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ANNUAL NATIONAL ASSESSMENT
GRADE 9
MATHEMATICS
MEMORANDUM
SET 1: 2012 EXEMPLAR
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Mathematics Grade 9 ANA Exemplars 2012 Memo (English)
Grade 9 Mathematics memorandum from the ANA Exemplars 2012 (English), 33 pages. Read online or get the free PDF download, question paper linked when available.
Mathematics · Grade 9 · ANA Exemplars · 2012 · English. Memorandum, 33 pages. Read online or download the PDF.
- Subject
- Mathematics
- Grade
- Grade 9
- Language
- English
- Document type
- Memorandum
- Year
- 2012
- Exam period
- ANA Exemplars
- Curriculum
- CAPS
- Pages
- 33
- File size
- 343.8 KB
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ANNUAL NATIONAL ASSESSMENT 2012
GRADE 9 MATHEMATICS
EXEMPLAR MEMORANDUM
REAL NUMBER SYSTEM
1.
1.1.1 4 - rational (1)
1.1.2 √2 - irrational (1)
1.1.3 0,2- rational (1)
1.2
NUMBERS REAL NON - REAL UNDEFINED
0 √
7
7 √
0
√7 √
√−7 √
(4)
1.3.1 7 (1)
0,7 =
10
1.3.2 13 (1)
0,13 =
100
1.3.3 201 1 (1)
2,01 = =2
100 100
1.4.1 2,5 × 10 × 7 = 1,75 × 10 (1)
1.4.2 0,007 (2)
2
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1.4.3 1,12 × 10 × 3 × 10 = 3,36 × 10 (2)
1.5.1 1,6 (1)
1.5.2 −√5 (1)
1.6.1 0,152, 0,153, 0,154 anyone of the three (1)
1.6.2 0,71; 0,72; 0,73; 0,74; 0,75; 0,76; 0,77 anyone of the seven (1)
1.7.1 2 and 3 (2)
1.7.2 4 and 5 (2)
1.7.3 1 and 2 (2)
1.7.4 4 and 5 (2)
1.8
NATURAL WHOLE INTEGER RATIONAL IRRATIONAL REAL
NUMBER NUMBER
7 √ √
15
√ √
1
2
8
0,081 √ √
2 √ √
−√16 √ √ √
0,528 √ √
2,6 √ √
6 √ √ √ √ √
2
3
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1.9.1 0,6 < 0,625 < 0,75 < 0,8 (1)
1.9.2 0,2 < 0, 2̇ 0̇ < 0, 2̇ < 0,24 (1)
1.9.3 √0,36 < √0,366 < 0, 6̇ < 0,69 (1)
1.9.4 −3 5 1 −2 (1)
< − <− <
2 6 6 3
1.9.5 −0, 12 < −0,11 < ̇−0,1 < −0,01̇ (1)
FINANCIAL MATHEMATICS
1. (3)
1.1 Selling price=22 × 1,50 = 33
Profit= Selling price – cost price = 33 − 20 = 13
1.2.1
a) 18% 75 000 = 13 500 (2)
b) Tax = 12 000 + 20% ( 97 500 − 80 000) (2)
= 12 000 + 2 100
= 14 100
c) Tax= 20 000 + 25% 30 000 (2)
= 27 500
d) Tax= 42 000 + 35% 80 000 (2)
= 70 000
1.2.2
Taxable income R140 000 R100 000 R230
000
Tax R25 000 R16 000 R45
4
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500
1.3 Increase: 1 740 − 1500 = 240
% increase = × 100 = 16 (2)
1.4 = (1 + ) (2)
= 1500(1 + 2(0,11))
= 1 830
or
× ×
SI= = = 330
Total amount = 1500 + 330 = 1 830
1.5.1 Investment = (1 + ) (3)
11,5
= 2 750 1 +
100
= 5 891,92
1.5.2 Interest earned= 5 891,92 − 2 750 = 3 141,92 (1)
1.6.1 = (1 + ) (4)
×
12
= 9500 1 +
100 × 4
= 24 463,29
Interest = 24 463,29 − 9 500
= 14 963,29
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1.6.2 = (1 + ) (4)
×
8
= 9500 1 +
100 × 2
= 17 793,32
Interest= 17 793,32 − 9 500
= 8293,32
1.6.3 = (1 + ) (4)
×
6,5
= 9500 1 +
100 × 12
= 15 956,86
Interest= 15 956,86 − 9 500
= 6456,86
1.6.4 = (1 + ) (4)
7,25
= 9500 1 +
100
= 16630,37
Interest= 16630,37 − 9 500
= 7130,37
6
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RATIO AND PROPORTION
1 (3)
1.1 Speed= km/h = 300 km/h
1.2 (3)
MINUTES 2 4 6 8 20
COST R1,60 R3,20 R4,80 R6,40 R16,00
1.3 (3)
NO OF MEN TIME
5 45
9
45 × 5
= = 25ℎ
9
OR
Time taken by 5 men = 45ℎ
Time taken by 1 man = (45 × 5)ℎ
×
Time taken by 9 men = = 25ℎ
1.4 (3)
TIME NO OF HOURS
8 3
2
2 =8×3
= 12
OR
In 8 hours no of workers needed = 3
In 1 hour no of workers needed = 3 × 8
In 2 hours no of workers needed = 3 × 8 ÷ 2
= 12
7
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1.5 (2)
NO OF KG COST IN R
3 24
7
7
=
24 3
7 × 24
=
3
Cost = R56
OR
Cost of 3 kg = 24
Cost of 1 kg = 24 ÷ 3
Cost of 7 kg = 8 × 7
= 56
1.6.1 Direct proportion (1)
1.6.2 =5 (1)
1.6.3 = 5(25) = 125 (1)
1.7 COST IN R NO OF (3)
APRICOTS
5,60 15
10,08
15 10,08
=
5,60
10,08 × 15
=
5,60
= 27
8
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1.8 , (3)
Cost per ℓ = = 8,65
1.9 (3)
NO OF KG COST IN R
0,35 25,10
1
1
=
25,10 0,35
2510
=
35
= 71,714 …
~ 71,71
1.10 (3)
NO OF KG COST IN R
12,5 90
17,2
7,2
=
90 12,5
7,2 × 90
=
12,5
Cost= R51,84
1.11 (3)
Distance in km No of ℓ
130 6,5
80
80
=
6,5 130
80 × 6,5
=
130
=4
No of ℓ = 4
1.12 (3)
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No of ℓ No of m
5 40
1
1
=
40 5
1 × 40
=
5
=8m
SPEED/TIME /DISTANCE
1. × =
1.1 120 × = 600
= h
= 5h
(2)
∴time taken is 5 hours.
1.2 × =
× 8 = 600
600
= /ℎ
8
= 75 /ℎ
(2)
10
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2. S (km/h) T (h) D (km) (3)
car 100 x 1 0x
air 300 6-x 300(6-x)
100 + 300(6 − ) = 1200
100 + 180 = 1200
−200 = −600
=3
Time by air = (6 − 3)ℎ
= 3ℎ
3. = (120 × 1,5) km = 180 km (3)
,
km
,
= = 125 km/h
h
660
z= h = 3h
220
4. d =s×t (3)
= (5 × 0,5) km
= 2,5 km
s × 0,25 = 2,5
2,5
s=
0,25
s = 10 km/h
5.1 Average speed = km/h (3)
= 7,5 km/h
5.2 Distance = × (1)
= (7,5 × 6) km
= 45 km
6. s×t=d (3)
(p + 20)t = q
t= h
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EXPONENTS
1.
1.1 99 × 1001 = 99 × 1 = 99 (1)
1.2 (1)
× = =
1.3 ( ) (2)
= =
1.4 (8 )(−4 ) −32 (3)
=
(2 ) 4
= −8
1.5 2 8 (3)
2
2 2
=
2
2
=
2
=2
1.6 15 . 3 . 25 (4)
9 . 125
. . .
=
.
=
=3
1
=
3
1.7 5 5 10 10 (3)
(2 )× = (2 )× = = 10 =
1.8 1 (1)
( )( )= = =
1.9 1 (2)
( . . )÷ = ÷ =
1.10 (3 ) ( ) 27 ×1 (3)
= = 27 = 27
( )
12
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PATTERNS, FUNCTIONS AND ALGEBRA
1
1.1 90 and 69 (2)
1.2 36; 72; 108; 144; 180 ; 216 (2)
1.3 39; 48 (2)
2
2.1 Add 6. (1)
2.2 =6 −3 (1)
2.3 6 − 3 = 45
6 = 48
=8 (1)
The 8th term = 45
3.1 Number 1 2 3 4 5 6 7 (2)
of
panels
Number 2 3 4 5 6 7 8
of poles
Number 6 12 18 24 30 36 42
of slabs
3.2 = +1 (2)
3.3 =6 (2)
3.4 ( ) = 10 + 1 = 11 (2)
( ) = 6(10) = 60
4 (3)
Input Rule Output
1 −2 + 3 1
2 −1
3 −3
5
Input Rule Output
−2 2 −1 7
13
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−1 1
0 −1
1 1
2 7 (5)
6
6.1 = −2(0) + 6(0) − 8 = −8 (1)
6.2 = −2(−1) + 6(−1) − 8 = −2 − 6 − 8 = −16 (1)
6.3 = −2 +6 −8
6.4 = −2( + 2) + 6( + 2) − 8 (4)
= −2( + 4 + 4) + 6 + 12 − 8
= −2 −8 −8+6 +4
= −2 −2 −4
7
7.1 =3 =1 (1)
7.2 = 3 = 27 (1)
7.3 1 (1)
=3 =
9
PRODUCTS
1
1.1 6 −3−5=6 −8 (2)
1.2 5 −6 +3= − +3 (3)
1.3 5 −6 + 3 = −6 +8 (3)
1.4 −9 (2)
1.5 9 −4 (2)
1.6 3( + 2 − 3) = 3 +6 −9 (3)
1.7 −5(2 + − ) = −10 −5 +5 (3)
1.8 − 8 + 16 (3)
1.9 4 − 12 + 9 (3)
14
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1.10 3(16 − 24 + 9) = 48 − 72 + 27 (3)
1.11 −5 +6−( + 4 + 4) = −9 + 2 (5)
1.12 3 − 6 − 12 + 12 − 5 + 10 = −14 + 6 + 10 (6)
1.13 2 1 2 2
(9 − 1) − ( − 1) = 3 − − +
3 3 3 3
7 1
= + (6)
3 3
1.14 1 1 1 1 1 1
4( −
)−( + + )=4 − − − −
16 2 16 4 2 16
1 5
=3 − −
2 16
(6)
1.15. 1 2 1
− − ( − + )
1 1
= − − +2 −
2 (6)
=2 −
FACTORS
1.
1.1 4(2 + 3 ) (2)
1.2 −5(2 + 3 ) (2)
1.3 + (1)
1.4 4(6 − 3 + 4) (2)
1.5 ( − 1) (2)
1.6 −2 (1 + ) (2)
1.7 2 (4 − 5) (2)
1.8 6 ( −2 + 3) (2)
1.9 4 (2 −3 − 5) = 4 (2 − 5)( + 1) (2)
1.10 ( − ) − ( − ) = ( − )( − ) (3)
1.11 ( − )( − ) = ( − )( − )( + ) (3)
1.12 − )( + ) (2)
1.13 ( − ) = ( − )( + ) (3)
15
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1.14 ( + )( − )=( + )( + )( − ) (3)
1.15 (4 + )(4 − ) (2)
1.16 3( −9 ) = 3( − 3 )( + 3 ) (3)
1.17 1 1 (3)
( − 4) = ( − 2)( + 2)
2 2
1.18 1 1 (3)
(4 − ) = (2 − )(2 + )
8 8
OR
4 − (2 − )(2 + )
=
8 8
2
2.1 (97 − 3)(97 + 3) = (94)(100) = 9400 (2)
2.2 (101 − 100)(101 + 100) = 1(201) = 201 (2)
2.3 (55 − 25)(55 + 25) = (30)(80) = 2400 (2)
2.4 (20,5 − 15,5)(20,5 + 15,5) = (5)(36) = 180 (2)
EQUATIONS
1. (2)
1.1 6 = 24 (3)
=4
1.2 ( − 3) = 0
= 0 or =3 (2)
1.3 2 + 4 = 9 − 12
7 = 16 or 16 = 7
16 16 (3)
= =
7 7
1.4 × 36: 9( − 5) − 4(3 − 4 ) = 468
9 − 45 − 12 + 16 = 468
25 = 525
= 21 (4)
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1.5 × 15: 45 + 3 + 195 = 5(2 + 1)
48 + 195 = 10 + 5
38 = −190
= −5 (4)
1.6 5 =5
∴ =4 (2)
1.7 2 =2
∴− =5 (2)
= −5
1.8 3 =3 (2)
∴ +1 =4
=3
2.1 + +1+ + 2 = 78
3 + 3 = 78
3 = 75
= 25
The numbers are 25; 26; 27 (4)
2.2
AGE NOW AGE IN 4 YEARS
Ben +4
Patrick 5 5 +4
5 + 4 = 3( + 4)
5 + 4 = 3 + 12
2 =8
=4
Ben’s age is 4 years and Patrick’s is 20 years (4)
2.3 Area = ×
−12 +6 = −6 ×
−12 + 6
=
−6
−6 (2 − 1)
=
−6
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= (2 − 1) (3)
2.4 Area = ×
2 − 8 = × ( + 2)
2 −8
=
+2
2(− 4)
=
+2
2( + 2)( − 2)
=
+2
= 2( − 2) (5)
2.5 =2
Area = × = 32
2 × = 32
2 = 32
= 16
= ±4
=4 (because > 0)
∴ =4 and = 2(4) = 8 (4)
INEQUALITIES
1.
1.1
(2)
1.2
•
(2)
18
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1.3
• •• (2)
1.4
(2)
1.5
•••
(2)
2
2.1 <2
(2)
2.2 3≤ ≤ 7and ∈
(2)
2.3 √3 ≤ < 2,1and ∈
(2)
2.4 −3 ≤ ≤ 0 and ∈ or −4 < < 1 and ∈ (2)
2.5 < −2 or ≥ 0 and ∈
(2)
ALGEBRAIC FRACTIONS
1.
1.1 2 (2)
1.2 − (2)
1.3 3 2 6 (5)
× × =
19
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1.4 ( − 4) 4 ( − 1) ( − 4)( − 1) (5)
× × =
4 ( − ) ( − )
1.5 2 (1)
+ = =
2 2 2
1.6 3(7 − 5) − 2(10 − 7) 21 − 15 − 20 + 14 −1 (3)
= =
6 6 6
1.7 5(6 − 10) + 6(3 − 5) − 180 30 + 50 + 18 − 30 − 180 (4)
=
30 30
48 − 60
=
30
16(3 − 10)
=
30
8(3 − 10)
=
15
1.8 3( − 1) − ( + 2)( + 1) 3 − 3 − − 3 − 2 (4)
=
( − 1)( + 1) ( − 1)( + 1)
− −5
=
( − 1)( + 1)
20
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GRAPHS
1
1.1
2
3
(4)
21
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1.2
y 23 x 1
3
2
(4)
2.
2.1 = −2 − 2 (2)
2.2 = −2 + 3 (2)
2.3 1 (2)
= +1
3
22
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3. (3)
Y
8
7
x 3
6
5 y x 1
4
3
3y x 3
2
B
1
A
-5 -4 -3 -2 -1 0 1 2 3 4 X
C
4. =2 +2 (3)
23
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SPACE AND SHAPE
1.
1.1 A.Trapezium
B. Octagon
C. Triangle
D. Rectangle
E. Parallelogram (5)
1.2 A. cylinder (4)
B. rectangular-based pyramid
C. cube
D. triangular pyramid
2.1 (1)
2.2 70° East of North or 20° North of East or 70° North (2)
3. In ∆ and ∆ (5)
= = 90°(given)
is common
= (given)
∴∆ ≡∆ (90°, ℎ , )
4. + = (Pyth) (5)
+ (3) = (5)
= 25 − 9
= 16
= 4unit
= −
= 4 − √7 unit
= + (Pyth)
= 7 + 9 unit
= 16unit
= 4 units
24
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5. 2 + = 180° (co-int< ; || ) (4)
+ 120° = 180°
2 = 60°
2 + + = 180° ( sum of < of ∆)
but = (< opp equal sides)
2 +2 = 180°
60° + 2 = 180°
2 = 120°
∴ = 60°
2 + + = 180°(sum of < of ∆)
but = (< oppequal sides)
2 +2 = 180°
120° + 2 = 180°
2 = 60°
∴ = 30°
2 + + = 180° ( suppl< on a str line
or is a str. <)60° + + 30° = 180°
2 = 180° − 90°
= 90°
6. = (< opp. sides)
40° + + = 180° ( sum of < of ∆) (6)
2 = 180° − 40°
2 = 140°
= 70°
= = 70° ( alt < , || )
+ = 180° (suppl< on a str line or K is a
str. <)
= 180° − 70°
2 = 110°
25
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7.
7.1 In ∆ and ∆ (4)
= ( common side )
= (alt. < , || )
= ( alt. < , || )
∴∆ ≡∆ (<, <, )
∴ =
= ( . ∆)
7.2 In ∆ and ∆ (4)
= (proved above)
= (alt. < , || )
= ( alt. < , ||)
∴∆ ≡∆ ( <, <, )
∴ = = ( . ∆)
7.3 The diagonals of a parallelogram bisect each other. (1)
8. = = = (prop. sides of similar quads) (6)
7
=
9 28
9×7
= = 2,25
28
6 7
=
28
6 × 28
= = 24
7
26
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9. In ∆ and ∆ (6)
= (alt < , || )
= (alt < , || )
= (vert. opp< )
∆ ||| ∆ (<, <, <)
∴ = = (prop. sides of similar ∆s)
2 3
∴ =
6
=9
,
and =
= 7,5
= +
= 2,5 + 7,5 = 10
TRANSFORMATIONS
1.
1.1 (−1; −4) (1)
1.2 (3; −6) (1)
1.3 (−3; −4) (1)
1.4 (−4; 3) (1)
2. Reflection about the X-axis (4)
3. Vertical translation
3.1 Reflection about the Y-axis (2)
3.2 Reflection about the Y-axis and then reflection about
the X-axis or reflection about the X-axis and then
reflection about the Y-axis (4)
4. ′(−6; 9) ; ′(3; 15) ; ′(15; −6) (3)
27
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5.1
(1)
5.2
(1)
28
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MEASUREMENT
1.
1.1 Area of ∆ = ℎ = × 3 × 4 cm = 6cm (2)
1.2 Area of ∆ = ℎ = × 6 × 4 cm = 12cm (3)
1.3 Area of ADEG= ( + ) = (9 + 6)4 cm = 30cm (3)
1.4 Area of BDEG= × ℎ = 6 × 4 cm = 24cm (3)
1.5 = + (Pyth)
= 3 + 4 cm
= 9 + 16cm
= 25cm
∴ BE = 5cm
1.6.1 Perimeter of ∆ = (5 + 5 + 6)cm = 16cm (2)
1.6.2 Perimeter of ADEG = (9 + 5 + 6 + 4)cm = 24cm (4)
2. = + (Pyth)
= 49 + 120 unit
= 169 unit
= 13 units
Area =
1 13
=( × × )unit
2 2
= 66,36 unit (3)
3.
3.1 Total surface area= 2(8 × 12) + 2(12 × 18) + 2(8 × 18)cm
= 912cm (6)
3.2 Volume= × ×ℎ
= (8 × 12 × 18)cm
= 1728cm (3)
29
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4. Volume of cylinder= ℎ
22
550 =
× (5) × ℎ
7
550 × 7
ℎ=
22 × (5)
(4)
ℎ=7
5. Volume = × ×ℎ
50 × 30 × ℎ = 24000
24000
ℎ=( )
50 × 30
ℎ = 16 (3)
DATA HANDLING
1.
1.1 20 30 30 40 50 50 60 70 70 70 70 80 80 90 100 (1)
1.2
1.2.1 Median = 70 (1)
1.2.2 Mode = 70 (1)
1.2.3 Range = 100 − 20 = 80 (1)
1.2.4 Mean = = 60,67 to decimal places (2)
2.
2.1 24 passengers (1)
2.2 Range = 70 − 1 = 69 (1)
2.3 Mean= = 26,7 = 27 (3)
2.4 AGES IN YEARS TALLY FREQUENCY
(INTERVAL) MARKS
1 - 13 IIIIII 7
14 - 26 III 3
27 - 39 IIIIIII 8
40 - 70 IIIII 6 (4)
30
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2.5 27-39 (1)
2.6 18 passengers (1)
3.
3.1 22 (1)
3.2 range= 15 − 2 = 13 (1)
3.3 SHOE SIZES TALLY FREQUENCY
(INTERVAL) MARKS
1-5 IIII 4
6 - 10 IIIIIIIIII 12
11 - 15 IIIII 6 (4)
4.
4.1 18 babies (2)
4.2 15 babies (1)
4.3 11 – 15 months (1)
5.1
DIE COIN COMBINED
OUTCOMES
1 H 1H
T 1T
2 H 2H
T 2T
3 H 3H
T 3T
4 H 4H
T 4T
5 H 5H
T 5T
6 H 6H
T 6T
(6)
31
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5.2
(2)
5.3
5.3.1 1 (1)
12
5.3.2 3 1 (1)
=
12 4
6.1 1 (1)
10
6.2 1 (1)
10
6.3 2 1 (1)
=
10 5
7.
7.1 18 3 (1)
=
30 5
7.2 12 2 (1)
=
30 5
8.
8.1 3 (1)
32
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8.2.1 3 (1)
10
8.2.2 2 1 (1)
=
10 5
8.2.3 1 (1)
10
H
H
T
H H
T
T
H
H
T
T
H
T
T
33
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