (ii) 45° – x°
25

(ii) QR = 16 cm, tan y = 4 PRAKTIS 7
3

2. (a) 291° 48′ Bahagian A

(b) 97° 9′ 1. C 2. D 3. A 4. B 5. C
8. A
(c) (i) 5 cm
CONTOH 6. B 7. D
(ii) kos/cos y = 7
25
Bahagian B
3. (a) (i) tan x = 5 1. (a) ✗
12
(b) ✓
(ii) PT = 10 cm, sin y = 15 (c) ✓
17 (d) ✗

(b) (i) 4.607 m 2. (a)
(ii) 28° 45′
(c) x = 9.238 cm, Luas/Area = 36.952 cm2

PRAKTIS 6

Bahagian A (b)
1. B 2. D 3. C 4. A 5. D 3. (a)
6. C 7. A 8. B 9. A 10. C
11. B 12. D 13. A 14. B 15. A (b)

Bahagian B
1. (a) (i) ✓

(ii) ✗
(iii) ✗
(b) (i) BENAR/TRUE
(ii) BENAR/TRUE
2. (a) (i) ✓
(ii) ✗
(b) (i) ✗
(ii) ✓

Bahagian C Bahagian C
1. (a) x = 80°, y = 55°, z = 25° 1. (a) E/D
F G/C
(b) (i) x = 90°
(ii) y = 96° 4 cm
(iii) z = 108°
K/A 3 cm J 3 cm H/B
(c) (i) x = 64°
(ii) y = 107°
(iii) z = 43°

2. (a) (i) x = 56°
(ii) y = 62°
(iii) z = 28°

(b) 31.66 cm
(c) (i) x = 35°

(ii) y = 22°
(iii) z = 33°

5

(b) (i) M 4 cm L (b) (i) SR
2 cm (ii)
H/G
4 cm P Q

K/E J/F B/C
3 cm 2 cm

A/D 2 cm N 4 cm S R
(c) R
(ii) L/M
P
CONTOH H G
1 cm 6 cm
E
K F T
1 cm
QS
J
Q
2 cm

B/A 4 cm C/N/D 2 cm

2. (a) 2 cm
2 cm
3 cm 2. (a) P

O

5 cm 5 cm

6 cm S R
K
(b) 102.5 cm3 (b) (i) KM
(c) 738 g (ii)

J

PRAKTIS 8

Bahagian A C 4. D 5. B
1. B 2. C 3. A 9. A 10. D
6. D 7. C 8. A 14. B
11. C 12. 13. B ML

Bahagian B PRAKTIS 9

1. (a) (i) ✓ Bahagian A 4. C 5. B
(ii) ✓ 1. B 2. D 3. A 9. A 10. C
6. A 7. D 8. B 14. A 15. C
(b) (i) Pembahagi sudut/Angle bisectors 11. D 12. B 13. D
(ii) Garis selari/Parallel lines
Bahagian B
2. (a) (i) HF
(ii) DB 1. (a) (i) m = –2, c = 8

(b) F

Bahagian C (ii) m=– 1 , c = 3
1. (a) P 4
Q
(b) (i) 5

(ii) –7

2. (a) (–2, 13); (5, –1)
(b) (i) PALSU/FALSE

(ii) BENAR/TRUE

3. (a) (i) 3x + 4y = 12

SR (ii) y= 3 x – 3
4 8

6

(b) (i) RS 25. (a) (i) 649 000
(ii) PQ (ii) 0.0230

4. (a) (i) x = 5 (b)

(ii) y = –3
(b) (i) 8x + 2y = 13
x y
(iv) 2 + 8 =1

Bahagian C

1. (a) y = –2x + 6

(b) (i) y =– 1 x +8 Bahagian C
2

(ii) 16
CONTOH 26. (a) (i) 32 cm
(iii) (8, 4) (ii) 8 : 3

2. (a) (i) k = 6 (b) (i) x = 3
(ii) y = –2x + 6 (ii) x = 5

(iii) 9 unit2 (c) y = –3, –2, –1, 0

(b) (i) 9 km 27. (a) A
(ii) 4y = 3x – 15

3. (a) (i) 2y = x + 5

(ii) 5
2
(b) (i) y = –3
(ii) 2y + 5x = 34
B C
(iii) k = 6.8
(b) (i) 30 cm2

(ii) 78 cm2

PENTAKSIRAN PT3 (c) (i) 9
(ii) (a) C ⊂ A
Bahagian A (b) B ⊂ A′

1. C 2. B 3. A 4. D 5. C 28. (a) (i) 5x + 7y = 55
6. D 7. B 8. C 9. A 10. C
11. D 12. B 13. C 14. D 15. A (ii) y = 55 – 5x
18. D 19. B 20. C 7
16. A 17. C
(b) (i) 360 cm2

Bahagian B (ii) 4 504.5 cm3
21. (a) (i) (a4)3 (c) k = 30.5

(ii) a4 29. (a) y
a–8
10

(b) x = 58° + 83° 8
= 141°
6
y + 62° + 58° = 180°
y = 180° – 120°
= 60°

22. (a) (i) BENAR/TRUE 4
(ii) PALSU/FALSE
2
(b) (i) ✓
(iii) ✓

23. (a) (i) ✓ –1 0 1 2 3 4 5x
(ii) ✓ –2
–4
(b) (i) Prisma/Prism –6
(iii) Kon/Cone

24. (a) P, Q
(b) (i) L
(ii) Badminton

7

(b) (i) (0, 4) 31. (a) (i) 66°
(ii) 90° (ii) 42°

(iii) Lawan arah jam/Anticlockwise (iii) 31°
  (b) (i) 2y = x + 11
(iv) Translasi/Translation 4
–3 (ii) Pintasan-x/x-intercept = –11
(c)
(c) (i) x = 8
(ii) JK
1
10 ML

30. (a) RM170.73 H/G
(b) (i) 25 cm

(ii) 1 012.5 cm2
(c) I/J

CONTOH 4 cm

K/L 5 cm F/E
B/A 2 cm

C/D

8

CONTOH