# Mathematics Grade 9 GEC Test 2022 Memo (English)

> Grade 9 Mathematics memorandum from the GEC Test 2022 (English), 12 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/grade-9-maths-english-memo/d26c869f-99cc-4b8f-b847-b46164e3a73f
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 9
- Language: English
- Document type: Memorandum
- Year: 2022
- Exam period: GEC Test
- Curriculum: CAPS
- Pages: 12

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Mathematics · Grade 9 · GEC Test · 2022 · English. Memorandum, 12 pages. Read online or download the PDF.

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                                        GEC PILOT STUDY
                                     MARKING GUIDELINE 2022
                                     MATHEMATICS: ENGLISH
                                           GRADE 9

SECTION A

         One mark per answer.
         There are no half marks.
         Tick () only the correct answer and underline the incorrect one.
         No marks should be credited for multiple answers. Underline them all.
         If there is no answer indicated, skip the question

                                                                                  Option
 No.                                      Expected answer

 1.       An irrational number                                                    D

 2.         14 = 2 × 7                                                            B
            16 = 2 × 2 × 2 × 2
          ∴LCM = 2 × 2 × 2 × 2 × 7
                = 112
 3.          𝑑                                                                    D
          𝑠=
             𝑡
              180
          𝑠=      = 90 km/h
                2
              210
          𝑡 = 60 = 3,5 h
          𝑑 =𝑠×𝑡
          𝑑 = 90 × 3,5 = 315 km
 4.       = 3[−(14)] + 4 × 2                                                      D
          = −42 + 8
          = −34

 5.         3(−2(2) × 3) + 6(2)2 (3)2                                             B
          =
                     2×3
            3(−12) + 216
          =
                 6
            180
          =
             6
          = 30

Grade 9 Mathematics: English Marking Guideline                  1

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                                                                           Option
 No.                                      Expected answer

 6.        −64 8                                                           A
        =         + −4
          2×2 2
        = −16 + 4 − 4
        = −16
 7.     = 𝑎12−3                                                            B
        = 𝑎9
 8.       1 1 11                                                           D
            +           11 30
        = 5 6 = 30 =       ×   = 11
          1 1        1  30   1
            ×
          5 6 30
 9.     = 23 (𝑥 5 )6                                                       D
        = 8𝑥 30
 10.    = 1 − 1 + 56 ÷ 56                                                  A
        =1−1+1=1
 11.    25                                                                 C

         Figure                  1               2               3    4
         Dots                    7               13              19   25

 12.                                      𝑇5 𝑇4 𝑇3 𝑇2                      C
                                            = = =
                                          𝑇4 𝑇3 𝑇2 𝑇1
                            1
        𝑎 = 16 and 𝑏 = 16

 13.                                                                       B
         Arrangement        1    2    3    4     5    6     7
         No of tables       1    2    3    4     5    6     7
         No of people       4    8    8    16    12   24    16

        Sequence 1: Arrangement 1; Arrangement 3 Arrangement 5; …

        Sequence 2: Arrangement 2; Arrangement 4; Arrangement 6;…

        The rule for the 7th arrangement is 4𝑛.
 14.    10 handshakes                                                      A

 15.      1                                                                C
        −
          4
 16.    7𝑥𝑎𝑦 + 3𝑎𝑥𝑦                                                        D

 17.    𝑥 2 + 7𝑥𝑦 − 15𝑦 2                                                  A

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                                                                Option
 No.                                      Expected answer

 18.      16𝑥 3 + 8𝑥 2                                          D
        =
            −2 𝑥 2
            2
          8𝑥 (2𝑥 + 1)
        =
             −2𝑥 2
        = −4 (2𝑥 + 1)
        = −8𝑥 − 4
 19.    = (1 − 3𝑎)(1 + 3𝑎)                                      B

 20.    = 3𝑥(𝑥 2 + 3𝑥 − 10)                                     D
        = 3𝑥(𝑥 + 5)(𝑥 − 2)
 21.    L = (𝑥 + 1) and W = (𝑥 − 2)                             D

 22.      6(𝑥 2 + 3𝑥 − 10)          8𝑥                          C
        =                     × 2
                 48𝑥             𝑥 −4
          6(𝑥 + 5)(𝑥 − 2)              8𝑥
        =                    ×
                48𝑥             (𝑥 − 2)(𝑥 + 2)
          (𝑥 + 5)
        =
          (𝑥 + 2)
 23.    = 15(𝑠 4 − 𝑧 4 )                                        C
        = 15(𝑠 2 − 𝑧 2 )(𝑠 2 + 𝑧 2 )
        = 15(𝑠 − 𝑧)(𝑠 + 𝑧)(𝑠 2 + 𝑧 2 )
 24.    𝑥 − 5 = −7                                              C
            𝑥 = −2
 25.    3𝑥 + 1 = 10                                             D
            3𝑥 = 9
             𝑥=3
 26.    𝑥 = 0 or 𝑥 = 4                                          B

 27.       6𝑥−1 = 62                                            B
        ∴ 𝑥−1=2
              𝑥=3
 28.    −6 = 12𝑥 − 18                                           B
        12 = 12𝑥
          𝑥=1
 29.    𝑦 = 𝑥2 + 𝑐                                              A
        𝑎 = (−1)2 + 2
        𝑎=3
 30.    3𝑥(6𝑥 − 8) = 0                                          D
        3𝑥 = 0 or 6𝑥 − 8 = 0
                      4
         𝑥 = 0 or 𝑥 = 3
 31.     4𝑥 2 − 4𝑥 − 48 = 0 … (÷ 4)                             D
           𝑥 2 − 𝑥 − 12 = 0
        (𝑥 + 3)(𝑥 − 4) = 0
           𝑥 = −3 or 𝑥 = 4

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                                                                                       Option
 No.                                     Expected answer

 32.                    𝑑2 = 𝑙 2 + 𝑤 2 …Theor. of Pyth.                                A
                   (𝑤 + 9)2 = (𝑤 + 7)2 + 𝑤 2
             𝑤 2 + 18𝑤 + 81 = 𝑤 2 + 14𝑤 + 49 + 𝑤²
              𝑤 2 − 4𝑤 − 32 = 0
            (𝑤 − 8)(𝑤 + 4) = 0
              𝑤 ≠ −4 or 𝑤 = 8
                          𝑙 = 8 + 7 = 15
                          A= 𝑙 ×𝑏
                            = 15 × 8
                            = 120 cm2
 33.    𝑦 = −2𝑥 − 4                                                                    D

 34.                                                                                   D

             1
        𝑦 =− 𝑥−1
             3
        𝑐 = −1
              1
        𝑚=−
              3

 35.    Rule: 𝑦 = 𝑥 2 + 2                                                              C
           258 = 𝑧 2 + 2
             𝑧 2 = 256
              𝑧 = 16
 36.                                                                                   A

        𝑦-values are all equal to 4, which implies that the horizontal line is above
        the X-axis.

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                                                                                            Option
 No.                                       Expected answer

 37.     𝑦 = 𝑚𝑥 + 𝑐                                                                         C
         𝑦 = 𝑚𝑥 + 4                Substitute 𝑐 = 4 from the graph
         0 = −3𝑚 + 4               Substitute (−3; 0) from the graph
        3𝑚 = 4
             4
         𝑚=3
                 4
         ∴ 𝑦 = 3𝑥 + 4
                 4
           𝑦 = 3 (−2) + 4
               −8
           𝑦=     +4
                3
               4
           𝑦=
               3
               4
        ∴ (−2; 3) lies on the graph (The learner may test different values to check which
        points lie on the graph. 𝑦 −intercept eliminates options B and D)

 38.      𝑦 = 2𝑥 − 8 (Line AB)                                                              D
               1                           1
        ∴ 𝑦 = −2𝑥 + 𝑐            m ⊥ AB= − 2
                1
          4 = − 2 (−2) + 𝑐       Substitute (−2; 4)
           𝑐=3
                     1
         ∴ 𝑦 = −2𝑥 + 3
 39.    𝐴′(−3; −5)                                                                          B

 40.    (𝑥; 𝑦)           (−𝑥; 𝑦)                                                            A

 41.    Translated 5 units up                                                               A

 42.    90°                                                                                 B

 43.    80°                                                                                 A

 44.    Alternate angles AB‖CD                                                              A

 45.       AÔ C = BÔ D Vertically Opp. ∠𝑠 =                                               D
        3𝑥 − 5 = 𝑥 + 250
               0

        3𝑥 − 𝑥 = 250 + 50
             2𝑥 = 300
              𝑥 = 150
        ∴ AÔ C = 3𝑥 − 50
                 = 3(15°) − 5°
                 = 45° − 5°
                 = 40°

 46.           ̂ D = 90° AD⊥BE,
              BO                                                                            D
        𝑥 +5° +35° = 90° …∠𝑠 on a str. line
                ∴ 𝑥 =50°

Grade 9 Mathematics: English Marking Guideline                   5

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                                                                                        Option
 No.                                     Expected answer

 47.    Two angles opposite equal sides are equal.                                      D

 48.     ̂H = GM
        MG     ̂ H = GH
                      ̂ M = 60° ∠𝑠 of            OR
        equilat. Δ                                ̂M = GM
                                                 𝐴G     ̂ H = 60° Alt. ∠𝑠 =, AB//CD
         ̂ D + GH
        GH      ̂ M = 180° ∠𝑠 on a str. line            ̂H = AG
                                                       AG     ̂M + MG
                                                                    ̂H°
                 ̂ D = 120°
                GH                                            = 60° + 60°
                 ̂B = GH
                EG     ̂ D Corr. ∠𝑠, AB//CD                  = 120°
           4𝑦 + 20° = 120°                              ̂H = EG
                                                       AG     ̂B        Vert. opp. ∠𝑠
                 4𝑦 = 100°                         4𝑦 + 20° = 120°
                       100                                 4𝑦 = 100°
                ∴𝑦=
                        4                                         100
                                                        ∴𝑦=
                     = 25°                                         4
        OR                                                    = 25°
         ̂H = GM
        MG     ̂ H = GH
                      ̂ M = 60° ∠𝑠 of
        equilat. Δ
         ̂H = GM
        BG     ̂ H = 60° … Alt. ∠𝑠, AB//CD
        4𝑦 + 20° + 60° = 180° … ∠𝑠 on a str
           4𝑦 + 20° = 120°
                 4𝑦 = 100°
                       100
                ∴𝑦=
                        4
                     = 25°

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                                                                                                Option
 No.                                      Expected answer

 49.    In ΔEFG and ΔABC                             OR                                         A
        EF FG EG                                     In ΔEFG and ΔABC
             =    =
        AB BC AC                                     EF FG EG
         12 cm   4 cm   8 cm                              =    =
               =      =      =2                      AB BC AC
         6 cm   2 cm   4 cm                           12 cm   4 cm   8 cm
        ∴ ΔEFG||| ΔABC         Corr. Sides are        6 cm
                                                            = 2 cm = 4 cm = 2
        proportional                                ∴ ΔEFG||| ΔABC Corr. Sides are
        𝐴𝐶̂ 𝐵 = 𝐸𝐺̂ 𝐹 = 86°                         proportional
           𝐵̂1 = ̂F = 58°                           𝐵̂1 = F̂ corr. … ∠𝑠 of similar △ 𝑠
           ̂
        ∴A =E    ̂2            corr.                 𝐵̂1 = 58°
        ∠𝑠 of similar △ =                           ̂ + 𝐵̂1 + Ĉ = 180° … sum of ∠𝑠 of △
                                                    A
           = 36°                                    ̂
                                                    A + 58° + 86° = 180°
        𝐸̂1 = ̂A               alter. ∠𝑠 DE∥AC      ̂
                                                    A = 180° − 144°
             =36°                                   ̂
                                                    A = 36°
                                                    𝐸̂1 = ̂
                                                          A alter…. ∠𝑠 = DE∥AC
                                                    𝐸̂1 =36°
 50.           𝐵̂2 = 𝐵̂1 = 420         diagonals of rhombus bisect ∠𝑠                           C
        𝐴𝐵̂𝐶 = 840
        𝐴𝐵̂𝐶 + 𝐵𝐶̂ 𝐷 = 1800                  Co-int ∠𝑠 AB∥CD
              0        ̂
          84 + 𝐵𝐶 𝐷 = 180        0

                     𝐵𝐶̂ 𝐷 = 960
                     𝐵𝐶̂ 𝐷 = 𝐵𝐶̂ 𝐴 + 𝐴𝐶̂ 𝐷
                      960 = 𝐵𝐶̂ 𝐴 + 𝐴𝐶̂ 𝐷
               But 𝐵𝐶̂ 𝐴 = 𝐴𝐶̂ 𝐷             diagonals of rhombus bisect ∠𝑠
                    ∴ 96 = 2𝐴𝐶̂ 𝐷
                          0

                       𝐴𝐶̂ 𝐷 = 480
 51.      𝐹̂2 = 𝐵̂1 = 680           Alt ∠𝑠 AB∥CD          OR                                    A
          𝐸̂2 + 𝐵̂1 = 180      0
                                   Co-int ∠𝑠 ED∥BF           𝐵̂1 = 𝐸̂1 = 680 … Corr. ∠𝑠 ED∥BF
        𝐸̂2 + 680 = 1800                                  𝐴̂ + 𝐷̂1 + 𝐸̂1 = 1800 …. ∠𝑠 of ∆
                  𝐸̂2 = 1120                             𝐴̂ + 𝐷̂1 + 680 = 1800
            𝐴̂ + 𝐷 ̂1 = 𝐸̂2        Ext ∠ of ∆                    𝐴̂ + 𝐷
                                                                      ̂1 = 1120
           ̂     ̂
          𝐴 + 𝐷1 = 112        0
                                                         But 𝐴̂ = 𝐷 ̂1 …∠𝑠 opp equal sides
              But 𝐴̂ = 𝐷   ̂1    ∠𝑠 opp equal sides
                                                            ∴ 𝐴̂ = 560
               ∴ 2𝐴̂ = 1120
                    𝐴̂ = 560
 52.    AC 2 = AB 2 + BC 2 … Theor. Of Pyth.                                                    C
             = (15 cm)2 + (8 cm)2
             = 225 cm2 + 64 cm2
             = 289 cm2
        AC = 17 cm

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                                                                    Option
 No.                                     Expected answer

 53.       𝑃 = 2(𝑙 + 𝑏) = 40                                        C
        𝑙 + 𝑏 = 20 For maximum area: 𝑙 = 𝑏 = 10
        ∴ 𝐴 = 𝑙2
            = 102
           = 100
        Using different combinations other 10 gives smaller area.
        e.g. 𝐴 = 9 × 11 = 99 cm2
             𝐴 = 8 × 12 = 96 cm2
 54.                                             1                  B
        For 2 smaller rectangles AB = (5 ÷ 10 = ) cm
                                                 2
                                 BC = 4 cm
                                         1
                               ∴area = 2 × 4 = 2 cm2

 55.    𝐴 = 𝜋𝑟 2 = 36𝜋                                              A
           ∴ 𝑟 2 = 36
              𝑟=6
              𝐶 = 2𝜋𝑟
                =2×6×𝜋
                = 12𝜋
 56.    𝐶 = 2𝜋𝑟                                                     B
          =2×𝜋×3
          = 6𝜋
                                     2
        For 2 out of 3 children: 𝐶 = × 6𝜋
                                     3
                                  = 12,57 cm
 57.         1                                                      A
        V= 2 × 𝑏 × ℎ × 𝑙
           1
        = ( × 8 × 6 × 14) cm3
           2
        = (4 × 6 × 14) cm3
        = 336 cm3
 58.    𝑆𝐴 = 2𝜋𝑟ℎ + 2𝜋𝑟 2                                           D
               22                22
        = (2 ×    × 7 × 16 + 2 ×    × 72 ) cm2
                7                7
        = (2 × 22 × 16 + 2 × 22 × 7) cm2
        = (704 + 308) cm2
        = 1012 cm2

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                                                                                Option
 No.                                     Expected answer

 59.     300 ml = 300 cm3                                                       B
              𝑉 = 𝜋𝑟 2 ℎ
                   22
        300 cm3 =      × 𝑟 2 × 13 cm
                    7
        300 cm3     7
                 ×     = 𝑟2
         13 cm     22
            1050
                  cm2 = 𝑟 2
             143
                     𝑟 = 2,71 cm
                     𝑑 = 2 × 2,71 cm
                       = 5,42 𝑐m
 60.    𝑆𝐴 = 2(𝑙𝑏 + 𝑏ℎ + 𝑙ℎ)                                                    C
        ℎ = 140 mm= 14 cm
        𝑏 = 2ℎ
        𝑏 = 2(14) cm= 28 cm                      OR
           3640 = 2(𝑙(28) + 28(14) + 𝑙(14))            3640
                                                            = 28𝑙 + 392 + 14𝑙
        3640 = 2(28𝑙 + 392 + 14𝑙)                        2
        3640 = 2(42𝑙 + 392)                      1820 − 392 = 42𝑙
        3640 = 84𝑙 + 784                              1428 = 42𝑙
                                                       1428
          84𝑙 =784                                          =𝑙
            𝑙 = 34 cm                                   42
                                                      34 cm = 𝑙

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SECTION B

       Marking guideline for Section B

               Do not penalise the learner for the same mistake more than once.
               There are no half marks.
               Underline errors committed by learners do not place a cross (X).
               In instances where learners have used different but mathematically sound
                strategies to solve problems, they (learners) must be credited.
               Credit learners for providing the correct answer without showing calculations.

       M                a mark for a correct method
       A                a mark for accurate calculation
                 key
       CA               a mark for consistent accuracy

 No.                                Expected answer                        Rationale/Clarification    Mark

 61.        2𝑦 = 2𝑥 + 4                                                   1 mark for the
             𝑦 = 𝑥 + 2 A                                                 standard form
            𝑦 −intercept if 𝑥 = 0
            𝑦 =0+2                                                        1 mark for plotting both
            𝑦=2                                                           intercepts
            𝑥 −intercept if 𝑦 = 0
            0=𝑥+2                                                         1 mark for the
            𝑥 = −2                                                        shape/gradient

                                                                          Full marks for the
                                                                          correct graph plotted

                                                                          OR                          3
                                                                          1 mark for plotting
                                                                          𝑥 −intercept
                                                                          1 mark for plotting
                                                                          𝑦 −intercept
                                                                          1 mark for the
                                                                          shape/gradient

                                                                          NB:
                                                                          1 mark if the learner has
                                                                          plotted incorrect
            CACA                                                        intercepts but the
                                                                          gradient of the graph is
                                                                          positive.

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 No.                         Expected answer                      Rationale/Clarification      Mark

 62.    A (1 ; 2), B (−2 ; 3) and C (−4 ; −1)                    1 mark for correct plotting
        𝐴′ (1 ; −2)                                              of 𝐴′ .
        𝐵′ (−2 ; −1)                                             1 mark for correct plotting
        𝐶 ′ (−4 ; −5)                                            of 𝐵′ .
                                                                 1 mark for correct plotting
                                                                 of 𝐶 ′ .

                                                                 Full marks for the
                                                                 correct plotted image
                                                                                               3

        AAA

 63.    MN ̂ P = MP̂N = 54°  M     ∠𝑠 opp equal sides (MN=MP)   1 mark for statement and
        ̂ + 54° + 54° = 180°
        M                           sum of int ∠𝑠 of△            reason
        ̂ = 72°CA
        M
        ̂ +Q
        M     ̂ = 180°              co-int ∠𝑠 (MN∥RQ)
            ∴Q ̂ = 108° CA                                      1 mark for 𝑄̂
        OR
         ̂P1 = 54°                 ∠𝑠 opp equal sides (MN=MP)    OR
         ̂
         N2 = 54°                   Alt ∠𝑠 (NR∥MQ)
           ̂ = 108° CA                                          1 mark for 𝑄̂
        ∴Q                         Opp ∠𝑠 of ∥gram

        In ∆QPR
        ∴Q̂+R ̂2 + P̂3 = 180°       sum of int ∠𝑠 of△
                    ̂     ̂
                But R 2 = P3     ∠𝑠 opp equal sides (PQ=RQ)
             108°+2P ̂3 = 180°
                   2P̂3 = 72°                                               ̂3
                                                                 1 mark for P
                     ̂3 = 36°  CA
                     P
           ̂        ̂
           P1 + 𝑦 + P3 = 180°
                but P̂1 = 54°      ∠𝑠 opp equal sides (MN=MP)    1 mark for 𝑦
         54° + 𝑦 + 36° = 180°       ∠𝑠 on a straight line
                      𝑦 = 90°CA

        OR
         ̂P1 = N
               ̂ 2 = 54° … alt ∠𝑠 (NR∥MQ)                                   ̂2
                                                                 1 mark for N
         ̂P3 = R
               ̂ 1 = 36° … alt ∠𝑠 (NR∥MQ)                                   ̂
                                                                 1 mark for R1

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 No.                         Expected answer                        Rationale/Clarification   Mark

        In ∆NPR
          𝑦+N ̂2 + R
                   ̂ 1 = 180°   … sum of int ∠𝑠 of △
        𝑦 + 54° + 36° = 180°
                     𝑦 = 90°CA                                    1 mark for 𝑦

                                                                   Full marks for answer      4
                                                                   only.

 64.    Angle at B + 45° + 45° = 180°                              1 mark for angle at B
        ∴ Angle at B           = 90° A
        AB 2 = (Home to 𝐴)2 − (Home to 𝐵)2 … Theorem of            1 mark for statement and
        PythM                                                     reason
              = (2 km)2 − (1 km)2
             = 4 km2 − 1 km2
              = 3 km2
        AB = √3 km2
              = 1,7 km
           Angle at the shop = angle at school  given
                              = 45°
        B to school = B to shop … sides opp. equal ∠𝑠              1 mark for distance from
                                                                   B to the school
                     = 3 km A

        (Shop to school)2 = (B to shop)2 + (B to school)2
                                                …Theorem of Pyth
                          = ( 3 km)2 + (3 km)2
                          = 9 km2 + 9 km2                                                     5
                          = 18 km2

        Shop to school     = √18 km2
                           = 4,2 km

        (School to Home)2 = (Home to B)2 + (B to school)2
        Theorem of Pyth
                          = ( 1 km)2 + (3 km)2
                          = 1 km2 + 9 km2
                          = 10 km2                                 1 mark for distance from
        School to home    = √10 km2                                school to home
                          = 3,2 km CA
                                                                   1 mark for the answer
        The total distance = 2 km + 1,7 km+ 3 km +4,2 km+3,2 km
                           = 14,1 km CA
                                                                   Full marks for answer
                                                                   only.

Grade 9 Mathematics: English Marking Guideline              12

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