# Mathematics Grade 9 P2 November Mock 2026 Memo (English)

> Grade 9 Mathematics Paper 2 memorandum from the November Mock Exam 2026 (English), 7 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-9-mathematics-2026-november-mock-exam-memorandum-paper-2/5fbec12a-33f0-47a8-bdc2-65573d99d0db
- Updated: 2026-10-29
- Subject: Mathematics
- Grade: Grade 9
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: November Mock Exam
- Paper: 2
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 7

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## About this document

Mathematics · Grade 9 · November Mock Exam · 2026 · English. Memorandum, 7 pages. Read online or download the PDF.

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## Document text

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### Page 1

Mock

                           hlayiso mock exams
                           Practice Series:
                           Grade 9 Exam Preparation
                           WWW.HLAYISO.COM

                           SENIOR PHASE
                     END-OF-YEAR EXAMINATION

                                    GRADE 9

                               MATHEMATICS P2
                           NOVEMBER 2026 (MOCK)
                            MARKING GUIDELINES

MARKS: 75

                     These marking guidelines consist of 7 pages.

Copyright reserved                                                  Please turn over

### Page 2

Mathematics/P2                                        2                               Hlayiso/Mock 2026
                                      Senior Phase – Marking Guidelines

NOTE:

    • If a learner answers a question TWICE, only mark the FIRST attempt.
    • Consistent accuracy (CA) applies in ALL aspects of the marking guidelines, unless stated
      otherwise. Stop marking at the second calculation error.
    • Accept answers rounded correctly to TWO decimal places. Penalise units only where the question
      asks for them.
    • Accept any other correct method and award marks accordingly.

GEOMETRY

   S A mark for a correct statement (a statement mark is independent of a reason).
   R A mark for the correct reason (a reason mark may only be awarded if the statement is correct).
 S/R Award a mark if the statement AND reason are both correct.

Accept the usual abbreviations, e.g. ∠s on a str line; vert opp ∠s; corresp ∠s, AB∥CD; alt ∠s; co-int
∠s; sum of ∠s in △; ext ∠ of △; ∠s opp = sides; opp ∠s of ∥m.
QUESTION 1

 1.1         B       (2x + 30◦ = 110◦ , vert opp ∠s, so x = 40◦ )         ✓B                        (1)

 1.2         C       (square)                                             ✓C                        (1)

 1.3         A (∠∠∠ gives similar, not necessarily congruent,             ✓A                        (1)
             triangles)

 1.4         B       P′ (−3 ; −4)                                         ✓B                        (1)

 1.5         D       2,4 kℓ = 2,4 m3 = 2 400 000 cm3                      ✓D                        (1)

                                                                                                    [5]

QUESTION 2

 2.1.1       B̂1 + B̂2 + B̂3 = 180◦         ∠s on a str line              ✓ 180◦
                                                                          ✓ reason
             x + 2x + 3x − 30◦ = 180◦       Substitution                  ✓ 210◦
             6x = 210◦                                                    ✓ x = 35◦                 (4)

             ∴ x = 35◦

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### Page 3

Mathematics/P2                                    3                                 Hlayiso/Mock 2026
                                  Senior Phase – Marking Guidelines

 2.1.2       B̂3 = 3(35◦ ) − 30◦ = 75◦                                  ✓ answer (CA from 2.1.1)(1)

 2.2         a = GĤK = 68◦               [corresp ∠s; AB ∥ CD]         ✓S ✓R (a)
             b = GK̂H = 40◦                    [alt ∠s; AB ∥ CD]        ✓S ✓R (b)
                            ◦     ◦    ◦     ◦                          ✓S ✓R (c)                 (6)
             c = HĜK = 180 − 68 − 40 = 72 [sum of ∠s in △]
             OR
             BĜH = 180◦ − 68◦ = 112◦ [∠s on a str line];
             c = 112◦ − 40◦ = 72◦

 2.3.1       3x + 40◦ + 5x − 20◦ = 180◦            [∠s on a str line]   ✓S
             8x + 20◦ = 180◦                                            ✓R
             8x = 160◦                                                  ✓ x = 20◦                 (3)
             x = 20◦

 2.3.2       PV̂W = 3(20◦ ) + 40◦ = 100◦ and                            ✓ both angles = 100◦
             VŴS = 4(20◦ ) + 20◦ = 100◦                                ✓R                        (2)
             ∴ PV̂W = VŴS
             ∴ PQ ∥ RS                            [alt ∠s equal]
             OR
             QV̂W + VŴS = 80◦ + 100◦ = 180◦ ; PQ ∥ RS [co-int
             ∠s suppl]

                                                                                                 [16]

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### Page 4

Mathematics/P2                                    4                                Hlayiso/Mock 2026
                                  Senior Phase – Marking Guidelines

QUESTION 3

 3.1.1       Trapezium                                                 ✓ answer                   (1)

 3.1.2       Kite                                                      ✓ answer                   (1)

 3.1.3       Rhombus                                                   ✓ answer                   (1)

 3.2         L̂ = M̂1                               [∠s opp = sides]   ✓ S/R L̂ = M̂1
                       ◦            ◦        ◦
             M̂1 = 180 − (3x − 20 ) = 200 − 3x [∠s on a str line]      ✓ S/R M̂1 = 200◦ − 3x
             3x − 20◦ = x + 10◦ + (200◦ − 3x)           [ext ∠ of △]   ✓ S/R ext ∠ / sum of ∠s
             5x = 230 ◦                                                ✓ x = 46◦               (4)
             x = 46◦
             OR
             x + 10◦ + 2(200◦ − 3x) = 180◦ [sum of ∠s in △]
             ⇒ 5x = 230◦ , x = 46◦
             (Check: K̂ = 56◦ , L̂ = M̂1 = 62◦ , KM̂N = 118◦ )

 3.3         Â + B̂ = 180◦                   [co-int ∠s; AD ∥ BC]     ✓S
             3y − 10◦ + 2y + 15◦ = 180◦                                ✓R
             5y = 175◦                                                 ✓ y = 35◦                  (3)
             y = 35◦

 3.4.1       In △AEC and △BED:                                         ✓ AE = EB
             AE = EB                                         [given]   ✓ CE = ED
             CE = ED                                         [given]   ✓ S/R
             Ê1 = Ê2                                [vert opp ∠s]    ✓ R (SAS)                  (4)
             ∴ △AEC ≡ △BED                             [SAS / s∠s]

 3.4.2       BD = AC                                        [≡ △s]     ✓ BD with reason           (1)

 3.5.1       In △MNP and △RST:                                         ✓ both pairs of angles
             M̂ = R̂                                       [given]     ✓ S/R P̂ = T̂
             N̂ = Ŝ                                       [given]     ✓ R (∠∠∠)                  (3)
             P̂ = T̂                      [sum of ∠s in △ / 3rd ∠]
             ∴ △MNP ||| △RST                               [∠∠∠]

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### Page 5

Mathematics/P2                                            5                                Hlayiso/Mock 2026
                                          Senior Phase – Marking Guidelines

             ST     RS
 3.5.2          =       [||| △s]                                              ✓ correct proportion
             NP MN
             ST 8                                                             ✓ ST = 12 cm               (2)
                =
              9    6
             ST = 12 cm

                                                                                                        [20]

QUESTION 4

 4.1                                  y                                       ✓ correct position and shape
                                                                              of image
                            C′                                                ✓ labels
                                                                              ✓ all three coordinates (3)

                     B′          A′

                                                       x

             A′ (−1 ; 2), B′ (−4 ; 2), C′ (−1 ; 5)

 4.2         (x ; y) → (−x ; y)                                               ✓ answer                   (1)

 4.3.1       (x ; y) → (x − 2 ; y − 6)                                        ✓ answer                   (1)

 4.3.2       B′′ (2 ; −4)                                                     ✓ answer                   (1)

 4.4         C′′′ (5 ; 1)   ((x ; y) → (y ; x))                               ✓ answer                   (1)

 4.5.1       D(2 ; 4), E(8 ; 4), F(2 ; 10)                                    ✓ any two correct
                                                                              ✓ all three correct        (2)

 4.5.2       4 times (22 = 4; area of △ABC = 4,5 units2 , area of             ✓ answer                   (1)
             △DEF = 18 units2 )

                                                                                                        [10]

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### Page 6

Mathematics/P2                                     6                                   Hlayiso/Mock 2026
                                   Senior Phase – Marking Guidelines

QUESTION 5

 5.1.1       EC = 15 − 9 = 6 cm and BE = AD = 8 cm                       ✓ EC = 6 and BE = 8
             BC2 = BE2 + EC2 [Pythagoras]                                ✓ substitution in Pythagoras
             BC2 = 82 + 62 = 100                                         ✓ BC = 10 cm              (3)
             BC = 10 cm

 5.1.2       P = 9 + 10 + 15 + 8 = 42 cm                                 ✓ answer (CA from 5.1.1)(1)

 5.1.3       A = 12 (AB + DC) × AD = 12 (9 + 15) × 8                     ✓ substitution
             = 96 cm2                                                    ✓ answer                    (2)
             OR
             rectangle 9 × 8 = 72 plus triangle 12 × 6 × 8 = 24; total
             96 cm2

 5.2.1       Two semicircles = one full circle with d = 40 m             ✓ 2 × 60
             P = 2 × 60 + π d = 120 + 3,14 × 40                          ✓ 3,14 × 40
             = 120 + 125,6 = 245,6 m                                     ✓ 245,6 m                   (3)

 5.2.2       Field area = 60 × 40 + π r2 = 2 400 + 3,14 × 202            ✓ 60 × 40
             = 2 400 + 1 256 = 3 656 m2                                  ✓ 3,14 × 202 (1 256)
             Grass = 3 656 − 22 × 3 = 3 656 − 66 = 3 590 m2              ✓ 3 590 m2                  (3)

                                                                                                    [12]

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### Page 7

Mathematics/P2                                     7                               Hlayiso/Mock 2026
                                   Senior Phase – Marking Guidelines

QUESTION 6

 6.1.1       h2 = 52 + 122 = 169     [Pythagoras]                      ✓ substitution
             h = 13 cm                                                 ✓ 13 cm                   (2)

 6.1.2       V = area of base × height = 12 × 5 × 12 × 20              ✓ 21 × 5 × 12 (= 30)
             = 600 cm3                                                 ✓ 600 cm3                 (2)

 6.1.3       TSA = 2 × 12 × 5 × 12 + (5 + 12 + 13) × 20                ✓ two triangles (60)
             = 60 + 600                                                ✓ three rectangles (600) (CA
             = 660 cm2                                                 from 6.1.1)
                                                                       ✓ 660 cm2                 (3)

 6.2.1       r = 0,7 m                                                 ✓ r = 0,7
             V = π r2 h = 3,14 × 0,72 × 1,5                            ✓ substitution
             = 2,31 m3 (2,3079)                                        ✓ answer                  (3)

 6.2.2       2,3079 × 1 000 = 2 307,9 ℓ                                ✓ answer (CA from 6.2.1)(1)
             Accept 2 307,9 ℓ / 2 308 ℓ / 2 310 ℓ (CA from 6.2.1)

 6.2.3       4 times bigger   (π (2r)2 h = 4π r2 h)                    ✓ answer                  (1)

                                                                                               [12]

                                                                                         TOTAL: 75

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