Grade 7 Revision Memorandum– CAPS

Section 1: Numbers, Operations and Relationships.

1.1 Whole Numbers:

1. 1.e 2.g 3.a 4.f 5.d 6.b 7.c

2. a) 285 348 + 136 581 b) 256 841 – 85 654

+1 +1 +1 1 1 7 13 1
2 8 5 3 4 8 2 5 6 8 4 1
+ 1 3 6 5 8 1 - 8 5 6 5 4
= 4 2 1 9 2 9 = 1 7 1 1 8 7

c) 284 968 + 613 750 d) 210 400 - 172 528

+1 +1 1 10 9 13 9
2 8 4 9 6 8 2 1 10 4 0 10
+ 6 1 3 7 5 0 - 1 7 2 5 2 8
= 8 9 8 7 1 8 = ∙ 3 7 8 7 2

e) 2 825 x 16 f) 1 293 ÷ 26

4 1 3 ∙ ∙ 4 9 r 1 9
2 8 2 5 2 6 1 2 9 3
x 1 6 - 1 0 4
45 13
11 16 19 5 0 ∙ 2
+ 2 8 2 5 0 - 2 3 4
4 5 2 0 0 ∙ 1 9
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g) 95 711 x 86 h) 81 037 ÷ 31

4 5
∙ 2 6 1 4 r 3
3 4 78 11
3 1 0 3 7
9 5 7 1 1
- 6 2
x 8 6 89 10
1 15 17 14 12
1
6 6 - 1 8 6
+ 7 6 5 6 8 8 0 ∙ ∙ 4 3
8 2 3 1 1 4 6 3 1
1 2 7
- 1 2 4
∙ ∙ 3
3.
a) 37 = 37, 74, 111, 148, 185 b) 122 = 122, 244, 366, 488, 610
c) 768 = 1 536, 2 304, 3 072, 3 840, 4 608
d) 791 = 791, 1 582, 2 373, 3 164, 3 955

4. a) 891 = (1, 3, 9, 11, 27, 33, 81, 99, 297, 891)

b) 274 = (1, 2, 137, 274)
c) 630 = (1, 2,3, 5, 6, 7, 9, 10, 14, 15, 18, 21, 30, 35, 42, 45, 63, 70, 90, 105,
126, 210, 315, 630)
d) 558 = (1, 2, 3, 6, 9, 18, 31, 62, 93, 186, 279, 558)

5. a) 578 = 2 x 17 x 17 b) 132 = 2 x 2 x 3 x 11
c) 600 = 2 x 2 x 2 x 3 x 5 x 5 d) 234 = 2 x 3 x 3 x 13

6. a) 48 and 972
Factors of 48 = 2 x 2 x 2 x 2 x 3
Factors of 972 = 2 x 2 x 3 x 3 x 3 x 3 x 3
Highest common factor = 2 x 2 x 3 = 12
Lowest common multiple = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 = 3 888

b) 175 and 255

Factors of 175 = 5 x 5 x 7
Factors of 255 = 3 x 5 x 17

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Highest common factor = 5

Lowest common multiple = 5 x 5 x 7 x 3 x 17 = 8 925

c) 41 and 100
Factors of 41 = 41
Factors of 100 = 2 x 2 x 5 x 5
Highest common factor = 1
Lowest common multiple = 41 x 2 x 2 x 5 x 5 = 4 100

d) 333 and 666

Factors of 333 = 3 x 3 x 37
Factors of 666 = 2 x 3 x 3 x 37
Highest common factor = 3 x 3 x 37 = 333
Lowest common multiple = 2 x 3 x 3 x 37 = 666

1.2. Exponents

1. a) 92 = 81 b) 82 = 64
c) 33 = 27 d) 63 = 216

e) √9 = 3 f) √144 = 12
3 3
g) √125 = 5 h) √27 = 3

2. a) 43 = 4 x 4 x 4 b) 102 = 10 x 10
c) 119 = 11 x 11 x 11 x 11 x 11 x 11 x 11 x 11 x 11
d) b7 = b x b x b x b x b x b x b

3 3 3 2
3. a) 52 + √16 + 9 b) ( √27 + √9) + ( √125 + √49)

= 25 + √25 = (3 + 3)3 + (5 + 7)2

= 25 + 5 = (6)3 + (12)2
= 30 = 216 + 144 = 360

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3 2
c) 10 × (√36 + √64) d) (33 + 23 ) − (32 + 22 )

= 10 × (6 + 4)2 = (27 + 8) − (9 + 4)
= 10 × (10)2 = 35 − 13
= 10 × 100 = 22
= 1 000

1.3. Integers

1. a) -10 < 10 b) 2 x 5 ____ (- 2) x (- 5)

10 = 10
c) 12 – 3 ____ -12 + 3 d) 18 ÷ 2 ____ -18 ÷ 2
9 > -9 9 > -9

2. a) -70 – (-8) + 16 b) -5 x -8 + 90 x -1 – (-34)

= -70 + 8 + 16 = 40 – 90 + 34
= - 46 = - 16

c) 3 x 7 + (6 x -10) d) 80 ÷ (-10) + (-13) – (- 15)

= 21 + (-60) = -8 – 13 +15
= -39 = -6

1.4. Common Fractions

1 1 1 2
1. a) 32 + 15 b) 23 4 − 4 5
7 6 93 22
=2+5 = −
4 5
35 12 465 88
= + 10 = − 20
10 20
47 7 377 17
= 10 = 4 10 = = 18 20
20

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3 7
c) 𝑜𝑓 60 d) 𝑜𝑓 88
4 8
3 60 15 7 88 11
=4 × =8 ×
1 1
3 15 7 11
=1 × =1 ×
1 1

= 45 = 77

5 4 1 2
e) 58 × 45 f) 14 4 × 10 3
9 45 24 3 19 57 32 8
= × = ×
8 5 4 3
9 3 19 8
=1 ×1 = ×1
1

= 27 = 152

85÷5 17 65 ÷5 13
2. a) = b) = 12
15÷5 3 60÷5
36 ÷36 1 49 ÷7 7
c) =2 d) = 11
72 ÷36 77÷7

78 1 93 5
3. a) = 1 77 b) = 8 11
77 11
63 3 16 1
c) = 4 15 d) = 53
15 3

1.5. Decimal Fractions

1. a) 0,495 = 0,50 b) 0,673 = 0,67

c) 0,759 = 0,76 d) 0,133 = 0,13

2. a) 7,11 + 0,941 b) 2,491 + 2,564

71 , 1 1 21 , 41 9 1
+ 0 , 9 4 1 + 2 , 5 6 4
= 8 , 0 5 1 = 5 , 0 5 5

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c) 3,321 – 2,291 d) 5,070 – 4,135

3 , 3 2 1 4 6
- 2 , 2 9 1 5 , 10 7 10

= 1 , 0 3 0 - 4 , 1 3 5
= 0 , 9 3 5

e) 6,43 x 5 f) 8,819 x 2
2 1 1 1
6 , 4 3 8 , 8 1 9
x 5 x 2
32 , 1 5 17 , 6 3 8

g) 1,23 x 0,6 h) 8,77 x 0,7

5 4
1 1
1 , 2 3 8 , 7 7
x 6 x 7
7 , 3 8 61 , 3 9

7,38 ÷ 10 = 0,738 61,39 ÷ 10 = 6,139

Divide by 10 for one decimal place. Divide by 10 for one decimal place.

i) 7,280 ÷ 0,5 j) 9,32 ÷ 0,9

1 , 4 5 6 1 , 0 3 5
5 7 , 2 8 0 9 9 , 3 2 0
- 5 - 9
2 , 2 0 , 3
- 2 , 0 - 0 , 0
0 , 2 8 0 , 3 2
- 0, , 2 5 0 , 2 7
0 , 0 3 0 0 , 0 5 0
- 0 , 0 3 0 0, - 0 4 5
0 , 0 0 0 5 0 etc

1,456 × 10 = 14,56 1,0355555 x 10 = 10,3555555

Remembering it is recurring because
of the division by 9 which always
gives recurring numbers.

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3. a) 0,166666 ≈ 0,17 b) 16,66666% ≈ 17%

84 21
c) = 25 d) 84 %
100
45 9
e) = 20 f) 0,45
100

g) 0,375 h) 37,5%
98 49
i) = 50 j) 98%
100
2 1
k) = 50 l) 0,02
100

m) 0,8 n) 80%
11
o) p) 11%
100
38 19
q) = 50 r) 0,38
100

4. a) 45% of 90 b) 95% of 200

45 90 95 200
= 100 × = 100 ×
1 1
45 9 95 2
= 10 × 1 = ×1
1
405
= = 40,5 = 190
10

c) 32% of 3 700 d) 15% of 3 210

32 3 700 15 3 210
= 100 × = 100 ×
1 1
32 37 15 321
= × = 10 ×
1 1 1
3 321
= 1 184 =2 × 1
963 1
= = 481,5 𝑜𝑟 481 2
2

5. a) from 94 to 90 b) from 20 to 50
𝑏𝑖𝑔−𝑠𝑚𝑎𝑙𝑙 𝑏𝑖𝑔−𝑠𝑚𝑎𝑙𝑙
dec = × 100 inc = × 100
𝑏𝑖𝑔 𝑠𝑚𝑎𝑙𝑙

94−90 50−20
dec = × 100 inc = × 100
94 20

dec = 4,26% inc = 150%

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c) from 42 to 50 d) from 70 to 52
𝑏𝑖𝑔−𝑠𝑚𝑎𝑙𝑙 𝑏𝑖𝑔−𝑠𝑚𝑎𝑙𝑙
inc = × 100 dec = × 100
𝑠𝑚𝑎𝑙𝑙 𝑏𝑖𝑔

50−42 70−52
inc = × 100 dec = × 100
42 70

inc = 19,05% dec = 25,71%

Section 2: Patterns, functions and Algebra

2.1 Numeric and Geometric Patterns

1. a) 4, 6, 8, … b) 7, 10, 13, …
i) 10 12 14 i) 16 19 22
ii) add 2 to the next term ii) add 3 to the next term
iii) T = 2 + 2n iii) T = 4 + 3n
iv) T10 = 2+ 2(10) = 22 iv) T10 = 4 + 3(10) = 34

c) 4, 12, 36, … d) 1, 4, 9, …
i) 108 324 972 i) 16 25 36
ii) multiply the previous term by 3 ii) square the position
4
iii) 𝑇 = 3 × 3𝑛 iii) T = n2
4
iv) 𝑇 = 3 × 310 = 78 732 iv) T = 102 = 100

e) 100, 94, 88, … f) 200, 100, 50 …

1 1
i) 82 76 70 i) 25 12 2 64

ii) subtract 6 from the last term ii) divide the last term by 2
iii) T = 106 – 6n iii) T = 400 ÷ 2n
25
iv) T = 106 – 6(10) = 46 iv) 𝑇 = 400 ÷ 210 = 64 = 0,390625

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g) i) Pattern in table:
Position 1 2 3 4 5 6 7
Number
of 1 3 5 7 9 11 13
crayons

ii) add 2 more crayons to the previous pattern

iii) T = 2n – 1
iv) T = 2(10) – 1 = 19

h) i) Pattern in table form:

Position 1 2 3 4 5 6
Number
1 4 7 10 13 16
of leaves
ii) Add three leaves to the previous pattern
iii) T = 3n – 2
iv) T = 3(10) – 2 = 28

i) i) Pattern in table form:

Position 1 2 3 4 5 6 7
Number
of lego 1 2 4 8 16 32 64
tips
ii) Double the previous blocks number of lego tips
1
iii) 𝑇 = 2 × 2𝑛
1
iv) 𝑇 = 2 × 210 = 512

j) i) Pattern in table form:

Position 1 2 3 4 5 6 7
Number
8 7 6 5 4 3 2
of petals
ii) Subtract one petal from the previous flower
iii) T=9–n

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iv) T = 9 – 10 = -1 So this flower doesn’t exist and the pattern ends at

position 9 with no petals on the flower.

2.2. Functions and Relationships

1. a) i) 15 ii) 10 iii) 16 iv) 25

b) i) 29 ii) 29 iii) 0 iv) 53

c) i) 18 ii) 10 iii) 108 iv) 6

d) i) 2 ii) 55 iii) 11 iv) 143

e) i) 165 ii) 5 iii) 41 iv) 3

2. a) i) -18 ii) 12 iii) 21

iv) 51 v) 𝑦 = 3𝑥 − 9

b) i) 24 ii) 8 iii) 24
iv) 99 v) T = n2 – 1

c) i) 14 ii) -10 iii) -26

iv) -58 v) T = 2 – 4n

d) i) -35 ii) 28 iii) 55

iv) 109 v) T = 9n + 1

e) i) -15 ii) -7 iii) -2

iv) 5 v) T = n – 10

3. a) 𝑦 = 3𝑥 + 7
i) y = 3(3) + 7 = 16 ii) y = 3(7) + 7 = 28

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iii) y = 3(-3) + 7 = -2 iv) -5 = 3x + 7

-12 = 3x
-4 = x
v) y = 22 = 3x + 7 vi) y = 40 = 3x + 7
15 = 3x 33 = 3x
5=x 11 = x

b) 𝑦 = 𝑥2 + 2
i) y = 32 + 2 = 11 ii) y = 02 + 2 = 2
iii) y = (-5)2 + 2 = 27 iv) y = 6 = x2 +2
4 = x2
x=±2

v) y = 38 = x2 + 2 vi) y = 102 = x2 + 2
36 = x2 100 = x2
x=±6 x = ± 10

c) 𝑦 = −5𝑥 + 1
i) y = -5(0) +1 = 1 ii) y = -5(4) + 1 = -19
iii) y = -5(-5) + 1 = 26 iv) y = -49 = -5x + 1
-50 = -5x
x = 10

v) y = 11 = -5x + 1 vi) y = -99 = -5x + 1

10 = -5x -100 = -5x
-2 = x 20 = x

d) 𝑦 = 𝑥2 + 𝑥
i) y = 02 + 0 = 0 ii) y = 32 + 3 = 12
iii) y = (-6)2 – 6 = 30

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2.3 Algebraic Expressions

1. a) 𝑥 3 + 4𝑥 2 − 9𝑥 + 12 b) 3𝑚 + 𝑚3 − 12
Variable = x variable = m
Constant = 12 constant = -12

c) 𝑡 5 − 100𝑡 d) 𝑝6 − 78𝑝 + 3
Variable = t Variable = p
Constant = 0 Constant = 3

2. a) Sam buys 4 candy bars and 12 packets of chips for a total of R100.
100 = 4 x cb + 12 x ch
b) There are 3 times as many girls in the class as there are boys.
Class = 3 x g + 1 x b
c) A Honda uses twice as much petrol as a Ford does.
Honda = 2 x petrol
d) A dog and a cat eat 40kg of food per month. The dog eats twice as much as
the cat.
40 = dog + cat and dog = 2 x cat so
40 = 2 x cat + cat
40 = 3 x cat
e) A tree grows 15cm each year
height = start + 15 x number of years.

2.4. Algebraic Equations

1. a) 3𝑥 = 30 b) 7 + 𝑥 = 10
x = 10 x=3

c) 45 = 5 + 2𝑥 d) 8 ÷𝑥 =2
40 = 2x x=4
x = 20

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e) 40 ÷ 𝑥 + 5 = 10 f) 20 + 𝑥 = 2𝑥 + 12
40 ÷ x = 5 20 – 12 = 2x - x
x=8 8=x

2. a) ab + c b) a2 – 3d
= (3) x (7) + 0 = (3)2 – 3 x -1
= 21 =9+3
= 12

c) ac – bd d) d3 – a3 +3b – 3c
= (3) x (0) – (7) x (-1) = (-1)3 –(3)3 +3 x 7 – 3 x 0
=0+7 = -1 – 27 + 21 – 0
=7 = -7

e) 7b – 3ad f) 45 – ab + ad – ac
= 7 x 7 – 3 x 3 x (-1) = 45 – 3 x 7 + 3 x (-1) – 3 x 0
= 49 + 9 = 45 – 21 – 3 – 0
= 58 = 21

2.5. Graphs

1. a) i) non-linear b) i) non-linear
ii) increasing ii) decreasing

c) i) linear i) linear
ii) increasing ii) increasing

e) i) linear
ii) decreasing

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2. a) A car travels at exactly 100km/h for 1 hour.

b) A car accelerates smoothly to 100km/h

c) A car has to do an emergency stop.

d) A sales graph for the year, where November, December and January have
high sales, February, March and April have low sales, and May – October
have medium sales.

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e) The amount of data used over the course of one month, if 1GB is purchased
on the first, and a second GB is purchased in the middle of the month.

Section 3: Space and Shape (Geometry)

3.1 Geometry of 2D Shapes

1. a) Equilateral triangle b) square

c) rhombus d) trapezium
e) isosceles triangle f) rectangle
g) kite h) paralelogram
i) right-angled triangle j) heptagon

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2.

3. a) congruent b) similar
c) neither d) congruent
e) similar

3.2. Geometry of 3D Objects

1. a) cube b) rectangular prism

i) 6 faces i) 6 faces
ii) 8 vertices ii) 8 vertices
iii) 12 edges iii) 12 edges

c) triangular prism d) square-based pyramid

i) 5 faces i) 5 faces
ii) 6 vertices ii) 5 vertices
iii) 9 edges iii) 8 edges

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e) triangular-based pyramid f) cylinder

i) 4 faces i) 3 faces
ii) 4 vertices ii) no vertices
iii) 6 edges iii) 2 edges

2. a) a cube

b) a rectangular prism.

3.3. Geometry of Straight Lines

1. a) line segment b) ray

one end point to another one end point to a continued point
point.

c) line d) parallel lines

continuous on both ends equally distant from each other for
every point.

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e) perpendicular lines
intersect each other at a right angle.

3.4. Transformation Geometry

1. a) translation – move a point or shape a certain distance up or down, and left or

right.
b) reflection – mirror the point or shape around a line
c) rotation – shift the shape around a certain point.
d) line of symmetry – the line that creates a mirror image within the shape.
e) enlargement - making the shape bigger by a certain factor
f) reduction – making the shape smaller by a certain factor.

2. a) either reflection or translation b) enlargement

c) translation d) rotation
e) reduction or enlargement

3. a) correct b) incorrect
c) correct d) incorrect

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Section 4: Measurement

4.1 Area and Perimeter of 2D Shapes

1. a) Perimeter = AB + BC + CD + DE + EA All sides are equal to each

= 6cm + 6cm + 6cm + 6cm + 6cm other, and given as 6cm
= 30cm

b) Perimeter = AB + BC + CA
= 8.48km + 7.98 km + 8.54 km
= 25km

c) Perimeter = AB + BC + CD + DE + EF + FA
= 0,61m + 0,76m + 0,59m + 0,63m + 0,85m +1,09m
= 4,53m

d) Perimeter = AB + BC + CD + DE + EF + FG + GH + HA
= 28,26mm + 22,43mm + 35,42mm + 39,57mm + 43,05mm + 34,8mm
+ 29,32 mm + 24,17mm
= 257,02mm

2. a) area of a square = 𝑙 × 𝑙 = 𝑙 2
b) perimeter of a square = 𝑙 + 𝑙 + 𝑙 + 𝑙 = 4𝑙
c) area of a rectangle = 𝑙 × 𝑏
d) perimeter of a rectangle = 𝑙 + 𝑙 + 𝑏 + 𝑏 = 2𝑙 + 2𝑏
1
e) area of a triangle = ×𝑏 ×⊥ℎ
2

f) perimeter of a triangle. = 𝑠𝑖𝑑𝑒 1 + 𝑠𝑖𝑑𝑒 2 + 𝑠𝑖𝑑𝑒 3

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3. a) Perimeter = 4𝑙 b) Perimeter = 2𝑙 + 2𝑏
= 4 × 120𝑐𝑚 = 2 × 40𝑚 + 2 × 20𝑚
= 480𝑐𝑚 = 120 𝑚

Area = 𝑙 2 Area = 𝑙 × 𝑏
= (120𝑐𝑚)2 = 40𝑚 × 20𝑚
= 14 400𝑐𝑚2 = 800𝑚2

c) Perimeter = 𝑠1 + 𝑠2 + 𝑠3 Perimeter = 𝑠1 + 𝑠2 + 𝑠3
= 15𝑐𝑚 + 25𝑐𝑚 + 29,15𝑐𝑚 = 125,24𝑐𝑚 + 127,58𝑐𝑚 + 117,61𝑐𝑚
= 69,15𝑐𝑚 = 370,43𝑐𝑚

1 1
Area = 2 × 𝑏 ×⊥ ℎ Area = 2 × 𝑏 ×⊥ ℎ
1 1
= 2 × 25𝑐𝑚 × 15𝑐𝑚 = 2 × 127,58𝑐𝑚 × 103,14𝑐𝑚

= 187,5𝑐𝑚2 = 6 579,3𝑐𝑚2

4.2. Surface Area and Volume of 3D Objects

1. a) Surface area of a cube ∴ 𝑆𝐴 = 6𝑙 2

b) Surface area of a rectangular prism ∴ 𝑆𝐴 = 2𝑙𝑏 + 2𝑙ℎ + 2𝑏ℎ
c) Volume of a cube ∴ 𝑉𝑜𝑙 = 𝑙 × 𝑙 × 𝑙 = 𝑙 3
d) Volume of a rectangular prism. ∴ 𝑉𝑜𝑙 = 𝑙 × 𝑏 × ℎ

2. Find the surface area and volume for each of these shapes:
a) 𝑆𝐴 = 2𝑙𝑏 + 2𝑙ℎ + 2ℎ𝑏
∴ 𝑆𝐴 = 2 × 30𝑐𝑚 × 20𝑐𝑚 + 2 × 30𝑐𝑚 × 25𝑐𝑚 + 2 × 25𝑐𝑚 × 20𝑐𝑚
∴ 𝑆𝐴 = 1 200𝑐𝑚2 + 1 500𝑐𝑚2 + 1 000𝑐𝑚2
∴ 𝑆𝐴 = 3 700𝑐𝑚2

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𝑉𝑜𝑙 = 𝑙 × 𝑏 × ℎ
∴ 𝑉𝑜𝑙 = 30𝑐𝑚 × 20𝑐𝑚 × 25𝑐𝑚
∴ 𝑉𝑜𝑙 = 15 000 𝑐𝑚3

b) 𝑆𝐴 = 6𝑙 2
∴ 𝑆𝐴 = 6 × 7𝑚 × 7𝑚 = 6 × (7𝑚)2
∴ 𝑆𝐴 = 294 𝑚2

𝑉𝑜𝑙 = 𝑙 3
∴ 𝑉𝑜𝑙 = 7𝑚 × 7𝑚 × 7𝑚 = (7𝑚)3
∴ 𝑉𝑜𝑙 = 343 𝑚3

c) 𝑆𝐴 = 2𝑙𝑏 + 2𝑙ℎ + 2ℎ𝑏

∴ 𝑆𝐴 = 2 × 19𝑚𝑚 × 15𝑚𝑚 + 2 × 19𝑚𝑚 × 17𝑚𝑚 + 2 × 17𝑚𝑚 × 15𝑚𝑚
∴ 𝑆𝐴 = 570𝑚𝑚2 + 646𝑚𝑚2 + 510𝑚𝑚2
∴ 𝑆𝐴 = 1 726 𝑚𝑚2

𝑉𝑜𝑙 = 𝑙 × 𝑏 × ℎ
∴ 𝑉𝑜𝑙 = 19𝑚𝑚 × 17𝑚𝑚 × 15𝑚𝑚
∴ 𝑉𝑜𝑙 = 4 845 𝑚𝑚3

d) 𝑆𝐴 = 2𝑙𝑏 + 2𝑙ℎ + 2ℎ𝑏

∴ 𝑆𝐴 = 2 × 5𝑐𝑚 × 5𝑐𝑚 + 2 × 5𝑐𝑚 × 20𝑐𝑚 + 2 × 20𝑐𝑚 × 5𝑐𝑚
∴ 𝑆𝐴 = 50𝑐𝑚2 + 200𝑐𝑚2 + 200𝑐𝑚2
∴ 𝑆𝐴 = 450𝑐𝑚2

𝑉𝑜𝑙 = 𝑙 × 𝑏 × ℎ
∴ 𝑉𝑜𝑙 = 5𝑐𝑚 × 5𝑐𝑚 × 20𝑐𝑚
∴ 𝑉𝑜𝑙 = 500𝑐𝑚3

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Section 5: Data Handling

5.1 Collect, Organise and Summarize Data

1. A sample is a small set/group taken from a larger set or group – usually randomly.
A population is the entire group with a specific characteristic.

2. a)
Intervals Tally Marks Frequency
10 – 19 || 2
20 – 29 || 2
30 – 39 |||| 5
40 – 49 |||| ||| 8
50 – 59 |||| 4

b) 5 1, 5, 6, 9
4 1, 2, 2, 3, 6, 6, 6, 8,
3 0, 2, 7, 8, 9
2 0, 5
1 6, 8

c) You could do intervals of 5, i.e. 15 – 19, 20 – 24, 25 – 29 and so on

You could also do intervals of 6 or 7 as well.

𝑠𝑢𝑚
d) Average = 𝑡𝑜𝑡𝑎𝑙 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛𝑠
16+18+20+25+30+32+37+38+39+41+42+42+43+46+46+46+48+51+55+56+59
Average = 21
830
Average = 21

Average = 39,52
e) Median = 42
f) Mode = 46
g) Maximum = 59
h) Minimum = 16

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i) Range = 59 – 16 = 43
j) Only certain how-to videos could have been sampled.
A non-random group of people were asked.
The sample is too small.
Any other relevant answer.

5.2. Representing data

1. a) a histogram, because the data is grouped into continuous intervals.

b)

Histogram of frequency of minute intervals of

YouTube videos watched
9
8
7
6
Frequency

5
4
3
2
1
0
0-9 10 – 19 20 – 29 30 – 39 40 – 49 50 – 59
Intervals of minutes

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Colours

5 13
10

32

50

Green Red Blue Yellow Pink

2.
Calculations:
Total observations = 13 + 32 + 50 + 10 + 5 = 110
13 13
Green angle = 110 × 360° = 43° Green %age = 110 × 100 = 12%

(Rounding to the nearest unit)

32 32
Red angle = 110 × 360° = 105° Red %age = 110 × 100 = 29%
50 50
Blue angle = 110 × 360° = 164° Blue %age = 110 × 100 = 45%
10 10
Yellow angle = 110 × 360° = 33° Yellow %age = 110 × 100 = 9%
5 5
Pink angle = 110 × 360° = 16° Pink %age = 110 × 100 = 5%

5.3. Interpret, Analyse and Report Data

1. a) Meat lover
b) Vegan
c) Pizza = 1 + 5 + 9 = 15 Pasta = 2 + 4+ 8 = 14
Burger = 3 + 3 +12 = 18 Salads = 4 + 7 + 5 = 16
Therefore, the burger is the most popular choice for all the diets.
d) Salads
e) Burgers

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f) Total = all pizza + all pasta + all burger + all salads

= 15 + 14 + 18 + 16
= 63 people were interviewed.

2. a) Nuts
b) Glazed
c) 30 people = 30% of total
By inspection we can see that 100 people were interviewed.
Or we could say 30 ÷ 30% = 100 people.
d) 150 x 18% = 27 people liked glazed doughnuts.

5.4 Probability

1 1
1. a) b)
6 6
1
c) 0 d) 6

1 1
2. a) b)
4 4
1
c) 0 d) 4

3. Possible outcomes: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
1
Probability of each: 10 for each one.

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