# Mathematics Grade 9 DEAF 2012 Memo (English)

> Grade 9 Mathematics memorandum from the DEAF 2012 (English), 10 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/grade-9-english-maths-memo-deaf/3175fcc9-9801-4d30-80b9-892b4b88d12d
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 9
- Language: English
- Document type: Memorandum
- Year: 2012
- Exam period: DEAF
- Curriculum: CAPS
- Pages: 10

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## About this document

Mathematics · Grade 9 · DEAF · 2012 · English. Memorandum, 10 pages. Read online or download the PDF.

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## Document text

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                                                               ANNUAL NATIONAL ASSESSMENT 2012
                                                                    GRADE 9 MATHEMATICS
                                                                  MEMORANDUM (DEAF LEARNERS)

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 (Deaf Learners)                                         1

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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: 1 mark
        = ( − ) ( − 2)  M/A                                          Common factor 2: 1 mark
                                                                  Common factor( − ): 2 marks
        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 (Deaf Learners)                         2

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2.5.3                                                                                                Multiplying by 6: 1 mark
                 –           =1                                                                           Expanding: 1 mark
                                                                                                              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
                      = 4 CA
                                                                                                                                 [30]

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
                                                     Year 1: 12 000 x 6,5% = 780                         Substitution: 1mark
         = 16 441,04 CA
                                                     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 (Deaf Learners)                                    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 3.4.2                             15 300 ÷ 24: 1 mark
                        = 637,50                                                        Answer: 1 mark      (2)
                        = 15 300

                                                                                                              [15]

QUESTION 4

4.1     18        23 A                                                                Answer: 2 marks
                                                                                         1 mark per term     (2)

4.2     Add 5 to the last term to get the next term. A                          Correct answer: 1 mark      (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 (Deaf Learners)                           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 (Deaf Learners)               5

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5.2.2   At point of intersection:−2 + 1 = − 2                                                  Equation:1 mark
                                    −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: 1 mark
                   = 60°  CA for 6.1.1                                                                     Reason: 1 mark
                                                                                                            Answer: 1 mark       (3)

         6.1.3        =     (corresponding Ls; PQ|| RS M
                      =2 °
                      = 40° A

                  OR

                      +     +   = 180°  (sum of Ls of ∆ = 180 ) M
                                             4 + + = 180°
                                            80° + 60° + = 180°
                                                 = 180° − 140
                                                 = 40° A
                  OR

                    +     = +            (ext L of = sum of int opp Ls) 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
                  OR

                                Grade 9 Mathematics Memorandum (Deaf Learners)                                     6

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                          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
      =                                                                                                     Answer: 1 mark     (4)
          (    )( )
      =
      = 9             
                                                                                                                                [27]

QUESTION 7

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 (Deaf Learners)                               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: 2 marks
           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)
           = (72 + 47,808 + 33,2)               CA                                                      Answer: 1 mark
                                                                                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 = 40,0268           ≈ 40                CA
                                                                                                                                   [15]
                                Grade 9 Mathematics Memorandum (Deaf Learners)                                       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      
                                      IIII IIII II
                      31 – 40                                                  9       
                                      IIII IIII
                      41 – 50                                                  7       
                                      IIII II
                                                                                       A                                                   (4)

9.1.2
                         14
                                                                                                      Labelling: 2 marks
                                                                                                   Correct bars: 2 marks
                         12

                         10
                                                                                                              
                          8                                                                                      A

        Frequency 
                          6

                          4

                          2

                          0
                                 1 - 10   11- 20     21 - 30     31 - 40   41 - 50
                                                                                                                                           (4)
                                             Number of marks 

9.2.1             a)          Range = 10 − 5 = 5            A                                                   Range: 1 mark
                                                                                                                                           (2)
                  b)          5 5 6 7 8 9 9 9 9 10                                                              Median: 1 mark
                              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 (Deaf Learners)                                    9

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   Grade 9 Mathematics Memorandum (Deaf Learners)     10

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