JAWAPAN LENGKAP TINGKATAN 1, MODEL 1 (BAHAGIAN C)
1
26 (a) Latihan Zul untuk satu hari / Zuls’ one day training  5 4 ÷ 3 =
1
4
7
4
1
4
21
4
1
7
× 3 = 4 jam / hours
Tambahan / Additional jam / hours  + = 2 jam / hours
Memenuhi keperluan jurulatih?  Tidak, kerana latihan Zul selama 2 jam < Keperluan jurulatih
3 jam
Satisfy coach’s requirement?  No, Zul’s training of 2 hours < Coach’s required 3 hours
3
(b) Melayu / Malay  × 60 = 45
4
1
Cina / Chinese  × 60 = 12
5
India / Indian  60 − 45 − 12 = 3 pekerja / staffs
(c)
5
7
× 210 = 150 paket / packets
150 paket / packets = 20 kg gula / sugar
20
2
= 15
150
1
= 2 kg
4
1 paket / packets =
𝑥 paket / packets
𝑥=
1
4
2
15
2
=
9
4
2
15
9
=4×
15
2
kg gula / sugar
= 16.875 paket / packets
𝑥 = 16.875 paket / packets < 30 paket / packets
Hafiz tidak dapat menjual lebih daripada 30 paket gula
Hafiz did not sell more than 30 packets of sugar
27 (a) Faktor bagi 15 / Factors of 15  1, 3, 5, 15
Faktor bagi 30 / Factors of 30  1, 2, 3, 5, 6, 10, 15, 30
Faktor bagi 45 / Factors of 45  1, 3, 5, 9, 15, 45
Faktor sepunya / Common factors  1,3,5,15
(b)
FSTB / HCF  2 × 2 × 5 = 20
(c)
2
8, 12, 18
2
4, 6, 9
2
2, 3, 9
3
1, 3, 9
3
1, 1, 3
1, 1, 3
Gandaan sepunya terkecil (GSTK) / Lowest common multiple (LCM)  2 × 2 × 2 × 3 × 3 = 72
28 (a) Luas bilik ketiga / Area of third room:
= (√32 × √8) − (√6 × √24)
= (√32 × 8) − (√6 × 24)
= √256 − √144
= 16 − 12
= 4 m2
Perbezaan luas bilik ketiga dan luas bilik pertama:
Difference between area of third and area of first room:
= √32 × √8 − 4
= √32 × 8 − 4
= √256 − 4
= 16 − 4
= 12 m2
(b) = 3√8000 + 3√1000
3
= √43 × 53 + 10
3
= √203 + 10
= 20 + 10
= 30
(c) Isi padu kubus C = Isi padu kubus A + Isi padu kuboid B
Volume of cube C = Volume of cube A + Volume of cuboid B
Panjang satu sisi kubus C / Length of a side of cube C = 3√512 = 8 cm
Nisbah merujuk kepada buku yang dimiliki oleh tiga orang itu
Ali : Bala
=2:5
The ratio reflect books owned by them
Bala : Chong
= 3:1
=2×3:5×3
Padankan nisbah Bala Ali : Bala
Bala : Chong =
3×5:1×5
Match Bala’s ratio
Ali : Bala : Chong = 6
: 15
:5
29 (a)
(b) (i) Peratus / Percentage =
24
40
× 100 = 60 %
𝑥
(ii) 80 = 40 × 100
𝑥=
80
× 40
100
𝑥 = 32
Markah perlu ditambah / Points to be added  32 − 24 = 8 mata / points
(c) (i) Peratus / Percentage =
90
×
1080
100 = 8.33 %
(ii) Jumlah baharu penumpang / New total number of passengers = 1080 – 90 + 60 = 1050
Nisbah / Ratio :
60 : 1050
2 : 35
Pecahan / Fraction :
60
=
1050
=
2
35
Perpuluhan / Decimal :
60
=
1050
=
2
35
= 0.0571
Peratus / Percentage :
60
=
1050
× 100
= 5.71%
30 (a) = 24𝑥𝑦 3 ÷ 6𝑗𝑥 4 × (−2𝑗𝑦)
=
24𝑥𝑦 3
6𝑗𝑥 4
=
4𝑦 3
𝑗𝑥 3
=
−8𝑦 4
𝑥3
× (−2𝑗𝑦)
× (−2𝑗𝑦)
(b)
Jarak AD / Distance AD = 5𝑝 + 11𝑞 + 7𝑝 = (12𝑝 + 11𝑞) m
(c) (i) Perimeter, P = 2(15 − 3𝑥𝑦) + 14
Perimeter, P = 30 − 6𝑥𝑦 + 14
Perimeter, P = (44 − 6𝑥𝑦) cm
1
(ii) Luas / Area A = × Lebar / Width × Tinggi / Height
2
1
Luas / Area A = × (14) × (9𝑥𝑦 − 2)
2
Luas / Area A = 7 (9𝑥𝑦 − 2)
Luas / Area A = (63𝑥𝑦 − 14) cm2
31 (a) Andaikan / Let :
−3𝑦 − 2𝑥 = 2 Persamaan 1 / Equation 1
3𝑥 + 5𝑦 = −2 Persamaan 2 / Equation 2
Untuk persamaan 2, ungkapkan 𝑦 dalam sebutan 𝑥
For equation 2, express 𝑦 in terms of 𝑥
−3𝑦 − 2𝑥 = 2
2 + 2𝑥
𝑦=
−3
Gantikan dalam persamaan 1
Substitute into equation 1
3𝑥 + 5𝑦 = −2
5(2 + 2𝑥)
3𝑥 +
= −2
−3
10 10𝑥
3𝑥 −
−
= −2
3
3
𝑥 4
− =
3 3
𝑥 = −4
Dari persamaan 1,
From equation 1,
−3𝑦 − 2𝑥 = 2
−3𝑦 − 2(−4) = 2
−3𝑦 = 2 − 8
𝑦=2
(b) Andaikan / Let :
𝑥 = Wang Amin / Amin’s money
𝑦 = Wang Ali / Ali’s money
Jika Amin memberi RM40 kepada Ali, mereka akan mempunyai jumlah wang yang sama
If Amin gives Ali RM40, they would have the same amount of money
𝑥 − 40 = 𝑦 + 40
𝑥 = 𝑦 + 80
Jika Ali memberikan RM35 kepada Amin, Amin akan mempunyai wang dua kali ganda daripada Ali
If Ali gives Amin RM35, Amin would have twice than Ali has
2(𝑦 − 35) = 𝑥 + 35
2𝑦 − 70 = 𝑥 + 35
2𝑦 = 𝑥 + 105
Selesaikan untuk 𝑥 dan 𝑦
Solving for 𝑥 and 𝑦
2𝑦 = 𝑥 + 105
2𝑦 = (𝑦 + 80) + 105
𝑦 = 185
𝑥 = 𝑦 + 80
𝑥 = 185 + 80
𝑥 = 265
Ya. 𝑥 > 𝑦, Amin mempunyai lebih wang berbanding Ali
Yes. 𝑥 > 𝑦, Amin has more money than Ali
(c) Andaikan / Assume :
𝑥 = hijau / green
3𝑥 = kuning / yellow
𝑥 + 3𝑥 ≤ 480
4𝑥 ≤ 480
𝑥 ≤ 120
𝑥 + 3𝑥 > 240
4𝑥 > 240
𝑥 > 60
60 < 60 𝑥 ≤ 120
Bilangan minimum guli hijau ialah 61 biji.
The minimum number of green marbles is 61.