# Mathematics Grade 9 Paper 1 November EC 2016 Memo (English)

> Grade 9 Mathematics Paper 1 memorandum from the November EC 2016 (English), 9 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/maths-p1-memo-gr9-nov2016-english/df5f1ef6-e692-44f5-8c0c-01c9600204c8
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 9
- Language: English
- Document type: Memorandum
- Year: 2016
- Exam period: November EC
- Paper: 1
- Curriculum: CAPS
- Pages: 9

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## About this document

Mathematics · Grade 9 · November EC · 2016 · English. Memorandum, 9 pages. Read online or download the PDF.

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

## Matching paper and memo

- [MATHS P1 QP GR9 NOV2016 English](https://www.hlayiso.com/library/maths-p1-qp-gr9-nov2016-english/6e1834ea-8b7d-4d04-bfdd-245d7a9b7357) — QP · 2016 · Eng

## Document text

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### Page 1

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                    SENIOR PHASE

                         GRADE 9

                  NOVEMBER 2016

                   MATHEMATICS
                   MEMORANDUM

MARKS:   100

               This memorandum consists of 9 pages.

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2                                    MATHEMATICS                            (EC/NOVEMBER 2016)

 NOTE:

      This is marking guideline. In instances where learners have used different
       mathematically sound strategies to solve the problems, they should be credited.
      Underline errors committed by learners and apply Consistent Accuracy (CA)
       marking.

                                              KEY
 M                Method mark
 CA               Consistent Accuracy mark
 A                Accuracy mark
 S                Statement
 R                Reason
 S/R              Statement and Reason

 QUESTION 1 [10 marks]

 Ques.                                          Mark Allocation                    Total
 1.1        B                                      1 mark for each correct answer   (1)

 1.2        C                                                                            (1)

 1.3        A                                                                            (1)

 1.4        B                                                                            (1)

 1.5        A                                                                            (1)

 1.6        D                                                                            (1)

 1.7        C                                                                            (1)

 1.8        A                                                                            (1)

 1.9        C                                                                            (1)

 1.10       C                                                                            (1)
                                                                                         [10]

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(EC/NOVEMBER 2016)                          MATHEMATICS                                               3

 QUESTION 2 [26 marks]

 Ques.      Solution                                      Mark Allocation                Total
                            −7
 2.1        6,74 × 10  A                                               Answer: 1 mark    (1)
            3
 2.2.1      √𝑥 3 + 𝑥 0                                                        𝑥 : 1 mark
            AA                                                             +1: 1 mark
            𝑥+1                                                                           (2)
 2.2.2      √0,03 𝑥 8 + 0,01𝑥 8                                               8
                                                                       √0,04 𝑥 : 1 mark
                       A                                               Answer: 1 mark
            √0,04 𝑥 8

            0,2 𝑥 4     A                                                                       (2)
 2.2.3        (2𝑑 2 𝑒)2                                                   2 4 2
                                                                         2 𝑑 𝑒 : 1 mark
            (4𝑑 −3 𝑒 −2 )−1                                             22 𝑑 −3 𝑒 −2 : 1 mark
               M           M                                            Answer: 1 mark
                2 4 2        2 −3 −2
            2 𝑑 𝑒 ×2 𝑑 𝑒
            16𝑑  A                                                                              (3)
                        2                                           2
 2.2.4      2(𝑥 + 2) − 2(𝑥 + 1)(𝑥 + 2)                             𝑥 + 4𝑥 + 4 : 1 mark
                  M                    M                          𝑥 2 + 3𝑥 + 2 : 1 mark
            2(𝑥 2 + 4𝑥 + 4) − 2( 𝑥 2 + 3𝑥 + 2)
                                                          2𝑥 2 + 8𝑥 + 8 − 2𝑥 2 − 6𝑥 − 4
            2𝑥 2 + 8𝑥 + 8 − 2𝑥 2 − 6𝑥 − 4      A                               : 1 mark
            2𝑥 + 4  CA
                                                                          Answer: 1 mark         (4)
 2.3.1      𝑥 2 + 5𝑥 − 24                                                  𝑥 + 8: 1 mark
                 A          A                                            𝑥 − 3: 1 mark
            (𝑥 + 8)(𝑥 − 3)
                                                                                                 (2)
 2.3.2      2(𝑎 − 𝑏) − 𝑏 + 𝑎                                           +1(𝑎 − 𝑏: 1 mark
            2(𝑎 − 𝑏) + 1(𝑎 − 𝑏)  M                               (2 + 1)(𝑎 − 𝑏): 1 mark
            (2 + 1)(𝑎 − 𝑏)     M                                       Answer : 1 mark
            3(𝑎 − 𝑏)  A
                                                                                                 (3)
 2.4.1      4𝑥 − 10 = 6                                                   4𝑥 = 16: 1 mark
            4𝑥 = 16  M                                                   Answer : 1 mark
            𝑥=4     A                                                                           (2)
 2.4.2      3𝑥 − 10 2𝑥 − 5                                              × LCD: 6: 1 mark
                   =
               2      3
                                                              9𝑥 − 30 = 4𝑥 − 10: 1 mark
                     3𝑥−10             2𝑥−5                             Answer: 1 mark
            6×(              ) =6×(         )    M
                        2               3

            9𝑥 − 30 = 4𝑥 − 10               A
            5𝑥 = 20
            𝑥=4        CA                                                                       (3)

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4                                   MATHEMATICS                           (EC/NOVEMBER 2016)

    2.4.3   𝑥2 = 4                                    (𝑥 + 2)(𝑥 − 2) = 0: 1 mark
            (𝑥 + 2)(𝑥 − 2) = 0  M                               Answer: 1 mark
            𝑥 = −2 or 𝑥 = 2   A                                                       (2)
               5                                                      5
    2.4.4   3𝑥 = 96                                                 2 : 1 mark
            𝑥 5 = 32                                            Answer: 1 mark
            𝑥 5 = 25 M
            𝑥=2       A                                                                (2)
                                                                                       [26]

    QUESTION 3 [8 marks]

    Ques.   Solution                              Mark Allocation             Total
    3.1.1   16  A                                             Answer: 1 mark  (1)
    3.1.2                                                          5𝑛: 1 mark
             AA
            5𝑛 + 1                                                  +1: 1 mark         (2)
    3.2.1   𝑦=𝑥        A                                        Answer: 1 mark        (1)
    3.2.2   𝑥 = −2     A                                        Answer: 1 mark        (1)

                               
                            𝑦 = −2𝑥 + 1

                                              

                                                  

            𝑦 intercept : 𝑦 = +1   A
                              1
            𝑥 intercept : 𝑥 = 2    A
            label    A                                                                 (3)
                                                                                        [8]

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(EC/NOVEMBER 2016)                  MATHEMATICS                                           5

  Question 4 [12 marks]

  Ques.     Solution                              Mark Allocation               Total
  4.1       𝑝. 𝑛. 𝑖 = 𝑆𝐼  M                                   Formula: 1 mark
              5 000 × 𝑛 × 0,12 = 1 800    M               Substitution: 1 mark
                                                                Answer: 1 mark
            𝑛=3       A

            OR

            𝐴 = 𝑃(1 + 𝑛𝑖)   M

            6 800 = 5 000(1 + 0,12𝑛)     M

            1.36 = 1 + 0,12𝑛

            0,36 = 0,12𝑛

            𝑛 = 3 A                                                               (3)
  4.2       Let the numbers be 𝑎 and 𝑏  M                 Any method: 1 mark
            𝑎 + 𝑏 = 143
            𝑎−𝑏 =7                                                  75: 1 mark
            𝑎 =𝑏+7
            𝑏 + 7 + 𝑏 = 143                                         68: 1 mark
            2𝑏 = 136
            𝑏 = 68
            𝑎 = 68 + 7
            𝑎 = 75           A      A
            The numbers are 75 and 68

            OR
            𝑎 + 𝑏 = 143
            𝑎−𝑏 =7
            ∴ 2𝑎 = 150 (adding the 2 equations)
               𝑎 = 75
            75 + 𝑏 = 143
                 𝑏 = 68                                                            (3)

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6                                    MATHEMATICS                  (EC/NOVEMBER 2016)

    4.3                                    S = 10         Answer: 2 marks
                     PENS     PENCILS

            10 − 7 = 3 boxes A                                                   (2)
    4.4     𝑑 = 𝑠 × 𝑡 M                            Formula/method: 1 mark
            𝑑 = 100 𝑘𝑚/ℎ × 3ℎ             A
            𝑑 = 300 𝑘𝑚                                      300 𝑘𝑚: 1 mark
                 𝑑                                           300 𝑘𝑚
            𝑆=𝑡                                                       : 1 mark
                                                              2 ℎ𝑟𝑠
                            300𝑘𝑚
            Av. Speed =             M                     Answer: 1 mark
                            2ℎ𝑟𝑠
            = 150 𝑘𝑚/ℎ      CA
            OR
             Speed      100 km           x km
             Time       3 hrs            2 hrs
            2x = 300
             x = 150 km/h                                                          (4)
                                                                                  [12]

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(EC/NOVEMBER 2016)                 MATHEMATICS                                              7

  QUESTION 5 [16 marks]

  Ques.     Solution                                         Mark Allocation       Total
  5.1.1     ∠𝐴𝐶𝐵 = ∠𝐷𝐶𝐹 = 32° (Vert. opp. ∠′𝑠)                    Statement and
            ∠𝐸𝐵𝐶 = ∠𝐴𝐶𝐵 = 32° (Alt. ∠′𝑠, EB ∥ DA)            reason: 1 mark each    (2)
                                          ′
  5.1.2     ∠𝐶𝐴𝐵 + ∠𝐴𝐵𝐸 = 180° ( Co int. ∠ 𝑠 ∶ 𝐸𝐵||𝐷𝐴 ) S/R      Statement and
            ∠𝐶𝐴𝐵 = 180° − 65° M                                  reason: 1 mark
            ∠𝐶𝐴𝐵 = 115°A                                     Substitution: 1 mark
                                                                 Answer: 1 mark
            OR
            ∠𝐶𝐴𝐵 + ∠𝐴𝐶𝐵 + ∠𝐴𝐵𝐶 = 180° (∠′𝑠 of a ∆) S/R
            ∠𝐶𝐴𝐵 = 180° − (32° + 33°) [∠𝐴𝐵𝐶 = 65° − 32°]M
            ∠𝐶𝐴𝐵 = 180° − 65°
            ∠𝐶𝐴𝐵 = 115° A                                                          (3)
  5.2.1     ∠𝐴 + ∠𝐴𝐵𝐶 = ∠𝐵𝐶𝐸 ( Ext ∠ of a ∆) S/R                 Statement and
                                                                  reason: 1 mark
            (2𝑥 − 48°) + (𝑥 + 14°) = 116°
                                            M
            3𝑥 − 34° = 116°                                   Substitution: 1 mark
            3𝑥 = 150°           A
            𝑥 = 50°                                              Answer: 1 mark

            OR                                 S/R
            ∠𝐴 + ∠𝐴𝐵𝐶 + + ∠𝐴𝐶𝐵 = 180° (∠′𝑠 of a ∆)
            (2𝑥 − 48°) + (𝑥 + 14°) + 64° = 180° M
            3𝑥 + 30° = 180°
            3𝑥 = 150°
            𝑥 = 50°          A

                                                                                     (3)
  5.2.2     ∠𝐴= 2𝑥 − 48°                                     Substitution: 1 mark
            = 2(50°) − 48° M
            = 100° − 48°                                         Answer: 1 mark
            = 52°     A                                                             (2)
  5.2.3     ∠𝐴𝐵𝐶 = 50° + 14° = 64°                             Correct statement:
            ∠𝐴𝐶𝐵 = 180° − 116° = 64°                                       1 mark
                            S                 R
             ∆𝐴𝐵𝐶 is an isosceles triangle (∠𝐴𝐵𝐶 = ∠𝐴𝐶𝐵)        Correct Reason:
                                                                           1 mark    (2)
  5.3.1       S            R                                 Correct statement:
            ∠𝐴𝐵𝐶 = 40° (Complementary ∠′𝑠)                                 1 mark
                                                                Correct Reason:
                                                                           1 mark
                                                                                     (2)
  5.3.2       S             R                                Correct statement:
            ∠𝐴𝐷𝑂 = 32° (AO =OD / radii)                                    1 mark
                                                                Correct Reason:
                                                                           1 mark     (2)
                                                                                     [16]

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8                                   MATHEMATICS                                   (EC/NOVEMBER 2016)

    QUESTION 6 [11 marks]

    Ques.   Solution                                            Mark Allocation                      Total
    6.1     STATEMENT                                     REASON             Correct
            𝐾𝐿 = 𝐾𝑁                                       Given A           statement
            𝐿𝑀 = 𝑁𝑀                                       Given  A          with
                                                                             reason: 1
            𝐾𝑀 = 𝐾𝑀                                       Common A
                                                                             mark each
            ∴ ∆KLM ≡ ∆KNM.                                𝑆𝑆𝑆A                                        (4)

    6.2.1   STATEMENT                                     REASON                Correct
            𝐴̂ = 𝑃̂                                       Alt .∠’s, AB ∥ PQ A statement
            𝐵̂ = 𝑄̂                                       Alt .∠’s, AB ∥ PQ A with
                                                                                reason: 1
            𝐴𝑂̂𝐵 = 𝑃𝑂̂𝑄                                   Vert. opp. ∠’s  A
                                                                                mark each
            ∴ ∆ ABO ||| ∆PQO.                             𝐴𝐴𝐴  A                                      (4)
    6.2.2   𝑂𝑄   𝑂𝑃 (Corr. sides are proportional)             Statement and reason: 1 mark
               =                                S/R
            𝑂𝐵   𝐴𝑂
                                                                        𝑥         12 𝑐𝑚 :
              𝑥    12 𝑐𝑚                                                      =             1 mark
                                                                       5 𝑐𝑚       6 𝑐𝑚
                 =             A
            5 𝑐𝑚   6 𝑐𝑚
                                                                              Answer: 1 mark
            𝑥 = 𝑂𝑄 = 10 𝑐𝑚     CA                                                                      (3)
                                                                                                       [11]

    QUESTION 7 [8 marks]
    Ques. Solution                                                Mark Allocation         Total
    7.1                                                                    14 𝑐𝑚 : 1 mark
            𝑑 = 7 × 2 = 14𝑐𝑚      M                                               𝜋𝑟 2
                                                                        𝑠 2 − : 1 mark
                                                                             2
                                              𝜋𝑟 2             196 𝑐𝑚2 − 77𝑐𝑚2 : 1 mark
            Area of the shaded part = 𝑠 2 −                                   Answer: 1 mark
                                               2     M
                               22
                                  × 49
                  = 14 × 14 −  7
                                2
                  = 196 𝑐𝑚2 − 77 𝑐𝑚2      A
                  = 119 𝑐𝑚2 CA                                                                        (4)
    7.2     2𝑙 + 2 𝑏 = 16        M                                 2𝑙 + 2 𝑏 = 16 : 1 mark
            𝑙+ 𝑏=8
             𝑏 =8−𝑙                                                    𝑙 × 𝑏 = 15 : 1 mark
            𝑙 × 𝑏 = 15     M                                                   5 :1 mark
            𝑙(8 − 𝑙) = 15
            8𝑙 − 𝑙2 = 15                                                            3 : 1 mark
            𝑙2 − 8𝑙 − 15 = 0
            (𝑙 − 5)(𝑙 − 3) = 0

            𝑙 = 5 or 𝑙 = 3   A
            𝑏 = 3 or 𝑏 = 5   A                                                                        (4)
                                                                                                       [8]

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(EC/NOVEMBER 2016)                  MATHEMATICS                                             9

  Question 8 [9 marks]

  Ques.     Solution                                   Mark Allocation               Total

  8.1.1
                                                           H                 GH
                                   G                       T
                                                                             GT

                                                                             RH
                                       R                  H
                                                          T                  RT

                                                       H                      YH
                                    Y                  T                     YT

                                                       H                    WH
                                     W                                      WT
                                                       T

                                                      H                     BH
                       A                            T
                                       B                                    BT         (2)
           Key:
           G− Green, R − Red, Y − Yellow, B − Black, W − White, H − head,
           T − tail

           2 marks for correct tree diagram.
  8.1.2    5    1                                               Answer: 1 mark
           10
              = 2  A
                                                                                       (1)
  8.1.3     2    1                                              Answer: 1 mark
               =   A
           10 5                                                                        (1)
  8.2.1    5                                                    Answer: 1 mark
              A
           8                                                                           (1)
  8.2.2    A A                                                      1
                                                                      : 1 mark
           1                                                          4
              = 25%                                               25% : 1 mark         (2)
           4
  8.2.3     A A                                                   5 : 1 mark
           5:1                                                      1 : 1 mark         (2)
                                                                                       [9]

                                                                          TOTAL:      100

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