My Nilam Maths Year 5 (DLP)

133
January

February

March

represents 20 mangoes [1 mark]
Based on the pictograph above, [2 marks]
(a) what is mode? [1 mark]
(b) find mean. [1 mark]
(c) find range.
(d) in which month the median is located?

134 Number of pupils

9 234 5
8 Number of boxes
7 [1 mark]
6 [2 marks]
5 [2 marks]
4
3
2
1
01

(a) Find the mode.

(b) Calculate the range.

(c) What is the mean?

Mathematics • Year 5 148 © Nilam Publication Sdn. Bhd.

ANSWERS

SKILL-BASED ACTIVITIES

UNIT 1 NUMBERS UP TO 1 000 000 B 1 718 492 2 189 816 3 369 853 6.3 3 1 554
4 263 292 5 395 481 6 296 638 A 1 146 4 917 910
7 285 323 8 276 793 2 74 984
1.1 2.2
A 1 921 181 eggs
A 1 735 204; Seven hundred thirty-five thousand 2 22 193 stamps 3 1 037 books

two hundred and four UNIT 7 FRACTIONS
2 486 136; Four hundred eighty-six thousand
7.1
one hundred and thirty-six
3 530 592; Five hundred thirty thousand five UNIT 3 SUBTRACTION A 1 2 2 3 1 3 11 7
3 4 8
hundred and ninety-two
4 907 406; Nine hundred seven thousand four 3.1 4 8 1 5 4 3 6 7 3
2 5 4
hundred and six A 1 27 095 2 342 675 3 531 216
354 713 6 266 000
B 1 Thousands, 8 000 4 95 559 5 404 200 7 6 19 8 7 9 11
576 000 3 617 000 24 8 18
2 Hundred thousands, 300 000 7 302 059 8 401 274 6 469 989 10 8
530 146
3 Hundreds, 800 B 1 302 000 2 7.2

4 Ten thousands, 50 000 4 400 517 5 3 3 7
4 4 8
5 Tens, 20 7 465 067 8 A 1 2 2 3 3

C 1 2 hundred thousands + 6 ten thousands + 3.2 1 2 1
8 3 2
3 thousands + 4 hundreds + 1 tens + 8 ones; A 1 25 762 2 144 700 3 575 300 4 8 5 2 6 3
200 000 + 60 000 + 3 000 + 400 + 10 + 8 425 072 6 307 025
2 3 hundred thousands + 7 ten thousands + 4 172 299 5 376 322 1 3 1
2 5 8
7 428 008 8 7 3 8 2 9 2

5 thousands + 6 hundreds + 4 tens + 9 ones; 3.3 7.3
300 000 + 70 000 + 5 000 + 600 + 40 + 9
3 8 hundred thousands + 0 ten thousands + A 1 84 272 people 3 5 700 nails 1 3
4 8
2 32 704 apples A 1 8 2 6 3 5

4 thousands + 3 hundreds + 2 tens + 1 ones; UNIT 4 MULTIPLICATION 4 9 11 5 6 9 6 5 1
800 000 + 4 000 + 300 + 20 + 1 12 10 2
4 6 hundred thousands + 5 ten thousands +

9 thousands + 7 hundreds + 6 tens + 3 ones; 4.1 7 6 1 8 6 1 9 6 11
600 000 + 50 000 + 9 000 + 700 + 60 + 3 7.4 8 5 12
5 3 hundred thousands + 0 ten thousands + A 1 12 408 2 18 639 3 16 642

4 34 344 5 24 785 6 78 848 A 1 3 marbles 5 52
6 RM1 850
7 thousands + 4 hundreds + 2 tens + 6 ones; 7 47 558 8 24 372 9 33 828 2 4 buttons 7 2 520 minutes
300 000 + 7 000 + 400 + 20 + 6 8 2 200 litres
6 7 hundred thousands + 1 ten thousands + 10 80 408 11 68 460 12 91 236 3 45 pieces

13 609 360 14 448 208 15 628 418 4 240 apples

4 thousands + 3 hundreds + 8 tens + 0 ones; B 1 40 326 2 52 600 3 134 720
700 000 + 10 000 + 4 000 + 300 + 80
7 1 hundred thousands + 9 ten thousands + 4 137 727 5 133 408 6 280 016 UNIT 8 DECIMALS

7 513 595 8 159 012 9 562 975

4 thousands + 9 hundreds + 6 tens + 5 ones; 10 362 204 11 763 851 12 769 365 8.1

100 000 + 90 000 + 4 000 + 900 + 60 + 5 13 49 670 14 92 740 15 36 900 A 1 40.58 2 19.171 3 230.28
6 963.103
D 1 < 3 > 16 75 000 17 592 000 18 809 000 4 550.783 5 29.262 9 902.782
12 370.451
2 > 4 > 19 428 544 20 712 744 7 381.407 8 177.196
3 0.98
E 1 784 412, 898 523, 988 317, 998 321 4.2 10 120.495 11 500.19 6 2.109

2 854 209, 845 920, 843 126, 834 612 A 1 584 740 bricks 8.2 3 23.62
6 498.873
3 284 963, 428 936, 482 369, 824 396 2 53 670 stamps A 1 3.78 2 9.76
3 9.468
1.2 & 1.3 3 Twelfth day; 21 856 rambutans 4 2.3 5 604.524 6 286.02
9 258.408
A R = 10 000 , S = 5 000 UNIT 5 DIVISION 7 3.044 8 679.82 12 18 098
B 1 Even number pattern 15 9 735
2 Odd number pattern 8.3
3 Even number pattern 3 3.05
4 Odd number pattern 5.1 A 1 8.25 2 12.118 6 10.55
5 Odd number pattern 9 0.712
C 1 77 915; 78 465 A 1 1 932 4 7 339 remainder 6 4 67.528 5 24.091 12 0.95
2 10 773; 14 481; 15 408
3 2 299; 3 349; 4 399 2 32 584 5 75 660 remainder 2 7 268.69 8 217.277 3 1.2 kg
4 4 796; 6 296; 7 796
1.4 3 68 724 6 114 079 remainder 3 8.4

B 1 3 708 4 19 203 remainder 18 A 1 4.96 2 17.52

2 4 792 5 33 769 remainder 21 4 135.975 5 66.96

3 9 268 6 11 027 remainder 52 7 507.52 8 77.3

C 1 51 228 2 100 3 324 700 10 3.9 11 57

A 1 200 000; 240 000; 244 000; 243 700 4 864 remainder 927 5 1 000 13 729.3 14 230
2 400 000; 400 000; 399 000; 398 700
3 600 000; 650 000; 650 000; 649 900 5.2 8.5
4 200 000; 150 000; 152 000; 152 400
5 800 000; 830 000; 832 000; 831 700 A 1 15 280 irons A 1 6.7 2 1.41
6 600 000; 580 000; 584 000; 584 300
7 600 000; 600 000; 604 000; 604 200 2 16 718 boxes 4 0.13 5 0.719
8 900 000; 930 000; 935 000; 934 800
B 428 856, 428 865, 428 874, 428 3 86 745 residents 7 0.06 8 0.57

10 9.017 11 0.017

UNIT 6 MIXED OPERATIONS 8.6

A 1 6.92 km 2 130.20 km

885, 6.1 UNIT 9 PERCENTAGE
A 1 144
428 921, 428 935, 428 939, 428 948 or other 4 175 076 2 2 300 3 529 970
7 336 120 5 532 6 1 440
reasonable answers. B 1 5 046 8 224 721 9.1
4 291 2 1 300 3 135 608
UNIT 2 ADDITION 6.2 5 159 440 6 223 252 A 1 9 2 4 3 17
A 1 210 fishes 20 5 25
2 121 beads
2.1 4 1210 5 3 6 23
A 1 704 146 4 25
4 627 865 2 669 729 3 746 785 3 2 770 coins
5 682 889 6 719 603 7 30% 8 40% 9 25%
10 84% 11 94% 12 65%

© Nilam Publication Sdn. Bhd. A1 Mathematics • Year 5

B 1 250% 2 340% 3 590% 11.4 14.2
6 319%
4 434% 5 716% A 1 4 decades 5 20 centuries 80 years A 1 0.65 l 2 820 ml 3 1 120 ml

7 1 1 8 2 1 9 4 3 2 9 decades 8 years 6 23 centuries 65 years 4 3.8 l 5 4 860 ml 6 11 640 ml
5 4 4
3 7 decades 2 years 7 58 centuries 80 years 7 14.15 l 8 59.165 l

10 3 4 11 1 11 12 7 1 4 56 decades 4 years 8 145 centuries 20 years 14.3
5 20 2
11.5 A 1 8.78 l 2 670 ml 3 6 800 ml

C 1 RM12 2 114 ml 3 20 kg A 1 2 decades 1 year 5 5 centuries 8 years 4 3.82 l 5 1 250 ml 6 13 460 ml
4 378 km 5 162 m 6 162 candies
7 50 8 36 2 4 decades 9 years 6 1 century 35 years 7 141.4 l 8 1.87 l

3 1 decade 6 years 7 1 century 63 years 14.4

4 1 decade 8 years 8 1 century 10 years A 1 3.24 l 2 2 400 ml 3 650 ml

UNIT 10 MONEY 11.6 4 30 000 ml 5 14 l 6 74.7 l

A 1 45 years 7 6 000 ml 8 10 400 ml

10.1 2 67 decades 2 years 14.5

A 1 RM588.40 5 RM909 200.60 3 6 decades 7 years A 1 0.3 l 2 400 ml 3 450 ml

2 RM6 318.40 6 RM760 078.70 4 0.121l 5 82 ml 6 889 ml

3 RM4 140.20 7 RM210 011.15 UNIT 12 LENGTH 7 850 ml 8 82 ml

4 RM384 040.50 8 RM956 349.65 14.6

10.2 12.1 A 1 2.05 l 2 425 ml 3 3.89 l
A 1 25 mm
A 1 RM7 229.80 5 RM893 909.45 4 3 120 cm 2 4.8 cm 3 7.6 cm
5 6.2 m 6 8.75 m
2 RM579.50 6 RM44 579.05 7 2 900 m 9 8.07 km UNIT 15 SPACE
12 6 400 m
3 RM720 164.50 7 RM770 602.65 10 95 mm 8 25 4 km 3 25.95 m 15.1
12.2 5 6 160 m A 1 46 cm
4 RM550 610.60 8 RM653 496.90 A 1 10.8 mm 4 40 cm 2 68 cm 3 92 cm
4 96 mm 11 1 080 cm 3 115.6 m B 1 100 cm2 5 34 cm 6 59.4 cm
10.3 7 167 mm 6 7 090 m 2 224 cm2
12.3 3 304 cm2 3 110°
A 1 RM148 159.20 4 RM408 165.50 A 1 67.3 mm 2 15.2 cm 3 8 320 m 4 84 cm2 6 60°
4 71 mm 5 1 730 cm 6 22 m C 1 81 cm3 2 115°
2 RM453 695.15 5 RM684 226.85 7 40 mm 8 17 670 m 2 584 cm3 5 285° T
12.4 3 187 mm 3 372 cm3
3 RM34 315.20 6 RM401 109.90 A 1 7.2 cm 6 84.05 cm 4 198 cm3 2
4 13.6 km 15.2
10.4 7 5 760 cm 3 1.36 m A 1 70°
12.5 4 80°
A 1 RM336 140 8 RM150 765 A 1 3.4 mm 2 4.1 cm B 1 M
4 30.7 m 5 1 220 cm
2 RM267 435 9 RM424 477.35 7 40.18 m 8 1 217 m
12.6
3 RM95 700 10 RM154 216.50 A 1 9.85 km

4 RM291 212 11 RM347 845

5 RM115 609.95 12 RM230 500 2 351 mm
5 32 500 cm
6 RM301 132 13 RM68 500 8 9.5 m

7 RM412 800

10.5

A 1 RM267 100 6 RM10 215.50 2 6.2 cm
5 25 cm
2 RM125 300 7 RM8 007 8 0.89 km

3 RM43 080.17 8 RM83.765 N

4 RM30 290 9 RM920.15 S

5 RM145.10 2 7.45 m 4

10.6 3 B P
Q
A 1 RM25 608 3 RM1 620.50 A

2 RM560 245 4 RM162 112.50 UNIT 13 MASS

10.7 13.1
A 1 5 000 g
A 1 RM48 241.60 4 6 008 g 2 6 450 g 3 700 g
7 0.5 kg 5 9 070 g 6 7 620 g
2 RM2 400 10 8 750 g 8 8.7 kg 9 19.06 kg
13.2 11 12 500 g 12 6 200 g
3 RM270 054.48 A 1 2.4 kg UNIT 16 COORDINATES
4 1.08 kg 2 2 180 g 3 3 950 g
10.8 7 8.025 kg 5 5 100 g 6 17 060 g
13.3 8 12.758 kg
A 1 (a) RM180 A 1 2.7 kg 3 1 070 g 16.1
4 12.12 kg 2 4 990 g 6 6 620 g
(b) RM185.40 7 2.8 kg 5 3 600 g A 1 y-axis
13.4 8 0.84 kg 3 2 000 g
(c) RM6 365.40 A 1 2.0 kg 6 16.2 kg 2 x-axis
4 4 000 g 2 900 g
(d) Balance at beginning of year: RM6 365.40; 7 2 400 g 5 30 kg 3 80 g 3 Origin
13.5 8 62 500 g 6 2.7 g
Interest rate = 3%; A 1 1.05 kg 4 Cartesian plane
4 0.075 kg 2 13 g 3 2 560 g
Interest value = RM190.96; 7 525 g 5 0.65 kg B 1 (0 , 0) 2 (3 , 2) 3 (4 , 4)
13.6 8 62 g 6 (0 , 5)
Balance at the end of year A 1 3.85 kg 4 (1 , 5) 5 (6 , 0)
2 25 kg
= RM6 556.36 Cy J

10

UNIT 11 TIME 9 F
8 G
11.1 7A

A 1 20 years 6 805 years 6

2 700 years 7 8 decades 9 years 5
4E
3 9 centuries 8 12 decades 4 years 3H

4 4 decades 9 4 centuries 82 years

5 85 years 10 6 centuries 48 years 2
1 IC
11.2 D Bx
1 2 3 4 5 6 7 8 9 10
A 1 8 decades 7 years 5 6 centuries 29 years 0

2 4 decades 3 years 6 9 centuries 60 years

3 11 decades 5 years 7 7 centuries 44 years UNIT 14 VOLUME OF LIQUID UNIT 17 RATIO

4 13 decades 9 years 8 7 centuries 9 years 14.1
A 1 7 000 ml
11.3 4 9 020 ml 2 400 ml 3 2 300 ml 17.1
7 0.56 l A 1 1 : 3
A 1 2 decades 2 years 5 1 century 85 years 10 9 700 ml 5 3 008 ml 6 23 040 ml 4 1 : 2 2 1 : 100 3 1 : 5
B 1 1 : 1 000 5 1 : 10 6 1 : 4
2 2 decades 6 years 6 1 century 58 years 8 7.5 l 9 14.12 l 2 1 : 1 000

3 1 decade 6 years 7 2 centuries 72 years 11 13 500 ml 12 4 750 ml

4 4 decades 9 years 8 5 centuries 77 years

Mathematics • Year 5 A2 © Nilam Publication Sdn. Bhd.

3 Pra-UPSR MODEL PAPER 10 (a) Height of cuboid Q = 1 × 20 cm
2

Paper 1 = 10 cm
1 C
K L or other 6 C 2 B 3 A 4 A 5 A So, value of x = 10 cm – 6 cm
reasonable 11 A 7 D 8 A 9 D 10 A
16 B 12 C 13 C 14 A 15 D = 4 cm
21 B 17 D 18 D 19 C 20 D
answers 26 B 22 B 23 C 24 D 25 D (b) Area of base of the composite shape
31 C 27 A 28 D 29 A 30 B
36 C 32 C 33 D 34 A 35 C = 3 × (6 cm × 6 cm)
37 C 38 C 39 A 40 C
= 108 cm2

1 cm (c) Volume of the whole diagram

1 cm = (6 × 6 × 20) + (6 × 6 × 10) + (6 × 6 × 6)

UNIT 18 DATA HANDLING = 720 + 360 + 216

18.1 Paper 2 = 1 296 cm3
A 1 Mode = 2, Median = 2, Mean = 3, Range = 5 1 (a) Digit 4
2 Mode = 13, Median = 13, Mean = 12.5, (b) 749 925  750 000 11 (a) Each square = 1 760 ÷ 22
Range = 5 (rounded off to the nearest thousand)
3 Mode = 40 kg, Median = 35 kg, = 80
Mean = 33.5 kg, Range = 25 kg
4 Mode = 60, Median = 65, Mean = 67, Mode of the sales = 7 × 80
Range = 30
18.2 2 (a) Percentage of coloured parts = 560
A Horizontal pictograph:
Title: Grade Obtained in Science Test = 3 × 100% (b) Mean = 1 760
12 4
Gred A
= 25% = 440
Gred B (c) Mak Minah’s earning in the four days
(b) 25% = 25
Gred C 100 = 1 760 × RM2.50

Gred D = 1 = RM4 400
4
Gred E 12 (a) Total number of pupils = 40
3 (a) Number of apples his friend receives Ratio of the number of male pupils to the
Key: represent 5 people = 1 × 36
number of female pupils = 1 : 4
Vertical pictograph: 4
Title: Grade Obtained in Science Test =9 Male pupil Female pupil
(b) Number of apples that Mr Ali still has 5  40
= 36 – 9
= 27  40 ÷ 5 = 8

4 (a) RM12 000 ÷ 8 × 6 So, the number of female pupils
(b) Price of 6 bicycles
= RM12 000 ÷ 8 × 6 =4
=4×8
= RM9 000 = 32

5 (a) 1 century 15 years = 115 years

Mr Haikal’s age = 115 + 3 (b) Total number of pupils now = 40 + 20
2
= 60

= 59 years Ratio of the number of male pupils to the

(b) Mr Haikal’s wife age number of female pupils = 1 : 5

= 115 years – 59 years

= 56 years

Gred A Gred B Gred C Gred D Gred E 6 (a) Mass of the watermelon = 5.6 kg Male pupil Female pupil
(b) Mass of one slice of watermelon 6  60
Key: represent 5 people = 5.6 kg ÷ 8
= 0.7 kg  60 ÷ 6 = 10
B Horizontal bar graph: = 700 g
Title: Number of Pupils in 5 Classes So, the total number of male pupils now
7 (a) Mode score = 5
Class =
Data: 1, 2, 3, 3, 5, 5, 5, 8
= 1 × 10
Exora Median = 3 + 5
2 = 10
Satria =4 Number of male pupils that entered the

(b) Mean score class = 10 – 8
= 2
Wira = (5 + 3 + 3 + 2 + 1 + 5 + 8 + 5) ÷ 8

= 32 ÷ 8 13 (a) Simple interest = 2 × RM10 000
100
Waja =4

Perdana 8 (a) Life span of the turtle = RM200
Number of = (0.6 × 100) years
= 60 years (b) Simple interest at the first year = RM200
0 4 8 12 16 20 24 28 32 36 40 pupils (b) Period of a turtle will live after it matures Total Mr Mahesh’s savings at the end of the
= 60 years – 15 years
Vertical bar graph: = 45 years first year
Title: Number of Pupils in 5 Classes = RM10 000 + RM200
9 (a) Volume of water in the bucket = 4.2 l = RM10 200
Number of pupils Compound interest at the second year
= 2 × RM10 200
40 = 4 200 ml
36 100
32 (b) Volume of water left
28 = RM204
24 = (1 – 1 ) × 4.2 l Total Mr Mahesh’s savings at the end of
20 3
16 = 2.8 l second year
12 = RM10 200 + RM204
= 2 800 ml = RM10 404
8
4 (c) Volume of water left = 2 800 ml

0 Perdana Waja Volume of water in container G

= Volume of water in container E

Class = 2 800 ml ÷ 2
Satria Exora
Wira = 1 400 ml

© Nilam Publication Sdn. Bhd. A3 Mathematics • Year 5

(c) First year: 102 × RM10 000 14 (a) Percentage of magazine 15 (a) Total number of pens
100 = 105 + 95 + 120 + 130
= 100% – 40% – 35%

= RM10 200 = 25% = 450

Second year: 102 × RM10 200 (b) Number of magazines = 25 × 10 000 Number of blue pens = 2 × 450
100 100 5

= RM10 404 = 2 500 = 180

Third year: 102 × RM10 404 Number of reference books = 40 × 10 000 (b) Number of red pens
100 100
= 2 × ( 3 × 450)
= 4 000 3 5
= RM10 612.08
2
Fourth year: 102 × RM10 612.08 Number of storybooks = 35 × 10 000 = 3 × 270
100 100
= 180
= RM10 824.32 = 3 500
(c) Number of black pens
Title: Number of Reading Materials At The
Fifth year: 102 × RM10 824.32 Library = 450 – 180 – 180
100
= 90

= RM11 040.81 Magazine

So, the amount of Mr Mahesh’s savings at

the end of the fifth year is RM11 040.81. Reference
book

Storybook

Key: represent 500 copies

TOPICAL DRILL PAPER 1

1 D 2 B 3 C 4 B 5 B 6 B 7 C 8 C 9 D 10 B 11 A 12 B 13 A 14 B
15 C 16 C 17 C 18 D 19 A 20 D 21 C 22 B 23 D 24 B 25 B 26 C 27 A 28 A
29 C 30 D 31 A 32 A 33 B 34 D 35 A 36 C 37 C 38 C 39 A 40 A 41 B 42 D
43 A 44 C 45 C 46 A 47 A 48 B 49 A 50 A 51 B 52 A 53 B 54 A 55 A 56 B
57 C 58 A 59 D 60 B 61 B 62 B 63 C 64 B 65 C 66 C 67 A 68 C 69 B 70 D
71 B 72 D 73 B 74 B 75 C 76 D 77 C 78 B 79 A 80 D 81 C 82 C 83 D 84 B
85 C 86 A 87 C 88 B 89 C 90 C 91 C 92 A 93 B 94 B 95 D 96 A 97 D 98 A
99 C 100 D 101 B 102 A 103 D 104 C 105 B 106 B 107 B 108 B 109 B 110 A 111 C 112 A
113 B 114 D 115 C 116 B 117 A 118 B 119 A 120 C 121 B 122 D 123 D 124 C 125 A 126 A
127 B 128 A 129 C 130 B 131 B 132 A 133 A 134 C 135 D 136 B 137 A 138 D 139 B 140 C
141 A 142 B 143 D 144 B 145 C 146 B 147 D 148 C 149 D 150 C 151 A 152 D 153 A 154 B
155 C 156 C 157 C 158 B 159 C 160 D 161 C 162 C 163 D 164 D 165 D 166 D 167 B 168 C
169 B 170 B 171 D 172 C 173 B 174 D 175 C 176 A 177 B 178 C 179 C 180 D 181 D 182 A
183 A 184 D 185 B 186 B 187 A 188 D 189 C 190 A 191 C 192 A 193 B 194 C 195 C 196 B
197 C 198 D 199 A 200 B 201 B 202 A 203 D 204 A 205 C 206 A 207 B 208 B 209 C 210 B
211 A 212 C 213 D 214 C 215 C 216 C 217 B 218 A 219 D 220 A 221 C 222 B 223 D 224 B
225 B 226 D 227 B

TOPICAL DRILL PAPER 2

1 (a) 40 579 (b) Even number pattern (b) 579 707 or 707 597 or 570 797 or other
(b) Ten thousands 4 (a) The suitable symbol is <. The smaller number
(c) Odd number reasonable answers.
2 (a) 909 909 is 597 320. 597 320 rounded off to the 7 (a) Descending order:
(b) Nine hundred nine thousand nine hundred 2 270, 1 960, 1 650, 1 340
nearest thousand becomes 597 000. (b) Even number pattern
and nine (b) 5 001 and 5 003 or other reasonable 8 (a) 398 764 and 397 846 or other reasonable

3 (a) Number sequence = 3 450 – 3 442 answers. answers
5 (a) 700 000 + 3 000 + 800 + 9 + 50 (b) thousands
=8 9 (a) 70 000 + 90 = 70 090
Y = 3 458 + 8 = 703 859 (b) 70 090 → 70 000
X = 3 450 + 8 (b) 703 859 rounded off to the nearest thousand
(rounded off to the nearest ten thousand)
= 3 458 = 3 466 becomes 704 000.
6 (a) The smallest digit is digit 3, and the place
Number sequence: 3 442, 3 450, 3 458,
value is ten thousands.
3 466, 3 474

Mathematics • Year 5 A4 © Nilam Publication Sdn. Bhd.

10 (a) 913 450 + 77 935 = 991 385 (b) Target of number of cars that must be sold in (b) Number of stamps in one album
(b) 900 000 + 90 000 + 1 000 + 300 + 80 + 5 = 18 125 ÷ 5
11 (a) 7 069 + 432 921 + 63 932 + 209 732 October = 3 625
= 713 654 38 (a) Number of magazines in one box
= 350 000 – 273 637 = 600 000 ÷ 2
(b) 713 654  700 000 = 76 363 = 300 000
(rounded off to the nearest hundred thousand) 24 (a) Number of school C’s pupils (b) Number of magazines that each bookstore
12 X + Y = 1 000 000 – 348 928 = 16 684 – 7 493
= 651 072 = 9 191 receives = 300 000 ÷ 32
Value of card X = (651 072 + 105 832) ÷ 2 Difference between the number of pupils in
= 9 375
= 378 452 school B and C
Value of card Y = (651 072 – 105 832) ÷ 2 = 9 191 – 7 493 39 (a) Number of baskets that each lorry carries
= 1 698
= 272 620 (b) Number of female participants = 8 131 = 1 380 ÷ 6
13 (a) Number of bottles of soy sauce produced by Number of male participants
= 16 684 – 8 131 = 230
factory Q = 8 553
= 73 583 + 8 065 Difference = 8 553 – 8 131 (b) Number of baskets that each wholesaler gets
= 81 648
(b) Total number of bottles of soy sauce = 422 = 230 ÷ 5
25 (a) 8 000 × 70 = 560 000
(b) 86 572 × 7 = 606 004 = 46

40 (a) Number of marbles in box Q = 583 200 ÷ 4

= 145 800

produced by the 3 factories (b) Number of marbles in each container

= 81 648 + 81 648 + 73 583 = 145 800 ÷ 8
= 236 879
14 (a) Number of Amir’s marbles = 1 550 + 490 26 (a) 5 × = 110 = 18 225

= 2 040 (b) 5 × = 110 41 (a) 32 800 ÷ 8 – 100
(b) Number of Gopal’s marbles = 2 040 + 340
= 22 (b) 32 800 ÷ 8 – 100 = 4 100 – 100
= 2 380
Total number of marbles of the 3 pupils 27 (a) Y × 100 = 935 300 = 4 000
= 1 550 + 2 040 + 2 380
= 5 970 Y = 9 353 42 (a) 6 × 7 083 ÷ 9 = 42 498 ÷ 9
15 (a) Points obtained by Saiful
(b) 7 × 54 038 = 378 266 = 4 722
= 138 379 + 7 592
= 145 971 28 (a) 23 × 23 074 = 530 702 (b) 4 722 → 5 000
Points obtained by Lina = 145 971 + 908 (rounded off to the nearest thousand)
(b) 6 591 × 100 = 659 100
= 146 879 43 (a) Number of marbles that one of his friends
(b) Total of their points 29 (a) The number of erasers in 7 boxes
= 138 379 + 145 971 + 146 879 receives
= 431 229 = 7 × 50 742
16 (a) 4Q + Q = 500 000 = (308 – 26) ÷ 6
(b) 4Q + Q = 500 000 = 355 194 = 282 ÷ 6
= 47
5Q = 500 000 (b) Number of new erasers in a box (b) Subtract and divide
Q = 100 000 44 (a) Number of bottles of orange juice can be
(c) P = 4Q = 50 742 + (50 742 ÷ 2)

= 4 × 100 000 = 76 113
= 400 000
17 (a) 350 127 – 93 548 = 256 579 Number of new erasers in 13 boxes produced in 2 weeks = 8 240 ÷ 5 × 14
(b) 2 hundred thousands + 5 ten thousands +
= 13 × 76 113 = 23 072

= 989 469 (b) Number of bottles of orange juice can be

30 (a) Number of apples in a basket produced in August = 8 240 ÷ 5 × 31

= 4 × 540 = 51 088

= 2 160 45 (a) Number of packets of marble obtained

(b) Number of apples in 25 identical baskets = 6 × 16 204 ÷ 12

= 25 × 2 160 = 8 102

= 54 000 (b) Number of marbles in 9 packets = 9 × 12

31 (a) Number of pupils that attend the motivational = 108

6 thousands + 5 hundreds + 7 tens + 9 ones course in February 46 (a) Number of passengers that cannot get in the
18 (a) 9 hundred thousands – 15 ten thousands –
= 2 × 3 012 rescue boats
270 hundreds = 6 024
(b) Number of pupils that attend the motivational = 2 150 – 30 × 65
= 900 000 – 150 000 – 27 000 = 2 150 – 1 950
course in March = 200
= 723 000 (b) Number of rescue boats that should be
= 3 × 6 024
(b) 50 thousands – 860 tens = 50 000 – 8 600 = 18 072
32 (a) Residents in Town Q = 3 × 19 740
= 41 400 added
= 59 220 = 200 ÷ 65
19 (a) 90 000 – 700 = 89 300 (b) Residents in Town R = 7 × 59 220 = 3 remainder 5
So, the number of rescue boats that should
(b) 100 000 – 60 = 99 940 = 414 540
(c) Town R = 4 × Town S
20 (a) 734 891 – 43 092 – K = 13 976
414 540 = 4 × Town S
(b) 734 891 – 43 092 – K = 13 976 103 635 = Town S be added is 4 more.
So, the number of residents in Town S is
734 891 – 43 092 – 13 976 = K 47 (a) Total number of biscuits she bought
= 6 × 75
677 823 = K = 450
(b) Number of packets that Aina obtained
21 (a) R – 518 698 = 345 923 = 6 × 75 ÷ 10
= 45
R = 345 923 + 518 698 103 635. (c) Number of biscuits that each of Yaya’s

= 864 621 33 (a) 105 161 ÷ 7 = 15 023
(b) 15 023 → 15 000
S – R = 127 332
(rounded off to the nearest thousand)
S – 864 621 = 127 332 34 (a) 9 000 ÷ 4 = 2 250
(b) 2 250 ÷ 25 = 90
S = 991 953 35 (a) 723 180 ÷ 18 = 40 176 remainder 12 siblings receives = 7 × 10 ÷ 5
(b) 723 180 ÷ 18
(b) Possible values of J: = 40 176.66667 = 14
= 40 176.67 (rounded off to 2 d.p.)
J = 703 749 ± 198 213 36 (a) 203 765 ÷ 5 = 40 753 48 (a) Total number of coins both of them own
(b) T ÷ 1 000 = 409 remainder 5
J = 703 749 + 198 213 @ 703 749 – = 12 688 + 2 × 12 688
T = (409 × 1 000) + 5
198 213 = 409 005 = 38 064
37 (a) Number of Hana’s stamps = 72 500 ÷ 4
= 901 962 @ 505 536 (b) The number of Zaid’s coins in each money
= 18 125
22 (a) Badrul’s score = 932 – 120 box
= 2 × 12 688 ÷ 8
= 812 = 3 172

(b) Badrul’s score = 812

Yusof’s score = 700 49 (a) 7 (b) 1 3
4 4
Difference of scores = 812 – 700

= 112 50 (a) 1 part = 2 – 1
3 2
23 (a) Sales of cars in August

= 273 637 – 56 431 – 76 823 – 54 957 = 1
6
= 85 426

© Nilam Publication Sdn. Bhd. A5 Mathematics • Year 5

R = 1 + 1 59 (a) 13 7 = 13 + 0.007 (b) Percentage of participants that went home
6 6 000
1 early = 100% – 74%

= 2 = 1 = 13.007 = 26%
6 3
(b) 13 7 + 237.935 = 13.007 + 237.935 70 (a) Percentage of balance of money
1 000
R 5 1 5 = 250.942 = (1 – 1 ) × 100%
(b) + 1 9 = 3 + 1 9 4
60 (a) 7 × 2.374 = 16.618
8 = 75%
= 1 9 21 × 2.374 = 3 × (7 × 2.374)
75
51 (a) 17 = 3 × 16.618 (b) Balance of money = 100 × RM1 060
5
= 49.854 = RM795

(b) 3 3 + 1 1 = 3 12 + 1 5 (b) 49.854 = 40 + 9 + 0.8 + 0.05 + 0.004 71 (a) Number of pens in box Q = 30 × 350
5 4 20 20 100
61 (a) Total mass of the curry powder she bought

= 4 17 = 6 × 0.534 kg = 105
20
= 3.204 kg (b) Number of pens in box R

1 1 1 2 (b) Mass of curry powder in each container 20
8 4 8 8 100
52 (a) 5 – 3 = 5 – 3 = 3.204 kg ÷ 4 = 105 + ( × 105)
= 126
7 = 0.801 kg
8 Total number of pens in 3 boxes
= 1 62 (a) Rope Q = 3.75 m + 1.745 m
= 350 + 105 + 126
(b) 15 = 5.495 m
8 = 581
Rope R = 5.495 m + 0.35 m

53 (a) Total mass of the fruits bought = 5.845 m 72 (a) 15% = 15
100
2 1 2 (b) Rope P = 3.75 m
5 3 3 3
= 1 kg + 3 kg + 2 kg Rope R = 5.845 m = 20

6 5 10 Difference in length = 5.845 m – 3.75 m 15
15 15 15 100
= 1 kg + 3 kg + 2 kg = 2.095 m (b) Number of children viewers = × 3 200

2 63 (a) Distance from K to L = 7.945 km + 9.475 km
5
= 7 kg = 17.42 km = 480

1 2 (b) Nearest route to go to L from K (c) Number of adult viewers
3 5
(b) Difference in mass = 3 kg – 1 kg = 1.53 km + 7.945 km = (100% – 15%) × 3 200

5 6 = 9.475 km = 85 × 3 200
15 15 100
= 3 kg – 1 kg Farthest route to go to L from K = 2 720

14 = 17.42 km 73 (a) Sixty-seven thousand fifty-four ringgit and
15
= 1 kg Difference distance = 17.42 km – 9.475 km seventy sen

= 7.945 km (b) RM67 054.70 → RM67 055
(rounded off to the nearest ringgit)
54 (a) 2 2 kg – 1 6 kg = 2 14 kg – 1 30 kg (c) Total distance travelled
5 7 35 35 74 (a) Tania’s money
= 17.42 km + 7.945 km = RM50 + (2 × RM20) + 20 sen + 50 sen

= 19 kg = 25.365 km = RM90.70
35 (b) RM90.70 → RM91
64 (a) Mass of box J = 5.69 kg – 1.785 kg
(rounded off to the nearest ringgit)
(b) Balance of meat = 1 6 kg + 2 2 kg – 1 kg = 3.905 kg 75 (a) RM73.80 = RM73.50 + RM0.30
7 5 2
(b) Mass of box L = (3.905 + 5.69) kg + 2.98 kg RM73.50 = 147 pieces of 50 sen coins
RM0.30 = 3 pieces of 10 sen coins
60 28 35 = 12.575 kg
70 70 70 So, RM73.80 equals to 147 pieces of 50 sen
= 1 kg + 2 kg – kg (c) Total mass of the three boxes

53 = 3.905 kg + 5.69 kg + 12.575 kg
70
= 3 kg = 22.17 kg

55 (a) Number of parents = (1 – 1 ) × 2 275 65 (a) 75% = 75 = 3 coins dan 3 pieces of 10 sen coins or other
5 100 4
reasonable answers.
75
= 4 × 2 275 (b) 75% = 100 (b) RM73.80 → RM74
5 (rounded off to the nearest ringgit)
= 0.75
= 1 820 76 (a) Simple interest of the first year

(b) Number of guests that cannot enter the hall 66 (a) 170% = 170 = 2.5 × RM30 000
100 100
3
= (1 – 7 ) × 2 275 = 1 7 = RM750
10
= 4 × 2 275 (b) First year = 102.5 × RM30 000
7 3 3 × 20 100
(b) 5 = 5 × 20
= 1 300 = RM30 750

56 (a) Total number of male pupils and female = 60 Second year = 102.5 × RM30 750
100 100
pupils in year 2018
= RM31 518.75
[ ( )] 1 = 60%
= 2 270 – 5 × 2 270 + 102.5
67 (a) Percentage of the shaded area Third year = 100 × RM31 518.75

[ ( )] 1 = 6 × 100% = RM32 306.72
2 088 + 4 × 2 088 20

= 1 816 + 2 610 = 30% Fourth year = 102.5 × RM32 306.72
100
= 4 426 30
(b) 30% = 100 = RM33 114.39
(b) Female gender of pupils
= 0.3 Fifth year = 102.5 × RM33 114.39
57 (a) Thirty-five point zero eight nine 68 (a) Percentage of female pupils = 100% – 35% 100

(b) 35.089 = 35 + 89 = 65% = RM33 942.25
1 000
89 So, Mr Farid’s savings in the fifth year is
000 65
= 35 1 (b) Number of female pupils = 100 × 3 000 RM33 942.25.

58 (a) 1 part = 0.495 – 0.345 = 1 950 77 (a) Price of 4 refrigerators = 4 × RM2 450
= RM9 800
= 0.15 69 (a) Number of participants left = 100 – 26
Price of a television
K = 0.495 + 0.15 + 0.15 = 74 = (RM12 890 – RM9 800) ÷ 2
= RM1 545
= 0.795 Percentage of participants left = 74 (b) Difference in price = RM2 450 – RM1 545
100
(b) K × 4 = 0.795 × 4 = RM905

= 3.18 = 74%

Mathematics • Year 5 A6 © Nilam Publication Sdn. Bhd.

78 (a) Value of deposit = 10 × RM122 000 87 (a) Time taken for each phase 97 (a) 1.25 kg = (1.25 × 1 000) g
100
= 3 decades 6 years ÷ 3 = 1 250 g

= RM12 200 = 1 decade 2 years (b) Mass of 7 identical packets of sugar

(b) Installment payment for each month (b) Time taken to finish the research = 7 × 1 250 g

( ) 90 = 3 decades 6 years + 1 decade 2 years = 8 750 g
= 100 × RM122 000 ÷ (5 × 12)
= 4 decades 8 years 4
= RM109 800 ÷ 60 98 (a) 580 g + 16 kg + 17 5 kg
= 48 years
= RM1 830
88 (a) Age of building K = 0.58 kg + 16 kg + 17.8 kg
79 (a) Ying-Ying’s savings
( ) 5 = 34.38 kg
= RM360 450 ÷ 6 = 300 years – 100 × 300 years
(b) 15 kg 40 g = 15 kg + 40 g
= RM60 075 = 300 years – 15 years
= 15 kg + (40 ÷ 1 000) kg
(b) Zulin’s savings = 2 × RM60 075 = 285 years
= 15.04 kg
(b) Age of building J = 29 decades 10 years
= RM120 150 99 (a) 1 000 × 350 g = 350 000 g
Age of building K = 28 decades 5 years
Dewi’s savings = 3 × RM60 075 = 350 kg
Total age
= RM180 225 (b) 5 7 kg ÷ 100 = 5.7 kg ÷ 100
= 29 decades 10 years + 28 decades 5 years 10
Total savings of Zulin and Dewi = 0.057 kg
= 58 decades 5 years
= RM120 150 + RM180 225 = 57 g
89 (a) 153 cm = (153 ÷ 100) m
= RM300 375 100 (a) Mass of 4 identical books
= 1.53 m
80 (a) Price of one smartphone = 21 kg 420 g ÷ 7 × 4
(b) 153 cm = (153 × 10) mm
= RM30 456 ÷ 8 = 21.42 kg ÷ 7 × 4
= 1 530 mm
= RM3 807 = 12.24 kg

(b) Total amount of money to be paid 90 (a) 6.8 m + 7 1 m + 974 cm (b) Mass of 5 books
4
= RM30 456 ÷ 8 × 3 = 680 cm + 725 cm + 974 cm = 21 kg 420 g ÷ 7 × 5

= RM11 421 = 2 379 cm = 21.42 kg ÷ 7 × 5

(c) Mr Gopal’s money (b) 321 cm = (321 × 10) mm = 15.3 kg

= (100 × RM100) + (30 × RM50) = 3 210 mm Mass of bag = 19 kg 150 g – 15.3 kg

= RM11 500 3
5
Balance of money = RM11 500 – RM11 421 91 (a) 21 km – 12 760 m – 7.45 km = 19.15 kg – 15.3 kg

= RM79 = 21.6 km – 12.76 km – 7.45 km = 3.85 kg

81 (a) 3 centuries 5 years = 1.39 km 101 (a) Mass of flour available now

= 3 centuries + 5 years (b) 5 km 329 m = 5 km + 329 m = 3 kg 350 g + 2 kg 400 g

= (3 × 100) years + 5 years = 5 000 m + 329 m = 3 350 g + 2 400 g

= 305 years = 5 329 m = 5 750 g

(b) 3 centuries 5 years 92 (a) Length of each part of the ribbon (b) Mass of flour left

= 3 centuries + 5 years = 320 cm ÷ 4 = 5 750 g – 2 kg 650 g

= (3 × 100) years + 5 years = 80 cm = 5.75 kg – 2.65 kg

= (300 ÷ 10) decades + 5 years 1 = 3.1 kg
4
= 30 decades 5 years (b) of the length of one part of the ribbon 102 (a) Mass of box Q = 5.6 kg – 2.55 kg

82 (a) 85 years = 80 years + 5 years 1 = 3.05 kg
4
= (80 ÷ 10) decades + 5 years = × 80 cm Mass of box R = 3.05 kg – 0.95 kg

= 8 decades 5 years = 20 cm = 2.1 kg

(b) 2 decades 3 years = 2 decades + 3 years = 0.2 m (b) Total mass of the three boxes

= (2 × 10) years + 3 years 93 (a) Distance from P to Q = 1 = 5.6 kg + 3.05 kg + 2.1 kg
5
= 23 years × 15.75 km = 10.75 kg

83 (a) 137 years + 2 centuries 18 years + 1 = 10 750 g
5
8 decades = × 15 750 m 103 (a) Mass of sugar for each container

= 137 years + 218 years + 80 years = 3 150 m = 7 × 1.2 kg ÷ 4

= 435 years (b) Distance from P to R through Q = 2.1 kg

= 4 centuries 35 years = 3 150 m + 15.75 km Mass of sugar needed for one cake

(b) 9 centuries 30 years = 3 150 m + 15 750 m = 2.1 kg ÷ 5

= 9 centuries + 30 years = 18 900 m = 0.42 kg

= (9 × 100) years + 30 years 94 (a) Length of red ribbon = 4.25 m = 420 g

= 930 years = 425 cm (b) Mass of flour needed for one cake

84 (a) 9 centuries 30 years – 45 decades 7 years Length of blue ribbon = 20 × 425 cm = 2 × 420 g
100
= 90 decades 30 years – 45 decades 7 years = 840 g

= 47 decades 3 years = 85 cm = 0.84 kg

(b) 7 × 6 decades 8 years (b) Length of yellow ribbon = 85 cm + 130 cm 104 (a) Mass of flour used to make biscuits

= 476 years = 215 cm = 1 × [(6 × 3.2 kg) – 13.65 kg]
3
= 470 years + 6 years Total length of 3 ribbons
1
= 47 decades 6 years = 425 cm + 85 cm + 215 cm = 3 × 5.55 kg

85 (a) Salmah’s age = 42 years old = 725 cm = 1.85 kg

= 4 decades 2 years 95 (a) Total amount of ribbons that Rani and Siti = 1 850 g

(b) Rahila’s age = 42 years old – 9 years need (b) Mass of flour left

= 33 years old = 65.7 cm + 17.9 cm + 65.7 cm 1
3
Total of their ages = 149.3 cm = (1 – ) × 5.55 kg

= 42 years old + 33 years old (b) Length of ribbon that Aida needs = 3.7 kg

= 75 years old = 4 × 149.3 cm 105 (a) Volume of water in the beaker = 2.5 l

= 7 decades 5 years = 597.2 cm (b) Volume of water in the beaker

86 (a) Age of Idris’s grandfather = 84 years old = 5.972 m = 2.5 l

Idris’s age = 84 years old ÷ 6 96 (a) The length of cloth used = 14.8 m – 6.7 m = (2.5 × 1 000) ml

= 14 years old = 8.1 m = 2 500 ml

(b) Age of Idris’s grandfather = 84 years old = 810 cm 106 ml l 2 l
5
Age of Idris’s father = 4 × 14 years old (b) The length of cloth used to make one children (a) 420 + 23 + 15

= 56 years old baju kurung = 420 ml + 23 000 ml + 15 400 ml

Difference in age = 810 cm ÷ 6 = 38 820 ml

= 84 years old – 56 years old = 135 cm

= 28 years old = 1.35 m

© Nilam Publication Sdn. Bhd. A7 Mathematics • Year 5

(b) 27 l 19 ml = 27 l + 19 ml (b) Volume of the whole diagram The ratio of length of sides of cube L to the
length of sides of cube M
= 27 l + (19 ÷ 1 000) l = (6 cm × 6 cm × 14 cm) +
= 4 : 12
= 27.019 l (6 cm × 6 cm × 6 cm) =1:3
(b) Volume of cube L = 4 cm × 4 cm × 4 cm
107 (a) 1 000 × 835 ml = 835 000 ml = 720 cm3
= 64 cm3
= 835 l 118 (a) Area of triangle HJK Volume of cube M
= 12 cm × 12 cm × 12 cm
(b) 11 3 l ÷ 100 = 11.3 l ÷ 100 = 1 × 6 cm × 6 cm = 1 728 cm3
10 2 Ratio of volume of cube L to the volume of
= 0.113 l
= 18 cm2 cube M
= 113 ml (b) Area of triangle HJF = 64 : 1 728
= 1 : 27
108 (a) Volume of water in container B = 1 × 6 cm × 12 cm 130 (a) The real distance from Ain’s house to
2
= 3 × 4 800 ml school
= 6 × 1 000 m
= 14 400 ml = 36 cm2 = 6 000 m
(c) Triangle HJK (b) The real distance from Ain’s house to the
= 14.4 l 119 (a) Perimeter of the shaded region
= (6 + 15 + 5 + 20 + 9) cm shop
(b) Volume of water in 2 bottles = 55 cm = 2 × 1 000 m
(b) Area of the shaded region = 2 000 m
= 14.4 l ÷ 8 × 2 Ratio of real distance from Ain’s house to

= 3.6 l the shop to the real distance from Ain’s

= 3 600 ml house to school
= 2 000 : 6 000
109 (a) Volume of orange juice bought 1 =1:3
2 131 (a) Mode = 5 kg
= 4 × 1 250 ml = (20 cm × 9 cm) – ( × 4 cm × 14 cm) (b) Median:
1 kg, 2 kg, 3 kg, 3 kg, 4 kg, 5 kg, 5 kg, 5 kg, 8 kg
= 5 000 ml = 180 cm2 – 28 cm2

=5l = 152 cm2

(b) Volume of orange juice in each bottle 120 (a) (2 , 4)

= 4 × 1 250 ml ÷ 8 (b) Point A is 2 units to the right from origin and

= 625 ml 4 units above on the plane

= 0.625 l 121 (a) Value of X = 6

110 (a) Volume of soy milk sold on Sunday (b) Coordinate of point H is (3 , 6)

= 42 7 l + 6 l 230 ml (c) Difference in distance = 4 – 3
10
=1
= 42.7 l + 6.23 l
122 (a) (2, 4)
= 48.93 l
(b) Total distance of Leoh’s movement
(b) Total volume of soy milk sold
= 5 × 2.5 km middle
7
= 42 10 l + 48.93 l = 12.5 km
Median = 4 kg
= 42 700 ml + 48 930 ml (c) Town T
(c) Range = 8 kg – 1 kg
= 91 630 ml 123 (a) Point P and point R
= 7 kg
111 (a) Volume of honey in each small container (b) Possible coordinates of point K:
(d) Mean
Point K = (r , 3r)
4 (3 + 5 + 8 + 5 + 2 + 4 + 3 + 1 + 5) kg
= 6 × 4 5 l ÷ 36 = (1 , 3) or (2 , 6) or = 9

= 0.8 l (3 , 9) or other reasonable = 4 kg

= 800 ml answers 132 (a) Mode = 163 cm

(b) Volume of honey in 7 small containers 124 (a) 1 : 6 (b) Median

= 7 × 800 ml (b) 6 : 1 = 9th data + 10th data
2
= 5 600 ml 125 (a) 1 : 2

= 5.6 l (b) Syafiq’ s money = RM10 156 cm + 163 cm
2
112 (a) Volume of cooking oil that she bought Adam’s money = 3 × RM10 =

= 5 × 3.5 l = RM30 = 159.5 cm

= 17.5 l 126 (a) Number of male pupils = 36 ÷ 4 (c) Range = 163 cm – 145 cm

= 17 500 ml =9 = 18 cm

(b) Volume of cooking oil in 3 containers (b) Number of female pupils (d) Mean

= 17 500 ml ÷ 7 × 3 = (36 ÷ 4) × 3

= 7 500 ml = 27 (2 × 145) + (3 × 150) +

= 7 l 500 ml Difference = 27 – 9 = (4 ×156) + (9 ×163) cm
18
113 (a) Number of perpendicular lines = 8 = 18

(b) Number of parallel lines = 4 127 (a) Ratio of number of female pupils to the = 2 831 cm
18
114 (a) 90° number of male pupils = 1 : 3
= 157.28 cm
(b) 120° Number of female pupils = 30
Number of male pupils = 3 × 30 133 (a) Mode = January
115 (a) Perimeter of the shaded region
= 90
= (14 + 24 + 25 + 25 + 24) cm (b) Total number of pupils in the group (b) Mean = (6 × 20) + (5 × 20) + (4 × 20)
= 30 + 90 3
= 112 cm = 120
128 (a) Number of roosters = 120
(b) Area of the shaded region Number of hens = 120 ÷ 4 300
3
= 2× ( 1 × 7 cm × 24 cm) = 30 =
2 (b) New number of roosters
= 168 cm2 = 120 – 10 = 100
= 110 (c) Range = 120 – 80
116 (a) Volume of cube P = 4 cm × 4 cm × 4 cm New number of hens
= 30 – 8 = 40
= 64 cm3 = 22 (d) February
New ratio of hens to roosters 134 (a) Mode = 3 boxes
Volume of cube Q = Length × Width × = 22 : 110 (b) Range = 5 – 1

Height = 1 : 5 =4
129 (a) Length of sides of cube M (c) Mean
7 × 64 cm3 = 7 cm × 4 cm × Height = 3 × 4 cm
= 12 cm
448 cm3 = 28 cm2 × Height

16 cm = Height (5 × 1) + (7 × 2) + (8 × 3) +

(b) Volume of 3 cuboids Q (5 × 4) + (5 × 5)
30
= 3 × 448 cm3 =

= 1 344 cm3

117 (a) Area of shaded plane = 88
30
= (8 cm × 6 cm) + (6 cm × 6 cm)

= 48 cm2 + 36 cm2 = 2.9

= 84 cm2

Mathematics • Year 5 A8 © Nilam Publication Sdn. Bhd.