# Mathematics Grade 9 June Common 2011 Memo (English)

> Grade 9 Mathematics memorandum from the June Common 2011 (English), 9 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/gr-9-maths-eng-memo/385e0538-f1e5-4f22-8830-0b1748710cce
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 9
- Document type: Memorandum
- Year: 2011
- Exam period: June Common
- Curriculum: CAPS
- Pages: 9

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## About this document

Mathematics · Grade 9 · June Common · 2011. Memorandum, 9 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Document text

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### Page 1

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                    Province of the
                    EASTERN CAPE
                    EDUCATION

                         SENIOR PHASE

                           GRADE 9

                           JUNE 2011

                   MATHEMATICS
                 MARKING GUIDELINE

MARKS: 100

                This marking guideline consists of 9 pages.

### Page 2

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2                     MATHEMATICS                 (JUNE 2011)

SECTION A

1.    D√                                               (1)

2.    C√                                               (1)

3.    C√                                               (1)

4.    C√                                               (1)

5.    A√                                               (1)

6.    A√                                               (1)

7.    C√                                               (1)

8.    B√                                               (1)

9.    B√                                               (1)

10    A√                                               (1)

11.   D √√                                             (2)

12.   A√                                               (1)

13.   C√                                               (1)

14.   B√                                               (1)

15.   A√                                               (1)

16.   A√                                               (1)

17.   A√                                               (1)

18.   C√                                               (1)

19    B√                                               (1)

20.   B√                                               (1)

21.   B√                                               (1)

22    C√                                               (1)

                                    TOTAL SECTION A:    23

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     (JUNE 2011)                                    MATHEMATICS                                                       3

SECTION B

QUESTION           NO    SOLUTION                                                         MARK         REASON
                                                                                          ALLOCATION
1.                 1.1   1.1.1      √25 ; 2,3 ; 16 ; 4          √                                      If all four numbers
                                                                                             (1)       mentioned.
                         1.1.2                                                                         If both numbers
                                    π ; √19                         √                        (1)       mentioned.

                   1.2   Real numbers                           √                            (1)       Answer only.

                   1.3   0,005 629 √                                                         (1)       Answer only.

                   1.4       (2a2b4)3 x 4ab3                                                           Product rule
                                 8a-2b7
                           = 8a b x 4ab3
                               6 12
                                                  √                                                    Simplification
                                 8a-2b7           √
                                9 8
                           = 4a b                 √                                          (3)       Answer

                   1.5   Let the unknown distance be x
                          480 km : 4 hours
                           x travelled km : 31/3 hours

                           480           4
                                    =      x3 √                                                        Direct proportion
                            x           10

                           12x       = 4800

                                   4800
                          X =
                                    12

                                 x = 400      √                                                        Answer

                                 The family is left with 80 km to its destination √

                                                  OR

                         Average speed for 4 hours:          = 120 km/h           √                    Average speed

                         Distance travel for 3 hours 20 minutes: 120 x 3                               Distance travelled
                                                                                                       in 3 hours 20
                                                                                                       minutes
                                                                        = 120 x
                                                                        = 400 km √

                         Distance left = 480 km – 400 km
                                       = 80 km                                        √       (3)      Difference
                                                                                             [10]

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  4                                               MATHEMATICS                                         (JUNE 2011)

QUESTION 2

2.1   2.1.1            a 2  b2      x(3a  4b)
                                  X
                       y ( a  b)   9a 2  16b 2

                   (a  b)(a  b)       x (3a  4b)
                                  X
                      y ( a  b)    (3a  4b )(3a  4b )                   √

                            ab      x
                                X                                                                     Simplification
                             y    3a  4b √

                           x(a  b)
                                                       √
                         y (3a  4b )                                                           (3)   Answer

      2.1.2                                                                                           Removing of
               (5x – 1)(2x + 3)                                                                       brackets
               = 10x 2 + 15x − 2x – 3             √
               = 10x 2 + 12x − 3                  √                                             (2)   Answer

                                                      OR
2.2           p 3 − p 2 r − 9p + 9r                          p 3 − 9p – p2r + 9r
          = (p 3 − p 2 r) – (9p − 9r)         √                 = p(p2 − 9) – r(p2 − 9) √             Correct grouping

          = p 2 ( p – r) - 9( p – r)          √                 = (p2 – 9)(p – r) √                   Common factor

          = ( p – r)( p 2 − 9)                                  = ( p – 3)(p + 3) )(p – 3 ) √         Answer

          = ( p – r )( p – 3 )( p + 3)        √                                                 (3)

2.3   2.3.1        6(x + 2) = 3( 3x – 4) – 3
                   6x + 12 = 9x – 12 – 3                           √
                   6x + 12 – 12 = 9x − 15 − 12
                   6x − 9x = - 27                                   √
                     - 3x = - 27
                         3x = 27
                         3x    27
                             =
                          3     3
                         x = 9                                         √                        (3)

      2.3.2            4x               x7
                               =                       OR
                       5                 3
                                                            4x        x7                             Cross
                                                               X 15 =     X 15                        multiplication
                                                            5          3
                  12x = 5x − 35                            12x = 5x – 35       √

                   7x = -35                                7x   = -35
                   7x     35                                7x      35
                      = -                                      = -
                   7       7                                7       7
                                                             x = -5            √

                   x    = -5                                                                    (2)   Answer

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 (JUNE 2011)                                  MATHEMATICS                            5

      2.3.3     3.5 x 1 = 75
                            3.5 x 1 = 75
                             3         3
                               x 1                                      Simplification
                             5       = 25            √
                             5 x 1 = 5 2                                Equal basis
                             x+1 = 2                 √
                          x+1–1=2–1
                             x = 1                  √            (3)    Answer

2.4          1
      y    =    x − 4
             2
       To find the y-intercept, let x = 0
                      1
                y =     (0)− 4
                      2
                                                                         Co-ordinate of y
                y = -4                                                   intercept
      Co-ordinate of y intercept is (0; -4)    √

      To find the x-intercept , let y = 0
                1
                     −4 = 0
                2
                1
                     = 4
                2
                  x
              2x      = 2x4
                  2
                    = 8
                                                                         Co-ordinate of x
      Co-ordinates of x-intercept (8;0 )      √                          intercept

                    y
                     4-
                                                         1
                    3-                              y=       −4
                                                         2
                      2-
                      1-             √
                                                                         x-intercept in the
      -x       | | | | | | | | | | | | | | | | | | | | | | x             graph
                  -2 -1-   2 4   6   8 10 12
                     -2-                                                 x-intercept in the
                     -3-                                                 graph
                     -4- √
                     -5-
                     -6-
                     -7-

                   -y                                              (4)
                                                                  [20]

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      6                                 MATHEMATICS                                  (JUNE 2011)

QUESTION 3

3.1

                                                            Structure 5
                                                                                 Correct drawing
                                                              √

                                                                           (1)

3.2       3.2.1   25 – 1 =   12 or any other relevant method
                     2                      √                              (2)   Answer

          3.2.2   Let the number of triangles be represented by n
                  The difference in the matchsticks sequence is 2. √
                  Hence the number of matchsticks is 2 times number of
                  triangles plus 1 i.e. 2n + 1 or any relevant method. √   (2)   Answer

          3.2.3   No. of n matchsticks = 2n + 1
                             n = 40
                    No. of matchsticks = (2 x 40) +1

                                         = 80 + 1

                                         = 81                   √
                   There are 81 matchsticks                                (1)   Answer

3.3       3.3.1   Let Linda’s age be x
                          Then Thami’s age is x + 12
                          But x + (x + 12) = 82        √                         Setting up the
                                                                                 equation
                                x + x + 12 = 82
                                       2x = 82 − 12
                                       2x = 70
                                       2       2
                                      x = 35        √
                       Linda’s age is 35 years                             (2)   Answer

          3.3.2   Thami’s age is x + 12
                           i.e.    35 + 12 = 47
                  Thami’s age is 47 years.                                 (1)   Answer
                                                                           [9]

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(JUNE 2011)                                         MATHEMATICS                                                 7

QUESTION 4

4.1
                                     •B      4                               ●A

                                             3--
                                                                                 √                  Position of C

                                             2--

                                      •C      1--
                                       (-1;1)
                |    |   |     |   | | |            |   |   |    |       |   |       |    |
                    -3       -2      -1                 1       2            3           4

                                                                                              (1)

4.2     C (-1;1)                                                                              (1)   Answer

4.3     Translation                                                                           (1)   Answer

4.4     4.4.1       Trapezium                                                                 (1)   Answer

        4.4.2       By rotating the shape all four angles meet a point, thus
                    forming 360º. The sum of the angles of any quadrilateral is
                    360º or any other reasonable logical argument√√                           (2)
                                                                                              [6]

QUESTION 5

5.1     5.1.1       R                √                                                       (1)   Answer

        5.1.2       Is similar                                       √                        (1)   Answer

        5.1.3       All angles of ∆ PQR = All angles of ∆ ABC                     √           (1)   Answer

        5.1.4       QR                                               √                        (1)   Answer

        5.1.5       RP                                               √                        (1)   Answer

5.2     5.2.1                 Statement                                          Reason
                    In ∆s ABO & PQO
                    AB = PQ                                      given √                            Reason
                     ABO =  OPQ                                alternate angles √                 Statement/Reas
                                                                                                    on
                     AOB =  POQ                                vertically opposite angles         Statement/Reas
                                                                 √                                  on
                    ∆ABC  ∆ RST                                  ,  ,S             √       (4)
                                                                                                    Condition of
                                                                                                    congruency

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  8                                        MATHEMATICS                                       (JUNE 2011)

      5.2.2   Side, Side, Side (SSS) ;Side, Angle, Side (S,A,S); Right
              Angle, Hypotenuse and Side (RHS)        (Any one of these) √             (1)   Answer
                                                                                      [10]

QUESTION 6

6.1   6.1.1      Volume of can     =  r2 h
                                   =     x (32,5 mm)2 x 75 mm                 √              Substitution

                                    =       x 1 056,25 mm3 x 75 mm
                                                                                             Answer
                                   =       24 8973,21 mm 3                        √

               Convert 24 8973,21 mm 3 to cm 3            (1 000 mm 3 = 1 cm 3 ) √           Conversion

                                                      1
                             24 8973,21 mm 3 X            = 248,97 cm 3         √
                                                     1000                             (4)    Answer

      6.1.2    Convert 248,97 cm 3 to mℓ (1 cm 3 = 1 mℓ)
                              :. 248,97 cm 3 = 248,97 mℓ       √                             Conversion

                           Space = Volume of can − liquid in the can
                                   = 248,97 mℓ – 200 mℓ
                        :. Space = 48,97 mℓ                                     √     (2)    Answer
                                                                                      [6]

QUESTION 7

7.1   First arrange the numbers in ascending order

      16; 18; 19; 23; 24; 26; 27; 29; 30; 33; 34; 35; 37; 38; 39; 40; 41; 42;
      45; 49.

              Stem                                  Leaf
                                                                                             1 mark per stem
                1                           6;   8; 9                √
                2                           3;   4; 6; 7; 9          √
                3                           0;   3; 4; 5; 7; 8; 9     √
                4                           0;   1; 2; 5; 9           √               (4)

7.2                                 √                         √
          Interval                Tally                               Frequency
                                                                                             Tally
           0 − 9                                                          0
         10 − 19                    ///                                   3                  frequency
          20 − 29                 ////                                    5
         30 − 39                 //// //                                  7
         40 − 49                   ////                                   5
                                             Total             20                     (2)

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 (JUNE 2011)                                    MATHEMATICS                                                9

7.3     7.3.1                             Sum of marks obtained
                Average Mark          =    No. of Marks obtained
                                                                                                Correct addition
                                        645
                                      =                                √
                                         20
                Average Mark          = 32,25                          √                 (2)    Answer

        7.3.2   Median       − Middle Value

                16; 18; 19; 23; 24; 26; 27; 29; 30; 33; 34; 35; 37; 38; 39; 40; 41;
                42; 45; 49.
                                           33  34
                             Median =
                                               2
                                            67
                                        =
                                             2

                                            =   33,5                       √             (1)    Answer

        7.3.3   Mode – value that appears most frequently/ often in a set of data.
                There is no mode                                       √                 (1)    Answer

        7.3.4   Range = Highest mark − Lowest mark
                      = 49 – 16                                                √

                             =   33                                                      (1)    Answer

7.4     7.4.1                                                                                       Labels (1)
                                      TEMPERATURES TAKEN ONE A DAY             √                    All 12 bars
                  T                                                                                 (3)
                  E     35                                                                          Max. 9 (2)
                                                                                                    Max.6 (1)
                  P                                                                                 Less than 6
                  E
                        30
                                                                                                    (NO
                  R     25                                                                          MARKS)
                  A
                  T     20
                  U
                  R
                        15
                  E     10

                  I     5                                                          √√√
                  N
                        0
                  º
                  c

                                                  TIME IN HOURS

                                                                                          (4)

        7.4.2   16:00                                                                     (1)       Answer
                                                                                         [16]

                                                                   TOTAL SECTION B:       77

                                                                      GRAND TOTAL:       100

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