jawapan-modul-aktiviti-pintar-bestari-matematik-tingkatan-2

JAWAPAN

BAB 1: POLA DAN JUJUKAN C 3. ✓
Muka Surat 1 1. a = 1 + 3 = 4
A
1. Menolak 4 daripada nombor sebelumnya. b=3+3=6
c=3+1=4
Subtract 4 from the previous number. 2. a = 5 – 3 = 2
b = 5 + 8 = 13
2. Mendarab nombor sebelumnya dengan 2. c = 8 + 13 = 21
Multiply the previous number by 2.
Muka Surat 2
3. Membahagi nombor sebelumnya dengan 3. A
Divide the previous number by 3. 1. ✗ 2. ✗

4. Menambah 2 segi tiga kepada bentuk B 86, 80
sebelumnya. 44, 38
Add 2 triangles to the previous shape. 1. Dua nombor sebelum:
Two numbers before: – 1 , –1
B 2
Urutan nombor genap/Sequence of even numbers: Dua nombor selepas:
8, 22, 36, 50 Two numbers after: –64, –128
Pola/Pattern:
Menambah 14 kepada nombor sebelumnya. 2. Dua nombor sebelum:
Add 14 to the previous number. Two numbers before:

Urutan nombor ganjil/Sequence of odd numbers: Dua nombor selepas:
15, 29, 43, 57 Two numbers after:
Pola/Pattern:
Menambah 14 kepada nombor sebelumnya. C 2. 13, 19 3. – 214, 1
Add 14 to the previous number. 1. 75, 375 48

Muka Surat 3 Nombor Perkataan Ungkapan algebra
A Number Word Algebraic expression

Jujukan nombor +2 Menambah 2 kepada 4 = 2 + 2(1)
Number sequences nombor sebelumnya. 6 = 2 + 2(2)
Add 2 to the previous 8 = 2 + 2(3)
1. 4 , 6 , 8 , 10 , … number. 10 = 2 + 2(4)
2 + 2n , n = 1, 2, 3, 4, …
+2 +2 +2

2. 10 , 7 , 4 , 1 , … – 3 Menolak 3 daripada 10 = 13 – 3(1)

−3 −3 −3 nombor sebelumnya. 7 = 13 – 3(2)

Subtract 3 from the 4 = 13 – 3(3)

previous number. 1 = 13 – 3(4)

13 – 3n , n = 1, 2, 3, 4, …

3. 5 , 10 , 15 , 20 , … + 5 Menambah 5 kepada 5 = 5(1)

+5 +5 +5 nombor sebelumnya. 10 = 5(2)

Add 5 to the previous 15 = 5(3)

number. 20 = 5(4)

5n , n = 1, 2, 3, 4, …

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B BAB 2: PEMFAKTORAN DAN
PECAHAN ALGEBRA
1. 25 2. Bukan/No
Muka Surat 8
3. Sebutan ke-9/9th term
A
Muka Surat 4
B 1. 2p2 – 2pq 2. 2p2 – 4pq
3. 2p2 – 4p2q
1. (a) 1, 3, 5, 7, …
(b) 2n – 1, n = 1, 2, 3, 4, … B
(c) 11 buah segi tiga/11 triangles
(d) n = 10 1. 3n – 18m 2. –2h2 – 18hk
3. –9e + 3 e2f
2. 150
2

Praktis PT3 C
1. (a)
1. n2 + 8n + 15 2. 4r2 – 23r – 6
–1 81 3. 6u2 – 31uv + 5v2 4. 16t2 – 4
5. 25r2 – 9s2 6. 4h2 + 28hk + 49k2
7. m2 – 16mn + 64n2

4 27 Muka Surat 9
1. 3m2 – n2
9 3. h2 + k2 2. 5f 2 – 3ef – 14e2
5. 9uw – 4u2 – w2 4. p2 – p + 16
14 3 7. –5r2 – 64s2 + 9r 6. 46h2 + 24h + 9
8. –8xy + 20y2

19 1 Muka Surat 10
1. (3x2 + 2x – 1) cm2 2. RM(12pq – 2q2)
3. (25y2 – 20y + 4) cm2

PQ Muka Surat 11
A
(b) (i) 10

(ii) Menambah/Add 2, 3, 4, … 1. 1, 5, y, 5y; 5y 2. 1, 7, k, 7k; 7k
3. 1, s; s 4. 1, 3x, 3y, 3xy, xy; 3xy
(c) (i) 28 (ii) 16 (iii) 9

2. (a) (i) + 5, + 4 (ii) 31 B

(b) (i) Nombor Fibonacci/Fibonacci Numbers 1. 3(a + 3) 2. h(5 + k)
4. 4(4m2 – 5)
(ii) x = 13, y = 144 3. r(s + 1)
6. 3e(3 + 7f )
(c) (i) Perimeter segi empat tepat: 5. 6v(u – 2v)
7. 4q2(3p + r) 8. 5wy(2w – 5xy)
Perimeter of rectangles:

14, 16, 18, 20, …

Pola: Menambah 2 kepada nombor Muka Surat 12

sebelumnya. 1. (n + 2)(n – 2) 2. (5 + y)(5 – y)
3. (r – 7s)(r + 7s)
Pattern: Add 2 to the previous number. 5. (a + 2)2 4. (6m – 1)(6m + 1)
7. (x + 5)2 6. (r – 3s)2
Luas segi empat tepat/Area of rectangle: 9. (2p + 1)(p – 6) 8. (3d – 4e)2
12, 15, 18, 21, … 11. (a + 7)(b + 2)
Pola: Menambah 3 kepada nombor 13. (3r – 1)(7 + s) 10. (–5y + 3)(y – 1)
sebelumnya.
Pattern: Add 3 to the previous number. 12. (m + 3)(n – 2)

14. (5y – z)(2x – y)

Pernyataan itu tidak benar. Muka Surat 13 2. 3h – 5k
The statement is not true.
1. (4x – 5y) kg
(ii) Perimeter/Perimeter = 26 cm 3. (4xy – 16y2) m2
Luas/Area = 30 cm2
Muka Surat 14
FOKUS KBAT
A
(a) 15, 13, 11, 9, … 1. 5y 2. –2 – 5v 3. 2k – 4
Pola: Menolak 2 daripada nombor sebelumnya. 8w 6h
Pattern: Subtract 2 from the previous number. 9

(b) 4, 8, 16, 32, … B
Pola: Mendarab nombor sebelumnya dengan 2.
Pattern: Multiply the previous number by 2. 1. 12n + 5m 2. 8 – 5t
30 16
(c) (7, 64)
(d) (1, 512) 3. 9u – 6 4. 8p – 9q
u2 – 2u 21pq

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5. 4y + 3x – 3xy 6. 6k + m 3. (a) y = 5 (b) x = 0.8
9xy mnk
Muka Surat 22
20 – hk 10q – pq + 9p
7. 8h2k 8. 12pq2 1. (a) J = 30 + 45m + 40n
(b) RM245
Muka Surat 15 (c) p = 4

1. 2k – 3h 2. 4x – 1 2. (a) L = a2 + 2ab (b) L = 340
4 (c) a = 8
4. 50 – 5w
3. 2 5–w Praktis PT3
3m
6. 3d 1. (a) (i) ✗ (ii) ✓ (iii) ✓
5. 16s2 e(d + 3)
5t (b) (i) z = 1 x – y2
8. k + 3h 4
7. m –5h
n(3n + 2) (ii) z = y
2–x
9. x(x + y)
4y(x – y) (c) (i) L = x2 – 1 y2
2
Muka Surat 16

1. 13a + 15 2. 3u2 (ii) L = 215.5
5(a – 5)(a + 5) 5(1 + t)
2. (a) (i) ✓ (ii) ✗ (iii) ✓
3. h – 2k 4. –4w
24h 2(u – 1)(3u – 1) (b) (i) y = 270° – 3x (ii) y = 165°

5. –2e – 5 6. –q (c) (i) v = w – 1 u (ii) v = 11
3e (5 – q)(r + p) 5

3 (a) (i) A (ii) y = 25

Praktis PT3 (b) (i) n = –6 (ii) b = 25

1. (a) (i) ✗ (ii) ✓ (iii) ✓ (c) (i) T = 10 + 0.03x + 0.12y

(b) 4 (ii) 52 minit/minutes
n–6
FOKUS KBAT

(c) 2x2 + 2xy + x + y L = x2 + 3x – 18

2. (a) (i) A (ii) 8(u – 1)(u + 1) SOALAN-SOALAN BERORIENTASIKAN PISA

(b) (i) 6f + 24 (ii) 8 – 3s 1. 0.45 m
12t 2. 35 m per minit/35 m per minute

(c) RM(20xy + 5y2) 2.1 km per jam/2.1 km per hour

FOKUS KBAT BAB 4: POLIGON
Muka Surat 27
(a) (140 – 17x + x2) m2 A
(b) (140 + 17x – x2) m2

BAB 3: RUMUS ALGEBRA Poligon sekata/Regular polygons: Q, T
Muka Surat 19
Poligon tak sekata/Irregular polygons: P, R, S, U
1. L = 4ab + 2a2 2. J = 4p + q B

3. z = x + y + 6 1. 3, 3, 3 2. 7, 7, 7

Muka Surat 20 Muka Surat 28
A A

1. ✗ 2. ✓ 3. ✗ 4. ✓ 5. ✗ 1. 2.

B 60° 90°
1. y = 4x + 12
2. u = s – 1 at
t2

3. a = 4b 4. x = 3z2 – y2 3.
1 + 3b 2
40°
5. l = T 2g (b) n = 5
4π2 (b) p = 4

Muka Surat 21

1. (a) k = 47

2. (a) m = 102

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B Muka Surat 32
1. 2. A

3.5 cm 1. 45° 2. 80° 3. 30°
3. 140°
2 cm
3. 24°
B 3. 14
3. 15
1. 108° 2. 135° 6. 9

Muka Surat 33
A

1. 60° 2. 36°

Muka Surat 29 B
A
1. 5 2. 12
Sudut pedalaman/Interior angles: b, d, e, j, k
Sudut peluaran/Exterior angles: a, c, f, i, m, n Muka Surat 34

A 1. 5 2. 18
4. 12 5. 8
Hasil
Muka Surat 35
Poligon Bilangan Bilangan tambah 1. x = 150°, y = 120°
Polygons sisi segi tiga sudut 2. 34°
Number pedalaman
Number Sum of Praktis PT3
of sides of interior
triangles 1. (a) (i) ✗
(b) (i) 20°
angles (c) 20° (ii) ✗ (iii) ✓
(ii) 18 (iii) 5
1. 3 7 (7 – 2) × 2. (a) (i) 5
4 5 180° (b) 9 (ii) 5
12 (c) (i) 45°
= 900° (ii) 5

5

FOKUS KBAT

2. a+b+c+d+e+f+g+h
1
2 (8 – 2) × = (6 – 2) × 180°
8 6 180°
3 = 4 × 180° gh
= 1 080° = 720°
6 4
5 a b+g
f

3. ΂ n – 2 ΃ × 180° b a+h e
c d

Muka Surat 30 BAB 5: BULATAN
A Muka Surat 38
A
1. 360° 2. 720° 3. 1 440°
B

1. p = 130° q = 120° r = 110°

p + q + r = 130° + 120° + 110° 1. P • • Lilitan/Circumference

= 360°

2. a = 70° b = 50° c = 70°

d = 90° e = 80° 2. Q • • Jejari/Radius

a+b+c+d+e

= 70° + 50° + 70° + 90° + 80°

= 360° 3. R • • Perentas/Chord

Muka Surat 31 4. S • • Diameter/Diameter
A

1. 120° 2. 130° 3. 110°
4. 115° 5. 130°

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B OO 2. Lengkok minor (b) PNS ialah paksi simetri bulatan sebab PNS
Minor arc ialah diameter bulatan itu.
1. Lengkok major PNS is the axis of symmetry of the circle because
Major arc 3. Tembereng minor PNS is the diameter of the circle.
Minor segment
4. Sektor minor 2. (a) 8 cm
Minor sector O 5. Sektor major (b) 16 cm
Major sector (c) Panjang lengkok PQR dan lengkok STU
6. Sukuan bulatan adalah sama kerana panjang perentas PWR
Quadrant O dan perentas SVU adalah sama.
The length of arc PQR is equal to the length of
Muka Surat 39 7. Semibulatan arc STU because the length of segment PWR is
1. Semicircle equal to the length of segment SVU.

O 2. B

P O
2.5 cm
O
Muka Surat 41
1. SO = OR2 – RS2

= 102 – 82
= 6 cm

ST = SO + OT

= 6 + 10

= 16 cm

3. Q 2. 26 cm 3. 9 cm 4. 51 cm

4 cm Muka Surat 42
A
O
4 cm 1. 22 cm 2. 62.84 cm

(Terdapat dua jawapan yang mungkin.) B 2. 5 cm
(There are two possible answers.) 2. 113.112 cm2
4. 1. 21 cm

B C
1. 616 cm2
3. 254.502 cm2

Muka Surat 43
A

60° A 1. 7 cm 2. 3.5 cm 3. 10 cm
O 3 cm

B

1. 1 963.75 cm2 2. 2 827.8 cm2

C

Muka Surat 40 1. 88 cm 2. 22 cm
A
Muka Surat 44 2. 22
1. (a) N ialah pusat bulatan sebab N ialah titik A 2. 18
persilangan pembahagi dua sama serenjang
bagi dua perentas. 1. 14 2
N is the centre of the circle because N is the 3
point of intersection of perpendicular bisector
of two segments. B

1. 4.5

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C 2. 160° Muka Surat 53
1. 300° 1.
Muka Surat 45 2. 351.904 cm2
A 2.
1. 231 cm2 2. 14 cm

B 2. 15°
1. 21 cm (b) 185.5 cm2
(b) 72 cm
C (b) 64 1 cm2
1. 280°
6
Muka Surat 46 (b) 4 cm
1. (a) 7 cm
2. (a) 28 cm

Muka Surat 47
1. (a) 42 2 cm

3

2. (a) 25.136 cm
(c) 201.088 cm2

Praktis PT3

1. (a) (i) 8.38 cm (ii) 3.06 cm

(b) (i) 14 cm (ii) 13.89 cm Muka Surat 54
A
2. (a) P : Tembereng/Segment 1. 896 cm2

Q : Sektor/Sector B 2. 756 cm2
1. 1 240 cm2 2. 996 cm2
(b) x = 94.29 cm, y = 125.71 cm
Muka Surat 55 2. 2 776 cm2
(c) (i) 124 4 cm (ii) 650 2 cm2 A 2. 704 cm2
9 9 1. 1 881 cm2
2. 9 856 cm2
FOKUS KBAT (b) 479 1 cm2 B
9 1. 2 904 cm2
1. (a) 80°
2. (a) 100 2 cm (b) 462 cm2 Muka Surat 56
A
3 1. 1 386 cm2

BAB 6: BENTUK GEOMETRI TIGA B
DIMENSI

Muka Surat 51
A

Mempunyai bucu/With vertices 1. 5 2. 21 3. 17 4. 15
• Kon/Cone
• Prisma/Prism Muka Surat 57
• Piramid/Pyramid
1. 858 cm2 2. 1 996.5 cm2

Tidak mempunyai bucu/Without vertices Muka Surat 58
• Silinder/Cylinder 1. 294 cm3 2. 315 cm3 3. 1 280 cm3
• Sfera/Sphere 4. 1 971.2 cm3 5. 457.38 cm3 6. 38 808 cm3

B Muka Surat 59

1. Piramid/Pyramid 2. Sfera/Sphere 1. 1 480 cm3 2. 5
3. Prisma/Prism

Muka Surat 52 2. Piramid/Pyramid Praktis PT3
1. Kon/Cone 4. Silinder/Cylinder
3. Prisma/Prism 6. Piramid/Pyramid 1. (a) (i) prisma, hemisfera/prism, hemisphere
5. Silinder/Cylinder 8. Prisma/Prism (ii) silinder, piramid/cylinder, pyramid
7. Kon/Cone
(b) (i) Trapezium/Trapezium
(ii) 6

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(iii) PQ2 + QR2 = 18 + 32
= 50
= PR2

Maka, PQR ialah sebuah segi tiga
bersudut tegak.
Thus, PQR is a right-angled triangle.

(c) 616 cm3 (b) Luas segi tiga SQT 32)
2. (a) (i) ✗ Area of triangle SQT
(ii) ✓ (iii) ✗ = 1 × (2 × 18) × (2 ×
(b) (i) 6 cm 2
(c) 28 (ii) 10.5 cm = 48 unit2

FOKUS KBAT Praktis PT3

1. 47.6% 2. 3 465 cm2 1. (a) y P
R
BAB 7: KOORDINAT 7 Q
Muka Surat 63 6
A 5
4
1. 6 unit/units 2. 6 unit/units 3
3. 9 unit/units 4. 10 unit/units 2
1
B
O 1 234 567 x

1. 5 unit/units 2. 14 unit/units (b) (i) (10, 7) (ii) 1 km
3. 19 unit/units 4. 5 unit/units
5. 14.32 unit/units (c) (i) (4, –4) dan/and (8, –6)

(ii) 7 unit/units

Muka Surat 64 2. (a) 9
1. b = 9 atau/or b = 1
2. (2, 12) dan/and (2, 2) (b) (i) (13, 11) (ii) 10 unit/units
3. (a) (i) 5 unit/units (ii) 8.94 unit/units
(iii) 314 2 unit2
(iii) 5 unit/units 7
(b) Segi tiga sama kaki/Isosceles triangle
(c) (i) m = 5, n = 6 (ii) 5.385 unit/units

FOKUS KBAT

Muka Surat 65 1. R
A
2. Bucu S adalah salah.
1. (–4, 2) 2. (2, 4) The vertex S is incorrect.
3. (2, –2) 4. (9, 0)
B Titik tengah PR/Midpoint of PR

= ΂2 + 8, 6 + 2΃ = (5, 4)
2 2

1. (8, 4) 2. (–2, 5) 3. (1, 9) Katakan koordinat bagi titik S ialah (m, n).
4. (6, –10) 5. (5, 7) 6. (5, 8)
7. (7, 9) 8. (–1, 3) Let the coordinates of point S be (m, n).

Muka Surat 66 ΂ 3 + m , 1 + n΃ = (5, 4)
1. (6, 15) 2 2
3. (a) (3, 2)
2. m = –4, n = 7 3 + m = 5 , 1+n=4
(b) (15, 14) 2 2
1+n=8
Muka Surat 67 3 + m = 10

1. (a) Q(–1, 2), R(–1, 6), S(7, 6) m=7 n=7

(b) (3, 4) Maka, koordinat bagi titik S ialah (7, 7).
Thus, the coordinates of point S are (7, 7).
2. (a) (3, 2) (b) (5, 3) (c) 10 unit2

Muka Surat 68 BAB 8: GRAF FUNGSI
Muka Surat 72
1. (a) (5, 4) (b) m = 1, n = 8 A
1. ✓
(c) 20.396 unit/units
Hubungan satu kepada satu
2. (a) PQ = (7 – 4)2 + [2 – (–1)]2 = 18
One -to- one relation
QR = (3 – 7)2 + (6 – 2)2 = 32

PR = (3 – 4)2 + [6 – (–1)]2 = 50

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2. Muka Surat 74 0 123
Hubungan satu kepada banyak 3. x –2 –1 1 0 –7 –26

One -to- many relation y92 2 cm
2 cm
3. ✓ y
Hubungan banyak kepada satu
Many -to- one relation 10

4. –2 –1 0 x
Hubungan banyak kepada banyak –10
Many -to- many relation –20 123

B y = 1 – x3
1. {(1, 5), (2, 10), (3, 15)}
–30

2. P 1 2 3
Q 5 10 15
4. y = 5x
3. Set Q 4. x –15 –10 –5 5 10 15
f(x) = 5x y –2 –3 –6 63 2
15
0123 y y= —3x0 2 cm
10 5432 2 cm
5 6
Set P 2 cm
0 2 cm 4
123

Muka Surat 73 –1 2
6
1. x –3 –2 x
y87 5 10 15

y –15 –10 –5 0
–2

8

–4

6 –6
4 y=5–x

2

–3 –2 –1 0 1 23 x Muka Surat 75
1. (a) 20%
2. x –3 –2 –1 0 1 2 3
y 18 7 0 –3 –2 3 12 (b) Ya/Yes
(c) 3.2 jam/hours
y 2 cm 2. (a) 50 m
(b) 4 saat/seconds
2 cm (c) 0.5 saat dan 2.5 saat
20
0.5 second and 2.5 seconds
15
y = 2x2 – x – 3 Muka Surat 76

10 1. (a) x 0 5 10 15 20 25 30

A 0 50 200 450 800 1 250 1 800

5

–3 –2 –1 0 123 x
–5

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(b) 2 cm (b) A = 39 – 6h 0 1 23
A (cm2) 2 cm (c) (i) x –2 –1 –4 0 6 14
2 000
1 500 s –6 –6 2 cm

(ii) 2 cm
y

15

1 000 10
s = x2 + 3x – 4

5

500 x

–2 –1 0 123
–5
0 x (cm)
–10
10 20 30

(c) (i) A = 650 cm2 2. (a) (i) –2.6 (ii) 1.6 2 cm
(ii) A = 1 150 cm2 (b) (ii) 2 cm
y
(d) Kek yang lebarnya 24 cm x
A cake with width 24 cm 6 3
4
2. (a) y1 = 1 500 + 200x 2
y2 = 1 800 + 150x

(b) x 0 1 2 3 4 5

y1 1 500 1 700 1 900 2 100 2 300 2 500

y2 1 800 1 950 2 100 2 250 2 400 2 550

(c) 2 cm –3 –2 –1 0 12
–2
y (RM) –4

2 cm

3 000

2 500 y2 = 1 800 + 150x
2 000
y = 1 500 + 200x
1

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

1 500 I2 700 680 660 640 620 600

1 000 (ii)

I (cm3) 2 cm

500 2 cm

0 123456 800
x(tahun) I2 = 700 – 20x

(d) Tahun ke-6 600
6th year
400
Praktis PT3 I1

1. (a) (i) Pemboleh ubah bersandar 200
Dependent variable
y 0 2 4 6 7 8 x (minit/minute)
Pemboleh ubah tidak bersandar x
Independent variable
(iii) Minit ke-7
(ii) 5 7th minute

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FOKUS KBAT Muka Surat 87
(a) (i) x 0 1 2 3 4 5 A

y 50 65 80 95 110 125 1. 110 km/j, 100 km/j, 1 minit
Nyahpecutan/Deceleration
(b)
2. 0 m/s, 30 m/s, 20 saat
y Pecutan/Acceleration

B

200 1. Pecutan/Acceleration = 0.4 m/s2
150
100 2. Pecutan/Acceleration = 3 km/j per minit
50 3. Nyahpecutan/Deceleration = 315 km/j2

0 1234567 Muka Surat 88

(c) 7 buah beg/bags 1. 12 m/s 2. 50 km/j 3. 60
4. 8 5. 20

Praktis PT3

x 1. (a) (i) B (ii) 10:30 a.m.

(b) 25 km = 25 km
20 min 20 j

60

BAB 9: LAJU DAN PECUTAN = 75 km/j
Muka Surat 83
A ≠ 76 km/j

km/min RM/kg cm/s Maka, kereta itu bergerak dengan laju tak

seragam.

m/s2 Thus, the car moves with non-uniform speed.

(c) (i) 6.25 jam/hours

B (ii) 126 km

1. 38.4 m/s 2. 9.6 m/min 2. (a) (i) v < w < u
3. 80 km/j
C Jarak (ii) Nyahpecutan/Deceleration = 5 400 km/j2
Distance
Jarak (b) (i) 86.4 km/j (ii) 180 cm/s
Distance
(c) (i) 36 min (ii) 210 km

(iii) 292.5 km/j

FOKUS KBAT

Masa Masa 1
3
O Time O Time 1. 33 km

✓✓ 2. Jarak di antara bandar M dengan bandar N

Muka Surat 84 Distance between town M and town N
A
= 80 × 1.5
1. Laju seragam/Uniform speed
2. Laju tak seragam/Non-uniform speed = 120 km

Masa yang diambil untuk perjalanan pulang
Time taken for return journey

B = 120
80 + 20
1. 80 km/j 2. 15.875 m/min
C
= 1.2 j

1. 30 cm/s 2. 54 km/j 3. 240 cm/min Jumlah masa untuk seluruh perjalanan
Total time taken for whole journey
Muka Surat 85 = 1.5 j + 15 min + 1.2 j
= 1.5 j + 0.25 j + 1.2 j
1. 2.25 min 2. 140.4 km = 2.95 j
Jam 1100 – Jam 0800 = 3 jam
3. (a) 120 km 1100 hours – 0800 hours = 3 hours

(b) Jam 0848/0848 hours

4. 70 km

Muka Surat 86 Maka, David tiba di bandar M sebelum jam
1100.
1. 96 km/j 2. 88 km/j Thus, David reaches town M before 1100 hours.

3. (a) m = 1.5, n = 4 (b) 87.5 km/j

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SOALAN-SOALAN BERORIENTASIKAN PISA Muka Surat 96

(a) 12.00 tengah hari/noon 1. 10 2. (0, –15) 3. 312 cm

(b) Jarak laluan Cindy/Cindy’s route = 270 km Praktis PT3

Masa/Time y y
x x
= 3 jam 20 minit 10 2 0 1. (a) (i) (ii) 0 (iii) –
= 3 1 jam – 7 00
(b) (i) 9 (ii) 8
3 3 20
(c) Kecerunan/Gradient = –300
Laju purata/Average speed
Harga motosikal merosot RM300 setiap
= 270 ÷ 3 1
3 tahun.

= 81 km j–1 The price of the motorcycle decrease RM300
∴ Aida memandu dengan lebih laju.
every year.
Aida drives faster.
2. (a) P(0, 6), Q(–6, 6); T(8, 0), U(12, 0)

(b) (i) – 5 (ii) 6
4 (ii) 310.48 cm

BAB 10: KECERUNAN GARIS LURUS (c) (i) 15
Muka Surat 93 8
A
FOKUS KBAT
1. 2 2. 6 (a) mAB = 1.5, mBC = 2
3. mencancang, mengufuk
(b) Perbezaan laju/Difference in speed
vertical, horizontal
B (c) 11.00 a.m.

1. Kecerunan AB, mAB

Jarak mencancang y2 – y1 BAB 11: TRANSFORMASI ISOMETRI
Muka Surat 100
==
x2 – x1 A
Jarak mengufuk

2. Pintasan-y/y-intercept = a 1. Berubah/Changed
b Bukan transformasi/Not a transformation
Pintasan-x/x-intercept =
Kecerunan PQ, mPQ 2. Tidak berubah/Unchanged
Transformasi/Transformation
0–a
= B

b–0 1. Beza/Different, Beza/Different
Tidak kongruen/Not congruent

2. Sama/Same, Sama/Same
Kongruen/Congruent

a C

=– 1. A D
b

Pintasan- y B Q P
=– C

Pintasan- x

Muka Surat 94 R
S
1. 1 2. 0
2. R
3. tidak tertakrif/undefined
5. 4
4. –1 5

6. – 1 7. Benar/True Q
2 D
A
8. Palsu/False 9. Benar/True B P

10. Benar/True S

Muka Surat 95

1. 1 2. –2 3. – 1 C
2 3
1
4. 0 5. 3 6. 5

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Muka Surat 101 2. (a) Translasi/Translation ΂ 5 ΃
A 3

1. Translasi kerana bentuk, saiz dan orientasi sama. (b) y
A translation because the shape, size and orientation
are the same. 11

2. Bukan translasi kerana orientasi tidak sama. 10 U
Not a translation because the orientation is different.
9
B 8R
7S
6Q

1. ΂ 5 ΃ 2. ΂–53΃ 3. ΂––35΃ 5 P
3
4N
C 3

1. ΂103΃ 2. ΂ 0 ΃ 3. ΂ 8 ΃ 2M
–6 –4 1

O x
1 2 3 4 5 6 7 8 9 10 11

Muka Surat 102 (c) Garis PS ialah imej bagi garis MN di bawah
A
translasi ΂ 5 ΃.
1. 3

The line PS is the image of the line MN under

–5 a translation ΂35 ΃.
P
Muka Surat 104
–4
P' A

2. 1. Ya/Yes 2. Bukan/No 3. Ya/Yes
B
P
P' 1. Pantulan pada garis AB.
+3 A reflection in the line AB.
+6
2. Pantulan pada garis MN.
3. A reflection in the line MN.

3. Pantulan pada garis UV.
A reflection in the line UV.

4. Pantulan pada paksi-y.
A reflection in the y-axis.

5. Pantulan pada paksi-x.
A reflection in the x-axis.

Muka Surat 105
A

P' 1. Q U
+2
P
–6

4. V Q'

P +3 2. U
–3
Q' Q
P'
V
B
1. B 2. D 3. C 4. A 3. U
Muka Surat 103
Q
1. (a) Translasi/Translation ΂–64΃
Q'
(b) 48 cm2
(c) F' = (11, 5) V 2. B 3. D

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1. D

12 PINTAR BESTARI PT3 Matematik Tingkatan 2

Muka Surat 106 2. Q' P'
1. P

U B' QR R'
A A' C

D D'

B

C C' 3. y
E' E
5 P' Q'
F' F 4Q
R
G' G
3 R'
2

V 1P C
O1 67
2 3 4 5 8 9 10 11 x

2. (a) y A B
Q
12 U 1.
y
11
P T 7

10 6
5P
9
4
8 P'

7 RS 3
6 y=6
5 2

4 1
T'
O x
3 1 2 3 4 5 6 7 8 9 10 11 12

2

1 U' Putaran 90° ikut arah jam pada titik (6, 1).
B A clockwise rotation of 90° at the point (6, 1).

O x
1 2 3 4 5 6 7 8 9 10 11

(b) Pantulan pada garis y = 6. 2. y
A reflection in the line y = 6.
7 180°
(c) (9, 1)
6
5 P'

Muka Surat 107 4
A 3P

2

1

1. Bukan putaran/Not a rotation O 1 2 3 4 5 6 7 8 9 10 11 12 x
2. Putaran/A rotation
3. Putaran/A rotation Putaran 180° pada titik (6, 4).
A rotation of 180° at the point (6, 4).
B

1. Putaran 90° ikut arah jam pada titik (0, –2). Muka Surat 109
A clockwise rotation of 90° at the point (0, –2). A

2. Putaran 90° lawan arah jam pada titik (2, –4). 1. P'
An anticlockwise rotation of 90° at the point (2, –4).
P
3. Putaran 90° lawan arah jam pada titik (11, 3).
An anticlockwise rotation of 90° at the point (11, 3). O

4. Putaran 180° pada titik (7, 4). 2. P
A rotation of 180° at the point (7, 4).
P'
5. Putaran 180° pada titik (5, 1). O 270°

A rotation of 180° at the point (5, 1).

Muka Surat 108
A

1. C

P R'

Q P'
R
Q' PINTAR BESTARI PT3 Matematik Tingkatan 2
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13

3. Praktis PT3

P 1. (a) (i) Translasi/Translation ΂1–32΃

P' (ii) Objek/Object: P
Imej/ Image: Q
O
(b) (i), (ii), (iii)
4. P
A
P' O
180° EB
D

B'

C

B 4. A A' D' C'
1. B 2. C 3. B
(c) (i) 2 (ii) (4, 2)
Muka Surat 110 2. (a) (i) A
1. (a) (c) S
(ii) R
y
P
B
11 A

A C' B' QB
10 D'
CD
9
CD A' (b) (i) Translasi/Translation ΂68΃

8 (ii) M J' H' Q

7

6

5
P

4

3

2

1

x E' K' L'

O 1 2 3 4 5 6 7 8 9 10 11 F' G'

(b) Putaran 90° ikut arah jam pada titik P. N P
Putaran 270° lawan arah jam pada titik P. L
A clockwise rotation of 90° at the point P. JH
An anticlockwise rotation of 270° at the point P. EK

2. (a) (i) Putaran ikut arah jam FG
A clockwise rotation
(c) B, A, F, D
(ii) 11 12 1
FOKUS KBAT
10 2

9 3 (a) (6, 2) (b) (6, 0) (c) (0, 0)
O
8 BAB 12: SUKATAN KECENDERUNGAN
7 4 MEMUSAT

65 Muka Surat 116
1. Mod/Mode = 5
(iii) 10:40 a.m.
(b) Putaran 300° ikut arah jam pada titik O. Min/Mean = 3
Median/Median = 3
A clockwise rotation of 300° at the point O. 2. Mod/Mode = 5.7
Min/Mean = 6.5
Muka Surat 111 Median/Median = 6
1. (a) (i) Isometri/Isometry 3. Mod/Mode = 45
Min/Mean = 42
(ii) Isometri/Isometry Median/Median = 45
(iii) Isometri/Isometry 4. Mod/Mode = 70 dan/and 110
(b) Q, R dan/and S Min/Mean = 102.5
2. x = 13, y = 67.4 Median/Median = 110

Muka Surat 112 4. 6 5. 2
1. 2 2. 4 3. 3 9. 4 10. 2
6. 2 7. 4 8. 3

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Muka Surat 117 2. Mod/Mode = Tiada/None
1. Mod/Mode = 20 Min/Mean = 50
Median/Median = 50
Min/Mean = 21
Median/Median = 20 Mod baharu/New mode = Tiada/None
Mod baharu/New mode = 25 Min baharu/New mean = 375
Min baharu/New mean = 26 Median baharu/New median = 60
Median baharu/New median = 25
Kesimpulan: Apabila setiap nilai dalam data Kesimpulan: Terdapat nilai ekstrem dalam
ditambah dengan 5, mod, min, dan median
baharu bertambah sebanyak 5 . data ini, median sesuai digunakan
Conclusion: When each value in the data is
added with 5, the new mode, mean and median untuk mewakili data kerana nilai ekstrem

increases by 5 . mempengaruhi nilai min .

Muka Surat 118 Conclusion: There is an extreme value in the data,

median is suitable to be used to represent
the data because the extreme value affects the

mean value.

1. Pendapatan bulanan Titik tengah, x Gundalan Kekerapan, f fx
Monthly income (RM) Midpoint, x Tally Frequency, f

1 001 – 2 000 1 500.5 4 6 002

2 001 – 3 000 2 500.5 2 5 001

3 001 – 4 000 3 500.5 7 24 503.5

4 001 – 5 000 4 500.5 4 18 002

5 001 – 6 000 5 500.5 3 16 501.5
Σ f = 20 Σ fx = 70 010

2. RM3 001 – RM4 000 3. RM3 500.50 4. RM4 000
Muka Surat 119

1. Jisim betik (g) Titik tengah, x Gundalan Kekerapan, f fx
Mass of papaya (g) Midpoint, x Tally Frequency, f
724.5 2 173.5
700 – 749 3 3 098
774.5 2 473.5
750 – 799 4
824.5
800 – 849 3

850 – 899 874.5 5 4 372.5

900 – 949 924.5 6 5 547

950 – 999 974.5 4 3 898
2. 900 g – 949 g 3. 862.5 g Σ f = 25 Σ fx = 21 562.5
4. 52%

Muka Surat 120 2. Min/Mean
1. Median/Median Tiada nilai ekstrem dalam data.
There is no extreme value in the data.
Nilai ekstrem, 15 000, akan mempengaruhi
nilai min tetapi tidak akan mempengaruhi 3. Mod/Mode
nilai median. Bagi data kategori yang tidak mempunyai
The extreme value, 15 000, will affect the mean nilai berangka, mod digunakan.
value but not the median value. For the categorical data without numerical value,
the mode is used.

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4. Mod, min dan median/Mode, mean and median FOKUS KBAT
Taburan data adalah seragam dan tiada nilai
ekstrem. (a) Mod/Mode = 1 200
The data distribution is uniform and has no extreme Median/Median = 1 200
value. Min/Mean = 2 260

Muka Surat 121 (b) Mod dan median sesuai untuk mewakili data itu
1. Mod/Mode = RM6 kerana adanya nilai ekstrem iaitu 20 000, yang
mempengaruhi nilai min.
Min/Mean = RM4.60 Mode and median are suitable to be used to represent
Median/Median = RM5 the data because there is an extreme value, 20 000
that affects the mean value.

2. Mod/Mode = 2 dan/dan 3 BAB 13: KEBARANGKALIAN MUDAH
Min/Mean = 2.6 Muka Surat 127
Median/Median = 2.5
1. 9 2. 1 3. 4
Muka Surat 122 25 3 15
1. Mod/Mode = 3 jam/hours
Muka Surat 128
Min/Mean = 3.3 jam/hours A
Median/Median = 3 jam/hours

2. Mod/Mode = 38 tahun/years old 1. Mungkin/Possible
Min/Mean = 40 tahun/years old 2. Tidak mungkin/Impossible
Median/Median = 39 tahun/years old 3. Tidak mungkin/Impossible
B
Muka Surat 123
(a) P: Min/Mean = RM6 410 1. S = {Lelaki/Boy, Perempuan/Girl}
A = {Lelaki/Boy}
Q: Min/Mean = RM6 400 n(S) = 2
(b) P: Julat/Range = RM8 900 n(A) = 1

Q: Julat/Range = RM3 890 2. S = {P, E, R, A, K}
(c) Walaupun gerai P memperoleh min A = {E, A}
n(S) = 5
keuntungan yang sedikit lebih tinggi daripada n(A) = 2
gerai Q dalam tempoh lima bulan itu, tetapi
nilai julat gerai P adalah jauh lebih besar 3. S = {Rosman, Muthu, Wei Hou, Kamala,
daripada gerai Q. Oleh itu, keuntungan gerai Q Siew Ling, Nadia, Siti}
adalah lebih konsisten dan akan memperoleh
keuntungan yang lebih banyak daripada gerai P A = {Kamala, Siew Ling, Nadia, Siti}
dalam tempoh masa yang lebih panjang. n(S) = 7
Although the mean profit gained by stall P is bit n(A) = 4
higher than stall Q in the five months, the profit range
of stall P is larger than stall Q. Therefore, the profit Muka Surat 129 Kad Kesudahan
of stall Q is more consistent and stall Q can gain 1. Nombor
Card Outcome
more profit than stall P over a longer period of time. Number

M (1, M)

Praktis PT3 1 K (1, K)

1. (a) (i) ✗ (ii) ✓

(b) (i) Mac/March (ii) RM2 400 B (1, B)

(c) (i) 73 kg

(ii) Mod dan median adalah sesuai untuk M (2, M)

mewakili data itu kerana ada nilai

ekstrem yang mempengaruhi nilai min.

Mode and median are suitable to be used 2 K (2, K)
B (2, B)
to represent the data because there is an M (3, M)

extreme value that affects the mean value.

2. (a) (i) Teh/Tea (ii) RM70

(b) (i) 36

(ii) x = 11, y = 12

(Mana-mana nombor bulat yang lebih

daripada 10.) 3 K (3, K)
B (3, B)
(Any whole number which is greater than 10.)

(c) RM1 250

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(a) S = {(1, M), (1, K), (1, B), (2, M), (2, K), (d) Peristiwa mendapat vokal atau konsonan
(2, B), (3, M), (3, K), (3, B)} The event of getting a vowel or a consonant
= {A1, A2, E, B, C, D1, D2, F}
(b) {2B} Oleh sebab {A1, A2, E, B, C, D1, D2, F}
(c) {(1, M), (1, K), (1, B), (2, M), (3, M)} = S, maka peristiwa mendapat vokal
(d) {(1, M), (1, K), (1, B)} ialah pelengkap bagi peristiwa mendapat
konsonan.
Muka Surat 130 Since {A1, A2, E, B, C, D1, D2, F} = S, then the
event of getting a vowel is the complement of
1. 3 2. 2 3. 3
5 7 5 the event of getting a consonant.

4. (a) 1 (b) 1 1
8 4 4
2. (a) (b) 1 475

Muka Surat 131 Praktis PT3
A
1. (a) Label sebarang 4 sektor dengan A, 1 sektor
1. (a) A′ ialah peristiwa mendapat nombor selain dengan B dan 3 sektor dengan C.
daripada 5. Label any 4 sectors as A, 1 sector as B and
A′ is an event of getting numbers other than 5. 3 sectors as C.

(b) A′ = {1, 2, 3, 4, 6} (b) (i) S = {A, B, C, D, E}
2. (a) A′ ialah peristiwa memilih murid yang
(ii) 1 (iii) 4
menyertai kurang daripada tiga persatuan. 5 5
A′ is an event of choosing a student who join
less than three societies. (c) (i) 32 (ii) 4
(b) A′ = {0, 1, 2}
2. (a) (i) C
B
1. 5 2. 1 3. 3 4. 3 (ii) Papan/Board A = 8 = 2
20 5
7 3 10 8
Papan/Board B = 10 = 2
25 5
Muka Surat 132
1. (a) 5 12 (c) 14 Papan/Board C = 9
17 17 20
17 (b)
Oleh sebab pecahan kawasan berlorek

2. (a) S = {–2, –1, 0, 1, 2, 3} pada papan C adalah paling besar, maka

2 1 kebarangkalian mengena kawasan
3 2
(b) (c) {1, 2, 3}, berlorek pada papan C adalah paling

tinggi. Maka, papan C dipilih.

(d) Peristiwa mendapat nombor negatif atau Since the fraction of the shaded region
nombor positif = {–2, –1, 1, 2, 3}
Oleh sebab {–2, –1, 1, 2, 3} ≠ S, maka on board C has the largest value, then
peristiwa mendapat nombor negatif bukan
pelengkap bagi peristiwa mendapat nombor the probability of hitting the shaded area
positif dalam eksperimen ini.
The event of getting a negative number or on board C is the highest. Thus, board C
a positive number = {–2, –1, 1, 2, 3}
should be chosen.
Since {–2, –1, 1, 1, 2, 3} ≠ S, then the event of
getting a negative number is not the complement (b) (i) Biru/Blue (ii) 5
of the event of getting a positive number in this 8
experiment.
(c) (i) 1
3

(ii) Guli biru sebab bilangan guli biru

adalah paling kurang.

Blue marble because the number of blue

marbles is the least.

Muka Surat 133 (iii) 5
14
1. (a) S = {A1, A2, B, C, D1, D2, E, F}

(b) {B, C, D1, D2, F}, 5 FOKUS KBAT
8

(c) 225 192

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