# Mathematics Grade 9 P1 November Mock 2026 Memo (English)

> Grade 9 Mathematics Paper 1 memorandum from the November Mock Exam 2026 (English), 7 pages. Read online or download the free PDF.

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- Updated: 2026-10-29
- Subject: Mathematics
- Grade: Grade 9
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: November Mock Exam
- Paper: 1
- 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 P1
                           NOVEMBER 2026 (MOCK)
                            MARKING GUIDELINES

MARKS: 75

                     These marking guidelines consist of 7 pages.

Copyright reserved                                                  Please turn over

### Page 2

Mathematics/P1                                     2                                 Hlayiso/Mock 2026
                                   Senior Phase – Marking Guidelines

NOTE:

    • If a learner answers a question TWICE, only mark the FIRST attempt.
    • If a learner has crossed out an attempt and not redone the question, mark the crossed-out version.
    • Consistent accuracy (CA) applies in ALL aspects of the marking guidelines. Stop marking at the
      second calculation error.
    • Accept correct alternative methods. Answers only: award full marks ONLY where the guideline
      says so.
    • Assuming answers/values in order to solve a problem is NOT acceptable.

SECTION A
QUESTION 1

                                                 √
 1.1         C (12 is not a perfect square, so 12 is irrational;        ✓C                         (1)
             √
               49 = 7, 0,3̇ = 13 and − 58 are rational)

 1.2         A (4,72 × 10−5 )                                           ✓A                         (1)

 1.3         B (d = −4; T1 = −4(1) + 13 = 9)                            ✓B                         (1)

 1.4         D (− 23 × 32 = −1)                                         ✓D                         (1)

 1.5         C (4x2 − 20x + 25)                                         ✓C                         (1)

                                                                                                   [5]

SECTION B

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### Page 3

Mathematics/P1                                    3                               Hlayiso/Mock 2026
                                  Senior Phase – Marking Guidelines

QUESTION 2

 2.1         252 = 2 × 2 × 3 × 3 × 7                                  ✓ prime factors
             252 = 22 × 32 × 7                                        ✓ exponential form          (2)

 2.2         HCF = 22 × 32 = 36                                       ✓ HCF = 36
             LCM = 22 × 32 × 5 × 7 = 1 260                            ✓ LCM = 1 260
                                                                      (CA from 2.1)               (2)

               √                                                        √
 2.3         = 64 + (−5) − 9 × 2                                      ✓ 64 = 8
             = 8 − 5 − 18                                             ✓ −5 and −18
             = −15                                                    ✓ answer                    (3)

 2.4         −7 + 12 = 5 ◦ C at 14:00                                 ✓ 5 ◦C
             5 − 9 = −4 ◦ C                                           ✓ −4 ◦ C
                                                                      (Answer only: full marks) (2)

                                                                                                  [9]

QUESTION 3

 3.1.1       Deposit = 10% × R8 400 = R840                            ✓ R840
             Balance = 8 400 − 840 = R7 560                           ✓ R7 560                    (2)

 3.1.2       SI = P × n × i = 7 560 × 2 × 0,15                        ✓ substitution
             SI = R2 268                                              ✓ R2 268
             Total = 7 560 + 2 268 = R9 828                           ✓ R9 828
             OR A = 7 560(1 + 2 × 0,15)                               (CA from 3.1.1)
             A = 7 560 × 1,3 = R9 828                                 OR
                                                                      ✓ substitution into
                                                                      A = P(1 + ni)
                                                                      ✓ 1,3
                                                                      ✓ R9 828                    (3)

              9 828
 3.1.3              = R409,50 per month                               ✓ answer (CA)               (1)
                24

 3.2         A = 12 000(1 + 0,08)3                                    ✓ substitution
             A = R15 116,54                                           ✓ answer                    (2)

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### Page 4

Mathematics/P1                                    4                               Hlayiso/Mock 2026
                                  Senior Phase – Marking Guidelines

 3.3         Distance = 100 × 3 = 300 km                              ✓ 300 km / 100 × 3
                     300                                              ✓ 3 h 45 min                 (2)
             Time =      = 3,75 h = 3 h 45 min
                      80
             OR inverse proportion: 100 × 3 = 80 × t ⇒ t = 3,75
             h

                                                                                                  [10]

QUESTION 4

               −27a6 b3 × 2ab4
 4.1         =                                                        ✓ −27a6 b3
                    6a5 b2
                                                                      ✓ −54a7 b7 or −9
               −54a7 b7
             =                                                        ✓ a2 b5                      (3)
                6a5 b2
             = −9a2 b5

              2x+2 × (23 )x
 4.2         =                                                        ✓ prime base 2
                 (22 )2x+1
                                                                      ✓ 24x+2 in numerator and
              2x+2+3x 24x+2                                           denominator
             = 4x+2 = 4x+2
               2           2                                          ✓ answer                 (3)
               0
             =2 =1

 4.3         = −3x4 + 1                                               ✓ −3x4
                                                                      ✓ +1                         (2)

                                                                                                   [8]

QUESTION 5

 5.1         = 2x2 + 8x − 3x − 12 − (x2 − 4x + 4)                     ✓ 2x2 + 5x − 12
             = 2x2 + 5x − 12 − x2 + 4x − 4                            ✓ x2 − 4x + 4
             = x2 + 9x − 16                                           ✓ answer                     (3)

 5.2.1       = 5x(x2 − 9)                                             ✓ 5x(x2 − 9)
             = 5x(x − 3)(x + 3)                                       ✓ answer                     (2)

 5.2.2       = (x − 5)(x + 3)                                         ✓ (x − 5)
                                                                      ✓ (x + 3)                    (2)

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### Page 5

Mathematics/P1                                    5                                Hlayiso/Mock 2026
                                  Senior Phase – Marking Guidelines

 5.2.3       = a(x − 2y) + 3(x − 2y)                                  ✓ +3(x − 2y)
             = (x − 2y)(a + 3)                                        ✓ answer                   (2)

               (x − 4)(x + 4)
 5.3         =                                                        ✓ numerator factorised
               (x + 4)(x − 3)
                                                                      ✓ denominator factorised
               x−4                                                    ✓ answer                 (3)
             =
               x−3

                                                                                                [12]

QUESTION 6

 6.1.1       ×12: 4(x − 1) − 3(x + 2) = 12                            ✓ multiplying by LCD 12
             4x − 4 − 3x − 6 = 12                                     ✓ removing brackets
             x − 10 = 12                                              correctly
             x = 22
                                                                      ✓ answer                   (3)

 6.1.2       3x−1 = 34                                                ✓ 34
             x−1 = 4                                                  ✓ answer                   (2)
             x=5

 6.1.3       x2 − 3x − 10 = 0                                         ✓ standard form
             (x − 5)(x + 2) = 0                                       ✓ factors
             x = 5 or x = −2                                          ✓ both answers             (3)

 6.2         Naledi: x Sipho: 4x
             4x + 6 = 3(x + 6)                                        ✓ equation
             4x + 6 = 3x + 18
             x = 12                                                   ✓ x = 12
             Sipho is 4 × 12 = 48 years old.                          ✓ 48 years                 (3)

                                                                                                [11]

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### Page 6

Mathematics/P1                                    6                               Hlayiso/Mock 2026
                                  Senior Phase – Marking Guidelines

QUESTION 7

 7.1         24 ; 35   (add 3, 5, 7, 9, 11, … or Tn = n2 − 1)         ✓ both terms correct for the
                                                                      mark                       (1)

 7.2.1       p = 21                                                   ✓p
             q = 26                                                   ✓q                         (2)

 7.2.2       Constant difference = 5                                  ✓ 5n
             Tn = 5n + 1                                              ✓ +1                       (2)

 7.2.3       5n + 1 = 156                                             ✓ = 156
             5n = 155                                                 ✓ Figure 31 (CA)           (2)
             n = 31 ∴ Figure 31

 7.2.4       5n + 1 = 200                                             ✓ n = 39,8 / 196 and 201
             5n = 199
             n = 39,8, which is not a whole number.                   ✓ No, with reason          (2)
             No, he cannot use all 200 matchsticks for one figure.
             (Figure 39 uses 196 and Figure 40 uses 201
             matchsticks.)

                                                                                                 [9]

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### Page 7

Mathematics/P1                                               7                               Hlayiso/Mock 2026
                                             Senior Phase – Marking Guidelines

QUESTION 8

               x     −1      0   2       3
 8.1                                                                             ✓ any two correct
               y     6       4   0       −2                                      ✓ all four correct         (2)

                         y

              (0 ; 4)

                                 (2 ; 0)
                                           x

                                     f
 8.2                                                                             ✓ y-intercept (0 ; 4)
                                                                                 ✓ x-intercept (2 ; 0)
                                                                                 ✓ straight line with negative
                                                                                 gradient through the
                                                                                 intercepts, labelled f     (3)

                       3−0    3 1
 8.3.1       m=              = =                                                 ✓ substitution
                     4 − (−2) 6 2
                                                                                 ✓ answer                   (2)

                1
 8.3.2       y = x + c; substitute A(−2 ; 0):                                    ✓ substitute a point
                2
                1                                                                ✓ equation (CA)            (2)
             0 = (−2) + c ⇒ c = 1
                2
                1
             y = x+1
                2
                                                      1
 8.3.3       Perpendicular, because m f × mg = −2 ×     = −1                     ✓ perpendicular with reason
                                                      2
                                                                                 (CA)                     (1)
             (Also accept: the gradients −2 and 21 are negative
             reciprocals of each other.)

 8.3.4       y=3                                                                 ✓ answer                   (1)

                                                                                                          [11]

                                                                                                   TOTAL: 75

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