# Mathematics Grade 9 June Common 2010 Memo (English)

> Grade 9 Mathematics memorandum from the June Common 2010 (English), 6 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/grade-9-mathematics-2010-june-common-memorandum-english/33c7515f-7490-4142-a8dc-5537f110c640
- Updated: 2026-10-29
- Subject: Mathematics
- Grade: Grade 9
- Language: English
- Document type: Memorandum
- Year: 2010
- Exam period: June Common
- Curriculum: CAPS
- Pages: 6

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## About this document

Mathematics · Grade 9 · June Common · 2010 · English. Memorandum, 6 pages. Read online or download the PDF.

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

## Document text

Text extracted from the PDF. Layout, diagrams and tables may not survive extraction.

### Page 1

Province of the
                     EASTERN CAPE
                     EDUCATION

                          SENIOR PHASE

                            GRADE 9

                            JUNE 2010

                    MATHEMATICS
                  MARKING GUIDELINE

MARKS: 100

TIME: 2 hours

                This Marking Guideline consists of 05 pages.

### Page 2

2           MATHEMATICS   (JUNE 2010)

SECTION A

1.   D√

### Page 3

(JUNE 2010)                                  MATHEMATICS                                            3

  2.     C√
  3.     B√
  4.     C √√
  5.     A √√
  6.     A√
  7.     C√
  8.     B√
  9.     A √√
  10     A√
  11.    D√
  12.    A √√
  13.    C√
  14.    D √√
  15.    B√
  16.    C√
  17.    C√
  18.    D √√
  19     D √√                                                                                [26]

  SECTION B

   QUESTION     NO      SOLUTION                                          MARK
                                                                          ALLOCATION
                1.1.1                                                     2
   1                                                                      1 mark
                            3
                        = 3k m √√   -4                                    specifically for
                                                                          negative index
                1.1.2                                                     4
                                2        2
                        =-[(a -4a+4) +(a +a-6)] +8√
                        = -[2 a2 – 3a – 2]+8√
                        =-2 a2+3a+2+8√
                        ==-2a2 + 3a + 10√
                1.2     4(a + b) – x2(a+ b)                               3
                        = (a+b)(4 – x2) √
                        =(a+b) (2 – x)(2+ x) √√
                1.3     1112-992                                          2
                        = (111 – 99)(111 + 99) √
                        =12 x 210
                        = 2 520         √
                                                                          3
                1.4.1   3x + 15 – 6 = -2x – 2√
                        5x           = -11√
                        x            = -11/5√
                1.4.2   3x + 1 = 27                                       3
                        3x + 1 = 33√
                        x + 1 = 3√
                        x=2 √

   2            2.1.1   Pattern 5 =32 since the common difference is 7√   1

### Page 4

4                                          MATHEMATICS                                            (JUNE 2010)

        2.1.2   7x1–3 =7-3=4                   or any logical method                3
                7x2–3 = 14–3 =11                                                    1 mark for
                7x3–3 = 21–3 =18√                                                   exploration
                                                                                    1 mark for 7n
                7xn -3= 7n-3√√                                                      and 1mark for
                                                                                    -3
        2.1.3   7x200 -3 =1397                                                      1
        2.2.1         √                                                             1
        2.2.2                      √√ Since parallel lines have the same gradient   2
                                                                                    Give full marks
                                                                                    even without
                                                                                    reason
        2.2.3   At x-intercept, the value of y is zero.                             3
                                         √
                               √
                                     √
        2.2.4             √√                                                        2

    3   3.1.1   Sum of angles of a hexagon =4x 180º                                 1
                                            = 720º
        3.1.2   Number of triangles in any polygon = n-2√                           2
                Therefore sum of angles in any polygon =n–2 x 180º√
        3.1.3   1080= 180º x 6√                                                     2
                      =180º x (8–2)
                So this is an octagon√

        3.2.1                                                                       2

                                                                                    1st Quadrant
                                                                                    image

### Page 5

(JUNE 2010)                                   MATHEMATICS                                             5

   3.          3.2.2    See 3.1.1 4th quadrant√                                     3
                        A  (2;-1), B  (3;0), C  (2;0), D  (1;-1) √√         1 mark for
                                                                                    correct image
                                                                                    2 marks for
                                                                                    correct co-
                                                                                    ordinates

   4           4.1      S = d/ t                                                    4
                        20 km/h = 1√
                                    t
                        t = 0,05 h√
                          = 3 min√
                        The rear of the train will emerge at 1:06 pm or 13:06√
               4.2         8 - 5      =135 ℓ √                                      4
                          11 8
                           64-55      =135ℓ
                            88
                         9 = 135ℓ √
                        88
                        So full capacity =135ℓ x88√
                                               9
                                         =1320ℓ√
               4.3      Volume of a cylinder = πr2h√                                4
                                                =22x 64cm2x17cm√√                   1 mark
                                                    7                               specifically
                                                                                    for the correct
                                                 = 3 419,43 cm3√                    height

   5          5.1.1    72º        =1/5 √                                            1
                       360º
              5.1.2    Number of learners for English= 126º x 120 = 42 learners√√   2
                                                        360º
              5.1.3    180º -72º =108º (angles on a straight line) √                3
                       Number of learners for Maths=108º x120
                                                      360º
                                                        = 36√
                                  Percentage for Maths =36 x100 = 30%√
                                                          120
              5.2.1    Arranged data in ascending order:                            2
                       15; 18; 20; 20; 23; 23; 25√
                       Median is 20√
              5.2.3    Range is 25-15=10                                            1
              5.2.4    Mean = 15+18+20+20+23+23+25√                                 2
                                       7
                            = 20,57√
              5.2.5                                                                 2
                           1 5 8√
                           2 0 0 3 3 5√

### Page 6

6                                MATHEMATICS                                            (JUNE 2010)

    5.3.1                                                                  4

                                                                           1 mark for
                                                                           correct scaling

                                                                           1 mark for
                                                                           labelling axes

                                                                           2 marks for
                                                                           correct plotting

    5.3.2   This indicates a weak correlation between the results in the   2
            two subjects since the points are fairly scattered. √√         1 mark for
                                                                            conclusion
                                                                           1 mark for the
                                                                           reason

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