MODUL GALUS MATEMATIK TAHUN 5

(b) 20 × 9 321 = 186 420 (b) 465 378 ÷ 4 = 116 344 baki 2

9321 116344
× 20 4 465378

186420 ‒4
6
8 (a) 25 × 6 625 = 165 625
6625 ‒4
25
× 25
33125 ‒24
13
+13250
165625 ‒12
17
(b) 50 × 8 215 = 410 750
‒16
8215 18
× 50
‒16
410750 2

9 (a) 32 × 3 872 = 123 904 2 (a) 688 424 ÷ 11 = 62 584

3872 62584
× 32 11 6 8 8 4 2 4

7744 ‒66
+11616 28

123904 ‒22
64
(b) 10 × 40 382 = 403 820 2
0 ‒55
4038 0 92
×1
‒88
40382 44

10 (a) Bilangan peket epal yang dibeli oleh Encik Chong ‒44
(b) 15 × M = 120 0
(c) M = 120 ÷ 15
=8 (b) 246 776 ÷ 25 = 9 871 baki 1

8 9871
15 1 2 0 25 2 4 6 7 7 6

‒120 ‒225
0 217

Bab 5 Bahagi ‒200
177
1 (a) 260 888 ÷ 8 = 32 611
‒175
32611 26
8 260888
‒25
‒24 1
20
3 (a) 364 512 ÷ 100 = 3 645 baki 12
‒16
48 3645
100 3 6 4 5 1 2
‒48
8 ‒300
645
‒8
8 ‒600
451
‒8
0 ‒400
512

‒500
12

(b) Y ÷ 9 = 17
Y = 17 × 9
= 153

17
×9

153

99

4 (a) 644 544 ÷ 12 = 53 712 6 (a) 126 580 ÷ 30 = 4 219 baki 10
4219
53712
12 6 4 4 5 4 4 30 1 2 6 5 8 0
‒120
‒60 65
44 ‒60
58
‒36 ‒30
85 280
‒270
‒84 10
14
(b) 305 920 ÷ 32 = 9 560
‒12 9560
24
32 3 0 5 9 2 0
‒24 ‒288
0 179
‒160
(b) 588 562 kg ÷ 13 = 45 274 kg 192
‒192
45274 0
13 5 8 8 5 6 2 ‒0
0
‒52
68 7 (a) 276 503 ÷ 52 = 5 317 baki 19
5317
‒65
35 52 2 7 6 5 0 3
‒260
‒26 165
96 ‒156
90
‒91 ‒52
52 383
‒364
‒52 19
0
(b) 279 450 ÷ 6 = 46 575
5 (a) RM540 500 ÷ 5 = RM108 100 46575

RM 1 0 8 1 0 0 6 279450
5 RM 5 4 0 5 0 0 ‒24
39
‒5 ‒36
4 34
‒30
‒0 45
40 ‒42
30
‒40 ‒30
5 0

‒5 100
0

‒0
0

‒0
0

(b) 341 448 ÷ 8 = 42 681

42681
8 341448

‒32
21

‒16
54

‒48
64

‒64
8

‒8
0

8 (a) 432 912 ÷ 45 = 9 620 baki 12 10 (a) 834 722 ÷ 7 = 119 246 6
2
9620 11924
45 4 3 2 9 1 2 7 83472 2
2
‒405 ‒7 0
279 13

‒270 ‒7
91 64

‒90 ‒63
12 17

‒0 ‒14
12 32

(b) 329 118 ÷ 30 = 10 970 baki 18 ‒28
4
10970
30 3 2 9 1 1 8 ‒4

‒30 (b) RM329 180 ÷ 5 = RM65 836
29
RM 6 5 8 3 6
‒0 5 RM 3 2 9 1 8 0
291
‒ 30
‒270 29
211
‒25
‒210 41
18
‒40
‒0 18
18
‒15
9 (a) 445 125 ÷ 15 = 29 675 30

29675 ‒30
15 4 4 5 1 2 5 0

‒30 11 (a) K ÷ 45 = 1 350
145 (b) K = 1 350 × 45
= 60 750
‒135
101 1350
× 45
‒ 90
112 6750
+5400
‒105
75 60750

‒75 Bab 6 Operasi Bergabung
0
1 (a) 120 × 35 + 4 115 = 4 200 + 4 115
(b) 129 334 ÷ 10 = 12 933 baki 4 = 8 315

12933 (b) 62 354 + 512 × 82 = 62 354 + 41 984
10 1 2 9 3 3 4 = 104 338

‒10 2 (a) 232 161 – 9 564 × 12 = 232 161 – 114 768
29 = 117 393

‒20 (b) 76 × (3 256 – 2 022) = 76 × 1 234
93 = 93 784

‒90 3 (a) 11 467 + 35 205 ÷ 15 = 11 467 + 2 374
33 = 13 814

‒30 (b) (26 324 + 6 456) ÷ 44 = 32 780 ÷ 44
34 = 745

‒30 4 (a) 370 656 ÷ 24 – 6 971 = 15 444 – 6 971
4 = 8 473

101

(b) 25 946 – 11 256 ÷ 6 = 25 946 – 1 876 2 (a) 8– 1 = 72 – 46
= 24 070 5
99 9
5 (a) 20 × 126 + 220 = 2 520 + 220
= 2 740 = 26
9
(b) 551 ÷ 29 + 62 = 19 + 62
= 81 8
=2
6 (a) 350 ÷ 14 – 20 = 25 – 20
=5 9

(b) 358 × 10 – 358 = 3 580 – 358 (b) 73 – 1 – 4= 7 3 – 1 1× 5 – 4×10
= 3 222 1
10 2 10 2 × 5 1×10
7 (a) 13 × 185 – 425 = 2 405 – 425
= 1 980 = 7 3 – 1 5 – 40
10 10 10
(b) 180 + 5 × 38 = 180 + 190
= 370 = 73 – 15 – 40
10 10 10
8 (a) (1 265 + 2 735) ÷ 50 = 4 000 ÷ 50
= 80 = 18
10
(b) (616 – 343) × 5 = 273 × 5
= 1 365 = 1 8÷2
10 ÷ 2
9 (a) 3 × 5 439 + 5 439 = 16 317 + 5 439
= 21 756 =14
5
(b) (1 528 – 128) ÷ 40 = 1 400 ÷ 40
= 35 3 (a) 4 + 2 2 – 1 = 4 × 6 + 2 2×10 – 1×15
5 3 2 5× 6 3×10 2 ×15
10 (a) 18 × 30 – 115 = 540 – 115
= 425 = 24 + 2 20 – 15
30 30 30
(b) (987 + 1 106) × 5 = 2 093 × 5
= 10 465 = 24 + 80 – 15
30 30 30

= 89
30

Bab 7 Pecahan = 29
2
30

1 (a) 3 + 2 1 = 3 + 2 1× 2 (b) 15 2 – 6 1 + 3 = 15 2 × 4 – 6 1×6 + 3× 3
4 2 4 2×2 3 24 3×4 2×6 4×3

= 3 + 22 = 15 8 – 6 6 + 9
44 12 12 12

= 25 =92 + 9
4 12 12

41 = 9 11
=2+ + 12

44

1 4 (a) 5 1 + 11 + 5 1 = 5 1 + 1 1× 3 + 5 1
=2+1+ 6 2 6 6 2×3 6

4

= 31 = 51 + 13 + 51
4 66 6

(b) 61 + 5 + 1 1 + 1× 2
=6+5+
84 8 4×2 5
= 11 ℓ
= 11 + 1 + 2
88 6

(b) 5– 2 3 50 23
=–
= 11 3 10 10 10
8
27
=

10

=27ℓ
10

102

5 (a) 32 + 1 = 3 2× 2 + 2 1×3 =7+ 8 + 2
2 88
3 2 3×2 2×3

= 34 + 23 = 82÷2
66 8÷2

= 57 1
6 = 8 kg

4

=5+ 6 + 1 (b) 40 1 – 15 3 + 20 = 40 1× 4 – 15 3 + 20
66 28 2×4 8

= 6 1 km = 4 – 3 + 20
6 40 15
88

(b) 1 – 4= 200 – 9 × 5 – 4×2 = 1
20 – 4 45 ℓ
2 5 10 2 × 5 5× 2 8

= 200 – 45 – 8 8 (a) 4 + 41 + 4 = 3 4× 2 + 4 1×5 + 4
10 10 10 3

= 147 52 5×2 2×5
10
= 38 + 4 5 +4
= 14 7 kg 10 10
10
= 13
11
10

6 (a) 9 + 2 –11 = 9 + 7×2 11 10 3
1 – = 11 + +
10 5 10 10 5× 2 10
10 10

= 9 + 14 – 11 = 12 3 jam
10 10 10 10

= 12 ÷ 2 (b) 4 + 2 2 – 6 3 = 4 + 2 2× 2 – 6 3
10 ÷ 2 5 10 5× 2 10

=6 =4+ 2 4 – 6 3
5 10 10

=5+1 = 1 kg
55 10

= 1 m 9 (a) 5 1 – 2 – 14 = 21 – 2 – 9
1 4 54 5
5 = 21× 5 – 2× 20 – 9 × 4
4 × 5 1× 20 5 × 4
(b) 2 1 + 4 + 9 = 2 + 4 + 1× 2 + 9 = 105 – 40 – 36
5 10 5× 2 10 20 20 20
= 29
=6+ 2 + 9 20
10 10 9
=1 m
= 6 + 11 20
10
(b) 8 – 2 3 + 8 = 16 – 2 3
= 6 + 10 + 1 88
10 10 = 15 8 – 2 3
88
= 7 1 kg = 13 5 km
10 8

7 (a) 10 5 – 3 1 + 7 = 10 5 – 3 1× 2 + 7
8 48 8 4×2 8

= 10 5 – 3 2 + 7
8 88

= 73 + 7
88

= 7 10
8

103

10 (a) 12 + 2 7 + 1 = 3 + 2 × 2 + 7 + 1× 5 (b) 13 × 0.255 kg = 3.315 kg
5 10 2 5× 2 10 2 × 5 10 (a) 30.46 kg – 18.228 kg – 9.1 kg = 12.232 kg – 9.1 kg

=3+ 4 + 7 + 5 = 3.132 kg
10 10 10 (b) 3.132 kg ÷ 18 = 0.174 kg

= 3 + 16 Bab 9 Peratus
10
4
= 3 + 10 + 6 1 (a) 4% =
10 10
100
= 4 6÷2 = 4. ÷ 100
10 ÷ 2

= 4 3 jam = 0.04
5 (b) 75% = 75

(b) 20 2 – 10 1 + 12 = 22 + 2 × 2 – 1 100
36 3×2 6 = 7 5. ÷ 100

= 22 + 4 – 1 = 0.75
66
2 (a) 7×4 = 28
= 22 3 ÷ 3 25×4 100
6÷3 = 28%

= 22 1 kg (b) 3 × 25 = 75
2 4 × 25 100
= 75%

Bab 8 Perpuluhan

1 (a) 12.35 + 0.199 + 32.352 = 12.549 + 32.352 3 (a) 12 = 7 × 20
= 44.901 5 5× 20
= 140
(b) 62.301 + 14.8 + 25.897 = 77.101 + 25.897 100
= 102.998 = 140%

2 (a) 94.7 – 25.128 – 31.45 = 69.572 – 31.45 (b) 145% = 100% + 45%
= 38.122 = 100 + 45 ÷ 5
100 100 ÷ 5
(b) 50 – 19.8 – 11.327 = 30.2 – 11.327 =1+ 9
= 18.873 20

3 (a) 85.899 + 34.12 – 60.415 = 120.019 – 60.415 =19
= 59.604 20

(b) 125.66 – 98.567 + 47.301 = 27.093 + 47.301
= 74.394

4 (a) 9 × 12.584 = 113.256 4 (a) 24% × 350 = 24 × 350
(b) 34 × 6.981 = 237.354 100

5 (a) 102.34 ÷ 28 = 3.655 = 24 × 350
(b) 215.418 ÷ 42 = 5.129 100

6 (a) 5.128 km – 0.23 km + 5.128 km = 4.898 km + 5.128 km = 8 400
= 10.026 km 100

(b) 10.4 liter + 8.364 liter – 15.33 liter = 84
= 18.764 liter – 15.33 liter (b) 65% × 1 500 = 65 × 1 500
= 3.434 liter
100
7 (a) 6.18 m + 10.254 m – 12.4 m = 16.434 m – 12.4 m = 65 ×1 500
= 4.034 m
100
(b) 7.5 kg ÷ 6 = 1.25 kg = 97 500

8 (a) 10 × 8.65 kg = 86.5 kg 100
(b) 10.35 liter ÷ 25 = 0.414 liter = 975

9 (a) 26.5 liter ÷ 20 = 1.325 liter

104

5 (a) 15% × RM200 = 15 × RM200 = 228
100
Bab 10 Wang
= RM30
(b) 65% × 1 920 = 65 × 1 920 1 (a) RM43 129.40 + RM112 563 + RM29 822.50 +
RM645.50
100 = RM155 692.40 + RM29 822.50 + RM645.50
= 65×192 = RM185 514.90 + RM645.50
= RM186 160.40
10
(b) RM349 500 – RM128 543.60 – RM54 298
12 480 = RM220 956.40 – RM54 2986
= = RM166 658.40

10 2 (a) RM548 921.20 + RM109 567.80 – RM210 548.70
= 1 248 = RM658 489 – RM210 548.70
= RM447 940.30
6 (a) 26% × 650 = 26 × 650
100 (b) RM299 430 – RM49 348.20 + RM1 243.90
= RM250 081.80 + RM1 243.90
= 26 × 65 = RM251 325.70
10
3 (a) RM7 349.40 × 32 = RM235 180.80
= 1 690 (b) RM64 568.90 ÷ 11 = RM5 869.90
10
4 (a) RM3 625.60 × 34 ÷ 10 = RM123 270.40 ÷ 10
= 169 = RM12 327.04

(b) 40% × 325 = 40 × 325 (b) RM255 136 ÷ 25 × 52 = RM10 205.44 × 52
100 = RM530 682.88

= 4 × 325 5 (a) RM638 500 – RM149 730 = RM488 770
10 (b) RM30 500 + RM25 325 + RM49 362
= RM55 825 + RM49 362
= 1300 = RM105 187
10
6 (a) RM143 800 + RM59 320 + RM25 600
= 130 = RM203 120 + RM25 600
= RM228 720
7 (a) 6% × 150 = 6 × 150
100 (b) RM210 540 – RM55 900 – RM73 230
= RM154 640 – RM73 230
= 6 ×15 = RM81 410
10
7 (a) RM98 542 – RM7 285 + RM98 542
= 90 = RM91 257 + RM98 542
10 = RM189 799

=9 (b) RM230 465.70 + RM12 484 – RM6 432.20
= RM242 949.70 – RM6 432.20
(b) 1 25 × 84 = RM236 517.50
25% × 84 =
4 100 8 (a) RM25 635.25 × 8 = RM205 082
(b) RM80 436 × 3 = RM241 308
84 21
= 9 (a) RM154 320 ÷ 3 × 2 = RM51 440 × 2
41 = RM102 880

= 21 (b) RM654 325.50 ÷ 30 × 8 = RM21 810.85 × 8
= RM174 486.80
8 (a) Peratus bilangan muka surat yang tidak mempunyai
gambar 10 (a) RM36
= 100% – 50% (b) RM37.45
= 50% (c) RM1 910.17
501 112
× 112 = 105
1002 2
= 56

(b) Peratus bilangan pokok getah yang masih ada
= 100% – 5%
= 95%

95 × 240 = 95× 24
100 10

= 2 280
10

Bab 11 Masa dan Waktu 4 (a) 1 abad 5 tahun + 62 tahun
= 1 abad 5 tahun + 6 abad 2 tahun
= 7 abad 7 tahun

1 (a) 35 tahun = 30 tahun + 5 tahun abad tahun
= (30 ÷ 10) dekad + 5 tahun 1 5
= 3 dekad 5 tahun 2
+6 7
(b) 3 abad 72 tahun = (3 × 100) tahun + 72 tahun 7
= 300 tahun + 72 tahun
= 372 tahun (b) 56 tahun + 6 dekad 9 tahun
= 5 dekad 6 tahun + 6 dekad 9 tahun
2 (a) 6 dekad 8 tahun + 5 dekad 2 tahun + 18 tahun = 12 dekad 5 tahun
= 138 tahun
dekad tahun

dekad tahun 56

68 +6 9

52 11 15

+1 8 + 1 – 10

12 18 12 5

+ 1 – 10

13 8 5 (a) 4 dekad 5 tahun – 9 tahun = 3 dekad 6 tahun

13 dekad 8 tahun = (13 × 10) tahun + 8 tahun dekad tahun
= 130 tahun + 8 tahun
= 138 tahun 3 15
(b) 84 abad 38 tahun – 46 abad 18 tahun – 7 abad 56 tahun
= 30 abad 64 tahun 4 5
– 9

3 6

abad tahun (b) 3 dekad 3 tahun – 7 tahun = 26 tahun
84 38
– 46 18 dekad tahun
38 20
2 13
abad tahun
3 3
37 120 – 7

38 20 2 6
–7 56
2 dekad 6 tahun = (2 × 10) tahun + 6 tahun
30 64 = 20 tahun + 6 tahun
= 26 tahun

3 (a) 6 × 12 dekad 7 tahun = 762 tahun 6 (a) 3 × 3 abad 70 tahun = 25 abad 90 tahun

12 dekad 7 tahun 3 abad 70 tahun
×6 × 7

72 dekad 42 tahun 21 490
+4 – 400
72 dekad 42 tahun = (72 × 10) tahun + 42 tahun
= 720 tahun + 42 tahun 25 abad 90 tahun
= 762 tahun
(b) 5 × 24 tahun = 1 abad 20 tahun
(b) 3 abad 72 tahun ÷ 12 = 31 tahun 24 tahun

0 abad 3 1 tahun ×5
12 3 abad 7 2 tahun 120 tahun

–0 +300 120 tahun = 100 tahun + 20 tahun
3 372 = (100 ÷100) abad + 20 tahun
–36 = 1 abad 20 tahun
12
–12
0

106

7 (a) 1 abad 20 tahun ÷ 10 tahun = 3 (b) 9 abad 37 tahun – 6 abad 89 tahun = 2 abad 48 tahun

0 12 tahun abad tahun
1 0 1 abad 20 tahun
8 137
‒ 0 + 100
1 120 9 37
‒ 120 –6 89
0
2 48
(b) 6 dekad 4 tahun ÷ 4
= 1 dekad 6 bulan 10 (a) 1 dekad 4 tahun × 6 = 8 dekad 4 tahun
= (1 × 10) tahun + 6 tahun
= 10 tahun + 6 tahun dekad tahun
= 16 tahun
14
1 dekad 6 tahun
×6
4 6 dekad 4 tahun
‒ 4 + 20 6 24
2 24
‒ 24 + 2 – 20
0
84

(b) 55 tahun – 3 dekad 3 tahun = 55 tahun – 33 tahun
= 22 tahun

55 tahun
+ 33 tahun

22 tahun

8 (a) 15 tahun + 1 dekad 6 tahun Bab 12 Panjang
= 1 dekad 5 tahun + 1 dekad 6 tahun
= 3 dekad 1 tahun

dekad tahun 1 (a) 2 cm 8 mm = 2 cm + (8 ÷ 10) cm

15 = 2 cm + 0.8 cm

+1 6 = 2.8 cm

2 11 (b) 374 cm = (374 ÷ 100) m

+ 1 ‒ 10 = 3.74 cm

31 (c) 31 km = (3 × 1 000) m + 1 ×1  m
2  2 000 

(b) 3 dekad 8 tahun + 17 tahun = 3 000 m + 500 m

= 3 dekad 8 tahun + 1 dekad 7 tahun = 3 500 m

= 5 dekad 5 tahun 2 (a) 1.42 cm + 1 7 cm = 1.42 cm + 1.7 cm
10
= (5 × 10) tahun + 5 tahun = 3.12 cm

= 50 tahun + 5 tahun (b) 5 3 km + 12.65 km + 3 298 m
4
= 55 tahun
= 5.75 km + 12.65 km + 3.298 km
dekad tahun = 21.698 km

38

+1 7

4 15

+ 1 ‒ 10

55 3 (a) 120 mm – 76 1 cm = (120 ÷ 10) cm + (76.2 ÷ 10) cm
5

9 (a) 1 abad 50 tahun ÷ 5 = 30 tahun = 12 cm – 7.62 cm

0 abad 30 tahun = 4.38 cm
5 1 abad 50 tahun
(b) 65.82 m – 129 cm – 14 1 m
‒ 0 + 100 4
1 150
‒ 150 = (65.82 × 100) cm – 129 cm – [(14 × 100) cm +
0
 1 ×100  cm]
 4 

= 6 582 cm – 129 cm – [1 400 cm + 25 cm]

= 5 028 cm

4 (a) 16 × 1.025 m = 16.4 m
= (16.4 × 100) cm
= 1 640 cm

107

(b) 30 × 4 2 km = 30 × ( 22 × 1 000) m Bab 13 Jisim
55
= 30 × 4 400 m 1 (a) 7.05 kg = (7.05 × 1 000) g
= 132 000 m = 7 050 g

5 (a) 2 658 mm ÷ 15 = 177.2 mm (b) 5 640 g = 5 000 g + 640 g
= 17.72 cm  5 000   640 

(b) 60.38 km × 10 = 603.8 km =  1 000  kg + 1 000  kg
= (603.8 × 1 000) m
= 603 800 m = 5 kg + 16 kg
25
6 (a) 12 3 m ÷ 6 = [(12 × 100) cm + ( 3 × 100) cm] ÷ 6
44 = 5 16 kg
= [120 cm + 75 cm] ÷ 6 25
= 195 cm ÷ 6
= 32.5 cm (c) 6 3 kg = 6 kg + 3 kg
55
(b) 2 1 km ÷ 8 = [(2 × 1 000) km + ( 1 × 1 000) km] ÷ 8
55 = (6 × 1 000) g +  3 ×1  g
= [2 000 m + 200 m] ÷ 8  5 000 
= 2 200 m ÷ 8
= 275 m = 6 000 g + 600 g
= 6 600 g
7 (a) 12.65 cm – 12 mm = 126 cm 5 mm – 12 mm
= 114 cm 5 mm 2 (a) 80.9 kg + 65 320 g = 80.9 kg + (65 320 ÷ 1 000) kg
= 80.9 kg + 65.32 kg
(b) 520 m + 3 1 km = 146.22 kg
4
(b) 4 3 kg + 1.25 kg + 990 g
= 520 m + [(3 × 1 000) m + ( 1 × 1 000) m] 10
4
= [(4 × 1 000) g + ( 3 × 1 000) g] + (1.25 × 1 000) g +
= 520 m + [3 000 m + 250 m] 10
= 520 m + 3 250 m
= 3 770 m 990 g
= [4 000 g + 300 g] + 1 250 g + 990 g
8 (a) 15 × 4.524 km = 15 × (4.524 × 1 000) m = 4 300 g + 1 250 g + 990 g
= 15 × 4 524 m = 6 540 g
= 67 860 m
3 (a) 4 320 g – 11 kg
(b) 32 mm + 4 5 cm = (32 ÷ 10) cm + 4.625 cm 5
8
= 3.2 cm + 4.625 cm = 4 320 g – [(1 × 1 000) g + ( 1 × 1 000) g]
= 7.825 cm 5

9 (a) 20.5 m – 11 m = 20.5 m – 1.2 m = 4 320 g – [1 000 g + 200 g]
5 = 4 320 g – 1 200 g
= 19.3 m = 3 120 g

(b) 50 × 55 mm = 50 × (55 ÷ 10) cm (b) 20.875 kg – 9 126 g – 3 kg
= 50 × 5.5 cm 8
= 275 cm
= 20.875 kg – (9 126 ÷ 1 000) kg – 0.375 kg
10 (a) 9 759 m – 4 3 km = 20.875 kg – 9.126 kg – 0.375 kg
10 = 11.374 kg

= 9 759 m – [(4 × 1 000) m + ( 3 × 1 000) m] 4 (a) 7 × 4 3 kg = 7 × [(4 × 1 000) g + ( 3 × 1 000) g]
10 44

= 9 759 m – [4 000 m + 300 m] = 7 × [4 000 g + 750 g]
= 9 759 m – 4 300 m
= 5 459 m = 7 × 4 750 g
(b) 18 × 845 cm = 18 × (845 ÷ 100) m
= 33 250 g
= 18 × 8.45 m
= 152.1 m (b) 25 × 125 g = 25 × [(125 ÷ 1 000) kg]

= 25 × 0.125 kg

= 3.125 kg

= 3 kg +  125  kg
1 000 

= 3 1 kg
8

108

5 (a) 76 400 g ÷ 16 = (76 400 ÷ 1 000) kg ÷ 16 11 (a) 15 × 1 7 kg = 15 × [(1 × 1 000) g + ( 7 × 1 000) g]
= 76.4 kg ÷ 16 10 10
= 4.775 kg = 15 × [1 000 g + 700 g]
= 15 × 1 700 g
(b) 3 1 kg ÷ 8 = [(3 × 1 000) g + ( 1 × 1 000) g] ÷ 8 = 25 500 g
22
= [3 000 g + 500 g] ÷ 8 (b) 5 124 g ÷ 28 = (5 124 ÷ 1 000) kg ÷ 28
= 3 500 g ÷ 8 = 5.124 kg ÷ 28
= 437.5 g = 0.183 kg

6 (a) 1 4 kg + 3 210 g Bab 14 Isi Padu Cecair
5
1 (a) 42 150 mℓ = (42 150 ÷ 1 000) mℓ
= [(1 × 1 000) g + ( 4 × 1 000) g] + 3 210 g = 42.15 ℓ
5
(b) 3 4 ℓ = (3 × 1 000) mℓ + ( 4 × 1 000) mℓ
= [1 000 g + 800 g] + 3 210 g 55
= 1 800 g + 3 210 g = 3 000 mℓ + 800 mℓ
= 5 010 g = 3 800 mℓ
(b) 3 1 kg + 4.155 kg
(c) 16 ℓ 320 mℓ = (16 × 1 000) mℓ + 320 mℓ
5 = 16 000 mℓ + 320 mℓ
= [(3 × 1 000) g + ( 1 × 1 000) g] + (4 .155 × 1 000) g = 16 320 mℓ

5 2 (a) 620 mℓ + 2 1 ℓ
= [3 000 g + 200 g] + 4 155 g 4
= 3 200 g + 4 155 g
= 7 355 g = 620 mℓ + [(2 × 1 000) mℓ + ( 1 × 1 000) mℓ
4
7 (a) 5.08 kg – 1 250 g = (5.08 × 1000) g – 1 250 g
= 5 080 g – 1 250 g = 620 mℓ + [2 000 mℓ + 200 mℓ ]
= 3 830 g = 620 mℓ + 2 200 mℓ
= 2 820 mℓ
(b) 120.45 kg – 80 1 kg = 120.45 kg – 80..25 kg
4 (b) 16 754 mℓ + 10.06 ℓ + 20 1 ℓ
= 40.2 kg 8

8 (a) 18 × 850 g = 18 × (850 ÷ 1 000) kg = (16 754 ÷ 1 000) ℓ + 10.06 ℓ + 20.125 ℓ
= 18 × 0.85 kg = 16.754 ℓ + 10.06 ℓ + 20.125 ℓ
= 15.3 kg = 46.939 ℓ

(b) 55 × 0.25 kg = 55 × (0.25 × 1 000) g 3 (a) 60 ℓ ‒ 45.124 ℓ = 14.876 ℓ
= 55 × 250 g
= 13 750 g = (14.876 × 1 000) mℓ

9 (a) 30 kg ÷ 16 = (30 × 1 000) g ÷ 16 = 14 876 mℓ
= 30 000 g × 16
= 1 875 g (b) 5 1 ℓ ‒ 3 ‒ 0.255 ℓ= 5.1 ℓ ‒ 0.375 ℓ ‒ 0.255 ℓ
ℓ
(b) 20 5 kg ÷ 6
8 10 8

= [(20 × 1 000) g + ( 5 × 1 000) g] ÷ 6 = 4.47 ℓ
8
4 (a) 10 × 35.762 ℓ = 10 × (35.762 × 1 000) mℓ
= [2 000 + 625 g] ÷ 6 = 10 × 35 762 mℓ
= 2 625 g ÷ 6 = 357 620 mℓ
= 437.5 g
(b) 9 × 7 ℓ = 9 × 0.7 ℓ
10 (a) 6 655 g – 2.95 kg – 658 g 10
= (6 655 ÷ 1 000) kg – 2.95 kg – (658 ÷ 1 000) kg = 6.3 ℓ
= 6.655 kg – 2.95 kg – 0.658 kg
= 3.047 kg 5 (a) 1.72 ℓ ÷ 8 = (1.72 × 1 000) mℓ ÷ 8
= 1 720 mℓ ÷ 8
(b) 1 3 kg + 3 kg + 4.56 kg = 1.75 kg + 3 kg + 4.56 kg = 215 mℓ
4
= 9.31 kg (b) 1 3 ℓ ÷ 100 = [(1 × 1 000) mℓ + ( 3 × 1 000) mℓ ] ÷ 100
= (9.31 × 1 000) g 55
= 9 310 g = [1 000 mℓ + 600 mℓ ] ÷ 100
= 1 600 mℓ ÷ 100
= 16 mℓ

109

6 (a) 1 1 ℓ + 1.05 ℓ (b) 8 250 mℓ + 2 1 ℓ + 6.4 ℓ
10 4

= [(1 × 1 000) mℓ + ( 1 × 1 000) mℓ ] = 8 250 mℓ + [(2 × 1 000) mℓ + ( 1 × 1 000) mℓ ]
10 4

+ (1.05 × 1 000) mℓ + (6.4 × 1 000) mℓ
= [1 000 mℓ + 100 mℓ ] + 1 050 mℓ = 8 250 mℓ + [2 000 mℓ + 250 mℓ ] + 6 400 mℓ
= 1 100 mℓ + 1 050 mℓ = 8 250 mℓ + 2 250 mℓ + 6 400 mℓ
= 2 150 mℓ = 16 900 mℓ

(b) 5 5 ℓ + 1 1 ℓ + 5 5 ℓ = 5 5 ℓ + 1 1× 2 ℓ + 5 5 ℓ 11 (a) 30 × 1 3 ℓ = 30 × [(1 × 1 000) mℓ + ( 3 × 1 000) mℓ ]
8 4 8 8 4×2 8 44
= 55ℓ + 12ℓ + 55ℓ = 30 × [1 000 mℓ + 750 mℓ ]
88 8 = 30 × 1 750 mℓ
= 1112 ℓ = 52 500 mℓ
8
= 12 1 ℓ (b) 30 160 mℓ ÷ 10 = (30 160 ÷ 1 000) ℓ÷ 10
2 = 30.16 mℓ ÷ 10
= 3.106 ℓ
= (12 × 1 000) mℓ + ( 1 × 1 000) mℓ
2 Bab 15 Ruang

= 12 000 mℓ + 500 mℓ 1 (a) 6 cm + 6 cm + 6 cm + 8 cm + 8 cm + 8 cm
= 12 500 mℓ + (8 cm – 6 cm) = 44 cm

7 (a) 10 000 mℓ ‒ 8 3 ℓ = (10 000 ÷ 1 000) mℓ ‒ 8.6 ℓ (b) (6 × 6) cm2 + (8 × 8) cm2 = 36 cm2 + 64 cm2
5 = 100 cm2
= 10 ℓ ‒ 8.6 ℓ
= 1.4 ℓ 2 (a) Perimeter segi empat tepat
= 10 cm + 10 cm + 4 cm + 4 cm
(b) 40.945 ℓ‒ 25.08 ℓ = 15.865 ℓ = 28 cm
= (15.865 × 1 000) mℓ Panjang sisi segi empat sama = 28 cm ÷ 4
= 15 865 mℓ = 7 cm

8 (a) 25 × 3 160 mℓ = 25 × (3 160 ÷ 1 000) ℓ (b) (7 × 7) cm2 + (10 × 4) cm2 = 49 cm2 + 40 cm2
= 25 × 3.16 ℓ = 89 cm2
= 79 ℓ
3 (a) 4 cm + 4 cm + 4 cm + 3 cm + 5 cm = 20 cm
(b) 20 × 0.75 ℓ = 20 × (0.75 × 1 000) mℓ (b) (4 × 4) cm2 + ( 1 × 4 × 3) cm2 = 16 cm2 + 6 cm2
= 20 × 750 mℓ 2
= 15 000 mℓ = 22 cm2

9 (a) 9 7 ℓ ÷ 16 4 (a) [8 × 14 × 6] cm3 + [6 × 6 × 6] cm3
10 = [8 × 14 × 6] cm3 + 216 cm3
= 672 cm3 + 216 cm3
= [(9 × 1 000) mℓ + ( 7 × 1 000) mℓ ] ÷ 16 = 888 cm3
10
(b) [8 × 8 × (20 – 8)] cm3 + [8 × 8 × (20 – 8)] cm3
= [9 000 mℓ + 700 mℓ ] ÷ 16 = [8 × 8 × 12] cm3 + [8 × 8 × 12] cm3
= 9 700 mℓ ÷ 16 = 768 cm3 + 768 cm3
= 606.25 mℓ = 1 536 cm3
(b) 12 100 mℓ ÷ 22 = (12 100 ÷ 1 000) ℓ ÷ 22
5 (a) (12 × 32 × 12) cm3= 4 608 cm2
= 12.1 ℓ ÷ 22 (b) 2 × (13 × 8) cm2 = 2 × 104 cm2
= 0.55 ℓ = 208 cm2

10 (a) 15.9 ℓ‒ 3 1 ℓ ‒ 4 150 mℓ 6 (a) 10 cm + 20 cm + 10 cm + 20 cm = 60 cm
2 (b) 1 × 10 cm × 10 cm = 50 cm2
2
= (15.9 × 1 000) mℓ ‒ [(3 × 1 000) mℓ + ( 1 × 1 000)
2 7 (a) 120°
(b) 90°
mℓ ] ‒ 4 150 mℓ
= 15 900 mℓ ‒ [3 000 mℓ + 500 mℓ ] – 4 150 mℓ
= 15 900 mℓ ‒ 3 500 mℓ – 4 150 mℓ
= 8 250 mℓ

110

8 (a) (b) Q 4 (a) (b) y

P 6N M
RS 5
4
3
2
1

x

Bab 16 Koordinat 0 12 3 45 6
(c) J(3, 2)

1 (a) (0, 0) Bab 17 Nisbah
(b) Q(4, 5)
(c) P 1 (a) 1 : 4
(d) y
(b) 1 : 10
6Q
5T 2 (a) 1 : 6
4R
3 (b) 1 : 7
2P
3 (a) 1 : 3
1 S
1 2 345 6 (b) 1 : 5
0 x
4 (a) 3 bahagian → 15 buah kereta
2 (a) J(1, 4)
(b) L 1 bahagian → 15 ÷ 3 = 5
(c) K Bilangan kereta merah ialah 5 buah.
(b) 1 bahagian → 20 biji
(d) y 7 bahagian → 20 × 7 = 140 biji
Bilangan bola golf ialah 70 biji.
4 J K
R 5 (a) 1 bahagian → 20 cm
3
M 10 bahagian → 20 × 10 = 200 cm
2 Panjang benang ialah 200 cm.
L (b) 8 bahagian → 1.24 kg
1N 1 bahagian → 1.24 ÷ 8 = 0.155 g
Jisim sebiji oren ialah 0.155 kg.
0 12 34 x
6 (a) 7 bahagian → 14 buah buku
3 (a) P(1, 3)
(b) y 1 bahagian → 14 ÷ 7 = 2
Bilangan buku Asyida ialah 2 buah.
6 Q (b) 5 bahagian → 55 biji
1 bahagian → 55 ÷ 5 = 11
5 Bi;angan guli biru ialah 11 biji.

4 P Bab 18 Pengurusan Data
3
1 (a) Min = 11 + 12 + 9 + 12 + 6
2
5
1 x = 50

0 123 456 5
(c) Q(5, 5) = 10
(b) 12
(c) 6, 9, 11, 12, 12
Median = 11

111

2 (a) Median = 30.24 + 30.5 + 28.1 + 29.14 + 27.05 + 25.37 6 (4×1) + (1× 2) + (1×3) + (3× 4) + (1×5)

6 (a) Min =
10
= 170.4
6 = 26
10
= 28.4
(b) Julat = 30.5 – 25.37 = 2.6 kg
(b) 5 – 1 = 4 kg
= 5.13 (c) 1, 1, 1, 1, 2, 3, 4, 4, 4, 5
(c) 25.37, 27.05, 28.1, 29.14, 30.24, 30.5
Median = 3 + 2
Median = 28.1+ 29.14 2
2
=5
= 57.24 2
2
= 2.5
= 28.62
(d) 1 kg

3 (1×3) + (2×5) + (3×7) + (4× 4) 7 (1× 20) + (2× 40) + (3×10) + (4× 20)

(a) Min = (a) Min =
1+ 2 + 3+ 4
1+ 2 + 3+ 4

= 50 = 210
10 10

=5 = 21
(b) 7 orang adik-beradik
(c) 3, 5, 5, 7, 7, 7, 4, 4, 4, 4 (b) 10, 10, 10, 40, 40, 20, 20, 20, 20, 20

Median = 7 + 7 Median = 20 + 20
2 2

= 14 = 40
2 2

=7 = 20

(1×11) + (2× 4) + (3×7) + (4×10) (c) Tajuk: Botol Terpakai yang Dikumpul

4 (a) Min = 1
1+ 2 + 3+ 4

= 80 2
10

= 8 kg 3
(b) 11 – 4 = 7 kg

(c) 4, 4, 7, 7, 7, 10, 10, 10, 10, 11 4

Median = 7 +10 Petunjuk: mewakili 10 botol
2

= 17 8 (1×10) + (2×12) + (3×18) + (4×16)
2
(a) Min =
= 8.5 kg 1+ 2 + 3+ 4

5 (a) Min = 152
10
(4×50) + (1×100) + (3×150) + (1× 200) + (1× 250)
= 15.2 ℓ
= (b) 10, 12, 12, 16, 16, 16, 16, 18, 18, 18
10
Median = 16 +16
= 1 200 2
10
= 32
= 120 mililiter 2
(b) 250 – 50 mililiter = 200 mililiter
(c) 50, 50, 50, 50, 100, 150, 150, 150, 200, 250 = 16 ℓ

Median = 100 +150
2

= 125
(d) 50 mililiter

112

(c) Tajuk: Isi Padu Air Minuman 10

1 Bilangan peserta 7
Bilangan buku
2 6

3 5

4 4
3
Petunjuk: mewakili 2 ℓ
2
9
1
35
30 0 Buku
25 Sains Sejarah Matematik English
20
15
10

5
0

A BC D
Kumpulan

113

NOTA

114