Standard Form - pelangibooks.com RESOURCE_MATHS 2017/Top... · Mathematics Form 4 Chapter 1...

1 © Penerbitan Pelangi Sdn. Bhd.

1 significant figure

1 angka bererti

2 significant figures

2 angka bererti

3 significant figures

3 angka bererti

1. (a) 700 740 744

(b) 50 52 52.0

(c) 4 4.1 4.06

(d) 0.3 0.28 0.282

(e) 0.08 0.080 0.0804

2. (a) 46.7 + 1.8 × 0.025 = 46.7 + 0.045 = 46.745 = 47 (2 significant figures/angka bererti)

(b) 29.3 – 5.88 ÷ 1.2 = 29.3 – 4.9 = 24.4 = 20 (1 significant figure/angka bererti)

(c) 0.432 ÷ 15 × 0.9 = 0.0288 × 0.9 = 0.02592 = 0.0259 (3 significant figures/angka bererti)

(d) 108 × (1.074 – 0.938) = 108 ÷ 0.136 = 14.688 = 14.7 (3 significant figures/angka bererti)

(e) 0.0257 + 37.121.6

= 0.0257 + 23.2 = 23.2257 = 23 (2 significant figures/angka bererti)

3. (a) Surface area of the sphere Luas permukaan sfera = 4 × 3.142 × 0.152

= 0.28278 m2

= 0.283 m2 (3 significant figures/angka bererti)

(b) Volume of the right pyramid Isi padu piramid tegak

= 13

× 4.7 × 5.8 × 6

= 54.52 cm3

= 55 cm3 (2 significant figures/angka bererti)

(c) The total amount that Muthu had to pay Jumlah wang yang perlu dibayar oleh Muthu = (7 × RM1.80) + (5 × RM12.65) = RM12.60 + RM63.25 = RM75.85 = RM80 (1 significant figures/angka bererti)

4. (a) 461 000 = 4.61 × 105

(b) 823 = 8.23 × 102

(c) 795.2 = 7.952 × 102

(d) 36.31 = 3.631 × 101 (e) 0.1107 = 1.107 × 10−1

(f) 0.00249 = 2.49 × 10−3

(g) 0.000056 = 5.6 × 10−5

5. (a) 2.663 × 102 = 266.3(b) 8.04 × 101 = 80.4(c) 7.89 × 104 = 78 900(d) 5.21 × 10−1 = 0.521(e) 4.431 × 10−5 = 0.00004431(f) 7 × 10−4 = 0.0007(g) 3.5 × 10−3 = 0.0035

6. (a) 2.34 × 104 − 10 900 = 2.34 × 104 − 1.09 × 104

= (2.34 – 1.09) × 104 = 1.25 × 104

(b) 7.1 × 105 × 6.5 × 10–2 = (7.1 × 6.5) × 105 × 10–2 = 46.15 × 105 + (–2)

= 4.615 × 101 × 103 = 4.615 × 104

(c) 1.8 × 10–3 ÷ 9.6 × 102

= (1.8 ÷ 9.6) × 10–3 – 2

= 0.1875 × 10−5

= 1.875 × 10–1 × 10–5

= 1.875 × 10–6

(d) 9.15 × 10–3 + 0.000326 = 9.15 × 10–3 + 3.26 × 10–4 = 9.15 × 10–3 + 0.326 × 10–3

= (9.15 + 0.326) × 10–3 = 9.476 × 10–3

(e) 0.00057 − 4.8 × 10−5 = 5.7 × 10−4 − 4.8 × 10−5

= 5.7 × 10–4 – 0.48 × 10–4

= (5.7 − 0.48) × 10−4 = 5.22 × 10−4

(f) (3.2 × 103)2 = 3.22 × (103)2 = 10.24 × 106 = 1.024 × 107

(g) 4.71 × 105 ÷ 0.003

= 4.71 × 105

3 × 10–3

= 1.57 × 105 – (–3)

= 1.57 × 108

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R 1 Standard FormBentuk Piawai

© Penerbitan Pelangi Sdn. Bhd. 2

Mathematics Form 4 Chapter 1 Standard Form

(h) 45 000

(6 × 104)2

= 4.5 × 104

62 × 108

= 0.125 × 104 – 8 = 1.25 × 10–1 × 10–4

= 1.25 × 10–5

(i) 5.86 × 10–4

(2 × 10–2)3

= 5.86 × 10–4

8 × 10–6

= 0.7325 × 10−4 − (−6)

= 7.325 × 10–1 × 102

= 7.325 × 101

7. (a) The width of the land Lebar tanah

= 4.8 ÷ (150 × 10–2) = 3.2 m

(b) The volume of the empty space Isi padu ruang kosong = 24 × 15 × 50 − 450 = 1.8 × 104 − 0.045 × 104 = 1.755 × 104 cm3

(c) The perimeter of the surface of the table Perimeter permukaan meja = 2(1.9 × 102 + 8.5 × 10) = 2(1.9 × 102 + 0.85 × 102) = 2(2.75 × 102) = 5.5 × 102 cm

SPM Practice 1

Paper 1

1. 0.00315 − 2.7 × 10−4 = 3.15 × 10−3 − 0.27 × 10−3 = (3.15 − 0.27) × 10−3 = 2.88 × 10−3

Answer/ Jawapan: C

2. 0.01573 = 0.01573 = 0.016 (2 significant figures / angka bererti)

Answer/ Jawapan: D

3. 36 725 = 36 725 = 36 700 (3 significant figures / angka bererti)

Answer/ Jawapan: C

4. The remainder of the flour Baki tepung = 500 − (500 × 25%) = 375 kg

The mass in each plastic bag Jisim tepung dalam setiap beg

= 375 ×1 00060

g

= 6 250 g = 6.25 × 103

Answer/ Jawapan: D

5. 0.0000589 = 5.89 × 10−5

= m × 10n

m = 5.89, n = −5

Answer/ Jawapan: C

6. 420.5 = 4.205 × 102

Answer/ Jawapan: A

7. 0.000481 = 4.81 × 10−4

Answer/ Jawapan: A

8. Total number of erasers produced Jumlah bilangan pemadam yang dihasilkan = 6 × 103 × 60 × 14 = 5 040 000 = 5.04 × 106

Answer/ Jawapan: C

9. 0.074(2 ×10–2)2

= 7.4 ×10–2

4 ×10–4

= 7.44

×10–2 – (–4)

= 1.85 × 102

Answer/ Jawapan: D

10. Volume of water that can be stored Isi padu air yang boleh ditakung

= 227

× 1002

2

× 98

= 770 000 = 7.7 × 105

Answer/ Jawapan: A

11. 476 200 = 4.762 × 105

Answer/ Jawapan: A

12. The total volume of the metal cube Jumlah isi padu logam yang berbentuk kubus = 50 × 63 = 10 800 cm3

The volume of a sphere Isi padu sebuah sfera

= 10 8003 000

= 3.6 cm3

Mathematics Form 4 Chapter 1 Standard Form


The volume of a sphere is Isipadu sebuah sfera

3.6 = 43

× 227

×r3

r3 = 0.8591 r = 0.9506 = 9.506 × 10−1

Answer/ Jawapan: B

13. The number of basket balls produced Bilangan bola keranjang yang dihasilkan = 285 000 × 35% = 99 750 = 9.975 × 104

Answer/ Jawapan: B


1. (a) 2y2 − 9

Yes / Ya

(b) 3p(2 − 5p) = 6p − 15p2

Yes / Ya

(c) (x − 1)(x + 3) = x2 + 2x − 3

Yes / Ya

(d) 2n(1 − n2) + 3n = 2n − 2n3 + 3n = 5n − 2n3

No / Tidak

(e) m2 + 2m

= m3 + 2m

No / Tidak

(f) 10 + 7h No / Tidak

2. (a) q + 2q(4 − q) = q + 8q − 2q2

= 9q − 2q2

(b) –3m(m + 2) = –3m2 – 6m

(c) (y −1)(4y − 3) = 4y2 − 4y − 3y + 3 = 4y2 − 7y + 3

(d) (p + 2)2 = (p + 2)(p + 2) = p2 + 2p + 2p + 4 = p2 + 4p + 4

(e) (2m − 3)2 = (2m − 3)(2m −3) = 4m2 − 6m − 6m + 9 = 4m2 − 12m + 9

(f) (x + 3)(x − 4) = x2 – 4x + 3x − 12 = x2 − x − 12

(g) (y − 3)2 = (y − 3) (y − 3) = y2 – 3y – 3y + 9 = y2 − 6y + 9

3. (a) Total cost / Jumlah kos = (2m − 3)(m + 1) = 2m2 − 3m + 2m − 3 = RM(2m2 − m − 3)

(b) Area of the shaded region Luas rantau berlorek

= 12

(2p + 6 + 14)(p + 5) − 12

(6)(p)

= (p + 10)(p + 5) − 3p = (p2 + 12p + 50) cm2

(c) Total distance / Jumlah jarak = (2y − 5)(y + 3) = 2y2 − 5y + 6y − 15 = (2y2 + y − 15) m

4. (a) 4m + 14m2 = 2m(2 + 7m)

(b) −5t2 − 20t = −5t(t + 4)

(c) 25 − h2 = (5 + h)(5 − h)

(d) −16p2 + 49 = 49 − 16p2 = (7 + 4p)(7 − 4p)

(e) 4 − 100y2 = 4(1 − 25y2) = 4(1 + 5y)(1 − 5y)

(f) −12k2 + 75 = 3(−4k2 + 25) = 3(5 − 2k)(5 + 2k)

(g) 3s2 − 27 = 3(s2 − 9) = 3(s + 3)(s − 3)

(h) 15 – 6x2 = 3(5 – 2x2)

5. (a) 9 − 6m + m2 = (3 − m)(3 − m) = (3 – m)2

(b) 5t2 − 14t + 8 = (t − 2)(5t − 4)

(c) 6h2 + 11h − 10 = (3h − 2)(2h + 5)

(d) −9 − 5n + 4n2 = (−9 + 4n)(1 + n)

(e) 21 + 5p − 6p2 = (3 + 2p)(7 − 3p)

(f) 2y2 − 4y − 6 = 2(y2 −2y − 3) = 2(y − 3)(y + 1)

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R 2 Quadratic Expressions and EquationsUngkapan dan Persamaan Kuadratik


Mathematics Form 4 Chapter 2 Quadratic Expressions and Equations

(g) 12 + 3x − 15x2

= 3(4 + x − 5x2) = 3(4 + 5x)(1 − x)

6. (a) Yes / Ya(b) No / Tidak(c) No / Tidak(d) No / Tidak(e) Yes / Ya(f) No / Tidak

7. (a) 3(p + 2)2 = 9 + p2

3(p2 + 4p + 4) = 9 + p2

3p2 + 12p + 12 − 9 − p2 = 0 2p2 + 12p + 3 = 0(b) 2 − n = (n + 3)(2n − 5) 2 − n = 2n2 − 5n + 6n −15 2n2 + n − 15 + n − 2 = 0 2n2 + 2n − 17 = 0

(c) x + 2x – 1

= 2x + 1

x + 2 = (2x + 1)(x − 1) x + 2 = 2x2 − 2x + x − 1 2x2 − x − 1 − x − 2 = 0 2x2 − 2x − 3 = 0

8. (a) Let the first odd number Katakan nombor ganjil pertama = x

The second odd number Nombor ganjil kedua = x + 2

x2 + (x + 2)2 = 394 x2 + x2 + 4x + 4 = 394 2x2 + 4x − 390 = 0 x2 + 2x – 195 = 0

(b) Area / Luas = 24

12

× (2x + 5x – 3) ×(x + 1) = 24

(7x – 3)(x + 1) = 48 7x2 + 7x – 3x + 3 = 48 7x2 + 4x − 3 – 48 = 0 7x2 + 4x – 51 = 0

(c) The price of 1 kg oranges Harga 1 kg oren = RMx

The price of 1 kg apples Harga 1 kg epal = RM(x + 3)

2x(x + 3) + x(x + 1) = 50 2x2 + 6x + x2 + x – 50 = 0 3x2 + 7x – 50 = 0

9. (a) LHS = 2p2 − 3p − 9 = 2(3)2 − 3(3) − 9 = 18 − 9 − 9 = 0 = RHS

\ p = 3 is a root of the equation. p = 3 ialah punca bagi persamaan.

(b) LHS = k(k + 2) = 4(4 + 2) = 24

RHS = 3k − 2 = 3(4) − 2 = 10 LHS ≠ RHS

\ k = 4 is not a root of the equation. k = 4 bukan punca bagi persamaan.

(c) LHS = 3m2 − 5 = 3(−1)2 − 5 = −2

RHS = 4m + 2 = 4(−1) + 2 = −2

LHS = RHS

\ m = −1 is a root of the equation. m = –1 ialah punca bagi persamaan.

10. (a) (2q + 5)(q − 2) = 0

2q + 5 = 0 or / atau q − 2 = 0 2q = −5 q = 2

q = – 52

(b) 2p2 + 11p + 9 = 0 (2p + 9)(p + 1) = 0

2p + 9 = 0 or / atau p + 1 = 0 2p = −9 p = −1

p = – 92

(c) 3n − 6n2 = 0 3n(1 − 2n) = 0

3n = 0 or / atau 1 − 2n = 0 n = 0 2n = 1

n = 12

(d) h2 + h − 20 = 0 (h + 5)(h − 4) = 0

h + 5 = 0 or / atau h − 4 = 0 h = −5 h = 4

(e) t2 − 4t = 21 t2 − 4t − 21 = 0 (t − 7)(t + 3) = 0

t − 7 = 0 or / atau t + 3 = 0 t = 7 t = −3

(f) 4y2 − 9y + 2 = 0 (4y − 1)(y − 2) = 0

4y − 1 = 0 or / atau y − 2 = 0 4y = 1 y = 2

y = 14



(g) 6m2 = 11m − 4 6m2 − 11m + 4 = 0 (2m − 1)(3m − 4) = 0

2m − 1 = 0 or / atau 3m − 4 = 0 2m = 1 3m = 4

m = 12

m = 43

(h) 3x2 + x = 5(x + 3) 3x2 − 4x − 15 = 0 (3x + 5)(x − 3) = 0

3x + 5 = 0 or / atau x − 3 = 0 3x = –5 x = 3

x = – 53

11. (a) (2y + 1)(y − 2) = 4 + y

2y2 − 3y − 2 = 4 + y 2y2 − 4y − 6 = 0 y2 − 2y − 3 = 0 (y − 3)(y + 1) = 0 y − 3 = 0 or / atau y + 1 = 0

y = 3 y = −1

(b) 5p(2p + 1) = 6 + 9p

10p2 + 5p = 6 + 9p 10p2 − 4p − 6 = 0 5p2 − 2p − 3 = 0 (5p + 3)(p − 1) = 0

5p + 3 = 0 or / atau p − 1 = 0 5p = −3 p = 1

p = – 35

(c) m2 + 3

m = 4

m2 + 3 = 4m m2 − 4m + 3 = 0 (m − 3)(m − 1) = 0

m − 3 = 0 or / atau m − 1 = 0 m = 3 m = 1

(d) 2t – 3 = 30

t + 2

(2t − 3)(t + 2) = 30 2t2 + t − 6 = 30 2t2 + t − 36 = 0 (2t + 9)(t − 4) = 0

2t + 9 = 0 or / atau t − 4 = 0 2t = −9 t = 4

t = – 92

(e) –7

h + 4 = 2h – 1

–7 = (2h − 1)(h + 4) –7 = 2h2 + 7h − 4 2h2 + 7h + 3 = 0 (2h + 1)(h + 3) = 0

2h + 1 = 0 or / atau h + 3 = 0 2h = −1 h = −3

h = – 12

12. (a) Use Pythagoras Theorem: Menggunakan Teorem Pythagoras:

(p +12)2 = (2p – 5)2 + p2

p2 + 2p + 1 = 4p2 – 20p + 25 + p2

4p2 – 22p + 24 = 0 2p2 – 11p + 12 = 0 (p – 4)(2p – 3) = 0

p – 4 = 0 or / atau 2p − 3 = 0

p = 4 p = 32

(Rejected / Ditolak)

(b) Speed = distance

time

Laju = jarakmasa

Let the time taken by the champion = t Katakan masa yang digunakan oleh juara = t

6t

– 6

t + 3060

= 2

6t

– 2 = 6

t + 12

6 – 2t

t = 6

t + 12

(6 – 2t)(t + 12

) = 6t

2t2 – 5t − 3 + 6t = 0 2t2 + t − 3 = 0 (t – 1)(2t + 3) = 0

t – 1 = 0 or / atau 2t + 3 = 0

t = 1 t = – 32

(Rejected / Ditolak)

(c) Let the number of students in the beginning = x Katakan bilangan murid pada asalnya = x

160

x + 1 =

160x – 8

160 + x

x =

160x – 8

(160 + x)(x – 8) = 160x 160x – 8x + x2 – 1 280 = 160x x2 – 8x − 1 280 = 0 (x – 40)(x + 32) = 0 (x – 40)(x + 32) = 0

x – 40 = 0 or / atau x + 32 = 0 x = 40 x = –32 (Rejected / Ditolak)



SPM Practice 2

Paper 2

1. 3m(m − 2) = 2 − 5m 3m2 − 6m = 2 − 5m 3m2 − m − 2 = 0 (3m + 2)(m − 1) = 0 3m + 2 = 0 or / atau m − 1 = 0

m = – 23

m = 1

2. x(5 + 2x) = −2 5x + 2x2 = −2 2x2 + 5x + 2 = 0 (2x + 1)(x + 2) = 0 2x + 1 = 0 or / atau x + 2 = 0

x = – 12

x = −2

3. 4p(p + 3) = 8p + 3 4p2 + 12p = 8p + 3 4p2 + 4p − 3 = 0 (2p − 1)(2p + 3) = 0 2p − 1 = 0 or / atau 2p + 3 = 0

p = 12

p = – 32

4. 3n = –5 – 2n

n – 6

3n(n − 6) = −5 − 2n 3n2 − 18n = −5 − 2n 3n2 − 16n + 5 = 0 (3n − 1)(n − 5) = 0 3n − 1 = 0 or / atau n − 5 = 0

n = 13

n = 5

5. 2m2 − 5 = 3(7 + 3m) 2m2 – 5 = 21 + 9m 2m2 – 9m − 26 = 0 (2m – 13)(m + 2) = 0 2m – 13 = 0 or / atau m + 2 = 0

m = 132

m = –2

6. (x + 3)2 = 5 + x x2 + 6x + 9 = 5 + x x2 + 5x + 4 = 0 (x + 1)(x + 4) = 0 x + 1 = 0 or / atau x + 4 = 0 x = −1 x = −4

7. h = 10 + t − 2t2

When the ball reach the ground, h = 0 Apabila bola itu sampai di tanah, h = 0

10 + t − 2t2 = 0 (5 − 2t) (2 + t) = 0 5 − 2t = 0 or / atau 2 + t = 0

t = 52

t = −2 (Not accepted/ Tidak diterima)

\ The ball took 2.5 seconds to reach the ground. Bola itu mengambil masa 2.5 saat untuk sampai di tanah.

8. (2m + 1)(m − 3) = 7 − 4m 2m2 − 6m + m − 3 = 7 − 4m 2m2 − m − 10 = 0 (2m − 5)(m + 2) = 0 2m − 5 = 0 or / atau m + 2 = 0

m = 52

m = −2

9. x(3x – 2)

x + 2 = 5

3x2 − 2x = 5(x + 2) 3x2 − 2x = 5x + 10 3x2 − 7x − 10 = 0 (3x − 10)(x + 1) = 0 3x − 10 = 0 or / atau x + 1 = 0

x = 103

x = −1

10. Area of the triangle Luas segi tiga = 33

12

(4x − 1) (x + 3) = 33

4x2 − x + 12x − 3 = 66 4x2 + 11x − 69 = 0 (x − 3) (4x + 23) = 0

x − 3 = 0 or 4x + 23 = 0

x = 3 atau

x = – 234

(Not accepted/ Tidak diterima)

11. v = t + 21 − 2t2 = −2t2 + t + 21

When / Apabila v = 0,

(−2t + 7) (t + 3) = 0

−2t + 7 = 0 or t + 3 = 0

t = 72

atau

t = −3 (Not accepted/ Tidak diterima)

\ The ball took 3.5 seconds to hit the wall. Bola itu mengambil masa 3.5 saat untuk terkena dinding.

12. xz

y

Let the longest side = x cm Biarkan sisi terpanjang = x cm

y = (x – 2) cm z = (x – 4) cm

y2 + z2 = x2

(x – 2)2 + (x – 4)2 = x2

x2 – 4x + 4 + x2 – 8x + 16 = x2

x2 – 12x + 20 = 0 (x – 10)(x – 2) = 0

x – 10 = 0 or x – 2 = 0 x = 10 atau x = 2 (Not accepted/ Tidak diterima)



y = 10 – 2 = 8 z = 10 – 4 = 6

\ The length of the three sides are 10 cm, 8 cm and 6 cm.

Panjang bagi tiga sisi ialah 10 cm, 8 cm dan 6 cm.


1. (a) blue, red, green, black Biru, merah, hijau, hitam

(b) Japan, Singapore, Brunei, France Jepun, Singapura, Brunei, Perancis

(c) triangle, square, trapezium, rectangle segi tiga, segi empat sama, trapezium, segi empat tepat

2. (a) B is the first 10 odd numbers. / B is a set of odd numbers which are less than 20.

B ialah sepuluh nombor ganjil yang pertama. / B ialah satu set nombor ganjil yang kurang daripada 20.

(b) C is the first 6 prime numbers. / C is a set of prime numbers which are less than 15.

C ialah enam nombor perdana yang pertama. / C ialah satu set nombor perdana yang kurang daripada 15.

(c) D is a set of quadrilaterals. D ialah satu set sisi empat.

3. (a) {1, 4, 9, 16, 25, 36, 49, 64, 81}(b) {1, 2, 3, 6, 9, 18}(c) {F, A, C, T, O, R, S}

4. (a) (i) 2 N

(ii) 8 N

(iii) 5 N

(b) (i) 2 J

(ii) 20 J

(iii) 64 J

(c) (i) pentagon K

(ii) square

K

segi empat sama

(iii) kite

K

lelayang

5. (a) Q1• 3•

12•

4• 2•

6•

(b) blue

red

white

yellow

•

• •

•

R

(c) P

RM

S•

• •

6. (a) L = {1, 4, 9, 16, 25, 36} n(L) = 6

(b) T = {I, E, A} n(T) = 3

(c) Y = {52, 61, 70} n(Y) = 3

7. (a) G = f

(b) H f

(c) I = f

8. (a) E = {3, 5}, F = {2, 3}, E F(b) G = {square, rectangle, rhombus, trapezium, …} G = {segi empat sama, segi empat tepat, rombus, trapezium, …} H = {square, rectangle, rhombus, trapezium, …} H = {segi empat sama, segi empat tepat, rombus, trapezium, …} G = H

9. (a) a = –3, b = –4 (b) a = 26, b = 39

10. (a) {288}

{multiples of 24} {288} {gandaan bagi 24}

(b) {trapezium}

{regular polygon} {trapezium} {poligon sekata}

(c) {factors of 8}

{factors of 16} {faktor bagi 8} {faktor bagi 16}

11. (a) A B False / Palsu

A = {3, 6, 9, …} and / dan B = {6, 12, 18, …}, so / jadi A B.

(b) D E True / Benar

D = {1, 2, 5, 10} and / dan E = {1, 2, 5, 7, 9, 10}, so / jadi D E.

(c) F H True / Benar

12. (a) JH • p

• r• q

• s

• t

(b) R

• 1

• 16

PQ

• 2 • 4

• 8

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R 3 SetsSet


Mathematics Form 4 Chapter 3 Sets

13. (a) S = {1, 2, 4, 8} Number of subsets / Bilangan subset = 24 = 16 Subsets of S / Subset bagi S: { }, {1}, {2}, {4}, {8}, {1, 2}, {1, 4}, {1, 8},

{2, 4}, {2, 8}, {4, 8}, {1, 2, 4}, {1, 2, 8}, {1, 4, 8}, {2, 4, 8}, {1, 2, 4, 8}

(b) T = {E, A} Number of subsets / Bilangan subset = 22 = 4 Subsets of T / Subset bagi T : { }, {E}, {A}, {E, A}

14. (a) • 15• 11

• 17

• 19

• 29

• 25

• 27

B

• 13

• 23

• 21

(b) E

F

15. (a) R = {17, 19, 23} Rʹ = {15, 16, 18, 20, 21, 22, 24, 25}

(b) S = {b, c, d, f, g, h, j} Sʹ = {a, e, i}

(c) T = {April, June, September, November} T = {April, Jun, September, November}

T ʹ = {January, February, March, May, July, August, October, December}

Tʹ = {Januari, Februari, Mac, Mei, Julai, Ogos, Oktober, Disember}

16. (a) (i) A = {12, 18, 24, 30} Aʹ = {10, 14, 16, 20, 22, 26, 28}(ii) n(Aʹ) = 7

(b) (i) Pʹ = {1, 2, 6, 7, 9, 10} Qʹ = {1, 2, 3, 7, 8, 9, 10}

(ii) n(Pʹ) = 6 n(Qʹ) = 7

17. (a) H = {1, 2, 4, 8} I = {2, 4, 6, 8, 10, 12} J = {3, 6, 9, 12}

(i) H I = {2, 4, 8}(ii) H J = { }(iii) I J = {6, 12}(iv) H I J = { }

(b) K = {13, 17, 19, 23, 29} L = {13, 23} M = {14, 15, 16, 17, 18, 19, 23, 24, 25, 26, 27, 28, 29}

(i) K L = {13, 23}(ii) L M = {23}

(iii) K M = {17, 19, 23, 29}(iv) K L M = {23}

18. (a) C D = {4, 7, 9}

C D

• 4• 8

• 11

• 12

• 13

• 1

• 3• 5

• 10

• 7• 9

(b) E = {1, 3, 5, 7, 9} F = {2, 3, 5, 7} G = {1, 2, 5, 10} E F G = {5}

E F

G

• 9

• 1

• 3• 7

• 5• 2

• 10

19. (a) P

Q

(b) A

B

C

(c) XY

Z

20. (a) A C = {a, c, h}

• b

BA

C

• e• i

• d• a

• h

• c

• f

• g

(b) A B C = {15}

A B

C

• 10

• 18

• 19

• 16

• 15• 20

• 11

• 17

• 13

• 14• 12



21. (a) (i) P Q P

(ii) P Q Q

(b) (i) P Q Q

(ii) P Q R

(c) (i) P R Q

(ii) Q R P

22. (a) A = {10, 12, 14, 16} B = {12, 15} A B = {12} (A B)ʹ = {10, 11, 13, 14, 15, 16}

(b) A = {a, n, g, l, e} B = {n, i, c, e} A B = {n, e} (A B)ʹ = {a, c, g, i, k, l}

23. (a) (A B)ʹ = {a, b, c, e, g, h}

• c

• e

• h

• d

• f• g

• b• a B

A

(b) (A B C)ʹ = {1, 3, 5, 9, 11, 13, 15, 17, 19}

•

•

• •

• 9

• 5

• 13•

311 7

• 1

15

17

• 19

A BC

24. (a) (i) X Y = {2, 5} (ii) n(Y Z) = 0 (iii) n(X Y)ʹ = 11(b) (i) 6 (ii) 4 (iii) 27(c) (i) Let the students who like both reading

and jogging = x Katakan murid yang suka kedua-dua membaca

dan berlari = x

The students who like reading only Murid yang suka membaca sahaja = 12 – x The students who like jogging only Murid yang suka berlari sahaja = 15 – x

12 – x + x + 15 – x + 3 = 24 30 – x = 24 x = 6

\ There are 6 students who like both reading and jogging.

Terdapat 6 murid yang suka membaca dan berlari.

(ii)

3

R

12 – x x 5 – x

J

25. (a) R S = {1, 2, 3, 4, 5, 6, 7, 10, 12, 16, 20, 28}

• 2

• 4

• 7• 12

• 5

13

6

2028

• •

• •

•

10

16

•

•

R S

(b) T = {1, 2, 3, 4, 6, 12} U = {2, 3, 5, 7, 11} V = {3, 6, 9, 12} T U V = {1, 2, 3, 4, 5, 6, 7, 9, 11, 12}

1•

4• 2•

5•

7•

11•

9•

3• 6• 12•

T U

V

26. (a) A B

A

B

C

(b) P Q R

PQ

R

27. (a) P Q = {a, b, d, e, j, c, f, g, h}

P

Q

i•

k•

I•

a• d•

b• e• j•

c•

g•

f•

h•

(b) X Y Z = {11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21,

22}



XY

Z11•

13•

18•

12•

16• 15•

14• 20•

21•

17•

22•

19•

28. (a) (i) P P Q

(ii) R P Q R

(b) (i) E F G

(ii) F E G

(iii) G E F G

29. (a) P = { j, a, d, e} Q = {c} P Q = {j, a, d, e, c} (P Q)ʹ = {b, f, g, h, i}

(b) P = {2, 4, 6, 8, 10, 12, 14} Q = {1, 3, 5, 15} P Q = {1, 2, 3, 4, 5, 6, 8, 10, 12, 14, 15} (P Q)ʹ = {7, 9, 11, 13}

30. (a) (A C)ʹ = {j, l, k, g, e, f }

b• i•

f•

e•

l• k•

a•

h•

j• g•

c•

d• A

BC

(b) (A B C)ʹ = {6, 12, 14, 20}

AB

C8•

22•

4•

2• 14•

20•

6•

16•

18• 12•

10•

31. (a) B

A

C

(b)

AB

C

32. (a) PQ

R

P R Q (b) P

Q

R

(P R) (P R)ʹ

33. (a) (i) 5 + 8 + 3 + 4 + x + x + 2x + 6 – x = 38 26 + 3x = 38 3x = 12 x = 4

(ii) n(E F)ʹ = 5 + 8 + 3 + 2(4) + 6 – 4 = 26

(iii) n[(E F) Gʹ] = 5 + 8 + 4 + x = 5 + 8 + 4 + 4 = 21

(b) (i) G H = {1, 5, 6, 8, 9}

(ii) G H I = {1, 4, 5, 6, 8, 9, 10, 13}

(iii) n(G H I)ʹ = 6(c) D = {s, t, a, r, e} E = {q, t} F = {o, u} (i) E F = {q, t, o, u} (ii) D E F = {s, t, a, r, e, q, o, u} (iii) n(D E F)ʹ = 2

(d) (i) A B

5

x24 – x 22 – x

(ii) 24 – x + x + 22 – x + 5 = 43 51 – x = 43 x = 8

(a) n(A B) = 24 – 8 + 8 + 22 – 8 = 38

(b) n(A B)ʹ = 24 – 8 + 22 – 8 + 5 = 35

(e) (i) Let x = number of librarians who wear spectacles

Katakan x = bilangan pustakawan yang memakai cermin mata

8 + 3 + 4 + x + 9 = 32 x = 32 – 24 x = 8



\ There are 8 librarians who wear spectacles.

Terdapat 8 pustakawan yang memakai cermin mata.

(ii) 8 + 3 + 8 + 9 = 28 students / murid (iii) n(P Q)ʹ = 9

SPM Practice 3

Paper 1

1. n(F G) = n(E) 1 + 4x + 18 + 2 + 12 = 5x + 18 4x + 33 = 5x + 18 33 – 18 = 5x – 4x x = 15

Answer/ Jawapan: B

2. Empty set = (X Y)ʹ Z ʹ Set kosong

Answer/ Jawapan: D

3. = {a, b, c, d, e, f, g, h} P = {c} Pʹ = {a, b, d, e, f, g, h}

Answer/ Jawapan: D 4. (P Q) R

Answer/ Jawapan: A

5. n(R) = 7 + x n(R Q)ʹ = 2 + 2x n(R) = n(R Q)ʹ 7 + x = 2 + 2x x = 5

n() = 2 + x + 7 + 5 + 2x + 2x = 14 + 5x = 14 + 5(5) = 39

Answer/ Jawapan: D

6.

ZX Y

Answer/ Jawapan: C

7. G

H

K

Shaded region: (G K) Hʹ Kawasan berlorek

Answer/ Jawapan: C

8. The number of female students who are not prefect is

Bilangan murid perempuan yang bukan pengawas ialah 18 = 3a a = 6

The number of students who are prefect is Bilangan murid yang menjadi pengawas ialah 5 + 2a = 5 + 2(6) = 17

Answer/ Jawapan: C

9. K = {1, 2, 3, 4, 5} L = {1, 2, 3} M = {5, 6, 7, 8} Mʹ = {1, 2, 3, 4}

Answer/ Jawapan: B

10.

7% 15%5%

3%

27%

2%3%

S

U

T

The percentage of U only Peratusan bagi U sahaja = 35 – 3 – 3 – 2 = 27%

The number of families that bought shampoo U only

Bilangan keluarga yang membeli syampu U sahaja = 27% × 9 000 = 2 430

Answer/ Jawapan: A

11. M = {2, 3, 5, 7, 11, 13} N = {1, 3, 9} Mʹ = {1, 4, 6, 8, 9, 10, 12} Nʹ = {2, 4, 5, 6, 7, 8, 10, 11, 12, 13} Mʹ Nʹ = (4, 6, 8, 10, 12}

Answer/ Jawapan: B

12.

A B CH

D

KL

Answer/ Jawapan: C



13.

xy

ComicKomik

FictionFiksyen

Non-fictionBukan-fiksyen

10

16

18

1125

Let the number of students who read all the three types of books = y

Katakan bilangan murid yang membaca ketiga-tiga jenis buku = y

The number of students who read comic is Bilangan murid yang membaca komik ialah

60 = 25 + 11 + 18 + y y = 6

Total number of students is Jumlah bilangan murid ialah

100 = 25 + 11 + 10 + 18 + 16 + 6 + x x = 100 – 86 = 14

Answer/ Jawapan: C

Paper 2

1. (a)

GE

F

(b)

GE

F

2. (a) A

CB

(b) A

CB

3. (a) P Q

(b) A

BC

4. (a) J

K

L

(b) J

K

L

5. (a)

PQ

R

(b)

PQ

R

6. (a)

Q

P

(b) Aʹ C

7. (a) P

33

7 612

8 8

7

15

Q R

(b) (i) n(Q) = 15 + 7 + 12 + 8 = 42 (ii) 8


CH

AP

TE

R 4 Mathematical ReasoningPenaakulan Matematik

1. (a) A statement Pernyataan(b) A statement Pernyataan(c) Not a statement; It is neither true nor false. Bukan pernyataan; Ia bukan benar atau palsu.(d) A statement Pernyataan(e) Not a statement; It is neither true nor false. Bukan pernyataan; Ia bukan benar atau palsu.

2. (a) True / Benar (b) False / Palsu

(c) False / Palsu(d) True / Benar(e) False / Palsu

3. True statementPernyataan benar

False statementPernyataan palsu

(a) 35

– 25

= 15

35

– 15

= 25

15

– 35

= 25

15

– 25

= 35

25

– 15

= 35

25

– 35

= 15

(b) {7, 8} {8, 10} = {8} {7, 8} {8} = {8, 10} {8, 10} {8} = {7, 8}

(c) 6 × 5 = 30 6 × 30 = 530 × 5 = 6

4. (a) Some prime numbers are odd numbers. Sebilangan nombor perdana ialah nombor ganjil.

(b) All trapeziums have a pair of parallel sides. Semua trapezium mempunyai sepasang sisi yang selari.

(c) Some positive numbers are integers. Sebilangan nombor positif ialah integer.

5. TrueBenar

FalsePalsu

(a) ✓

(b) ✓

(c) ✓

6. CanBoleh

CannotTidak boleh

ConclusionKesimpulan

(a)

✓

All positive numbers are greater than zero is true.Semua nombor positif adalah lebih besar daripada sifar adalah benar.

(b)

✓

All even numbers are divisible by 3 is false.Semua nombor genap boleh dibahagi tepat dengan 3 adalah palsu.

(c)

✓

All obtuse angles are greater than 90° is true.Semua sudut cakah adalah lebih besar daripada 90° adalah benar.

7. (a) Some quadrilaterals have two parallel lines. Sebilangan sisi empat mempunyai dua garis selari.

(b) All factors of 8 are factors of 16. Semua faktor bagi 8 ialah faktor bagi 16.

(c) Some perfect squares are odd numbers. Sebilangan kuasa dua sempurna ialah nombor ganjil.

8. (a) NegationPenafian

True / FalseBenar / Palsu

Multiples of 12 are not multiples of 6.Gandaan bagi 12 bukan gandaan bagi 6.

False / Palsu

(b) 34 43 True / Benar

(c) m { a, i, m} False / Palsu

9. Statement 1Pernyataan 1

Statement 2Pernyataan 2

(a) 100 = 10 102 = 100

(b) 100 is a perfect square.100 ialah kuasa dua sempurna.

100 has one significant figure.100 mempunyai satu angka bererti.

(c) 25 is greater than 5.25 adalah lebih besar daripada 5.

25 is greater than 15.25 adalah lebih besar daripada 15.

(d) 1 is a factor of 9.1 ialah faktor bagi 9.

3 is a factor of 9.3 ialah faktor bagi 9.


Mathematics Form 4 Chapter 4 Mathematical Reasoning

10. (a) 12

+ 13

= 56

and / dan 12

× 13

= 16

(b) 7 is a factor of 14 and 21. 7 ialah faktor bagi 14 dan 21.

(c) Set {x, y, z} has 23 subsets and 3 elements. Set {x, y, z} mempunyai 23 subset dan 3 unsur.

11. Statement 1Pernyataan 1

Statement 2Pernyataan 2

(a) There are 28 days in February.Terdapat 28 hari dalam bulan Februari.

There are 29 days in February.Terdapat 29 hari dalam bulan Februari.

(b) (am)n = amn am + n = am × an

(c) Rhombus is a polygon.Rombus ialah sebuah poligon.

Quadrilateral is a polygon.Sisi empat ialah sebuah poligon.

12. (a) 25 = 5 or/atau 52 = 25.

(b) 9 or 27 is a multiple of 3. 9 atau 27 ialah gandaan bagi 3.

(c) Rectangle or kite is a quadrilateral. Segi empat tepat atau lelayang ialah sebuah sisi empat.

13. (a) True / Benar (b) False / Palsu

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

14. (a) True / Benar (b) True / Benar

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

15. AntecedentAntejadian

ConsequentAkibat

(a) θ is an obtuse angle.θ ialah sudut cakah.

90° θ 180°

(b) x + y > 0 y –x

16. Implication 1Implikasi 1

Implication 2Implikasi 2

(a) If y = 9, then y = 3.Jika y = 9, maka y = 3.

If y = 3, then y = 9.Jika y = 3, maka y = 9.

(b) If E F = E, then F E.Jika E F = E, maka F E.

If F E, then E F = E.Jika F E, maka E F = E.

17. (a) (i) If x is an even number, then x is divisible by 2.

Jika x ialah nombor genap, maka x boleh dibahagi tepat dengan 2.

(ii) x is an even number if and only if x is divisible by 2.

x ialah nombor genap jika dan hanya jika x boleh dibahagi tepat dengan 2.

(b) (i) If A = f, then n(A) = 0. Jika A = f, maka n(A) = 0.

(ii) A = f if and only if n(A) = 0. A = f jika dan hanya jika n(A) = 0.

(c) (i) If m2 = 16, then m = ±4. Jika m2 = 16, maka m = ±4.

(ii) m2 = 16 if and only if m = ±4. m2 = 16 jika dan hanya jika m = ±4.

18. ConverseAkas

True / FalseBenar / Palsu

(a) If P Q = P, then P Q.Jika P Q = P, maka P Q.

True / Benar

(b) If x 5, then x 8.Jika x 5, maka x 8.

False / Palsu

(c) If q is a multiple of 5, then q is a multiple of 10.Jika q ialah gandaan bagi 5, maka q ialah gandaan bagi 10.

False / Palsu

19. Premise 1Premis 1

Premise 2Premis 2

ConclusionKesimpulan

(a) All reptiles lay eggs.Semua reptilia bertelur.

Snake is a reptile.Ular ialah reptilia.

Snake lays eggs.Ular bertelur.

(b) If R S, then R S = R.Jika R S, maka R S = R.

R S R R S

(c) If n is an even number, then n + 1 is an odd number.Jika n ialah nombor genap, maka n + 1 ialah nombor ganjil.

6 is an even number.6 ialah nombor genap.

6 + 1 is an odd number.6 + 1 ialah nombor ganjil.

20. (a) Haziq wears uniform to school.

Haziq memakai pakaian seragam ke sekolah.

(b) 8 is divisible by 2. 8 boleh dibahagi tepat dengan 2.

(c) m is an integer. m ialah integer.



21. (a) n 0

(b) All triangles have three sides. Semua segi tiga mempunyai tiga sisi.

(c) h is a negative number. h ialah nombor negatif.

(d) If PQR is a right-angled triangle, then ∠Q = 90°.

Jika PQR ialah segi tiga bersudut tegak, maka ∠Q = ∠90°.

22. Reasoning by deduction

Penaakulan secara deduksi

Reasoning by induction

Penaakulan secara aruhan

(a) ✓

(b) ✓

(c) ✓

23. (a) 4 is a factor of 24. 4 ialah faktor bagi 24.

(b) The sum of the interior angles of polygon x is (8 – 2) × 180 = 1080°.

Hasil tambah sudut dalaman bagi poligon x ialah (8 – 2) × 180 = 1080°.

(c) 25 = 32 Set P has 32 subsets. Set P mempunyai 32 subset.

24. (a) 4n + 3, n = 1, 2, 3, …

(b) 2n – n, n = 1, 2, 3, …

(c) 7(10 – n)2, n = 1, 2, 3, …

(d) n2

+ n, n = 1, 2, 3, …

(e) n–n, n = 1, 2, 3, …

25. (a) nth term / sebutan ke-n = 8 – 3(n – 1)2, n = 1, 2, 3, …

8 – 3(n – 1)2 = –139 3(n – 1)2 = 147 (n – 1)2 = 49 n – 1 = ±7 n 0, n = 7 + 1 = 8

The 8th term in the sequence is –139. Sebatan ke-8 dalam urutan itu ialah –139.

(b) Volume / Isi padu

= 13π(3)2(4)

= 12π cm3

SPM Practice 4

Paper 2

1. (a) (i) False / Palsu

(ii) If x = 4, then x = 2.

Jika x = 4, maka x = 2.

(b) or / atau

(c) 5 is a positive number. / 5 ialah nombor positif.

(d) 5 + 3n2, n = 1, 2, 3, 4, …

2. (a) False / Palsu

(b) x3

+ y4

= 1

(c) Volume / Isi padu = 13π(7)2(14)

= 6863

× 227

= 2 1563

= 718 23

cm3

Conclusion : The volume of the cone is

71823

cm3.

Kesimpulan: Isi padu kon ialah 718 23

cm3.

3. (a) True / Benar

(b) All / Semua

(c) If / Jika 2x + 3 5, then / maka x 1. If / Jika x 1, then / maka 2x + 3 5.

4. (a) Statement / Pernyataan

(b) EFGHK is a pentagon. EFGHK ialah pentagon.

(c) If the area of square KLMN is 25 cm2, then the side of square KLMN is 5 cm.

Jika luas segi empat sama KLMN ialah 25 cm2, maka sisi segi empat sama KLMN ialah 5 cm.

If the side of square KLMN is 5 cm, then the area of square KLMN is 25 cm2.

Jika sisi segi empat sama KLMN ialah 5 cm, maka luas segi empat sama KLMN ialah 25 cm2.

(d) (6 – 2) × 180° = 720° Sum of the interior angles of a regular

hexagon is 720°. Hasil tambah sudut pedalaman bagi suatu heksagon

sekala ialah 720°.

5. (a) (i) 8 is a factor of 16 or an even number. 8 ialah faktor bagi 16 atau suatu nombor genap.

(ii) True / Benar

(b) n 3



(c) If a number is an even number, then it is divisible by 2.

Jika suatu nombor ialah nombor genap, maka nombor itu boleh dibahagi tepat dengan 2.

If a number is divisible by 2, then it is an even number.

Jika suatu nombor boleh dibahagi tepat dengan 2, maka nombor itu ialah nombor genap.

6. (a) Some / Sebilangan

(b) PQR has two sides with the same length. PQR mempunyai dua sisi yang sama panjang.

(c) A set with n elements has 2n subsets where n = 0, 1, 2, 3, …

Satu set dengan n unsur mempunyai 2n subset dengan keadaan n = 0, 1, 2, 3, …

7. (a) (i) How to solve this problem? Bagaimanakah menyelesaikan masalah ini?

(ii) All multiples of 6 are multiples of 3. Semua gandaan bagi 6 adalah gandaan bagi 3.

(b) If 6 is a multiple of 3, then all multiples of 6 are multiples of 3.

Jika 6 ialah gandaan bagi 3, maka semua gandaan bagi 6 ialah gandaan bagi 3.

If all multiples of 6 are multiples of 3, then 6 is a multiple of 3.

Jika semua gandaan bagi 6, ialah gandaan bagi 3, maka 6 ialah gandaan bagi 3.

(c) 12 is a multiple of 3. 12 ialah gandaan bagi 3.


CH

AP

TE

R 5 The Straight LineGaris Lurus

1. (a) (b) (c)

Vertical distanceJarak mencancang

3 units / unit

3 units / unit

0 unit / unit

Horizontal distanceJarak mengufuk

2 units / unit

0 unit / unit

4 units / unit

(a) (b)

2. Ratio between A and BNisbah di antara A dan B

4— = 2 2

1—3

Ratio between B and CNisbah di antara B dan C

2— = 2 1

2 1— = — 6 3

Ratio between A and CNisbah di antara A dan C

6— = 2 3

3 1— = — 9 3

GradientKecerunan

21—3

3. (a) 24

= 12

(b) 32

(c) 05

= 0

4. (a) G(–2, 0) H(4, 3) 3 – 0 Gradient / Kecerunan = ———— 4 – (–2) 1 = –– 2

(b) E(–5, 2) F(3, –1) –1 – 2 Gradient / Kecerunan = ———— 3 – (–5) 3 = – –– 8

3 – 0 3 5. (a) ———— = –– 3 – (–1) 4

2 – 8 –6(b) ——–– = —– = 3 5 – 7 –2

–12 – (–5) –7(c) ————— = —– = 1 –8 – (–1) –7

3 – (–4) 7(d) ———–– = – — 2 – 6 4

6. (a) GradientKeceruman

Steepness (Steepest = 1)Kecuraman (Paling curam = 1)

(i) 96

= 32

2

(ii) 68

= 34

3

(iii) 39

= 13

4

(b) The greater the value of the gradient,

the steeper the slope of the line.

Semakin besar nilai kecerunan, semakin

curam garis lurus itu.

7.

(a) (b) (c)

GradientKecerunan

1 – (–1)———— –3 – 1 1= – — 2

3 – 2——–3 – 0 1= — 3

–1 – (–3)————— –2 – 3 2= – — 5

Sign of gradientTanda kecerunan

NegativeNegatif

PositivePositif

NegativeNegatif

Direction of inclinationArah kecondongan

Inclined downwards to the rightCondong ke kiri

Inclined upwards to the rightCondong ke kanan

Inclined downwards to the rightCondong ke kiri

8. x-interceptPintasan-x

y-interceptPintasan-y

(a) –2 2

(b) 2 4



GradientKecerunan

(a)4 2

2 1– — = – — 4 2

(b)–3 4

4 4– —– = — –3 3

(c)1 –3

–3–—– = 3 1


Mathematics Form 4 Chapter 5 The Straight Line

10. y-interceptPintasan-y

x-interceptPintasan-x

GradientKecerunan

(a)2 7 – 2

7

(b)–6 – –6

2 = 3 2

(c) –(8 × 3) = –24 8 3

(d)–12 6 –( –12

6) = 2



GradientKecerunan

(a) 5 –88—5

(b) 2 3 3– — 2

(c) 3– — 2

–4

3–3–4 ÷ – —4 2 8= – — 3

12. (a) (i) Gradient / Kecerunan –1 = –––– –3

= – 1—3

(ii) Coordinates of M / Koordinat M

= (–3, 0)

(b) (i) y-intercept / Pintasan-y = 3

(ii) Gradient / Kecerunan 3 = – —– –6 1 = — 2

(c) (i) x-intercept / Pintasan-x 4 = –34 ÷ – —4 3 = 3

(ii) m = 3 n = 0

13. (a) x 0 2 4

y –2 0 2

y

0x

–2 2

2

–2

–4

4 6

y = x – 2

(b) x 0 1 2

y 0 4 8

y

0x

–1 1 2 3

y = 4x

4

8

12

14.LHS

Sebelah kiri

RHSSebelah kanan

The point lie on the line

Titik itu terletak pada garis lurus tersebut

(a) 0 1—(–2) + 1 = 0 2

LHS = RHS Yes / Ya

(b) 6 1—(8) + 2 = 4 4

LHS ≠ RHS No / Tidak

15. (a) y = 3x – 2

(b) y = – 3—4

x + 7

16. Equation in the form y = mx + c

Persamaan dalam bentuk y = mx + c

GradientKecerunan


(a) y = 2x + 4 2 4

(b) 2y = x – 10

y = 1—2

x – 5

1—2

–5

17. (a) The equation is y = 1. Persamaan ialah y = 1.

(b) The equation is y = –3. Persamaan ialah y = –3.

(c) The equation is y = –1.5. Persamaan ialah y = –1.5.

18. (a) The equation is x = 2. Persamaan ialah x = 2.

(b) The equation is x = –3. Persamaan ialah x = –3.

(c) The equation is x = 1. Persamaan ialah x = 1.



19. (a) y = 12

x + c

At / Di A(0, –1),

–1 = 12

(0) + c

c = –1

\ The equation is y = 12

x – 1.

Persamaan ialah y = 12

x – 1.

(b) A12

, 5, m = –4

y = –4x + c

At / Di A12

, 5, 5 = –41

2 + c

c = 7

\The equation is y = –4x + 7. Persamaan ialah y = –4x + 7.

5 – 3 20. (a) Gradient, m = ——— Kecerunan, m 2 – 1

= 2

\ y = 2x + c

At / Di P(2, 5), 5 = 2(2) + c c = 1

\The equation is y = 2x + 1. Persamaan ialah y = 2x + 1.

–3 – ( –4)(b) Gradient, m = ————– Kecerunan, m 2 – 6 1 = – — 4

\ y = – 14

x + c

At / Di P(6, –4), –4 = – 14

(6) + c

c = – 52

\ The equation is y = – 14

x – 52

.

Persamaan ialah y = – 14

x – 52

.

–7 – 8(c) Gradient, m = ———— Kecerunan, m 3 – (–3)

= – 5—2

\ y = – 5—2

x + c

At / Di P(–3, 8), 8 = – 5—2

(–3) + c

c = 1—2

\ The equation is y = – 5—2

x + 1—2

.

Persamaan ialah y = – 5—2

x + 1—2

.

0 – 4 21. (a) Gradient of AB = ———— Kecerunan AB 2 – (–8) 2 = – — 5

2 \ y = – ––x + c 5 2 At / Di B(2, 0), 0 = – —(2) + c 5 4 c = — 5 2 4 \ The equation is y = – —x + —. 5 5 2 4 Persamaan ialah y = – —x + —. 5 5

(b) \ The equation is x = 3. Persamaan ialah x = 3.

6 – (–1) (c) Gradient of AB = ————– Kecerunan AB 4 – (–10) 1 = — 2 1 \ y = —x + c 2 1 At / Di B(4, 6), 6 = —(4) + c 2 c = 4 1 \ The equation is y = —x + 4. 2 1 Persamaan ialah y = —x + 4. 2

22. (a) x 0 1

y = 3x 0 3

x 1 2

y = –2x + 10 8 6

0 2

2

4

6

8

4 6 8

y = 3x

y = –2x + 10

P(2, 6)

x

y

(b) x 0 1

y = x + 3 3 4

x 0 3

2y = – —x – 2 3 –2 –4

0 2

2

4

4–2–4

–2

–4

y = x + 3

P(–3, 0)

2y = – –x – 2 3

x

y



(c) x 0 2

1y = —x + 1 2 1 2

x 0 –2

1y = – —x – 3 2 –3 –2

y

x0

2

2

4

–2

–2–4–6

–4

P(–4, –1)

1 y = ––x + 12

1y = – ––x – 32

23. (a) y + 2x = 3 ……..… 1 y – 5x = –4 ……… 2

1 – 2, 7x = 7 x = 1

Substitute x = 1 into 1. Gantikan x = 1 ke dalam a.

y + 2(1) = 3 y = 1

\ The point of intersection is (1, 1). Titik persilangan ialah (1, 1).

(b) 3y + x = 7 …… 1 2y – x = 3 …… 2

1 + 2, 5y = 10 y = 2

Substitute y = 2 into 1. Gantikan y = 2 ke dalam a.

3(2) + x = 7 x = 1


(c) 4y + 5x = 22 × 3, 12y + 15x = 66 …….. 1

5y + 3x = 21 × 5, 25y + 15x = 105 …… 2

2 – 1, 13y = 39 y = 3

Substitute y = 3 into 1. Gantikan y = 3 ke dalam a.

12(3) + 15x = 66 15x = 30 x = 2


24. (a) (i) Gradient of PQ Kecerunan PQ 4 = – –– 2 = –2

(ii) Gradient of RS Kecerunan RS 2 = – — 1 = –2

(b) (i) Gradient of PQ Kecerunan PQ 3 – 0 = ————– –1 – (–2) = 3

(ii) Gradient of RS Kecerunan RS 6 – (–3) = ———— 4 – 1 = 3

25. (a) 2y – 4x = 6 4y = 7 + 8x 7 y = 2x + 3 y = 2x + –– 4 Gradient, m

1 = 2 Gradient, m

2 = 2

Kecerunan, Kecerunan,

Since m1 = m

2 = 2, the lines are parallel.

Oleh sebab m1 = m

2 = 2, kedua-dua garis adalah selari.

(b) 5x = 2y + 8 3y = 4x – 1

5 4 1 y = —x – 4 y = —x – — 2 3 3 5 4 Gradient, m

1 = — Gradient, m

2 = —

Kecerunan, 2 Kecerunan, 3

Since m1 ≠ m

2, the lines are not parallel.

Oleh sebab m1 ≠ m

2, kedua-dua garis adalah tidak selari.

(c) 3 – 2x + y = 0 3y = 6x – 8 8 y = 2x – 3 y = 2x – — 3

Gradient, m1 = 2 Gradient, m

2 = 2

Kecerunan, Kecerunan,

\ Since m1 = m

2 = 2, the lines are parallel.

Oleh sebab m1 = m

2 = 2, kedua-dua garis adalah

selari.



26. (a) 4y – 8x = 10 5 y = 2x + –– 2

\ Gradient of LP = 2

Kecerunan bagi Lp

\ y = 2x + c

At / Di P(2, 5), 5 = 2(2) + c c = 1

\ The equation of Lp is y = 2x + 1.

Persamaan bagi Lp ialah y = 2x + 1.

(b) 2 – 3x + y = 0 y = 3x – 2

\ Gradient of LP = 3

Kecerunan bagi Lp

\ y = 3x + c

1 5 5 1 At / Di P—, —, –– = 3— + c 3 2 2 3 3 c = — 2 3 \ The equation of L

p is y = 3x + —.

2 3 Persamaan bagi L

p ialah y = 3x + —.

2

(c) 2x + 2y = 0 y = –x

\ Gradient of LP = –1

Kecerunan bagi Lp

\ y = –x + c

At / Di P(–2, –3), –3 = –(–2) + c c = –5

\ The equation of LP is y = –x – 5.

Persamaan bagi Lp ialah y = –x – 5.

8 – (–1) 27. (a) (i) Gradient of PQ = ———— Kecerunan PQ 4 – (–2) 3 = — 2

Let coordinates of T be (0, y). Katakan koordinat T ialah (0, y).

6 – y 3 Then / Maka ——– = — 6 – 0 2 6 – y = 9 y = –3

\ The coordinates of T are (0, –3). Koordinat bagi T ialah (0, –3).

6 – (–1) (ii) Gradient of PU = ———— Kecerunan PU 6 – (–2) 7 = — 8 7 \ y = —x + c 8 7 At / Di P(–2, –1), –1 = —(–2) + c 8 3 c = –– 4 7 3 \ The equation of PU is y = —x + —. 8 4

7 3 Persamaan bagi PU ialah y = —x + —. 8 4

(b) (i) 3x – y = 4 …….... 1 x – 2y = –7 × 3, 3x – 6y = –21 …… 2

1 – 2, 5y = 25 y = 5

Substitute y = 5 into 1. Gantikan y = 5 ke dalam a. 3x – 5 = 4 x = 3

\ The point of intersection is P(3, 5). Titik persilangan ialah P(3, 5). 1 (ii) Gradient of AP = — Kecerunan AP 2

1 \ y = —x + c 2 c = –4

\ The equation of the line passing through point B and parallel to AP

1 is y = —x – 4. 2 Persamaan bagi garis yang melalui titik B dan 1 selari dengan AP ialah y = —x – 4. 2

(c) (i) y = 3 – 2x

When / Apabila y = 5, 5 = 3 – 2x x = –1

\ The coordinates of A are (–1, 5). Koordinat bagi A ialah (–1, 5).

(ii) y = 3 – 2x

Let x-intercept = a. Katakan pintasan-x = a.

At / Di (a, 0), 0 = 3 – 2a 3 a = — 2 3 \ The equation of line PQ is x = —. 2 3 Persamaan garis PQ ialah x = —. 2



(iii) 3y – 2x = 6

y = 2—3

x + 2

2 Gradient of line passing through A = —. 3 2 Kecerunan garis yang melalui A = — 3 2 \ y = —x + c 3 2 At / Di A(–1, 5), 5 = —(–1) + c 3 17 c = —– 3

\ The equation of the line passing through point A and parallel to the

line 3y – 2x = 6 is y = 2—3

x + 17—–3

.

Persamaan bagi garis yang melalui titik A dan selari dengan garis

3y – 2x = 6 ialah y = 2—3

x + 17—–3

.

SPM Practice 5

Paper 1

1. 4x – 3y = 9 3y = 4x – 9

y = 4—3

x – 3

Gradient/ Kecerunan = 4—3

y-intercept / pintasan-y = –3

Answer/ Jawapan: B

2. Midpoint of XY, P / Titik tengah xY, P

= 6 + 2——–2

, –3 + 1———–2

= (4, –1)

Gradient of PZ / Kecerunan PZ

= 3 – (–1)————––2 – 4

= – 2—3

Answer/ Jawapan: D

3. When/ Apabila y = 0,

4x – 3(0) = 24 x = 6

Answer/ Jawapan: A

4. y = mx + c, m = – 2—3

, (–3, 5)

5 = – 2—3

(–3) + c

c = 3

y-intercept / pintasan-y = 3

Answer/ Jawapan: D

5. At / Di (0, 6), 6 = –2(0) + c c = 6 y = –2x + 6

When / Apabila y = 0, 0 = –2x + 6 2x = 6 x = 3

The point of intersection Titik persilangan = (3, 0)

Answer/ Jawapan: B

6. Gradient / Kecerunan

= 5 – (–1)2 – (–4)

= 6—6

= 1

Answer/ Jawapan: B

7. OS——OT

= 5—2

–10——OT

= 5—2

OT = – 4

The coordinates of T / Koordinat T = (0, –4)

Gradient of ST Kecerunan ST

–4 – 0

——–——0 – (–10)

= – 2—5

5y = – kx – 20

y = – k—5

x – 4

By comparing, Bandingkan,

– k—5

= – 2—5

k = 2

Answer/ Jawapan: C

8. 2y = mx + 12

y = m—2

x + 6

y-intercept/ Pintasan-y = 6 OL = 6 units / unit

The coordinates of K is (–8, 0). Koordinat K ialah (–8, 0).



Gradient of KL is Kecerunan bagi KL ialah

m—2

= 6 – 0

——–——0 – (–8)

m = 6—8

(2)

= 3—2

The value of m/ Nilai m = 3—2

Answer/ Jawapan: D

9. OT = 12 units / unit ST = 13 units / unit

OS = (ST)2 – (OT)2 = 132 – 122

= 5 units/ unit

The coodinates of S/ Koordinat S = (0, –5)

Gradient of ST/ Kecerunan ST

= –5 – 0

——–—0 – 12

= 5

—–12

Answer/ Jawapan: B

10. Gradient / Kecerunan = 2 – (–1)

——–——5 – 3

= 3—2

The equation of the straight line is Persamaan garis lurus ialah

y – (–1) = 3—2

(x – 3)

y + 1 = 3—2

x – 9—2

y = 3—2

x – 11—–2

y-intercept/ Pintasan-y = – 11—–2

Answer/ Jawapan: A

11. When/ Apabila y = 0,

12 – 3(0) + 4x = 0 4x = –12 x = –3

Answer/ Jawapan: C 12. 6 – 2y = 3x 2y = –3x + 6

y = – 3—2

x + 3

y-intercept / Pintasan-y = 3 When / Apabila y = 0,

0 = – 3—2

x + 3

3—2

x = 3

x = 2

x-intercept / Pintasan-x = 2

Answer/ Jawapan: B

Paper 2

1. (a) 3y = x − 9

y = x—3

– 3, m = 1—3

The equation of the straight line KL is Persamaan garis lurus KL ialah

y − 2 = 1—3

[x − (−3)]

= 1—3

(x + 3)

= 1—3

x + 1 + 2

y = 1—3

x + 3

(b) y = 1—3

x + 3

When / Apabila y = 0, 0 = 1—3

x + 3

−3 = 1—3

x

x = −9

Therefore, the x-intercept of straight line KL is −9.

Maka, pintasan-x bagi garis lurus KL ialah –9.

2. (a) x = −2(b) m

PQ= m

RS

mPQ

= 4 – 0

——––0 – 8

= – 1—2

mRS

= – 1—2

The equation of the straight line RS is Persamaan bagi garis lurus RS ialah

y − (−2) = – 1—2

[x − (−2)]

y + 2 = – 1—2

(x +2)

y = – 1—2

x – 1 – 2

y = – 1—2

x − 3

(c) y = – 1—2

x − 3

When / Apabila y = 0, 0 = – 1—2

x − 3

3 = – 1—2

x

x = −6



Therefore, the x-intercept of straight line RS is −6.

Maka, pintasan-x bagi garis lurus RS ialah –6.

3. (a) x = 3

(b) y = 1—3

x + 7

m = 1—3

y = mx + c

–4 = 1—3

(3) + c

c = –4 – 1 = –5

y = 1—3

x – 5

(c) When / Apabila y = 0,

1—3

x – 5 = 0

1—3

x = 5

x = 15


4. (a) y = – 1—2

x + 10

m = – 1—2

2y = kx + 9

y = k—2

x + 9—2

k—2

= – 1—2

k = –1

(b) 2y = kx + 9 2y = –x + 9 0 = –x + 9 x = 9


5. (a) mRS

= mPQ

= 1 – (–1)1 – (–3)

= 12

y = mx + c

3 = 12

(–2) + c

c = 4

y = 12

x + 4

(b) y = 0

0 = 12

x + 4

12

x = –4

x = –8 x-intercept / Pintasan-x = –8

6. (a) y = 2x + 12 y-intercept / Pintasan-y = 12

(b) y = 8

(c) m = 2 y = mx + c 8 = 2(5) + c c = 8 – 10 = –2 y = 2x – 2

7. (a) y = –3x – 2 k = –3(–2) – 2 = 4 y = 4

(b) 0 = –3x – 2 3x = –2

x = – 2—3

\ Q– 2—3

, 0

m = – 4——– 2—

3 = 6 y = 6x + 4

8. (a) 5 − 2y = − 2x − 2y = − 2x − 5

y = x + 5—2

\ m = 1

(b) 2y + 15 = 6x

When / Apabila y = 0, 15 = 6x

x = 5—2

\ x-intercept of straight line BC is = 5—2

.

Pintasan-x bagi garis lurus BC ialah 5—2

.

(c) 5 − 2y = −2x

y = x + 5—2

............a

2y + 15 = 6x

y = 3x − 15—–2

...........b

Substitute a into b :

Gantikan a ke dalam b:

x + 5—2

= 3x − 15—–2

2x + 5———2

= 6x + 15———–2

2x + 5 = 6x − 15 20 = 4x x = 5



When / Apabila x = 5, y = 5 + 5—2

= 7 1—2

\ B5, 7 1—2

9. (a) x + 2y − 2 = 0

When / Apabila y = 0, x − 2 = 0 x = 2

\ Coordinates of S is (2, 0). Koordinat bagi S ialah (2, 0).

(b) x + 2y − 2 = 0

y = − 1—2

x + 1

y-intercept of PS / Pintasan-y bagi PS = 1

Coordinates of P / Koordinat P = (0, 1)

Coordinate of Q / Koordinat Q = (2, 2).

mPQ

= 2 – 1———2 – 0

= 1—2

The equation of straight line QP is Persamaan garis lurus QP ialah

y − 1 = 1—2

(x − 0)

y = 1—2

x + 1

10. (a) mPQ

= mSR

= 2

The equation of straight line SR is Persamaan garis lurus SR ialah y − (−3) = 2(x − 6) y = 2x − 15

y-intercept / Pintasan-y = −15

(b) m = 2 – (–3)————1 – 6

= –1

The equation of straight line QR is Persamaan garis lurus QR ialah y − (−3) = −1(x − 6) y = −x + 3 (c) y = −x + 3

When/ Apabila y = 0, 0 = −x + 3 x = 3

Therefore, the x-intercept of straight line QR is 3.

Maka, pintasan-x bagi garis lurus QR ialah 3.


1. (a) Mass (kg)Jisim (kg)

(b) Length (cm)Panjang (cm)

51 – 60 21 – 25

61 – 70 26 – 30

71 – 80 31 – 35

81 – 90 36 – 40

2. (a) (b) (c)

Lower limitHad bawah

101 44 2.5

Upper limitHad atas

115 48 2.9

Lower boundarySempadan bawah

100.5 43.5 2.45

Upper boundarySempadan atas

115.5 48.5 2.95

Size of class intervalSaiz selang kelas

15 5 0.5

3. (a) Highest value / Nilai tertinggi = 99 Lowest value / Nilai terendah= 52

Size of class interval / Saiz selang kelas 99 – 52 = ———–– 5 = 9.4 ≈ 10 Use 10

The class intervals are / Selang kelas ialah: 52 – 61, 62 – 71, 72 – 81, 82 – 91 and / dan

92 – 101.

(b) Highest value / Nilai tertinggi = 80 Lowest value / Nilai terendah = 12

Size of class interval / Saiz selang kelas 80 – 12 = ———–– 7 = 9.7 ≈ 10 Use 10

The class intervals are / Selang kelas ialah: 12 – 21, 22 – 31, 32 – 41, 42 – 51, 52 – 61,

62 – 71 and / dan 72 – 81.

4. (a) Size of class interval / Saiz selang kelas = 10

Number of class interval / Bilangan selang kelas 98 – 25 = ———— 10 ≈ 8

\ Suitable / Sesuai.

(b) Size of class interval / Saiz selang kelas = 20

Number of class interval / Bilangan selang kelas 98 – 25 = ———— 20 ≈ 4

\ Not suitable / Tidak sesuai.

5. (a) Class intervalSelang kelas

TallyGundalan

FrequencyKekerapan

0.5 – 1.0 |||| || 7

1.1 – 1.6 || 2

1.7 – 2.2 |||| 5

2.3 – 2.8 |||| || 7

2.9 – 3.4 |||| |||| 9

3.5 – 4.0 || 2

(b) Wages (RM)Gaji (RM)

TallyGundalan

FrequencyKekerapan

400 – 499 ||| 3

500 – 599 |||| 5

600 – 699 |||| 4

700 – 799 |||| 4

800 – 899 || 2

900 – 999 || 2

6. (a) Modal class / Kelas mod = 30 – 34

(b) Modal class / Kelas mod = 3.7 – 4.5


MidpointTitik tengah

0.5 – 1.0 0.75

1.1 – 1.6 1.35

1.7 – 2.2 1.95

2.3 – 2.8 2.55

(b) Class intervalSelang kelas


31 – 40 35.5

41 – 50 45.5

51 – 60 55.5

61 – 70 65.5

CH

AP

TE

R 6 Statistics IIIStatistik III


Mathematics Form 4 Chapter 6 Statistics III

8. (a) MidpointTitik tengah

FrequencyKekerapan

Midpoint × Frequency Titik tengah × Kekerapan

MeanMin 88.3= —— 21= 4.20

3.3 7 23.1

3.8 3 11.4

4.3 2 8.6

4.8 5 24.0

5.3 4 21.2

Total / Jumlah 21 88.3

(b) MidpointTitik tengah

FrequencyKekerapan

Midpoint × FrequencyTitik tengah × Kekerapan

MeanMin 405= —— 30= 13.5

2 4 8

7 7 49

12 3 36

17 9 153

22 6 132

27 1 27

Total / Jumlah 30 405

9. (a) Mean / Min = 394——24

= 16.42

(b) (i) MidpointTitik tengah

FrequencyKekerapan


MeanMin 392= —— 24= 16.33

3 1 3

8 5 40

13 6 78

18 6 108

23 2 46

28 3 84

33 1 33


(ii) MidpointTitik tengah

FrequencyKekerapan


MeanMin 382= —— 24= 15.92

5.5 6 33

15.5 12 186

25.5 5 127.5

35.5 1 35.5


(c) The smaller the size of class interval, the more accurate the value of the mean to the actual mean. Semakin kecil saiz selang kelas, semakin tepat nilai min kepada min sebenar.



10. (a) Lower boundarySempadan bawah


1.95 2.45

2.45 2.95

2.95 3.45

3.45 3.95

3.95 4.45

01.95 2.45 2.95 3.45 3.95 4.45

2

FrequencyKekerapan

Height (m)Tinggi (m)

4

6

8

10

12

11. (a) 6 + 9 + 7 + 12 + 6 + 10 = 50 students / murid

(b) 25 – 29 minutes / minit

(c) 25 + 29————

2 = 27 minutes / minit

(d) 6 + 9 + 7———–—

50 × 100% = 44%

12. (a)

09.5 19.5 29.5 39.5 49.5 59.5

10

20

30

40

Frequency / Kekerapan

Length / Panjang (cm)

50




25.5

35.5

45.5

55.5

65.5

75.5

2

4

6

8

10

12

015.5 25.5 35.5 45.5

Temperature (°C)Suhu (°C)

55.5 65.5 75.5 85.5

FrequencyKekerapan

14. (a) 3.10 kg

(b) 2.93 – 2.97 kg and / dan 3.03 – 3.07 kg

(c) 8 + 14 + 8 + 20 + 10 = 60 watermelons / buah tembikai

(d) Mean / Min 2.95 × 8 + 3.00 × 14 + 3.05 × 8 + 3.10 × 20 + 3.15 × 10 = ———————————————— 60 183.5 = ——— 60 = 3.058 kg

15. (a) Cumulative frequency

Kekerapan longgokan

0

10

18

27

39

44

46



16. (a) Cumulative frequency

Kekerapan longgokan

0

20

35

42

47

50

52


FrequencyKekerapan


Cumulative frequencyKekerapan longgokan

2 0 2.5 0

3 4 3.5 4

4 5 4.5 9

5 6 5.5 15

6 7 6.5 22

7 7 7.5 29

8 2 8.5 31

02.5 3.5 4.5 5.5

Mass (kg)Jisim (kg)

6.5 7.5 8.5

5

10

15

20

25

30




18. (a) Range / Julat = 7.2 – 1.2 = 6.0

(b) Range / Julat = 11.4 – 2.1 = 9.3

90 + 99 50 + 59 19. (a) Range / Julat = ———— – —––—— 2 2 = 94.5 – 54.5 = 40.0

20. (a) (i) Median / Median = RM10

(ii) First quartile / Kuartil pertama = RM6

(iii) Third quartile / Kuartil ketiga = RM16

(iv) Interquartile range / Julat antara kuartil = 16 – 6 = RM10

(b) (i) Median / Median = 3.6 seconds

(ii) First quartile / Kuartil pertama = 3.28 seconds

(iii) Third quartile / Kuartil ketiga = 4.0 seconds

(iv) Interquartile range / Julat antara kuartil = 4.0 – 3.28 = 0.72 second

21. (a) (i) 40 books / buku

(ii) 23 books / buku

40 – 13 (iii) ——— × 100% = 67.5% 40

(b) (i) 37.5 seconds / saat

(ii) 51.5 – 26.5 = 25 seconds / saat

(iii) 150 – 140150

× 100% = 6.67%

SPM Practice 6

Paper 1

1.Number of

childrenBilangan anak

Cumulative frequency

Kekerapan longgokan

FrequencyKekerapan

1 4 4

2 14 10

3 21 7

4 28 7

5 30 2

Mode/ Mod = 2

Answer/ Jawapan: D

2. Age (years)

Umur (tahun)

Frequency, f

Kekerapan, f

Midpoint, x

Titik tengah, x

fx

20 – 29 2 24.5 49

30 – 39 6 34.5 207

40 – 49 10 44.5 445

50 – 59 7 54.5 381.5

60 – 69 5 64.5 322.5

Σf = 30 Σfx = 1 405

Mean age/ Min umur = 1 405———30

= 46.83 years/ tahun

Answer/ Jawapan: C

3. Mode / Mod = 4

4 has the highest frequency. 4 mempunyai kekerapan yang tertinggi.

Hence / Maka, y = 10

Answer/ Jawapan: B

4. Mean age / Min umur

= 15(4) + 16(8) + 17(6) + 18(5) + 19(2)—————————————————–

25

= 16.72

Answer/ Jawapan: B

5. 2—3 × 15 = 10

Angle of sector representing Thursday Sudut bagi sektor yang mewakili hari Khamis

= 10

———————————35 + 15 + 10 + 10 + 20

× 360°

= 40°

Answer/ Jawapan: B

6. Midpoint of the modal class = 649.5 Titik tengah bagi kelas mod

Size of the class interval = 100 Saiz selang kelas

Modal class = 600 – 699 Kelas mod

Answer/ Jawapan: B

7. Mean / Min

= 449.5(5) + 549.5(8) + 649.5(9) + 749.5(8)———————————————————–

30 = 616.17

Answer/ Jawapan: B



8. Mean / Min

= 2.45(7) + 3.45(9) + 4.45(3) + 5.45(6)—————————————————

25 = 3.77

Answer/ Jawapan: C

9. Modal class / Kelas mod = 25 – 28

Midpoint of the modal class Titik tengah bagi kelas mod

= 25 + 28————

2 = 26.5

Answer/ Jawapan: A

10. Mean / Min

= 14.5(8) + 18.5(9) + 22.5(7) + 26.5(15) + 30.5(11)—————————————————————–

8 + 9 + 7 + 15 + 11 = 23.46

Answer/ Jawapan: C

11. MarksMarkah

0 1 2 3 4 5

FrequencyKekerapan

4 6 1 x 2 3 Σf = 16 + x

Median = Value of 11th Nilai ke-11

(16 + x) + 1———–——

2 = 11

17 + x = 22 x = 5

Mean/ Min

= (1)(6) + (2)(1) + (3)(5) + (4)(2) + (5)(3)——————————————————–

16 + 5

= 46—–21

= 2.19

Answer/ Jawapan: D

12. 4, 4, 4, m, 8, 9

Median = 5

4 + m—–—–

2 = 5

m = 6

Another 2 data are added in, Dua lagi data ditambahkan 4, 4, 4, 6, 8, 9, 7, 8

Mean/ Min = 4 + 4 + 4 + 6 + 8 + 9 + 7 + 8

——————————————8

= 50—–8

= 6.25

Answer/ Jawapan: B

13. Mode/ Mod = 2

Frequency/ Kekerapan = 10 students/ murid

Therefore, the highest value of x is 9. Jadi, nilai maksimum x ialah 9.

Answer/ Jawapan: A

14. Range/ Julat = 71 + 80————

2 –

21 + 30————

2 = 75.5 – 25.5 = 50

Answer/ Jawapan: C

Paper 2

1. (a) DonationDerma

FrequencyKekerapan


0 – 4 0 2

5 – 9 5 7

10 – 14 6 12

15 – 19 9 17

20 – 24 5 22

25 – 29 3 27

30 – 34 2 32

35 – 39 0 37

(b) Mean / Min

=

0(2) + 5(7) + 6(12) + 9(17) + 5(22) + 3(27) + 2(32) + 0(37)

————————————————–0 + 5+ 6+ 9+ 5+ 3+ 2+ 0

= 515——30

= 17.17

(c)

0

1

2 7 12 17 22 27 32 37

2

3

4

5

6

7

8

9


Donation (RM) / Derma (RM)

(d) Number of people who donated more than RM20

Bilangan orang yang menderma lebih daripada RM20 = 5 + 3 + 2 = 10 person / orang



2. (a) FrequencyKekerapan


2 53

5 58

8 63

3 68

3 73

2 78

1 83

(b) Modal class / Kelas mod = 61 – 65

(c) Mean / Min

=

53(2) + 58(5) + 63(8) + 68(3) + 73(3)+ 78(2) + 83(1)

——————————————————–24

= 65.08

(d)

050.5 55.5 60.5 65.5 70.5 75.5 80.5 85.5

1

2

3

4

5

6

7

8


Marks / Markah

(e) 3 + 2 + 1 = 6 students / murid

3. (a) (i)MidpointTitik tengah

Upper boundary

Sempadan atas


32 34.5 0

37 39.5 2

42 44.5 9

47 49.5 19

52 54.5 31

57 59.5 37

62 64.5 40

(ii) Modal class / Kelas mod = 50 − 54

(b) Mean / Min

=

0(32) + 2(37) + 7(42) + 10(47) + 12(52) + 6(57) + 3(62)

———————————————0 + 2 + 7 + 10 + 12 + 63

= 1 990———

40 = 49.75

(c)

5

34.5 39.5 44.5 49.5 54.5 59.5 64.5Mass (kg) / Jisim (kg)

10

15

20

25

30

35

40

Cumulative frequency / Kekerapan longgokan

0

(d) Number of students whose mass are more than 52 kg

Bilangan murid yang jisim lebih daripada 52 kg = 40 − 25

= 15 students / orang


FrequencyKekerapan


12 – 16 3 14

17 – 21 4 19

22 – 26 7 24

27 – 31 9 29

32 – 36 5 34

37 – 41 2 39

(b) Modal class / Kelas mod = 27 – 31

(c) Mean / Min

=

3(14) + 4(19) + 7(24) + 9(29) + 5(34) + 2(39)

———————————————30

= 26.5 calculators / kalkulator

(d)

0

2

4

6

8

10

11.5

16.5

21.5

26.5

Number of calculators / Bilangan kalkulator


31.5

36.5

41.5

(e) 5 + 2 = 7 days / hari




Upper boundary

Sempadan atas


Kekerapan longgokan

17 19.5 0

22 24.5 8

27 29.5 22

32 34.5 39

37 39.5 61

42 44.5 74

47 49.5 80

(b) Mean score / Min skor

=

22(8) + 27(14) + 32(17) + 37(22) + 42(13) + 47(6)

8 + 14 + 17 + 22 + 13 + 6

= 2 740

80

= 34.25

(c)

19.5

10

0

20

30

40

50

6063

40

70

80


ScoreSkor24.5 29.5 34.5 39.5 44.5 49.5

(d) 80 – 63 = 17 Percentage of participants take part in final

stage Peratusan peserta yang mengambil bahagian dalam

peringkat akhir

= 1780 × 100%

= 21.25%


FrequencyKekerapan


46 – 50 3 48

51 – 55 4 53

56 – 60 7 58

61 – 65 4 63

66 – 70 3 68

71 – 75 4 73

76 – 80 5 78

(b) Size of the class interval Saiz selang kelas = 50.5 – 45.5 = 5 kg(c) Mean / Min

=

48(3) + 53(4) + 58(7) + 63(4) + 68(3) + 73(4) + 78(5)

30 = 63.33 kg

(d)

0 48 53 58 6863 73

1

2

3

4


Mass (kg) / Jisim (kg)

5

6

7

78

(e) Modal class / Kelas mod = 56 – 60 kg

7. (a) PointsMata


FrequencyKekerapan

45 – 49 47 5

50 – 54 52 7

55 – 59 57 8

60 – 64 62 6

65 – 69 67 5

70 – 74 72 5

75 – 79 77 4

(b) Mean / Min

=

47(5) + 52(7) + 57(8) + 62(6) + 67(5) + 72(5) + 77(4)

—————————————————–40

= 60.75



(c)

0

1

2

3

4

5

6

7

8


Points / Mata42 47 52 57 62 67 72 77 82

(d) Number of teams / Bilangan pasukan = 5 + 5 + 4 = 14 teams / pasukan

8. (a) (i) Modal class / Kelas mod = (16 − 20) (ii) Mean / Min

=

1(3) + 3(8) + 6(13) + 9(18) + 7(23) + 4(28) + 2(33)

————————————————–1 + 3 + 6 + 9 + 7 + 4 + 2

= 606——32

= 18.94

(b) Upper boundary

Sempadan atas


Kekerapan longgokan

0.5 0

5.5 1

10.5 4

15.5 10

20.5 19

25.5 26

30.5 30

35.5 32

(c)

0

4

0.5 5.5 10.5 15.5 20.5 25.5 30.5 35.5Score / Skor

8

12

16

20

24

28

32

Cumulative frequency / Kekerapan longgokan

(d) 25% × 32 = 8 m = 14


1. (a) (i) Possible / Mungkin

(ii) Possible / Mungkin

(iii) Impossible / Tidak mungkin

2. (a) 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20

(b) S, A, M, P, L, E

(c) 1, 2, 3, 6, 9

3. (a) S = {A, B, C, D, E}

(b) S = {male, female} S = {lelaki, perempuan}

4. (a) A = {a, e}

(b) B = {11, 13, 17, 19}

(c) C = {2, 4, 6, 8, 10}

(d) D = {b, c, d, f, g, h}

(e) F = {36}

5. (a) Possible / Mungkin (b) Impossible / Tidak mungkin

6. (a) (i) Ratio / Nisbah = 20—–50

= 2—5

(ii) Ratio / Nisbah = 50 – 20—–——50

= 3—5

(b) (i) Ratio / Nisbah = 500——800

= 5—8

(ii) Ratio / Nisbah = 300——800

= 3—8

7. (a) n(S) = 200

(i) n(2 siblings) = 90 n(2 orang adik-beradik) = 90

Probability / Kebarangkalian

n(2 siblings) n(2 orang adik-beradik) = —————— ————————— n(S) n(S) 90 = —— 200 9 = — 20

(ii) n(less than 2 siblings) n(kurang daripada 2 orang adik-beradik) = 17 + 50

= 67 P(less than 2 siblings) P(kurang daripada 2 orang adik-beradik)

n(less than n(kurang daripada 2 siblings) 2 orang adik-beradik) = —————— ————————– n(S) n(S) 67 = —— 200

(b) n(S) = 25 + 30 + 12 + 33 = 100

(i) n(black) / n(hitam) = 30

P(black) / P(hitam) = 30100

= 310

(ii) n(green) / n(hijau) = 33

P(green) / P(hijau) = 33100

(iii) n(red or blue) / n(merah atau biru) = 25 + 12 = 37 P(red or blue) / P(merah atau biru)

= 37100

8. (a) 2—5

× 250 = 100

(b) 7—–10

× 200 = 140

(c) 3—8

× 360 = 135

5 9. (a) (i) P(yellow) / P(kuning) = ————– 5 + 4 + 3 5 = — 12

(ii) Expected number of times Bilangan kali yang dijangkakan 5 = — × 600 12 = 250

10. (a) The most likely colour to be chosen is blue because the probability of getting a blue

150 1 marble is the highest which is —— = — 300 2

Warna yang paling mungkin terpilih ialah biru kerana kebarangkalian mendapat sebiji guli berwarna biru

adalah paling tinggi iaitu 150——300

= 1—2 .

CH

AP

TE

R 7 Probability IKebarangkalian I


Mathematics Form 4 Chapter 7 Probability I

(b) P(orange marble) / P(guli berwarna oren): x + 30 4 ———– = —– 300 + x 13 13(x + 30) = 4(300 + x) 13x + 390 = 1 200 + 4x 9x = 810 x = 90

(c) n(S) = 300 – y + 30 = 330 – y

n(green) / n(hijau) = 70 70 1 P(green) / P(hijau) : ———– = — 330 – y 4 280 = 330 – y y = 50

SPM Practice 7

Paper 1

1. Total number of balls / Jumlah bilangan bola = 36 + 52 = 88

Probability of choosing a white ball Kebarangkalian memilih sebiji bola putih

= 36—–88

= 9—–22

Answer/ Jawapan: D

2. Number of female passengers in the bus Bilangan penumpang perempuan di dalam bas = 60 – 24 = 36

Number of female passengers in the bus after 8 female passengers left the bus is

Bilangan penumpang perempuan di dalam bas selepas 8 penumpang perempuan turun dari bas ialah

= 36 – 8 = 28

Total number of passengers Jumlah bilangan penumpang = 60 – 8 = 52

Probability of choosing a female Kebarangkalian memilih seorang perempuan

= 28—–52

= 7—–13

Answer/ Jawapan: A

3. Let the number of red cards be x. Biarkan bilangan kad merah ialah x.

x———––

x + 7 +5 =

3—5

5x = 3x + 36 2x = 36 x = 18

Probability of picking a card that is not yellow Kebarangkalian memilih sekeping kad yang bukan berwarna

kuning = 1 – P(yellow / kuning)

= 1 – 5———–––

18 + 7 +5

= 1 – 1—6

= 5—6

Answer/ Jawapan: A

4. Number of days that Lim is late to school Bilangan hari Lim lewat ke sekolah

= 1—–32

× 192

= 6

Answer/ Jawapan: D

5. Number of red apples / Bilangan epal merah

= 1—6

× 150

= 25

Number of green apples / Bilangan epal hijau = 150 – 25 = 125

Answer/ Jawapan: D

6. Number of girls / Bilangan perempuan = 90 – 35 – 15 = 40

P(girl) / P(perempuan) = 40—–——

90 – 15

= 40—–75

= 8—–15

Answer/ Jawapan: B

7. P(vowel) / P(huruf vokal) = 2841000

Number of chances to get a vowel Bilangan peluang untuk mendapat huruf vokal

= 2841000

× 50

= 14.2 = 14

Answer/ Jawapan: A


Mathematics Form 4 Chapter 7 Probability I

8. Number of black toy / Bilangan mainan berwarna hitam

= 3—7

× 56

= 24

24 + x———–56 + x

= 1—2

48 + 2x = 56 + x 2x – x = 56 – 48 x = 8

Answer/ Jawapan: B

9. Number of boys / Bilangan lelaki = 60

Total number of pupils = 100 + 20 Jumlah bilangan murid = 120

\ P(boy) / P(lelaki) = 60—––120

= 1—2

Answer/ Jawapan: D

10. Factors of 32 / Faktor bagi 32 = {2, 4, 8}

Probability / Kebarangkalian = 3—9

= 1—3

Answer/ Jawapan: A

11. P(black) / P(hitam) = 17 + 3—–—————–

4 + 17 + 1 + 3

= 20—–25

= 4—5

Answer/ Jawapan: D 12. The number of green beans in the box Bilangan kacang hijau di dalam kotak = 20 – 5 = 15

The number of beans in the box Bilangan kacang di dalam kotak = 15 + 20 + 6 – 5 = 36

The probability of choosing a green bean Kebarangkalian memilih sebiji kacang hijau

= 15—–36

= 5—–12

Answer/ Jawapan: B

13. The probability of choosing a blue ball Kebarangkalian memilih sebiji bola biru

= 1 – 1—4

– 1—3

= 5—–12

Let the number of blue ball = x Biarkan bilangan bola biru = x

x—–

48 =

5—–12

x = 20

The number of blue balls that need to be taken out is

Bilangan bola biru yang perlu diambil keluar ialah

20 – n———–48 – n

= 3—–10

200 – 10n = 144 – 3n –7n = –56 n = 8

Answer/ Jawapan: C

14. The probability of choosing a student who wears S size T-shirt

Kebarangkalian memilih seorang murid yang memakai baju kemeja-T bersaiz S

1—4

= 80———–——–

80 + 140 + x 220 + x = 320 x = 100

Answer/ Jawapan: B


1. (a) No, it intersects the circle at two points. Tidak, ia menyilang bulatan itu pada dua titik.

(b) No, it does not touch the circle. Tidak, ia tidak menyentuh bulatan itu.

(c) Yes, it touches the circle at one point. Ya, ia menyentuh bulatan itu pada satu titik.

2. (a)

O

P

TangentTangen

3. (a)

A

B

O

T

4. (a) (i) ∆TAO is congruent to / kongruen dengan ∆TBO .

(ii) TA = 6 cm

(iii) x = 30°

(iv) y = 2(180° – 90° – 30°) = 120°

(b) (i) ∆TOB is congruent to / kongruen dengan ∆TOA .

(ii) x = 90°

(iii) y = 40°

(iv) TA = 4 cm

5. (a) (i) tan 35° = XY—–6

XY = 4.2 cm

(ii) OY = 4.22 + 62

= 7.32 cm

6. (a) ∠LBA

(b) ∠LBA

(c) ∠LBA

7. (a) x = ∠REF y = ∠RFE

(b) x = ∠RWU y = ∠RUW

(c) x = ∠QRF y = ∠RFE

(d) x = ∠MRQ or y y = ∠RLM or x

(e) x = ∠QRS y = ∠PRT

8. (a) x = 90° y = 70°

(b) x = 55° y = 180° – 55° – 38° – 40°

= 47°

(c) x = 90° – 32° = 58° y = 32°

(d) x = 100° ÷ 2 = 50°

y = 70° – 40° = 30°

(e) x = 62° y = 90° – 62° = 28°

9. (a) x = 74° – 34° = 40°

y = 180° – 40° – 40° – 34° – 35° = 31°

(b) x = 90° – 25° = 65° y = 25° + (110° – 65°) = 70°

180° – 88°(c) x = ————— 2 = 46°

y = 180° – 63° – 46° = 71°

CH

AP

TE

R 8 Circles IIIBulatan III


Mathematics Form 4 Chapter 8 Circles III

10. (a)

4 common tangents 4 tangen sepunya

(b)


(c)


11. (a) (i) EF = HG

(ii) SU = VT

(iii) SY = VY

(b) (i) PQ = KL (ii) PN = KN (iii) ∠PNO = ∠KNO

180° – 84° 12. (a) (i) ∠EXO = —————— 2 = 48°

(ii) EG = 4 + 6 = 10 cm

QS 13. (a) (i) tan ∠QRS = —– RQ QS tan 65° = —– RQ 14 RQ = ———– tan 65° = 6.53 cm

OP(ii) cos 65° = —— OS 2 OS = ——–— cos 65° = 4.73 cm 2 4.73 —— = ——– 6.53 RS RS = 15.44 cm RO = 15.44 – 4.73 = 10.71 cm

(b) (i) M

P N

UT

20°

13 cm

3 cm

PM sin 20° = —– 13 PM = 4.45 cm

\ MT = 4.45 + 3 = 7.45 cm

TU (ii) cos 20° = —– 13 TU = 12.22 cm

14. (a) OT = 72 – 252 = 25.96 cm

OP = 7 + 4 = 11 cm

PT = 25.96 – 11 = 14.96 cm

(b) ∠POR = ∠POX OX = 7 – 4 = 3 cm

cos ∠POX = 311

∠POX = cos–1 311

= 74.17°

(c) RS = PX = 112 – 32 = 10.58 cm

Area of OPSR / Luas bagi OPSR

= 12

× (4 + 7) × 10.58

= 58.19 cm2

Area of the shaded region Luas rantau berlorek

= 58.19 – 74.17°360°

× 227

× 72 –

180° – 74.17°360°

× 227

× 42 = 58.19 – 31.73 – 14.78 = 11.68 cm2

15. (a) cos ∠QOS = 520

∠QOS = cos–1 520

= 75.52

(b) ∠RPS = ∠QOS

tan 75.25 = SR3

SR = tan 75.52° × 3 = 11.62 cm

X

O

P

3 cm11 cm



(c) SQ = 202 – 52 = 19.36 cm

RQ = 19.36 – 11.62 = 7.74 cm

Area of OPRQ / Luas bagi OPRQ

= 12

× (3 + 5) × 7.74

= 30.96 cm2

Area of the shaded region Luas rantau berlorek

= (2 × 30.96) – 2(75.52°)360°

× 227

× 52 –

360° – 2(75.52°)360°

× 227

× 32 = 61.92 – 32.97 – 16.42 = 12.53 cm2

SPM Practice 8

Paper 1

1. ∠PRS = ∠UPS = 50°

∠RSP = 180° – 110° = 70°

x = 180 – 50° – 70° = 60°

Answer/ Jawapan: C

2. m = 180° – ∠URQ = 180° – 40° = 140°

∠XRW = ∠URW – ∠URX = 90° – 40° = 50°

n = 180° – ∠RXW – ∠XRW = 180° – 90° – 50° = 40°

m – n = 140° – 40° = 100°

Answer/ Jawapan: B

3. ∠QTS = 40°

∠TSQ = 180° – 40°—–————

2 = 70° ∠PQT = 70°

\ x = 180° – (68° + 70°) = 42°

Answer/ Jawapan: A

4. ∠QOS = 2 × 62° = 124°

∠QSO = 180° – 124°—–————

2 = 28°

x = 50°+ 28° = 78°

Answer/ Jawapan: D

5. The angle substended by the major arc RS, ∠RPS is more than 180°.

Therefore, ∠RPS = 240°. Sudut dicangkum oleh lengkok major RS. ∠RPS adalah lebih

besar daripada 180°. Oleh itu ∠RPS = 240°.

Answer/ Jawapan: A

6. Reflex ∠AOB / Refleks ∠AOB = 2 × 105° = 210°

Obtuse ∠AOB / Cakah ∠AOB = 360° – 210° = 150°

x = 360° – (90° + 90° + 150°) = 30°

Answer/ Jawapan: C

7. ∠PQM = ∠PMN = 74° ∠LQM = 180° – 74° = 106°

∠QML = 180° – 106°—–————

2 = 37°

x = ∠QML = 37°

Answer/ Jawapan: A

8. ∠TOQ = 2 × 30° = 60°

∠OTP = 180° – 40° = 140°

∠OQP = 90°

x = 360° – (60° + 140° + 90°) = 70°

Answer/ Jawapan: A

9. ∠QOS = 2 × 36° = 72°

∠OQR = 90°

∠OSR = 360° – (72° + 90° + 42°) = 156°

x = 180° – 156° = 24°

Answer/ Jawapan: A



10. OP = 52 + 122

= 13 cm

Answer/ Jawapan: B

11. ∠AYC = 46°

∠YAC = 180° – 46°—–————

2 = 67°

x = 67°

Answer/ Jawapan: B 12. ∠QOR = 180° – (90° + 24°) = 66°

∠OSQ = 66°——2

= 33°

x = 90° – 33° = 57°

Answer/ Jawapan: B

13. ∠POL = 2∠PQL = 2(50°) = 100°

∠OLP = 180° – 100°—–————

2 = 40°

∠QLM = ∠QPL = ∠QPO + ∠OPL = 20° + 40° = 60°

∠KLP = ∠PQL = 50° x = 180° – ∠KLP – ∠OLP – ∠QLM = 180° – 50° – 40° – 60° = 30°

Answer/ Jawapan: A

14. ∠SYX = ∠STY = 60°

x = 180° – ∠SXY – ∠SYX = 180° – 50° – 60° = 70°

Answer/ Jawapan: C

15. P

x Q

80� 12 cm

Let LM = XQ Biarkan LM = xQ

sin 80° = XQ——12

XQ = 11.82 cm LM = 11.82 cm

Answer/ Jawapan: D

16. ∠OQS = ∠OQR – ∠SQR = 90° – 35° = 55°

∠SOQ = 180° – ∠OQS – ∠OSQ = 180° – 55° – 55° = 70°

x = 180° – ∠SOQ – ∠OQR = 180° – 70° – 90° = 20°

Answer/ Jawapan: B


1. (a)

x

y

0

85°

Quadrant I / Sukuan I

(b)

x

y

0314°

Quadrant IV / Sukuan IV

(c)

x

y

0

168°

Quadrant II / Sukuan II

(d)

x

y

0

244°

Quadrant III / Sukuan III

2.

x-coordinateKoordinat-x

y-coordinateKoordinat-y

y-coordinatex-coordinate

Koordinat-yKoordinat-x

(a)–0.9 0.4

0.4–0.9

= – 49

(b)–0.9 –0.4

–0.4–0.9

= 49

(c)0.4 –0.9

–0.90.4

= – 94

(d)0.8 –0.6

–0.60.8

= – 34

3.

x-coordinateKoordinat-x

y-coordinateKoordinat-y

y-coordinatex-coordinate

Koordinat-yKoordinat-x

(a) (i)0.86 0.50

0.500.86

= 0.58

(ii)0.70 0.70

0.700.70

= 1.00

(iii)0.50 0.86

0.860.50

= 1.72

sin θ(y-coordinate)

(Koordinat-y)

cos θkos θ

(x-coordinate)(Koordinat-x)

tan θ

y-coordinatex-coordinate Koordinat-y

Koordinat-x (b) (i) 0.50 0.86 0.58

(ii) 0.70 0.70 1.00

(iii) 0.86 0.50 1.72

CH

AP

TE

R 9 Trigonometry IITrigonometri II


Mathematics Form 4 Chapter 9 Trigonometry II

4. sin θ

(y-coordinate)(Koordinat-y)

cos θkos θ

(x-coordinate)(Koordinat-x)

tan θ

y-coordinatex-coordinate Koordinat-y

Koordinat-x (a)

0.64 –0.760.64–0.76

= – 0.84

(b)–0.76 –0.66

–0.76–0.66

= 1.15

(c)–0.90 0.42

–0.900.42

= –2.14

(d)–0.36 0.98

–0.360.98

= –0.37

7. sin θcos θkos θ

tan θ

(a) 1 0Undefined

Tidak tertakrif

(b) 0 –1 0

(c)–1 0

UndefinedTidak tertakrif

(d) 0 1 0

5. QuadrantSukuan

sin θcos θkos θ

tan θ

(a)III

NegativeNegatif

NegativeNegatif

PositivePositif

(b)IV

NegativeNegatif

PositivePositif

NegativeNegatif

(c)II

PositivePositif

NegativeNegatif

NegativeNegatif

(d)III

NegativeNegatif

NegativeNegatif

PositivePositif

6. sin θ

cos θkos θ

tan θ

(a) 12

32

1

3

(b) 32

12 3

8. (a) sin 236° = –0.8290 cos 85° / kos 85° = 0.08716 tan 105° = –3.732

(b) sin 96° = 0.9945 cos 298° / kos 298° = 0.4695 tan 311° = –1.150

(c) sin 304.8° = –0.8211 cos 195.3° / kos 195.3° = –0.9646 tan 253.9° = 3.465

(d) sin 183°15ʹ = –0.05669 cos 168°42ʹ / kos 168°42ʹ = –0.9806 tan 271°56ʹ = –29.62

9. (a) θ = 55°26ʹ, 180° – 55°26ʹ = 55°26ʹ, 124°34ʹ

(b) θ = 180° + 21°22ʹ, 360° – 21°22ʹ = 201°22ʹ, 338°38ʹ

(c) θ = 180° + 35°25ʹ, 360° – 35°25ʹ = 215°25ʹ, 324°35ʹ

10. (a) θ = 180° – 59°11ʹ, 180° + 59°11ʹ = 120°49ʹ, 239°11ʹ

(b) θ = 37°53ʹ, 360° – 37°53ʹ = 37°53ʹ, 322°7ʹ

(c) θ = 81°20ʹ, 360° – 81°20ʹ = 81°20ʹ, 278°40ʹ

11. (a) θ = 9°, 180° + 9° = 9°, 189°

(b) θ = 180° – 67°23ʹ, 360° – 67°23ʹ = 112°37ʹ, 292°37ʹ

(c) θ = 180° – 42°51ʹ, 360° – 42°51ʹ = 137°9ʹ, 317°9ʹ



14. (a) x 0° 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360°

cos xkos x

1 0.87 0.5 0 –0.5 –0.87 –1 –0.87 –0.5 0 0.5 0.87 1

30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360°

y = cos x

0

–1.0

–0.5

0.5

1.0

y

x

(b) x 0° 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360°

tan x 0 0.58 1.73 UndefinedTidak

tertakrif

–1.73 –0.58 0 0.58 1.73 UndefinedTidak

tertakrif

–1.73 –0.58 0

y = tan x

30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360°0

–2

–1

1

2

y

x

12. (a) (i) tan x = 46

= 23

(ii) sin y = 45

(b) (i) sin x = 45

8

EC =

45

EC = 10 cm

\AC = 2EC = 2(10) = 20 cm

(ii) cos y / kos y = 1220

= 35

13. (a) tan 135° + 2 cos 360° tan 135° + 2 kos 360°

= –1 + 2(1) = 1

(b) sin 150° – 4 tan 180°

= 12

– 4(0)

= 12

(c) cos 60° + sin 210° – tan 225° kos 60° + sin 210° – tan 225°

= 12

+ – 12 – 1

= –1

(d) tan 0° – sin 180° + cos 270° tan 0° – sin 180° + kos 270° = 0 – 0 + 0 = 0



15. (a) y = cos x / kos x

(b) y = sin x

(c) y = cos x / kos x

(d) y = tan x

(e) y = tan x

16. (a) (i) y = tan x

(ii) tan θ = –1 θ = 45°

Angle in the second quadrant, Sudut dalam sukuan kedua, a = 180° – 45° = 135°

tan θ = 1 θ = 45°

Angle in the third quadrant, Sudut dalam sukuan ketiga, b = 180° + 45° = 225°

(b) (i) (90°, 1)

(ii) (270°, 0)

(iii) 45°, 1

2 cos 45° / kos 45° =

1

2

sin 45° = 1

2

The intersection point, C, Titik persilangan, C,

cos x / kos x = sin x

\ x = 45°

SPM Practice 9

Paper 1

1. tan y = 8—–15

= PT—–QS

PT = 16 cm QS = 30 cm

QR = QS – RS = 30 – 18 = 12 cm

TR = 122 + 162 = 20 cm

cos x = – QR—–TR kos x

= – 12—–20

= – 3—5

Answer/ Jawapan: C

2. At second quadrant 90° x 180°, Pada sukuan kedua 90° x 180°,

tan x = Negative value / Nilai negatif

At third quadrant 180° x 270°, Pada sukuan ketiga 180° x 270°,

tan x = Positif value / Nilai positif

Answer/ Jawapan: B

3. cos ∝/kos ∝= 15—–17

= SQ—–PQ

SQ = 15 cm, PQ = 17 cm

SP = 172 – 152 = 8 cm

tan θ = – SP—–SQ

= – 8—–15

Answer/ Jawapan: C

4. tan x = 4—3

= FH—–GH

FH = 8 cm, GH = 6 cm

FG = 82 + 62 = 10 cm

tan 35° = EF—–FG

EF = 10 tan 35° = 7 cm

Length of EFG / Panjang EFG = FG + EF = 10 + 7 = 17 cm

Answer/ Jawapan: C

5. sin 90° = 1 sin 180° = 0 sin 270° = –1

Answer/ Jawapan: B

6. tan ∠QPU = 5—–12

= UQ—–PU

UQ = 5 cm, PU = 12 cm,

PQ = 13 cm

PQ = QR, PR = 26 cm

TR = 262 – 102 = 24 cm

cos θ / kos θ = – TR—–PR

= – 24—–26

= – 12—–13

Answer/ Jawapan: D



7. y = sin 225° = –0.707

y = cos 225° / kos 225° = –0.707

sin 225° = cos 225° / kos 225°

Answer/ Jawapan: B

8. sin 90° = 1 sin 180° = 0

Answer/ Jawapan: B

9. sin x= 15—–17

= JK—–JL

JK = 45 cm, JL = 17 × 3 = 51 cm

KL = 512 – 452 = 24 cm

Answer/ Jawapan: C

10. cos ∠SRQ/kos ∠SRQ

= – 5—–13

= PR—–QR

PR = 2.5 m, QR = 6.5 m

QP = 6.52 – 2.52 = 6 m

Answer/ Jawapan: C

11. y = sin x 0.5 = sin x x = 30°, 150°,… The coordinates of the point H is (150°, 0.5). Koordinat bagi titik H ialah (150°, 0.5).

Answer/ Jawapan: C

12. tan m = 0.750.87

m = tan–1 0.750.87

m = 40°46’ θ = 180° + 40° 46’ = 220° 46’

Answer/ Jawapan: C

13. sin x = 0.34—––

1 = 0.34

cos y / kos y = 0.34—––

1 = 0.34

sin x + cos y = 0.34 + 0.34 sin x + kos y = 0.68

Answer/ Jawapan: B

14. cos 0° / kos 0° = 1 cos 90° / kos 90° = 0 cos 180° / kos 180° = –1 cos 270° / kos 270° = 0

Answer/ Jawapan: A

y

m 0

F(–0.87, –0.75)

θ

x

15. y

x0.5

0.87α

tan ∝ = – 0.87—––0.5

= –1.74

Answer/ Jawapan: D

16. cos x / kos x= 3—5

= QS—–PS

PQ = 8 cm, QS = 6 cm, PS = 10 cm

RQ—–QS

= 5—2

QS = 6 cm, RQ = 5—2

(6)

= 15 cm

PR = 82 + 152 = 17 cm

Answer/ Jawapan: C

17. cos y / kos y= – 15—–17

= – PQ—–UQ

PQ = 15 cm, UQ = 17 cm

US = UQ – QS = 17 – 9 = 8 cm

TU = ST2 – US2

= 102 – 82 = 6 cm

tan x = – TU—–US

= – 6—8

= – 3—4

Answer/ Jawapan: D

18.

yx

RQ

P

4 cm

5 cm3 cm

sin x= 3—5

= PQ—–PR

PQ = 3 cm, PR = 5 cm, QR = 4 cm

cos y = – QR—–PR kos y

= – 4—5

Answer/ Jawapan: A


1. (a) (i) Horizontal line: Garis mengufuk: AO(ii) Angle of elevation of B from A: Sudut dongakan bagi B dari A: ∠OAB(iii) Angle of depression of C from A: Sudut tunduk bagi C dari A: ∠OAC

2. (a)

62°

1.8 m

(b)

52°37′63°

10 m15 m

AB

(c)

30°

1 m1.6 m

25°

3. (a) 13°

25 cm

4. (a) tan ∠PSQ = 60—–QS

QS = 60

tan 28°

= 112.84 m

tan ∠PRQ = 60—–QR

QR = 60

tan 35°

= 85.69 m Distance between RS Jarak di antara RS = 112.84 − 85.69 = 27.15 m

(b) tan 12° = 25—–x

x = 25

tan 12° = 117.62 m

P

Q35°

28°

R S

30°

y

x12°

tan 30° = y

117.62

y = 117.62 × tan 30° = 67.91 m

The height of the lighthouse Tinggi rumah api = 67.91 − 25 = 42.91 = 43 m

(c) (i) ∠PSX = ∠SPT

tan 40° = STPT

ST = 12 × tan 40° = 10.07 m

(ii) tan ∠UPV = UV—–PV

∠UPV = tan–1 10.0712 + 6

= 29° 13’

SPM Practice 10

Paper 1

1. sin 60° = 80KM

KM = 92.38 m

Answer/ Jawapan: B

2. tan 46° = x

20

x = 20 × tan 46° = 20.71 m

The height of the kite from the ground Tinggi layang-layang dari tanah

= 20.71 + 1.4 = 22.11 m

Answer/ Jawapan: B

3. JN

Q

Angle of depression

Sudut tunduk

Angle of depression from point N is ∠JNQ. Sudut tunduk dari titik N ialah ∠JNQ.

Answer/ Jawapan: B

P T V

X S U

80 m

60°M L

K

1.4 m

20 m

x

CH

AP

TE

R 10 Angles of Elevation and DepressionSudut Dongakan dan Sudut Tunduk


Mathematics Form 4 Chapter 10 Angles of Elevation and Depression

4. tan 43°14ʹ = SP—–5

SP = 4.7

The depth of the water in the tank is 4.7 m. Kedalaman air dalam tangki ialah 4.7 m.

Answer/ Jawapan: D

5.

Y O

X75°

Horizontal planeSatah mengufuk

Answer/ Jawapan: A

6. tan ∠KML = 15—–25

∠KML = 30°58ʹ

Answer/ Jawapan: A

7.

15 m

1.7 m

h

35�

tan 35° = h—–25

h = 10.503 m

The height of the monument Tinggi tugu = 10.503 + 1.7 = 12.2 m

Answer/ Jawapan: C

8.

FE

H

M

N

G

15 m

12 m

58�

tan 58° = GE—–12

GE = 19.2 m

GN = GE – HF = 19.204 – 15 = 4.204 m

Angle of depression of H from G Sudut tunduk H dari G = ∠MGH

= ∠NHG

tan θ = GN—–NH

= 4.204——––

12

θ = 19°18ʹ

Answer/ Jawapan: A

9. In / Dalam ∆TQR,

TQ—–10

= tan 72°

TQ = 10 × tan 72° = 30.777 m

In / Dalam ∆SPR,

30.777———––

PQ + 10 = tan 34°

PQ + 10 = 30.777———––tan 34°

PQ = 45.629 – 10 = 35.629 m

Answer/ Jawapan: B

10.

Guard houseRumah jaga

50 m

Ph

k

Q58.4�

42�

tan 42° = k—–

50 k = 45.02 m

tan 58.4° = h—–50

h = 81.27 m

Height between the 6th floor and the 11th floor of the condominium

Tinggi di antara tingkat ke-6 dan tingkat ke-11 bagi kondominium

= h – k = 81.27 – 45.02 = 36.25 m

Answer/ Jawapan: A

11. 12.58�

3.8m0.88mHaris’s eyes levelParas mata Haris

Mr. Teeh

tan 12.58° = h—–

3.8 h = 0.848 m

Height of Mr. Tee / Tinggi Mr. Tee = 0.848 + 0.88 = 1.728 1.73 m


Answer/ Jawapan: B 1. (a) Yes / Ya

(b) Yes / Ya

(c) No / Tidak

2. (a)

Vertical plane Satah mencancang

(b)

Inclined plane Satah condong

(c)

Horizontal plane Satah mengufuk

3. (a) (i) Horizontal plane: Satah mengufuk: PQRS, TUVW

(ii) Vertical plane: Satah mencancang: – (iii) Inclined plane: Satah condong: PQUT, QRVU, RSWV, SPTW

4. (a) AS, BP, CQ, DR, PD AP, BP, CP, DP, PQ

PD, PB, DB PD, PB, DB, PQ

AB, AD, CD, CB, QP, RD, SP

AB, BC, CD, DA, PA, PC, AC

5. (a) (i) Normal to the plane SQR is Normal kepada satah SQR ialah

PS .(ii) Normal to the plane PSR is Normal kepada satah PSR ialah QS .

(b) (i) Normals to the plane SRCD are Normal kepada satah SRCD ialah

PS, QR, BC and / dan AD.

(ii) Normal to the plane PQCD is Normal kepada satah PQCD ialah

none / tiada .

6. Orthogonal projectionUnjuran ortogon

(a) DC

(b) PR

(c) AC

(d) BQ

(e) DP

7. (a) B C

GF

A D

HE

BG

(b) B C

GF

A D

HE

FH

8. Angle between line and planeSudut di antara garis dengan satah

(a) ∠HBG

(b) ∠GAI

(c) ∠HCG

(d) ∠IHG

CH

AP

TE

R 11 Lines and Planes in 3-DimensionsGaris dan Satah dalam Tiga Matra


Mathematics Form 4 Chapter 11 Lines and Planes in 3-Dimensions

(e) ∠AIB

9. (a) ∠VPO

(b) ∠CVO

(c) ∠DVO

10. (a) (i) PR = 82 + 152

= 17 cm

20 tan ∠WPR = —– 17 ∠WPR = 49°38'

(ii) TR = 202 + 152

= 25 cm 8 tan ∠QTR = —– 25 ∠QTR = 17°45'

(b) (i) HC = 82 + 42

= 8.94 cm

tan ∠BHC = 5

8.94 ∠BHC = 29°13'

B

CH8.94 cm

5 cm

(ii) BD = 82 + 52

= 9.43 cm

tan ∠HBD = 4

9.43 ∠HBD = 22°59'

D B

H

9.43 cm

4 cm

(c) (i) QN = 42 + 32

= 5 cm

tan ∠PNQ = 45

∠PNQ = 38°40'

(ii) KQ = 82 + 42

= 8.94 cm

tan ∠RKQ = 3

8.94 ∠RKQ = 18°33'

Q

R

K 8.94 cm

3 cm

(d) (i) tan ∠EGF = 56

∠EGF = 39°48'

20 cm

17 cmP R

W

8 cm

25 cmT

Q

R

Q

P

N

4 cm

5 cm

E

FG

5 cm

6 cm

(ii) tan ∠KEF = 65

∠KEF = 50°12'

11. Line of intersection between the two planes

Garis persilangan di antara dua satah

(a) VB

(b) AC

(c) VB

(d) VO

(e) BD

12. (a) E H

CD

GF

A B

(b) T W

R

VU

P Q

S

13. Angle between two planesSudut di antara dua satah

(a) ∠RLM

(b) ∠KSN

(c) ∠QUL or / atau ∠RTM

(d) ∠PSQ or / atau ∠KNL

14. (a) ∠WYZ

(b) ∠WVZ / ∠XUY / ∠LMN

(c) ∠MLN

15. (a) (i) tan ∠AHD = 54

∠AHD = 51° 20’

(ii) tan ∠CHD = 74

∠CHD = 60° 15’

(b) (i) tan ∠TVW = TW6

TW = 6 × tan 65° = 12.87 cm

E

FK 6 cm

5 cm

A

DH 4 cm

5 cm

HD

C

7 cm

4 cm

VW

T

6 cm



(ii) tan ∠WMV = 63

∠WMV = 63° 26’

(c) (i) tan ∠SQT = 42

∠SQT = 63° 26’

(ii) tan ∠UQT = 32

∠UQT = 56° 19’

SPM Practice 11

Paper 1

1. The angle between the plane PQR and the plane QSR is ∠PRS.

Sudut di antara satah PQR dengan satah QSR ialah ∠PRS.

Answer/ Jawapan: B

2. MN is the line of intersection between two planes. JM and MI are perpendicular to MN.

MN ialah garis persilangan antara dua satah. JM dan MI adalah berserenjang dengan MN.

\∠JMI

Answer/ Jawapan: D

3. TU is the normal to the base TUVW. QT is the orthogonal projection.

TU ialah normal kepada tapak TUVW. QT ialah unjuran ortogon.

\∠UQT

Answer/ Jawapan: B

4. PU is the normal to the base UVWT. UW is the orthogonal projection.

PU ialah normal kepada tapak UVWT. UW ialah unjuran ortogon.

\∠PWU

Answer/ Jawapan: A

5. UT is the line of intersection between the two planes.

QU and UP are perpendicular to UT. UT ialah garis persilangan antara dua satah. QU dan UP adalah berserenjang dengan UT. \∠PUQ

Answer/ Jawapan: B

W

VM 3 cm

6 cm

Q

TS 4 cm

2 cm

Q

UT

2 cm

3 cm

6. X

UT M

12

12.5

tan θ = 12.5—––12

θ = 46.17°

The angle between plane XUVY and plane XTWY. Sudut di antara satah xUVY dan satah xTWY = ∠UXT = 2(46.17°) = 92.3°

Answer/ Jawapan: D

7.

P Q

RS

WYZ

X

V

o

M

24 m M

80

V

P Q

��5824 VM = 802 – 242

= 5 824 cm

OM

V

��5824

24 m

Angle between planes WPQX and PQRS Sudut di antara satah WPQx dan PQRS = ∠VMO

cos θ = 24

5 824 kos θ θ = 71°40ʹ

Answer/ Jawapan: C

Paper 2

1. (a) ∠SMP

(b) tan ∠SMP = 75

∠SMP = 54°28’

M

SP

5 cm

7 cm

2. (a) P

A

X

B C

D

Q R

S

10 cm

6 cm

5 cmθ



(b) tan θ = 3—5

θ = 30°58ʹ

3. (a) ∠GCB

(b) tan ∠GCB = 9—8

∠GCB = 48°22ʹ

4. (a) ∠FGL

(b) tan ∠FGL = 5—6

∠FGL = 39°48ʹ

5. (a) ∠AED

(b) tan ∠AED = 6—–10

∠AED = 30°58ʹ

6. (a) A

E’F’2.5 m

1.5 m

CF’

F

1.5 m

42°

AFʹ = (1.5)2 + (2.5)2 tan 42° = 1.5—–FFʹ

= 2.915 m FFʹ = 1.666 m

A

F

F’

1.666 m

2.915 m

tan ∠FAFʹ = 1.666——––2.915

∠FAFʹ = 29°45ʹ

(b)

G

E

E’

1.666 m25°

EEʹ = FFʹ = 1.666 m

cos 25° / kos 25° = 1.666——––

EG EG = 1.838 m

G 6 cm

5 cm

F

L

A D

E

6 cm

10 cm

7. (a)

A

P Q

B

R

CD

S

O

E

18.25º

(b) tan 18.25° = x—–

18 x = 5.936 cm

EO = 5.936 + 5 = 10.936 cm

Let M be the midpoint of PS. Biarkan M ialah titik tengah PS.

Let N be the midpoint of QR. Biarkan N ialah titik tengah QR.

E

OM N18 cm

10.936 cmθ

tan θ = 18——–––

10.936 = 58°43ʹ

Angle between the two planes Sudut di antara dua satah = ∠MEN = 2 × 58°43ʹ = 117°26ʹ

E

Nx

18 cm18.25°


Paper 1

1. 0.03658 = 0.03658 = 0.037 (2 significant figures / angka bererti)

Answer/ Jawapan: B

2. 7 290 000 = 7.29 × 106

Answer/ Jawapan: C

3. 0.000213 − 3.45 × 10−5 = 2.13 × 10−4 − 3.45 × 10−5 = 2.13 × 10−4 − 0.345 × 10−4

= (2.13 − 0.345) × 10−4

= 1.785 × 10−4

Answer/ Jawapan: A

4. 0.054————–(3 × 10–3)3

= 5.4 × 10–2

————–27 × 10–9

= 0.2 × 10−2 − (−9)

= 2 × 10−1 × 107 = 2 × 106

Answer/ Jawapan: D

5. 1.8 × 106 × (100% − 12%) = 1.8 × 106 × 88% = 1.584 × 106

Answer/ Jawapan: A

6. ∠MNP = ∠KPN

(5 – 2) × 180°——————–

5 = 108°

∠SNP = 180° − 108° = 72°

Interior angle of NPQRS Sudut pedalaman NPQRS = (5 − 2) × 180° = 540°

x = 540° − 240° – ∠SNP – ∠NPQ – ∠PQR = 540° − 240° – 72° – 72° – 108° = 48°

Answer/ Jawapan: B

7. ∠PMN = 80° ∠MNR = 100° ∠MNQ = 100° − 35° = 65°

x = 360° − 130° − 80° − 65° = 85°

∠PLK = 100°

∠QPL = 180° − 130° = 50°

y = 180° − 100° − 50° = 30°

x + y = 85° + 30° = 115°

Answer/ Jawapan: D

8. ∠TUS = (180° − 120°) ÷ 2 = 30°

∠TSQ + ∠TUQ = 180°

∠TUQ = 180° – 100° = 80°

∠SUQ = 180° − 100° − 30° = 50°

x = ∠SUQ = ∠TUQ – ∠TUS = 80° – 30° = 50°

Answer/ Jawapan: A

9. ∠PMR is more than 90° and less than 180°. Therefore, ∠PMR = 140°.

∠PMR adalah lebih daripada 90° dan kurang daripada 180°. Oleh itu, ∠PMR = 140°.

Answer/ Jawapan: B

10. y

x

LB

KC

AD

0 2

2

4

6

8

4 6 8 10

Answer/ Jawapan: A

11. Area of image = k2 × Area of object Luas imej Luas objek 8 = k2 × 32

k2 = 8—–32

= 1—4

k = 1—2

Answer/ Jawapan: C

12. cos x = 12—–13

PQ—–13

= 12—–13

PQ = 12 cm

End-of-Year Assessment


Mathematics Form 4 End-of-Year Assessment

PQ : QR = 3 : 2

PQ—–QR

= 3—2

12—–QR

= 3—2

QR = 8 cm

TQ = 132 – 122

= 5 cm

TR = 52 + 82

= 9.43 cm

Answer/ Jawapan: D

13. y = cos x −1 = cos x x = 180°, … The coordinates of the point G is (180°, −1). Koordinat titik G ialah (180°, −1).

Answer/ Jawapan: C

14. y

m x0

θ

E(–0.82, 0.78)

tan m = 0.780.82

m = tan–1 0.780.82

= 43° 34' θ = 180° − 43° 34' = 136° 26'

Answer/ Jawapan: A

15.

T

S

P Q

R

U

Answer/ Jawapan: C

16. M

K

40°

Answer/ Jawapan: B

17. P

Q

60 m

50°

cos 50° = 60PQ

PQ = 60

cos 50° = 93.34 m

Answer/ Jawapan: B

18. Mode of the number of books is 2. Therefore the value of m is the highest which is 9. The possible value of n is less than 9 which is 8.

Mod bagi bilangan buku ialah 2. Oleh itu, nilai m adalah tertinggi, iaitu 9. Nilai yang mungkin bagi n ialah kurang daripada 9, iaitu 8.

m = 9 , n = 8

Answer/ Jawapan: D

19. (2p − q)(p + 3q) = 2p2 + 6pq − pq − 3q2

= 2p2 + 5pq − 3q2

Answer/ Jawapan: D

20. p – 2——–p2

÷ p2 – 4——––

3p =

p – 2——–p2

× 3p——–————

(p + 2)(p – 2)

= 3——–––

p(p + 2)

Answer/ Jawapan: C

21. y = x2

—3

+ 2

x2 = 3(y – 2)

x = 3(y – 2)

Answer/ Jawapan: B

22. 4 – 10x——–––2

= 3x – 2

4 – 10x = 6x – 4 16x = 8

x = 1—2

Answer/ Jawapan: B

23. 4023 = 4023

= 40pq

p = 2, q = 3

Answer/ Jawapan: D

24. (43 × 27)13 × (m

12)–2 = (43 × 33)

13 × m–1

= (4 × 3) × m–1

= 12—–m

Answer/ Jawapan: C



25. 2 x + 4 5 – 1—2

x

2 x + 4 , x + 4 5 – 1—2

x

−2 x x + 1—2

x 5 − 4

3—2

x 1

x 2—3

–2 x 2—3

Answer/ Jawapan: D

26. 2x—–3

−2 and / dan 4 − 3x −2

2x −6 − 3x −6 x −3 x 2

x = −2, −1, 0, 1, 2

Answer/ Jawapan: D

27.

P

K

Q

R

MN

S

L

The angle between the plane NPQ and the base PQRS is ∠NPS.

Sudut antara satah NPQ dan tapak PQRS ialah ∠NPS.

Answer/ Jawapan: D

28. Mean = 6.7

6.7 =

4(x) + 5(4) + 6(2x) + 7(6) + 8(7)

x + 4 + 2x + 6 + 7

6.7(3x + 17) = 16x + 118 20.1x + 113.9 = 16x + 118 4.1x = 4.1 x = 1 Answer / Jawapan : D

29. n(P Q) = n(R) x + 5 + (3x − 10) + 3 = 3 + (2x + 13) 4x − 2 = 2x + 16 2x = 18 x = 9

Answer/ Jawapan: A

30. S = {1, 2, 3, 4} T = {1, 2, 3, 4, 5, 6} U = {6, 7, 8, 9, 10} U’ = {1, 2, 3, 4, 5}

Answer / Jawapan : B

31.

12% 16%3%4%

23%

2%1%

K

M

L

The percentage of L only Peratusan bagi L sahaja = 25 − 3 − 4 − 2 = 16%

The number of students who bought pen L only Bilangan murid yang membeli pen L sahaja = 16% × 2 000 = 320

Answer / Jawapan : B

32. X

Y

Z

(X Z) Y

Answer/ Jawapan: C

33. 2y = px + 4

y = p—2

x + 2

y-intercept / Pintasan-y = 2 , K(0, 2)

OK—––OM

= 1—3

OK = 2, 2—––

OM =

1—3

OM = 6

OM = 6 units / unit

Coordinates M / Koordinat M = (6, 0)

Gradient of KM / Kecerunan KM

p—2

= 0 – 2——–6 – 0

p = –2—–3

Answer/ Jawapan: A

34. 4y + 11 = 20x

4y = 20x − 11

y = 5x – 11—–4

Gradient / Kecerunan = 5,

y-intercept / Pintasan-y = – 11—–4

Answer/ Jawapan: A



35. y = 3x + c, (1, −3) −3 = 3(1) + c c = −6 y = 3x − 6

When/ Apabila y = 0, 0 = 3x − 6 x = 2

The point of intersection / Titik persilangan = (2, 0)

Answer/ Jawapan: B

36. M

15 cmK

x cm

53° 8'

tan 53°8ʹ = x

—–15

x = 20 cm

Answer/ Jawapan: D

37. (3x − 4)2 + 5x − 9 = 9x2 − 24x + 16 + 5x − 9 = 9x2 − 19x + 7

Answer/ Jawapan: B

38. P(Restaurant C) P(Restoran C)

= 1 0802 000

Number of resident who likes Restaurant C Bilangan penduduk yang menggemari Restoran C

= 1 0802 000 × 850

= 459

Answer / Jawapan : A

39. Let the number of yellow cards = x Biarkan bilangan kad kuning

Probability of taking a yellow card = 1—4 Kebarangkalian memilih sekeping kad kuning

x————–

9 + 3 + x =

1—4

4x = 9 + 3 + x 3x = 12 x = 4

Probability of taking a red card Kebarangkalian memilih sekeping kad berwarna merah

= 9—–16

Answer/ Jawapan: B

40. Probability of choosing a white ball Kebarangkalian memilih sebiji bola putih

= 48———––

48 + 34

= 48—–82

= 24—–41

Answer/ Jawapan: D

Paper 2

1. (a)

M

N

P

(b)

M

N

P

2. x + y = 18 .................................. a 2x + 3y = 43 .................................. b a × 2 : 2x + 2y = 36 .................................. c b − c: y = 7

Substitute y = 7 into a, Gantikan y = 7 ke dalam a, x + 7 = 18 x = 11

\ x = 11 and / dan y = 7

3. When the toy car hit the block, v = 0. Apabila kereta mainan melanggar blok, v = 0.

15 + 7t − 2t2 = 0 (5 − t)(3 + 2t) = 0

5 – t = 0 , 3 + 2t = 0

t = 5 t = − 3—2

(Not accepted / Tidak diterima)

\ t = 5 s

4. (a) ∠NEH

(b) tan ∠NEH = 5—–12

∠NEH = 22°37'

5. (a) Gradient of PQ = Gradient of RS Kecerunan PQ = Kecerunan RS

= –1 – (–4)1 – (–3)

= 34



The equation of PQ is, Persamaan PQ ialah,

y − 8 = 34 (x − 2)

y = 34 x + 6

12

(b) When / Apabila y = 0 , y = 34 x + 6

12

0 = 34 x + 6

12

−612 =

34 x

x = –823

\ x-intercept / pintasan-x = −8 23

6. (a)

PM

G

H

8 cm

4 cmJ

K

N

Q

(b) GJ = 42 – 42

= 5.66 cm

tan ∠PGJ = 8——

5.66 ∠PGJ = 54°43’

7. 1—2

× 22—–7

×22 ×7 + 3 1—2

×(5 + BF) ×4 ×44 = 148

44 + 8 (5 + BF) = 148 44 + 40 + 8BF = 148

BF = 148 – 44 – 40——————–

8 = 8 cm

8. (a) (i) False / Palsu

(ii) If x = 2, then x3 = 8. Jika x = 2, maka x3 = 8.

(b) 3 is a multiple of 16 or 3 is an odd number.

3 ialah gandaan bagi 16 atau 3 ialah nombor ganjil.

(c) Premise 2 / Premis 2: x 5

(d) 3n2 − 1, n = 1, 2, 3, …

9. (a) Perimeter of the whole diagram Perimeter seluruh rajah = OB + Arc BA + AQ + Arc QR + RO

= 10 + 120—––360

×2 × 22—–7

×10 + (10 − 7) +

90—––360

×2 × 22—–7

×7 + 7

= 10 + 20.9524 + 3 + 11 + 7 = 51.95 cm

(b) Area of the shaded region = Area of AOB − Area of POQ + Area of QOR

Luas rantau berlorek = Luas AOB – Luas POQ + Luas QOR

= 120—––360

× 22—–7

×102 − 30—––360

× 22—–7

×72 +

90—––360

× 22—–7

×72 = 104.7619 − 12.8333 + 38.5 = 130.43 cm2

10. (a) x = 4

(b) Gradient of RS = Gradient of PQ Kecerunan RS = Kecerunan PQ

= 6 – 3——–4 – 0

= 3—4

The equation of RS is

Persamaan bagi RS

y − (−6) = 3—4

[x − (−2)]

y + 6 = 3—4

x + 3—2

y = 3—4

x − 9—2

(c) When / Apabila y = 0, y = 3—4

x − 9—2

0 = 3—4

x − 9—2

9—2

= 3—4

x

x = 6

\ x-intercept / Pintasan-x = 6

11. (a) Not a statement Bukan Pernyataan

(b) If the perimeter of equilateral triangle PQR is 15 cm, then the side of equilateral triangle PQR is 5 cm.

If the side of equilateral triangle PQR is 5 cm, then the perimeter of equilateral triangle PQR is 15 cm.

Jika perimeter segi tiga sama sisi PQR ialah 15 cm, maka sisi segi tiga sama sisi PQR ialah 5 cm.

Jika sisi segi tiga sama sisi PQR ialah 5 cm, maka perimeter segi tiga sama sisi PQR ialah 15 cm.

(c) Premise 2: 36 is an even number. Premis 2: 36 ialah nombor genap.

(d) 12 × 4 × 9 = 18 cm2

Area of triangle KLM is 18 cm2. Luas segi tiga KLM ialah 18 cm2.



12. (a) (i) Q = {10, 15, 20, 25} R = {14, 23}

(ii) (a.) P = {10, 12, 14, 16, 18, 20, 22, 24} (P R) = {14} n(P R) = 1

(b.) (Q R) = {10, 14, 15, 20, 23, 25} n(Q R) = 6

(b) (i) n(D) = 18 x + 3 + 1 + (3x – 2) = 18 4x + 2 = 18 x = 4 Number of students who like to eat

durians only Bilangan murid yang suka makan durian sahaja = 3x – 2 = 3(4) – 2 = 10

(ii) Number of students who like to eat the

three types of fruits, x = 4 Bilangan murid yang suka makan ketiga-tiga jenis

buah-buahan, x = 4 (iii) n(A) = 19 y + x + 1 + 6 = 19 y + 4 + 1 + 6 = 19 y = 8

Number of students who like to eat papayas and apples

Bilangan murid yang suka makan betik dan epal = y + x = 8 + 4 = 12

13. (a) (i) P

Y X10 m

4 mθ

tan θ = 4—–10

θ = 21°48’

(ii) P

Y Z

4 m

θ

tan 53°8’ = 4—–

YZ

YZ = 4—–———

tan 53°8ʹ YZ = 3 m

(b) (i) Number of yellow balls Bilangan bola kuning

= 1 – 1—3

– 1—2 × 48

= 8

(ii) Let the number of yellow balls added = x Biarkan bilangan bola kuning yang ditambah = x

1—5

= 8 + x—–——48 + x

48 + x = 5(8 + x) = 40 + 5x 8 = 4x x = 2

14. (a)

ST

UV

PQ

X

3 cm10 cm

P

Q

X

10 cm 80°

UQ = UX + XQ = TP + XQ = 3 + 1.7365 = 4.74 cm

(b) sin 80° = PX—–10

PX = 9.8481 cm

TU = PX = 9.85 cm

(c) Area of TUQP / Luas TUQP

= 1—2

× (3 + 4.74) × 9.85

= 38.12 cm2

15. (a) MarksMarkah

FrequencyKekerapan


10 – 14 0 12

15 – 19 3 17

20 – 24 6 22

25 – 29 8 27

30 – 34 7 32

35 – 39 4 37

40 – 44 2 42

45 – 49 0 47

cos 80° = XQ—–10

XQ = 1.7365 cm



(b) Mean mark / Min markah

= 3(17) + 6(22) + 8(27) + 7(32) + 4(37) + 2(42)

30

= 855—––30

= 28.5

(c)

0 12 17 22 27 32 37 42 47MarksMarkah

FrequencyKekerapan

1

2

3

4

5

6

7

8

(d) Number of students who passed the contest Bilangan murid yang lulus pertandingan = 7 + 4 + 2 = 13




27 24.5 6

32 29.5 21

37 34.5 40

42 39.5 62

47 44.5 76

52 49.5 84

57 54.5 88

(b) Mean age / Min umur = 6(27) + 15(32) + 19(37) + 22(42) + 14(47) + 8(52) + 4(57)

88

= 3 571—–—–

88

= 40.58



(c)

19.50

10

20

30

40

50

6064

70

80

90


Age (years)Umur (tahun)

24.5 29.5 34.5 39.5

40

44.5 49.5 54.5

(d) 88 – 64 = 24

Percentage of participants who get a hamper Peratusan peserta yang mendapat sebuah hamper

= 2488

× 100%

= 27.27%

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