Page 27 - Hybrid PBD 2022 Tg 4 - Matematik

Matematik Tingkatan 4 Bab 2 Asas Nombor
12. Hitung nilai bagi setiap yang berikut. SP 2.1.3 TP3
Calculate the value of each of the following. Video

(a) 6134 7 + 2163 7 = 11330 7
222 3 + 121 3 = 1120 3
1 1
1 1
6 1 3 4 7
2 2 2 3
+ 1 2 1 3 + 2 1 6 3 7
1 1 3 3 0 7
1 1 2 0 3
(c) 33342 5 + 1043 5 = 34440 5
(b) 1732 8 + 671 8 = 2623 8

1 1 1 1
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1 7 3 2 8 3 3 3 4 2 5
+ 6 7 1 8 + 1 0 4 3 5
2 6 2 3 8 3 4 4 4 0 5

13. Hitung nilai bagi setiap yang berikut. SP 2.1.3 TP3
Calculate the value of each of the following.

(a) 101101 2 + 1011 2 = 111000 2
312 4 + 123 4 = 1101 4
4 8 1 → 2 5 6
→ 4 2 0 – 1 101101 2 4 5 10
312 4 5 4 10 2 2 8 – 0
4 5 – 0 1011 2 → + 1 1 10
123 4 → + 2 7 10 2 1 4 – 0
4 1 – 1 5 6 10
8 1 10 2 7 – 0
0 – 1 2 3 – 1
Tukar setiap nombor kepada asas 10, lakukan pengiraan dan 2 1 – 1
tukar semula hasil itu kepada asas tersebut.
Convert each number to base 10, do the calculation and convert the 0 – 1
result back to the base.
(c) 2736 9 + 548 9 = 3385 9
(b) 1205 6 + 244 6 = 1453 6

→ 6 3 9 3 → 9 2 5 0 7
1205 6 2 9 3 10 2736 9 2 0 5 8 10
6 6 5 – 3 → + 9 2 7 8 – 5
244 6 → + 1 0 0 10 548 9 4 4 9 10
6 1 0 – 5 9 3 0 – 8
3 9 3 10 2 5 0 7 10
6 1 – 4 9 3 – 3
0 – 1 0 – 3

14. Hitung nilai bagi yang berikut. SP 2.1.3 TP3
Calculate the value of the following.

(a) 6561 7 − 2606 7 = 3652 7
313 5 – 244 5 = 14 5
5 5 12 5 8
2 0 8
6 5 6 1 7
3 1 3 5
– 2 6 0 6 7
– 2 4 4 5 3 6 5 2 7

1 4 5

(b) 110110 2 − 11101 2 = 11001 2 (c) 2313 4 − 1332 4 = 321 4
2 6
0 2 0 2 1 2 5
1 1 0 1 1 0 2 2 3 1 3 4
– 1 1 1 0 1 2 – 1 3 3 2 4
1 1 0 0 1 2 3 2 1 4

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