Tg4_praktis_2

2BAB Modul
Asas Nombor PBD
Number Bases

2.1 Asas Nombor/ Number Bases

1 Nyatakan tiga nombor dalam asas yang sama selepas nombor yang diberikan.
State three numbers in the same base after the given number. SP 2.1.1 TP 1

CONTOH (a) 324 (b) 4567
1002 = 334, 1004, 1014 = 4607, 4617, 4627

= 1012, 1102, 1112

(c) 1102 (d) 1223 (e) 314
= 1112, 10002, 10012 = 2003, 2013, 2023 = 324, 334, 1004

(f) 4225 (g) 1367 (h) 879
= 4235, 4245, 4305 = 1407, 1417, 1427 = 889, 1009, 1019

2 Nyatakan nilai bagi digit yang dibulatkan kepada nombor dalam asas sepuluh.
State the value of the digit that is circled to the number in the base ten. SP 2.1.1 TP 1

CONTOH (a) 20213 = (b) 4156 =
1101002 =
33 32 31 30 62 61 60
25 24 23 22 21 20 2021 415
110100
Nilai bagi digit 2/ Nilai bagi digit 5/
Nilai bagi digit 1/ Value of digit 2 = 2 × 33 Value of digit 5 = 5 × 60
Value of digit 1 = 1 × 22
= 54 =5
=4

(c) 1012 = (d) 10213 = (e) 5236 =

22 21 20 33 32 31 30 62 61 60
101 1021 523
Nilai bagi digit 1/
Nilai bagi digit 1/ Value of digit 1 = 1 × 33 Nilai bagi digit 2/
Value of digit 1 = 1 × 20 Value of digit 2= 2 × 61
= 27
=1 = 12

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Bab 2

3 Cerakinkan setiap nombor berikut mengikut nilai tempat digit-digitnya.
Write each of the following numbers according to the value of the digits. SP 2.1.1 TP 1

CONTOH (a) 1112 = 1 × 22 + 1 × 21 + 1 × 20
100112 = 1 × 24 + 0 × 23 + 0 × 22 + 1 × 21
+ 1 × 20 Nilai tempat 12 11 10
Place value = 111
Nilai tempat 24 23 22 21 20
Place value = 10011

(b) 3314 = 3 × 42 + 3 × 41 + 1 × 40 (c) 3526 = 3 × 62 + 5 × 61 + 2 × 60

Nilai tempat 42 41 40 Nilai tempat 62 61 60
Place value = 331 Place value = 352

4 Tukarkan setiap asas nombor kepada nombor asas sepuluh.
Convert each number given to a number in base ten. SP 2.1.2 TP 1

Tip Kalkulator

Boleh digunakan untuk menukar asas 2 dan 8 kepada asas 10.
Can be used to change basic 2 and 8 to base 10.

CONTOH (a) 10112 =1 × 23 + 0 × 22 + 1 × 21 + 1× 20
=8 + 0+2+1
110012 = 1×24 + 1 × 23 + 0 × 22 + 0 × 21 +
1× 20 = 1110

= 16 + 8 + 0 + 0 + 1

= 2510

(b) 2203 = 2 × 32 + 2 × 31 + 0× 30 (c) 30314 = 3 × 43 + 0× 32 + 3 × 31 + 1× 30
= 18 + 6 + 0 = 192 + 0 +9 +1

= 2410 = 20210

(d) 21234 = 2 × 43 + 1 × 42 + 2 × 41 + 3× 40 (e) 435 = 4 × 51 + 3× 50
= 128 + 16 + 8 + 3 = 20 + 3

= 15510 = 2310

5 Tukarkan setiap nombor dalam asas sepuluh kepada sebarang nombor asas.
Convert each number of base ten to the mention bases. SP 2.1.2 TP 1

CONTOH = 100012 (a) 1910 = ______2
1710 = ______2
2 19
2 17 2 9 --- 1
2 8 --- 1 2 4 --- 1
2 4 --- 0 2 2 --- 0
2 2 --- 0 2 1 --- 0
2 1 --- 0
0 --- 1
0 --- 1
= 00112

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Bab 2

(b) 2410 =______3 (c) 20910 = ______4

3 24 4 209
3 8 --- 0 4 52 --- 1
3 2 --- 2 4 13 --- 0
4 3 --- 1
0 --- 2
= 2203 0 --- 3
= 31014

6 Cari nilai p dalam persamaan yang berikut.
Find the value of p in the following equation. SP 2.1.2 TP 1

CONTOH (a) p436 = 11000112

23p45 = 4229 ×112000) 0=19192 1=0 (1 × 26) + (1 × 25) + (1 × 21) + (1

4229 = (4 × 92) + (2 × 91) + (2 × 90) = 34410

Menukar asas 10 Bandingkan Menukar asas 10 Bandingkan
kepada asas 6 jawapan
kepada asas 5 jawapan Convert base 10 to Compare the answer
base 6 =
Convert base 10 to base Compare the = 2436 dan/and p436
6 99
5 = answer

2 344 =an2d32334p54d5an/ 6 16 --- 3 Maka/Thus, p = 2
2 68 --- 4
2 13 --- 3 Maka/ Thus, 6 2 --- 4
2 2 --- 3 p=3
0 --- 2
0 --- 2
= 2436
= 23345

(b) 7758 = 40p45 (c) 11011102 = 2p57
7758 = (7 × 82) + (7 × 81) + (5 × 80) = 50910
×112021) 1+1(012 = (1 × 26) + (1 × 25) + (1 × 23) + (1
× 21) = 11010

Menukar asas 10 Bandingkan Menukar asas 10 Bandingkan
kepada asas 5 jawapan/ Compare kepada asas 7 jawapan/ Compare
Convert base 10 to the answer Convert base 10 to the answer
base 5 base 7
=404p041545 dan/and = 2157 dan/and 2p57
5 509 7 110
5 101 --- 4 Maka/Thus, p = 1 7 15 --- 5 Maka/Thus, p = 1
5 20 --- 1 7 2 --- 1
5 4 --- 0
0 --- 2
0 --- 4
= 2157
= 40145

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Bab 2

7 Selesaikan setiap yang berikut.
Solve each of the following. SP 2.1.2 TP 1

(a) Tunjukkan 3(74) + 72 + 7(4) sebagai nombor dalam asas enam.
Show 3(74) + 72 + 7(4) as number in base six.

3(74) + 72 + 7(4) = (3 × 74) + 72 + (4 × 7) Menukar asas 10 kepada asas 6/
= (3 × 74) + 72 + (4 × 71) Convert base 10 to base 6
= 728010
6 7280

6 1213 --- 2

6 202 --- 1

6 33 --- 4

6 5 --- 3

0 --- 5

= 534126

(b) Diberi bahawa 647 = x5 = y3, cari
G(iiv)enniltahiatb6a4g7i = x/5v=alyu3e, find
x, for x,

664477 = (x65 × Menukar asas 10 kepada asas 5/ Bandingkan/Compare
= 4610 Convert base 10 to base 5
= 71 + 4 × 70 ) 141x5 = x1541
5 46 =

5 9 --- 1

5 1 --- 4

0 --- 1

= 1415

(ii) nilai bagi y. / Value for y.

664477 = y(63 Menukar asas 10 kepada asas 3/ Bandingkan/ Compare
= Convert base 10 to base 3
× 71 + 4 × 70) 1201y3 = y3
3 46 = 1201
= 4610 3 15 --- 1
3 5 --- 0
3 1 --- 2

0 --- 1

= 12013

8 Cari penambahan nombor-nombor dalam asas dua yang berikut.
Find the addition of the base two numbers in the following. SP 2.1.3 TP 1

CONTOH (a) 1012 + 112 =

11012 + 10112 = 11111012 12 + 12 = 102 1101 12
+ 10112 + 112
12 + 12 + 12 = 112
110002 10002

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Bab 2

(b) 110102 + 101012 = (c) 1011112 + 110112 =

110102 11 01 11 11 11 12
+ 101012 + 110112

1011112 10010102

9 Cari penolakan nombor-nombor dalam asas dua yang berikut.
Find the subtraction of the numbers in the following two bases. SP 2.1.3 TP 1

CONTOH 1100 112 12 – 12 = 02 (a) 1012 – 112 =
10112 – 1012 = – 1012
1012
102 – 12 = 12 1102 – 112

102

(b) 11112 – 11012 = (c) 100112 – 1112 =

11112 11 1010112
– 11012 – 1112

102 11002

10 Hitung hasil tambah nombor-nombor berikut dalam asas yang sama.
Calculate the sum of the following numbers in the same base. SP 2.1.3 TP 1

CONTOH (a) 3214 + 2014 =

121202122333 + (2101×33=2) + (2 × 31) + (2 × 30) = 11791100 23021144 = (3 × 42) + (2 × 41) + (1 × 40) = 35371100
= (2 × 32) + (0 × 31) + (1 × 30) = = (2 × 42) + (0 × 41) + (1 × 40) =
=
Menukar asas 10
1 Menukar asas 10 51 710 kepada asas 4/
kepada asas 3/ + 3310 Convert base 10 to
1710 Convert base 10 to base 4
+ 1910 base 3 9010
4 90
3610 3 36 4 22 --- 2
3 12 --- 0 4 5 --- 2
3 4 --- 0 4 1 --- 1

1 --- 1 0 --- 1
0 --- 1
= 11224
= 11003

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Bab 2

(b) 40325 + 2105 = (c) 32125 + 14215 =
40325 = (4 × 53) + (3 × 51) + (2 × 50) = 51710
2105 = (2 × 52) + (1 × 51) = 5510 32125 = (3 × 53) + (2 × 52) + (1 × 51) + (2 × 50)
= 4×35213)0
Menukar asas 10 14215 = (1 + (4 × 52) + (2 × 51) + (1 × 50)
kepada asas 5/
Convert base 10 to = 23610
base 5
51710 Menukar asas 10
+ 5510 5 572 kepada asas 5/
5 114 --- 2 Convert base 10 to
57210 5 22 --- 4 base 5
5 4 --- 2
43210 5 668
0 --- 4 + 23610 5 133 --- 3
5 26 --- 3
66810 5 5 --- 1
5 1 --- 0
= 42425
0 --- 1
= 101335

11 Cari hasil tolak nombor-nombor berikut.
Find the subtraction of the following numbers. SP 2.1.3 TP 1

CONTOH (a) 3214 – 2014 =

33220044 – 2(311×4 = + (2 × 41) + (0 × 40) =5610 23021144 = (3 × 42) + (2 × 41) + (1 × 40) = 5710
= 42) = (2 × 42) + (1 × 40) = 3310

2114 = (2 × 42) + (1 × 41) + (1 × 40) = 3710 Menukar asas 10
kepada asas 4/
41 Menukar asas 10 5710 Convert base 10 to
kepada asas 4/ – 3310 base 4
5610 Convert base 10 to
– 3710 base 4 2410 4 24
4 6 --- 0
1910 4 19 4 1 --- 2

4 4 --- 3 0 --- 1

4 1 --- 0

0 --- 1 = 1204

= 1034

(b) 40325 – 2105 = (c) 32125 – 14215 =

241003525==(2(4××525)3)++(1(3××515)1=) +55(210× 50) = 51710 32125 = (3 × 53) + (2 × 52) + (1 × 51) + (2 × 50)
14215 = + (4 × 52) + (2 × 51) + (1 × 50)
Menukar asas 10 = 4×352130)
(1

kepada asas 5/ = 23610

Convert base 10 to Menukar asas 10

base 5 kepada asas 5/ Convert

51710 5 462 base 10 to base 5
– 5510 5 92 --- 2
5 18 --- 2 43210 5 196
46210 – 23610 5 37 --- 1
5 7 --- 4
5 3 --- 3 19610 5 1 --- 2

0 --- 3 0 --- 1

= 33225 = 12415

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Bab 2

12 Hitung hasil tambah nombor-nombor berikut dalam asas nombor yang dinyatakan.
Calculate the sum of the following numbers in the base given. SP 2.1.3 TP 1

CONTOH (a) 234 + 467 = _______ 5

4526 + 2378 = ______7 426374 = (2 × 41) + (3 × 40) = 13411100
= (4 × 71) + (6 × 70) =
24352768 = (4 × 62) + (5 × 61) + (2 × 60) = 1175961100
= (2 × 82) + (3 × 81) + (7 × 80) = Menukar asas 10
kepada asas 5/ Convert
11 Menukar asas 10 1110 base 10 to base 5
kepada asas 7/ + 3410
17610 Convert base 10 to 5 45
+ 15910 base 7 4510 5 9 --- 0
5 1 --- 4
33510 7 335 5 0 --- 1

7 47 --- 6 = 1405

7 6 --- 5

0 --- 6

= 6567

(b) 425 + 678 = _______ 6 (c) 1046 + 6839 = _______ 8

467258 = (4 × 51) + (2 × 50) = 52521100 61084396 = (1 × 62) + (4 × 60) = (430×10 90) = 56110
= (6 × 81) + (7 × 80) = = (6 × 92) + (8 × 91) +

Menukar asas 10 Menukar asas 10
kepada asas 6/ Convert kepada asas 8/ Convert
base 10 to base 6 base 10 to base 8

2210 6 77 4010 8 601
+ 5510 6 12 --- 5 + 56110 8 75 --- 1
6 2 --- 0 8 9 --- 3
7710 60110 8 1 --- 1
0 --- 2
0 --- 1
= 2056 = 113118

13 Hitung hasil tolak nombor-nombor berikut dalam asas nombor yang dinyatakan.
Calculate the subtraction of the following numbers in the base given. SP 2.1.3 TP 1

CONTOH (a) 1207 – 234 = _______ 5
1207 = (1 × 72) + (2 × 70) = 5110
7689 – 768 = _______6 234 = (2 × 41) + (3 × 40) = 1110
7766889==(7(7××819)2+) +(6(6××809)1)=+6(2810× 90) = 62910
Menukar asas 10
62910 Menukar asas 10 5110 kepada asas 5/ Convert
– 6210 kepada asas 6/ – 1110 base 10 to base 5
Convert base 10 to
56710 base 6 4010 5 40
5 8 --- 0
6 567 5 1 --- 3

6 94 --- 3 0 --- 1

6 15 --- 4 = 1305

6 2 --- 3

0 --- 2
= 23436

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Bab 2

(b) 436 – 179 = _______ 3 (c) 1059 – 11102 = _______ 5
436 = (4 × 61) + (3 × 60) = 2710 1059 = (1 × 92) + (5 × 90) = 8610
179 =(1 × 91) + (7 × 90) = 1610 11102 = (1 × 23) + (1 × 22) + (1 × 21) = 1410

2710 Menukar asas 10 8610 Menukar asas 10
– 1610 kepada asas 3/ Convert – 1410 kepada asas 5/ Convert
base 10 to base 3 base 10 to base 5
1110 7210
3 11 5 72
3 3 --- 2 5 14 --- 2
3 1 --- 0 5 2 --- 4

0 --- 1 0 --- 2

= 1023 = 2425

(d) 10325 – 589 = _______ 7 (e) 32106 – 2227 = _______ 8

150839 2=5 = (1 × 53) + (3 × 51) + (2 × 50) = 14210 232221706==(2(3××726)3)++(2(2××716)2)++(2(1××706)1=) =1174261010
(5 × 91) + (8 × 90) = 5310
Menukar asas 10
Menukar asas 10 kepada asas 8/ Convert
kepada asas 7/ Convert base 10 to base 8
base 10 to base 7
14210 72610 8 612
– 5310 7 89 – 11410 8 76 --- 4
7 12 --- 5 8 9 --- 4
8910 7 1 --- 5 61210 8 1 --- 1

0 --- 1 0 --- 1

= 1557 = 11448

14 Cari nilai bagi x bagi setiap soalan yang berikut.
Find the value of x for each of the following questions. SP 2.1.3 TP 1

CONTOH (a) 324 + x8 = 667
(a) x3 + 234 = 467
x8 = 667 – 324

x3 = 467 – 234 667 = (6 × 71) + Menukar asas 10
324 = (6 × 70) = kepada asas 8/ Convert
467 = (4 × 71) + Menukar asas 10 (3 × 41) + 4810 base 10 to base 8
kepada asas 3/
234 = ((26××417)0+) =(33×44100) Convert base 10 to (2 × 40) = 1410 8 34
= 1110 base 3 8 4 --- 2
4810
3410 3 23 – 1410 0 --- 4
– 1110 3 7 --- 2
3 2 --- 1 3410 = 428
2310
0 --- 2 Maka/thus, x = 42

Maka/thus, x= 212 = 2123

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Bab 2

(b) x5 – 237 = 346 (c) 789 – x4 = 278

x5 = 346 + 237 789 – 278 = x4

346 = (3 × 61) + 2210 Menukar asas 10 789 = (7 × 91) + 7110 Menukar asas 10
237 = (4 × 60) = 1710 kepada asas 5/ Convert 278 = (8 × 90) = 2310 kepada asas 4/ Convert
(2 × 71) + base 10 to base 5 (2 × 81) + base 10 to base 4
(3 × 70) = (7 × 80) =
5 39 4 48
2210 7110
+ 1710 5 7 --- 4 – 2310 4 12 --- 0

3910 5 1 --- 2 4810 4 3 --- 0

Maka/thus, x = 124 0 --- 1 Maka/thus, x = 300 0 --- 3
= 1245 = 3004

15 Selesaikan masalah yang berikut.
Solve the following problems. SP 2.1.4 TP 1

(a) Imswanainkdiitudiimalianhtamoalenhikibbuirnuy.aBuenratupkakmahembiblaenlig2a2n2m23abniijki manik biru dan manik putih. 40% daripada
putih yang di minta oleh ibu Iswandi?

Nyatakan jawapan dalam nombor asas lapan.

Iswandi was asked by his mother to buy a total of 22223 blue beads and white beads. 40% of the beads
are blue beads. What is the number of white beads requested by Iswandi’s mother? State the answer in

the number of base eight.

Maklumat /Information Penyelesaian / Solution Menukar asas 10
kepada asas 8/
Manik Biru (B) + Manik Putih 2TCu2ok2nav2re3kra=t n2(222×22232332)t3+ok(ne2up×ma3db2ae)+rnb(o2am×seb3o11)r0+a(s2a×s 3100) Convert base 10 to
= 8010
(P) = 22223 + White Beads base 8
Blue Beads (B) %Manik Putih /White beads
= 100% - 40% = 60% 8 48
M(Pa) n=ik22B2ir2u3(B) = 40% 8 6 --- 0
Bilangan manik Putih / Number of White
daripada 22223 beads 0 --- 6
Blue Beads (B) = 40% of 22223 = 60% × 8010 = 4810
Bilangan manik putih dalam asas lapan = 608
Soalan /Question / Number of White beads in base eight =
608 biji / seeds
Bilangan manik Putih = ?

Number of White beads = ?

Jawapan dalam nombor asas 8

Answer in number of base 8

(b) Cikgu Shanti bercadang hendak menghadiahkan tiga buah kalkulator kepada murid yang tidak

berkemampuan. Harga mbaegmi sbeabyuaarnhykaadlkeunlgaatonrdiualaahkeRpMing10w0a1n4g. Beliau telah membeli tiga buah
kalkulator tersebut dan kertas RM 12123. Berapakah baki

wangnya dalam nombor asas lima?

Teacher Shanti intends to present three calculators to poor students. The price for a calculator is RM

10014. She bought the three calculators and paid by two pieces of RM 12123 notes. How much is the
balance in the number of base five?

Maklumat /Information Penyelesaian / Solution Menukar asas 10
kepada asas 5/
Hadiahkan tiga buah kalkulator Tnuokmabrkoarnas1a0s01104 dan 12123 kepada Convert base 10 to
Give three calculators
Harga sebuah kalkulator = RM10014 Convert 10014 dan 12123 to number base 5
The price of a calculator = RM10014 base 10
Bayar dengan dua keping RM12123 5 35
Pay with two pieces of RM12123 1120102134 = (1 × 43) + (1 × 40) = 6510 5 7 --- 0
= (1 × 33) + (2 × 32) + 5010 5 1 --- 2
Soalan /Question (1 × 31) + (2 × 30) =
Baki wang / Remaining money = ? 0 --- 1
Jawapan dalam nombor asas 5 Baki / Balance = 2
Answer in base number 5 = 1205
Baki dalam = l1im00a10/ – 6510 = 3510
asas Balance in

base five = RM1205

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Praktis Pentaksiran 2

Kertas 1

1 Tentukan nilai bagi digit 4, dalam asas 9 B3A33124 00+23200144 4 = ______4

sepuluh, bagi nombor 35426. 4, in the base ten, C 3333330344
Determine the value of the digit D

of the number 35426.
A4 C 40 10 A478 7+0638 = ______8 C 132

B 24 D 42

B 100 D 1000

2 Nilai bagi digit 8 untuk nombor 784939 ialah
8 × 9m. Nyatakan nilai bagi m.
11 1A101171212 + 4325 = _____C_4 549
The value of digit 8 for number 784939 is 8 × 9m.
State the value of m. B 487 D 2230

A1 C3

B2 D4 12 1A3536 3– 289 = ______10 C 357

3 Nnoymatbaokar nas1a1s150.1102 sebagai asas dalam B 223 D 107

State 11101102 as the base in number in base 5. 13 Jika k10 = 5407, maka k =
A 314 C 1312
If k105=4 5407, then k =
B 433 D 226 A
C 273

4 Jika 46n07 = 21000003, cari nilai bagi n. B 400 D 540

If 462n07 = 21000003, find the value of n. 14 Diberi h4 + 345 = 768, cari nilai bagi h.
A C 5
Given h4 + 345 = 768, find the value of h.
B3 D6 A 42 C 162

B 53 D 223

5 Ungkapkan 2(64) + 24 + 6 sebagai satu 15 Cjaawriapnailnaidbaalagmi 3a5s6a+s t1ig0a112 dan tunjukkan

nombor dalam asas tujuh ialah

Express 2(64) + 24 + 6 as a number in the base Find the value of 356 + 10112 and show the
answer in base three
seven is

A 3524 C 7044 A 34 C 1000

B 5066 D 10423 B 36 D 1021

6 1101112 + 10112 = 16 UPSR ialah suatu nombor dalam asas empat.

A 11010011111022 C 1100000101100022 Jika UPSR = 7510, cari nilai bagi U.
B D UPSR is a number in base four. If UPSR = 7510,
find the value of U.
7 1AB10110100001102 00–22101112=
C 111111101022 A0 C2
D
B1 D3

8 A1B10111102 10–001021202 + 10012 = 17 Diberi t3 = 35 + 34 + 6 + 2, cari nilai bagi t.
C Given t3 = 35 + 34 + 6 + 2, find the value of t.
D 110000100122 A 110202 C 101122

B 110022 D 101122

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Kertas 2

1 Diberi bahawa 20020013 = 2 × 3h + 2 × 3k + 3l, cari nilai bagi
Given that 20020013 = 2 × 3h + 2 × 3k + 3l, find the value of

2M0a0k2a0/0T1h3e=n 2 × 36 + 2 × 33 + 1 × 30 =2× 3h + 2 × 3k + 1 × 3l
: h = 6, k = 3 dan / and l=0

(a) h – k

h–k=6–3=3

(b) h − l
k

h − l = 6 − 0 = 2
k 3

2 Nyatakan lima digit nombor pertama dalam
State the first four numbers in

(a) asas empat / base four
14, 24, 34 dan/and 104

(b) asas tiga / base three
13, 23, 103 dan/and 113

3 (a) Nyatakan nilai bagi digit 4, dalam asas sepuluh, untuk nombor 664789.
State the value for digit 4, in base ten, for number 664789.

94 93 92 91 90
66478

Nilai bagi digit 4 / Value for digit 4 = 4 × 92 = 324 Menukar asas 10 kepada asas 5
(b) Nyatakan nilai bagi 5278 dalam asas lima. Convert base 10 to base 5

State the value for 5278 in base five. 5 343
5 68 --- 3
5278 = (5 × 82) + (2 × 81) + (7 × 80) = 34310 5 13 --- 3
5 2 --- 3
4 Ungkapkan 3(44) + 2(42) + 4(3) + 1 sebagai satu nombor dalam
Express 3(44) + 2(42) + 4(3) + 1 as a number in 0 --- 2
= 23335

3(44) + 2(42) + 4(3) + 1 = 3 × 44 + 2 × 42 + 3 × 41 + 1 × 40 = 81310

(a) asas lima / base five (b) asas tujuh / base seven

Menukar asas 10 kepada asas 5 Menukar asas 10 kepada asas 7
Convert base 10 to base 5 Convert base 10 to base 7

5 813 7 813
5 162 --- 3 7 116 --- 1
5 32 --- 2 7 16 --- 4
5 6 --- 2 7 2 --- 2
5 1 --- 1
0 --- 2
0 --- 1 = 22417
= 112235

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5 Diberi bahawa 546 = m3 = n8, cari (b) nilai bagi n / value of n
Given that 546 = m3 = n8, find
546 = 5 × 61 + 4 × 60 = 3410 Menukar asas 10 kepada asas 8
Convert base 10 to base 8
(a) nilai m / value of m
8 34
Menukar asas 10 kepada asas 3 8 4 --- 2
Convert base 10 to base 3
0 --- 4
3 34
3 11 --- 1 = 428
3 3 --- 2 Maka / Then n = 42
3 1 --- 0

0 --- 1

= 10213
Maka / Then m = 1021

6 Cari nilai bagi 578 + 1100112 dan tunjukkan jawapan dalam asas lima.
Find the value of 578 + 1100112 and show the answer in base five.

J1517080=11(52 × 81) + (7 × (810)×=244)7+10 (1 × 21) + (1 × 20) = 5110 Menukar asas 10 kepada asas 5
= (1 × 25) + Convert base 10 to base 5

4710 5 98
+ 5110 5 19 --- 3
5 3 --- 4
9810
0 --- 3

= 3435

7 Cari nilai bagi 345 + 667 dan tunjukkan jawapan dalam asas yang berikut.
Find the value of 345 + 667 and show the answers in the following bases

364657 = (3 × 51) + (4 × 50) = 14981100 1910
= (6 × 71) + (6 × 70) = + 4810

(a) asas tiga / base three, 6710

(b) asas lapan / base eight.

Menukar asas 10 kepada asas 3 Menukar asas 10 kepada asas 8
Convert base 10 to base 3 Convert base 10 to base 8

3 67 8 67
3 22 --- 1 8 8 --- 3
3 7 --- 1 1 1 --- 0
3 2 --- 1
0 --- 1
0 --- 2
= 1038
= 21113

© Oxford Fajar Sdn. Bhd. (008974-T) 2019 26

8 Carikan nilai bagi y, dalam asas sepuluh, jika 647 + y4 = 889.
Find the value of y, in base ten, if 647 + y4 = 889.
y4 = 889 – 647

864897 = (8 × 91) + (8 × 90) = 84061100 Menukar asas 10 kepada asas 4
= (6 × 71) + (4 × 70) = Convert base 10 to base 4

8010 4 34
– 4610 4 8 --- 2
4 2 --- 0
3410
0 --- 2
Maka/then y= 202
= 2024

9 Diberi bahawa p adalah 8 bagi penyelesaian 2218 – 11p9 = 657, buktikan.
Given that p is 8 for solution 2218 – 11p9 = 657, prove.

11p9 = 2218 – 657

262517 8==(6(2××718)2+) +(5(2××708)1=) +47(110× 80) = 14510 Menukar asas 10 kepada asas 9
Convert base 10 to base 9

14510 9 98
– 4710 9 10 --- 8
9 1 --- 1
9810
0 --- 1
Maka/then p = 8
= 1189

10 Jadual di bawah menunjukkan taburan asegkruoumppoufla1n43174s3tu7doernatsngatmtenudriidngyaonneg menghadiri satu kursus.
The table below shows the distribution of course.

Murid Tingkatan 2 Tingkatan 3
Students Form 2 Form 3

Lelaki 1203 1134
Boys
337 100102
Perempuan
Girls

Seorang pelajar dipilih secara rawak dari kumpulan itu telah dipanggil ke hadapan untuk memberi
pendapat. Apakah kebarangkalian seorang pelajar lelaki dari tingkatan 3 akan dipilih dan berikan
jawapan dalam nombor asas sepuluh?

A student was randomly selected from the group and called forward to give an opinion. What is the probability

that a boys student from Form 3 will be selected and give the answer in a number of base ten.

Tukarkan kepada nombor asas sepuluh / Convert to number in base ten

1437 = 8010 1134 = 2310
1203 = 1510 100102 = 1810
337 = 2410

Kebarangkalian seorang pelajar lelaki tingkatan 3 dipilih

The probability of a Form 3 boy student is selected

= 2310
8010

© Oxford Fajar Sdn. Bhd. (008974-T) 2019 27

Genius

jBueszoarhenardgaanan1t0a4rage3l5asgealiarssijruaspoiraelnahdaRnM224 1g6e.laBseraiikrasnirahparigaala, hdaRlaMm1R2M6. Mdaalnaamkaalsaabs eszeapuhlaurhg,abbaaggiis1at0u5 gelas
gelas

air sirap dan segelas jus oren.

The price difference between 35 glass of orange juice and 24 glass of syrup water is RM 126. While the difference
price between 105 orange juice glasses and 104 glass of syrup water is RM 216. Give the price, in RM in the base
ten, for a glass of syrup water and a glass of orange juice.

Tukarkan kepada nombor asas sepuluh / Convert to number in base ten

35 = 310 105 = 510
24 = 210 104 = 410
126 = 810 216 = 1310

Andaikan jus oren = x dan air sirap = y.
Assume orange juice = x and syrup water = y.

Bina dua persamaan dan selesaikan dengan menggunakan persamaan serentak.
Form two equations and solve by using simultaneous equations.

351100xx – 241100yy = 810 -------(1) x31=0(933111000)–→–2–2122(011y11000)yyyy=====88101801.110500 1–0 910
– = 1310 -------(2)

(1) × 2
610x – 410y = 1610 -------(3)

(3) – (2)

x = 310 TMhaekna, ,11glgaeslsaos fjuosraonrgeenju=icReM=3R10Md3an10 aainrdsisryarpup=wRaMte0r.=51R0 M0.510

© Oxford Fajar Sdn. Bhd. (008974-T) 2019 28