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DEPARTMENT OF BASIC EDUCATION
REPUBLIC OF SOUTH AFRICA
ANNUAL NATIONAL ASSESSMENT 2012
GRADE 9 MATHEMATICS
TEST (BRAILLE)
MEMORANDUM
MARKS: 140
This memorandum consists of 9 pages.
Important Information
This is a marking guideline. In instances where learners have used different but mathematically sound strategies to solve
the problems they should be credited.
Unless stated otherwise, learners who give a correct answer only, should be awarded full marks.
Apply consistency accuracy and underline errors committed by learners.
KEY
M Method mark
CA Consistent Accuracy mark
A Accuracy mark
QUESTION 1
1.1 C 1.2 B 1.3 C 1.4 C 1.5 B Give 1 mark for each
1. correct answer.
1.6 B 1.7 C 1.8 C 1.9 C 1.10 B
[10]
QUESTION 2
2.1 3,56 × 10 A Answer: 1 mark (1 )
2.2.1 (3 ) + 2 27 +2 : 1 mark
= 27 + 2 A
= 29 CA Answer = 29 : 1 mark (2)
2.2.2 a²b² 4a²bc Answer: 2 marks
×
ac² 20b³ :: 1 mark
= M
: 1 mark
(2 )
=
Grade 9 Mathematics Memorandum (Braille) 1
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Mathematics Grade 9 BRAILLE 2012 Memo (English)
Grade 9 Mathematics memorandum from the BRAILLE 2012 (English), 9 pages. Read online or get the free PDF download, question paper linked when available.
Mathematics · Grade 9 · BRAILLE · 2012 · English. Memorandum, 9 pages. Read online or download the PDF.
- Subject
- Mathematics
- Grade
- Grade 9
- Language
- English
- Document type
- Memorandum
- Year
- 2012
- Exam period
- BRAILLE
- Curriculum
- CAPS
- Pages
- 9
- File size
- 215.3 KB
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BRAILLE 2012
- Other filesMemo · EngMemo · Afr
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2.3.1 4ab (5 ² ²+ 2 −3) Distributive law: 1 mark each term
= 20 +8 − 12 A 20 : 1 mark
8 : 1 mark
−12 : 1 mark (3 )
2.3.2 (2 − 1) − ( + 1)( − 1) Expanding: 1 mark
Simplifying: 1 mark
=4 − 4 + 1 (− + 1 M Answer: 1 mark (3)
=3 −4 +2 CA
2.4.1 8 ³+4 ² Common factor: 1 mark
(2 + 1): 1 mark (2)
= 4 ² (2 + 1) A
2.4.2 9 ²−36 ²
= 9( ²−4 ²) M
= 9( − 2 )( + 2 ) A
OR
= (3 − 6 )(3 + 6 ) A
= 3( − 2 )3( + 2 )
Common factor: 1 mark
= 9( − 2 )( + 2 ) A Difference of two squares: 1 mark (2 )
2.4.3 − −2 +2
= ( − ) − 2( − ) M/A Common factor t: 1mark
= ( − ) ( − 2) M/A Common factor 2: 1mark
Common factor( − ): 2marks
OR
= ( − 2) − ( − 2)
= ( − 2)( − )
(4)
2.5.1 3( + 6) = 12 Dividing by 3: 1 mark
+ 6 = 4 M Answer:1 mark
= −2 CA OR
OR −18 both sides:1 mark
Answer:1 mark (2 )
3 + 18 = 12
3 = −6 M
= −2 CA
2.5.2 −2 = 0 Common factor: 1 mark
( − 2) = 0 M Answer: 2 marks
(1 mark per factor) (3 )
= 0 or = 2 CA
Grade 9 Mathematics Memorandum (Braille) 2
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2.5.3 +1 −1 Multiplying by 6: 1 mark
– =1 Expanding: 1 mark
3 6 Answer: 1 mark (3 )
2( + 1) − ( − 1) = 6 M
2 + 2 − + 1 = 6 CA
= 3 CA
2.5.4 2 = 32 2 : 1 mark
Applying exponential law:1 mark
2 =2 M
Answer: 1 mark (3 )
+ 1 = 5 M/ CA
[30]
= 4 CA
QUESTION 3
3.1 3
Number of boys = 4 × 240 = 180 M /CA
3
: 1 mark
4
Answer: 1 mark (2)
3.2 Speed = M/ A Formula / Substitution: 1 mark
40 /ℎ =
× 60 M 0,225: 1 mark
(3)
OR 0,225 × 60 minutes Answer: 1 mark
= 13,5 CA
3.3 (1 + ) M/ A
= OR
Formula: 1 mark
= 12 000(1 + 0,065) M
= 16 441,04 CA Year 1: 12 000 x 6,5% = 780 Substitution: 1mark
Year 2: 12 780 x 6,5% = 830,70 (3)
Year 3: 13 610,7 x 6,5% = 884,70 Answer: 1 mark
OR
Year 4: 14 495,4 x 6,5% = 942,20
Year 5: 15 437,6 x 6,5% = 1 003,44
= 1+ There will be R16 441,04 in his account
,
= 12 000 1 +
= 16 441,04
3.4.1 15% cash = × 15 000 Answer: 1 mark (1)
= 2 250 A
3.4.2 = (1 + ) CA from 3.4.1 Formula: 1 mark
= 12 750[1 + 2(0,1)]
= 15 300 Substitution: 2 marks
OR Answer: 1 mark
OR
=
Formula: 1 mark
Grade 9 Mathematics Memorandum (Braille) 3
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Substitution: 1mark
( )( )
= Answer: 1 mark
(4)
Answer: 1 mark
= 2 250
Total amount= 12 750 + 2 250
= 15 300
3.4.3 Monthly payments = 15 300 ÷ 24 CA from 3.4.1 15 300 ÷ 24: 1mark
= 637,50 Answer: 1 mark (2)
[15]
QUESTION 4
4.1 18 23 A Answer: 2marks
1 mark per term (2)
4.2 Add 5 to the last term to get the next. A Correct answer: 1mark (1)
4.3 = 5 − 2 A Answer: 2 marks (2)
4.4 = 5 − 2 CA from 4.3 Substitution into the formula: 1 mark
38 = 5 − 2 CA Solving equation: 1 mark
40 = 5 M Answer: 1 mark (3)
∴8 = A
[ ]
Grade 9 Mathematics Memorandum (Braille) 4
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QUESTION 5
5.1.1 Parallel A Answer: 1 mark (1)
5.1.2 = −2 A Answer: 1 mark (1)
5.1.3 4 A Answer: 1 mark (1)
5.1.4 Non-linear A Answer: 1 mark (1)
5.2.1
X-intercept and Y-intercept: 2 marks per graph +
Gradient: 1 mark per graph+
Correct Labelling of graph: 1 mark per graph +(8)
Grade 9 Mathematics Memorandum (Braille) 5
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5.2.2 At point of intersection:−2 + 1 = − 2 Equation:1mark
−3 = −3 Answer: 1 mark (2)
=1
= (1) − 2 = −1
∴ (1; −1)
OR
= −2 + 1 = 1; = (−2(1) + 1) = −1
= −2 = 1; = 1− = −1
Therefore, this point lies on both graphs and it must be the point of intersection [14]
QUESTION 6
6.1 6.1.1 2 + 3 + 4 = 180° (TQ S is a straight angle) M/ A Statement: 1 mark
9 = 180° Reason: 1 mark
= 20° A (3)
Answer: 1 mark
6.1.2 = 3 (alternate Ls ;PQ|| RS) M Statement: 1mark
= 60° CA for 6.1.1 Reason: 1 mark
Answer: 1 mark (3)
6.1.3 = (corresponding Ls; PQ|| RS
=2 °
= 40° M
OR
+ + = 180° (sum of Ls of ∆ = 180 ) M
4 + + = 180°
80° + 60° + = 180°
= 180° − 140
= 40° A
OR
+ = + (ext L of = sum ofintoppLs) M
2 +3 = +
5 = 60° + Statement: 1 mark
100° = 60° + Reason: 1 mark
= 40° A Answer: 1 mark (3)
6.2 ∆ ≡∆ (S, <, S) A Statement with vertices in the correct order: 1 mark
Reason : 1 mark (2)
6.3.1 In ∆ ABD and ∆ ACD Correct statement with reason: 1 mark
AB = AC (given) A Correct statement with reason: 1 mark
Correct statement with reason: 1 mark
BD = CD (given) A Correct deduction and reason : 1 mark (4)
AD = AD ( common side) A
∆ ABD ≡ ∆ ACD (s,s,s) A
Grade 9 Mathematics Memorandum (Braille) 6
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OR
In ∆ ABD and ∆ ACD
ABD = ACD (opposite angles of a kite)
AB = AC (given)
BD = CD (given)
∆ ABD ≡ ∆ ACD (s, , s)
6.3.2 In ∆ ABE and ∆ ACE
AB = AC (given) A
AE = AE (common side) A
= corr. <s of congr. ∆s/ bisecting diagonals) A
∆ ABE≡ ∆ ACE (s,L,s) A
OR
In ∆ ABE and ∆ ACE
AB = AC (given)
Correct statement with reason: 1 mark
= (L opp = sides of ) Correct statement with reason: 1 mark
= corr. <s of congr. ∆s) or BE=EC (AD bisects BC) Correct statement with reason: 1 mark
Correct deduction and reason: 1 mark (4)
∆ ABE≡ ∆ ACE (L, L,s)
6.3.3 + = 180°( B C is str L) A Correct statement with reason: 1 mark
Correct statement with reason: 1 mark
But = (corr. <s of congr. ∆s) A Conclusion: 1 mark (3)
∴ = = 90° A
OR
BC AD (diagonals of a kite)
6.3.4 AE is perpendicular to BC A Answer: 1 mark (1)
6.4 = (proportional sides of similar∆ ) A Proportional sides and reason: 2 marks
substitution: 1 mark
= A Answer: 1 mark
(4)
(15)(6)
=
10
= 9 A
[27]
QUESTION7
7.1 (−1; 1) ′(1; 1) A Answer:2 marks
1 mark per pair of co-ordinates (2)
7.2 (1; 7) ′(1; 1) A Answer:2 marks
1 mark per pair of co-ordinates (2)
Grade 9 Mathematics Memorandum (Braille) 7
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7.3 The perimeter = (3 + 1,5 + 1 + 1 + 2 + 0,5) =9 Length of reduced sides: 1 mark
M/A Answer: 1 mark (2)
OR
(6+3+2+2+4+1)÷2= 9 units
7.4 Area of fig. P: Area of reduced fig = 4: 1 M/ A Answer: 2marks
OR (Dimensions halved:Area of original) (2)
Area of fig P to Area of reduced Fig = 1:0,25 or 1:
OR
Area of fig P to Area of reduced Fig = 10:2,5
[8]
QUESTION 8
8.1
3-D figure Name of figure No. of No. of vertices No. of 1 mark
faces edges for each
answer
Cylinder 3 faces No vertices 2
edges
Shape of the faces: Circles and a rectangle
(6)
8.2 The surface area of a rectangular prism is: Formula/ substitution into the correct
= 2( × ) + 2( × ℎ) + 2( × ℎ) formula: 3 marks
= 2(7,2 × 5) + 2(7,2 × 3,32) + 2(5 × 3,32) M/ A Calculation: 1 mark (5)
Answer: 1 mark
= (72 + 47,808 + 33,2) CA
Wrong formula: 0 marks
= 153, 01 CA
OR
2 + 2( + )ℎ
= 2(7,2 × 5) + 2(7,2 + 5)3,32
= 72 + 81,01
= 153,01
8.3 Volume of a cylinder = 1 000 Formula &substitution: 2 marks
ℎ = 1 000 M Calculation: 1 mark
(2,82) ℎ = 1 000 M Answer: 1 mark (4)
ℎ = 40
ℎ = 40,0268 ≈ 40 CA
[15]
Grade 9 Mathematics Memorandum (Braille) 8
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QUESTION 9
9.1.1 Class Interval Tally marks Frequency 1 mark per interval
1 – 10 ___ 0
11 – 20 2
II
21 – 30 12
IIIIIIIIII
31 – 40 9
IIII IIII
41 – 50 7
IIII II
A (4)
9.1.2 (a) Modal class is 11 – 20 answer (1)
9.1.2 (b) A=9 answer (1)
9.1.2 (c) B = 20 answer (1)
9.1.2 (d) 16 answer (1)
9.2.1 Range = 10 − 5 = 5 A Range: 1 mark
5 5 6 7 8 9 9 9 9 10 Median: 1 mark (2)
Median = 8 A
9.2.2 Mode = 5 Answer: 1 mark (1)
9.2.3 Mean= M Formula:1 mark
Answer: 1 mark (2)
= 6 A
[13]
TOTAL: 140
Grade 9 Mathematics Memorandum (Braille) 9
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