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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
CH
AP
TE
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
3 © Penerbitan Pelangi Sdn. Bhd.
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
© Penerbitan Pelangi Sdn. Bhd. 4
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)
CH
AP
TE
R 2 Quadratic Expressions and EquationsUngkapan dan Persamaan Kuadratik
5 © Penerbitan Pelangi Sdn. Bhd.
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
© Penerbitan Pelangi Sdn. Bhd. 6
Mathematics Form 4 Chapter 2 Quadratic Expressions and Equations
(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)
7 © Penerbitan Pelangi Sdn. Bhd.
Mathematics Form 4 Chapter 2 Quadratic Expressions and Equations
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)
© Penerbitan Pelangi Sdn. Bhd. 8
Mathematics Form 4 Chapter 2 Quadratic Expressions and Equations
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.
9 © Penerbitan Pelangi Sdn. Bhd.
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
CH
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TE
R 3 SetsSet
© Penerbitan Pelangi Sdn. Bhd. 10
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
11 © Penerbitan Pelangi Sdn. Bhd.
Mathematics Form 4 Chapter 3 Sets
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}
© Penerbitan Pelangi Sdn. Bhd. 12
Mathematics Form 4 Chapter 3 Sets
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
13 © Penerbitan Pelangi Sdn. Bhd.
Mathematics Form 4 Chapter 3 Sets
\ 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
© Penerbitan Pelangi Sdn. Bhd. 14
Mathematics Form 4 Chapter 3 Sets
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
15 © Penerbitan Pelangi Sdn. Bhd.
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.
© Penerbitan Pelangi Sdn. Bhd. 16
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.
17 © Penerbitan Pelangi Sdn. Bhd.
Mathematics Form 4 Chapter 4 Mathematical Reasoning
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
© Penerbitan Pelangi Sdn. Bhd. 18
Mathematics Form 4 Chapter 4 Mathematical Reasoning
(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.
19 © Penerbitan Pelangi Sdn. Bhd.
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
9. x-interceptPintasan-x
y-interceptPintasan-y
GradientKecerunan
(a)4 2
2 1– — = – — 4 2
(b)–3 4
4 4– —– = — –3 3
(c)1 –3
–3–—– = 3 1
© Penerbitan Pelangi Sdn. Bhd. 20
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
11. x-interceptPintasan-x
y-interceptPintasan-y
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
y-interceptPintasan-y
(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.
21 © Penerbitan Pelangi Sdn. Bhd.
Mathematics Form 4 Chapter 5 The Straight Line
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
© Penerbitan Pelangi Sdn. Bhd. 22
Mathematics Form 4 Chapter 5 The Straight Line
(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
\ The point of intersection is (1, 2). Titik persilangan ialah (1, 2).
(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
\ The point of intersection is (2, 3). Titik persilangan ialah (2, 3).
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.
23 © Penerbitan Pelangi Sdn. Bhd.
Mathematics Form 4 Chapter 5 The Straight Line
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
© Penerbitan Pelangi Sdn. Bhd. 24
Mathematics Form 4 Chapter 5 The Straight Line
(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).
25 © Penerbitan Pelangi Sdn. Bhd.
Mathematics Form 4 Chapter 5 The Straight Line
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
© Penerbitan Pelangi Sdn. Bhd. 26
Mathematics Form 4 Chapter 5 The Straight Line
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
x-intercept / Pintasan-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
x-intercept / Pintasan-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
27 © Penerbitan Pelangi Sdn. Bhd.
Mathematics Form 4 Chapter 5 The Straight Line
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.
© Penerbitan Pelangi Sdn. Bhd. 28
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
7. (a) Class intervalSelang kelas
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
MidpointTitik tengah
31 – 40 35.5
41 – 50 45.5
51 – 60 55.5
61 – 70 65.5
CH
AP
TE
R 6 Statistics IIIStatistik III
29 © Penerbitan Pelangi Sdn. Bhd.
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
Midpoint × FrequencyTitik tengah × Kekerapan
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
Total / Jumlah 24 392
(ii) MidpointTitik tengah
FrequencyKekerapan
Midpoint × FrequencyTitik tengah × Kekerapan
MeanMin 382= —— 24= 15.92
5.5 6 33
15.5 12 186
25.5 5 127.5
35.5 1 35.5
Total / Jumlah 24 382
(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.
© Penerbitan Pelangi Sdn. Bhd. 30
Mathematics Form 4 Chapter 6 Statistics III
10. (a) Lower boundarySempadan bawah
Upper boundarySempadan atas
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
31 © Penerbitan Pelangi Sdn. Bhd.
Mathematics Form 4 Chapter 6 Statistics III
13. (a) MidpointTitik tengah
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
© Penerbitan Pelangi Sdn. Bhd. 32
Mathematics Form 4 Chapter 6 Statistics III
16. (a) Cumulative frequency
Kekerapan longgokan
0
20
35
42
47
50
52
17. (a) Mass (kg)Jisim (kg)
FrequencyKekerapan
Upper boundarySempadan atas
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
Cumulative frequencyKekerapan longgokan
33 © Penerbitan Pelangi Sdn. Bhd.
Mathematics Form 4 Chapter 6 Statistics III
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
© Penerbitan Pelangi Sdn. Bhd. 34
Mathematics Form 4 Chapter 6 Statistics III
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
MidpointTitik tengah
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
Frequency / Kekerapan
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
35 © Penerbitan Pelangi Sdn. Bhd.
Mathematics Form 4 Chapter 6 Statistics III
2. (a) FrequencyKekerapan
MidpointTitik tengah
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
Frequency / Kekerapan
Marks / Markah
(e) 3 + 2 + 1 = 6 students / murid
3. (a) (i)MidpointTitik tengah
Upper boundary
Sempadan atas
Cumulative frequencyKekerapan longgokan
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
4. (a) Class intervalSelang kelas
FrequencyKekerapan
MidpointTitik tengah
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
Frequency / Kekerapan
31.5
36.5
41.5
(e) 5 + 2 = 7 days / hari
© Penerbitan Pelangi Sdn. Bhd. 36
Mathematics Form 4 Chapter 6 Statistics III
5. (a) MidpointTitik tengah
Upper boundary
Sempadan atas
Cumulative frequency
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
Cumulative frequencyKekerapan longgokan
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%
6. (a) Mass (kg)Jisim (kg)
FrequencyKekerapan
MidpointTitik tengah
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
Frequency / Kekerapan
Mass (kg) / Jisim (kg)
5
6
7
78
(e) Modal class / Kelas mod = 56 – 60 kg
7. (a) PointsMata
MidpointTitik tengah
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
37 © Penerbitan Pelangi Sdn. Bhd.
Mathematics Form 4 Chapter 6 Statistics III
(c)
0
1
2
3
4
5
6
7
8
Frequency / Kekerapan
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
Cumulative frequency
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
© Penerbitan Pelangi Sdn. Bhd. 38
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
39 © Penerbitan Pelangi Sdn. Bhd.
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
© Penerbitan Pelangi Sdn. Bhd. 40
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
41 © Penerbitan Pelangi Sdn. Bhd.
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
© Penerbitan Pelangi Sdn. Bhd. 42
Mathematics Form 4 Chapter 8 Circles III
10. (a)
4 common tangents 4 tangen sepunya
(b)
3 common tangents 3 tangen sepunya
(c)
1 common tangents 1 tangen sepunya
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
43 © Penerbitan Pelangi Sdn. Bhd.
Mathematics Form 4 Chapter 8 Circles III
(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
© Penerbitan Pelangi Sdn. Bhd. 44
Mathematics Form 4 Chapter 8 Circles III
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
45 © Penerbitan Pelangi Sdn. Bhd.
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
© Penerbitan Pelangi Sdn. Bhd. 46
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
47 © Penerbitan Pelangi Sdn. Bhd.
Mathematics Form 4 Chapter 9 Trigonometry II
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
© Penerbitan Pelangi Sdn. Bhd. 48
Mathematics Form 4 Chapter 9 Trigonometry II
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
49 © Penerbitan Pelangi Sdn. Bhd.
Mathematics Form 4 Chapter 9 Trigonometry II
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
© Penerbitan Pelangi Sdn. Bhd. 50
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
51 © Penerbitan Pelangi Sdn. Bhd.
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
© Penerbitan Pelangi Sdn. Bhd. 52
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
53 © Penerbitan Pelangi Sdn. Bhd.
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
© Penerbitan Pelangi Sdn. Bhd. 54
Mathematics Form 4 Chapter 11 Lines and Planes in 3-Dimensions
(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
55 © Penerbitan Pelangi Sdn. Bhd.
Mathematics Form 4 Chapter 11 Lines and Planes in 3-Dimensions
(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°
© Penerbitan Pelangi Sdn. Bhd. 56
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
57 © Penerbitan Pelangi Sdn. Bhd.
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
© Penerbitan Pelangi Sdn. Bhd. 58
Mathematics Form 4 End-of-Year Assessment
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
59 © Penerbitan Pelangi Sdn. Bhd.
Mathematics Form 4 End-of-Year Assessment
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
© Penerbitan Pelangi Sdn. Bhd. 60
Mathematics Form 4 End-of-Year Assessment
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.
61 © Penerbitan Pelangi Sdn. Bhd.
Mathematics Form 4 End-of-Year Assessment
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
MidpointTitik tengah
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
© Penerbitan Pelangi Sdn. Bhd. 62
Mathematics Form 4 End-of-Year Assessment
(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
16. (a) MidpointTitik tengah
Upper boundarySempadan atas
Cumulative frequencyKekerapan longgokan
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
63 © Penerbitan Pelangi Sdn. Bhd.
Mathematics Form 4 End-of-Year Assessment
(c)
19.50
10
20
30
40
50
6064
70
80
90
Cumulative frequencyKekerapan longgokan
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%