# Technical Mathematics Grade 10 P1 November Mock 2026 Memo

> Grade 10 Technical Mathematics Paper 1 memorandum from the November Mock Exam 2026 (English), 8 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-10-technical-mathematics-2026-november-mock-exam-memorandum-paper-1/e466b9ff-67e5-43e9-a132-d19295ea1223
- Updated: 2026-10-29
- Subject: Technical Mathematics
- Grade: Grade 10
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: November Mock Exam
- Paper: 1
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 8

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## About this document

Technical Mathematics · Grade 10 · November Mock Exam · 2026 · English. Memorandum, 8 pages. Read online or download the PDF.

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

## Matching paper and memo

- [Grade 10 Technical Mathematics P1 November 2026 Mock](https://www.hlayiso.com/library/grade-10-technical-mathematics-2026-november-mock-exam-question-paper-paper-1/46e79a1c-3b95-4b7d-b147-193250abf8de) — QP · 2026 · Eng

## Document text

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### Page 1

hlayiso mock exams
                             Practice Series:
                             Grade 10 Exam Preparation
                             WWW.HLAYISO.COM

                                  NATIONAL
                             SENIOR CERTIFICATE

                                     GRADE 10

                         TECHNICAL MATHEMATICS P1
                     NOVEMBER EXAMINATION 2026 (MOCK)
                              MARKING GUIDELINES

  MARKS: 100

                       These marking guidelines consist of 8 pages.

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### Page 2

Technical Mathematics/P1                        2                                 Hlayiso/Mock 2026
                                      NSC – Marking Guidelines

MARKING CODES
 A   Accuracy                        CA      Consistent accuracy
 M   Method                          R       Rounding
 NPR No penalty for rounding         NPU     No penalty for units omitted
 S   Simplification                  SF      Substitution in correct formula
 AO  Answer only
NOTE:

    • If a candidate answers a question TWICE, only mark the FIRST attempt.
    • Consistent accuracy (CA) applies in ALL aspects of the marking guidelines where indicated.
    • No penalty for rounding (NPR) for ALL questions, unless stated otherwise.
    • AO = answer only. Where AO is indicated, the stated marks are awarded for a correct answer
      without working.
    • Where a question says WITHOUT using a calculator, the working must be shown to earn full
      marks.

QUESTION 1

             √     √       √
 1.1           64 < 75 < 81                                           ✓8                           A
                 √
             8 < 75 < 9                                               ✓9                           A
             Between 8 and 9                                                                      (2)

 1.2.1       Non-real                                                 ✓ answer                     A
                                                                                                  (1)

 1.2.2       Irrational                                               ✓ answer                     A
                                                                                                  (1)

 1.3         45 = 32 + 8 + 4 + 1                                      ✓ powers of 2 / method       M
             = 1 × 25 + 0 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 1 × 20
             = 1011012                                                ✓ answer                   CA
                                                                                                 (2)

 1.4           10112                                                  ✓ method (carrying)          M
             + 1102
             = 100012                                                 ✓ answer                   CA
                                                                      AO: full marks
             (11 + 6 = 17)
                                                                                                  (2)

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### Page 3

Technical Mathematics/P1                         3                             Hlayiso/Mock 2026
                                       NSC – Marking Guidelines

 1.5         4,58 × 10−4 m                                        ✓ answer                      A
                                                                                               (1)

 1.6         (2 − i)(3 + 4i) = 6 + 8i − 3i − 4i2                  ✓ expansion                  A
             = 6 + 5i − 4(−1)                                     ✓ i2 = −1                    A
             = 10 + 5i                                            ✓ answer                    CA
                                                                                              (3)

                                                                                              [12]

QUESTION 2

 2.1         2x2 + 10x − 3x − 15 − x2 + x                         ✓ 2x2 + 7x − 15              A
             = x2 + 8x − 15                                       ✓ −x2 + x                    A
                                                                  ✓ answer                    CA
                                                                                              (3)

 2.2.1       3(x2 − 9)                                            ✓ common factor              A
             = 3(x − 3)(x + 3)                                    ✓ answer                    CA
                                                                                              (2)

 2.2.2       (2x + 1)(x − 3)                                      ✓ (2x + 1)                    A
                                                                  ✓ (x − 3)                     A
                                                                                               (2)

 2.2.3       y(x − 2) + 3(x − 2)                                  ✓ grouping                   A
             = (x − 2)(y + 3)                                     ✓ answer                    CA
                                                                                              (2)

 2.2.4       (2m − 1)(4m2 + 2m + 1)                               ✓ (2m − 1)                    A
                                                                  ✓ (4m2 + 2m + 1)              A
                                                                                               (2)

              (x − 2)(x + 2) x + 3
 2.3         =              ×                                     ✓ numerator factors          A
              (x + 3)(x − 2) x + 2
                                                                  ✓ denominator factors        A
             =1
                                                                  ✓ ÷ → × (reciprocal)         A
                                                                  ✓ answer                    CA
                                                                                              (4)

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### Page 4

Technical Mathematics/P1                   4                            Hlayiso/Mock 2026
                                 NSC – Marking Guidelines

               4a6 b2 · 3a−2
 2.4         =                                              ✓ 4a6 b2                  A
                   6a4 b
                                                            ✓ 12a4 b2                CA
               12a4 b2
             =                                              ✓ answer                 CA
                6a4 b                                                                (3)
             = 2b

                                                                                     [18]

QUESTION 3

               5x · 5 − 5x
 3.1.1       =                                              ✓ 5x · 5                  A
                  4 · 5x
               5 (5 − 1)
                x                                           ✓ common factor           A
             =                                              ✓ answer                 CA
                  4 · 5x
             =1                                                                      (3)

               2x · 22x+2
 3.1.2       =                                              ✓ 22x+2                   A
                   23x
                x+2x+2−3x                                   ✓ 23x                     A
             =2
                                                            ✓ answer                 CA
             = 22 = 4                                                                (3)

 3.2.1       log2 32 = 5                                    ✓ answer                   A
                                                                                      (1)

 3.2.2       log3 34 + log 102                              ✓ 4 and 2                 A
             = 4+2 = 6                                      ✓ answer                 CA
                                                                                     (2)

                                                                                      [9]

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### Page 5

Technical Mathematics/P1                       5                             Hlayiso/Mock 2026
                                     NSC – Marking Guidelines

QUESTION 4

 4.1.1       (x − 4)(x + 3) = 0                                 ✓ factors                   A
             x = 4 or x = −3                                    ✓x=4                       CA
                                                                ✓ x = −3                   CA
                                                                                           (3)

 4.1.2       ×6: 3(x + 1) − 2(x − 2) = 12                       ✓ multiplying by LCD        M
             3x + 3 − 2x + 4 = 12                               ✓ removing brackets        CA
             x=5                                                ✓ answer                   CA
                                                                                           (3)

 4.1.3       3x+2 = 3−2                                         ✓ 3−2                       A
             x + 2 = −2                                         ✓ equating exponents       CA
             x = −4                                             ✓ answer                   CA
                                                                                           (3)

 4.2         −4 ≤ 2x < 6                                        ✓ subtracting 1            A
             −2 ≤ x < 3                                         ✓ −2 ≤ x < 3             CA
                                                                ✓ number line (closed at −2,
                 −3   −2   −1   0    1    2    3   4            open at 3)               CA
                                                                                          (3)

 4.3         y = 3x − 7                                         ✓ y as subject              A
             x + 2(3x − 7) = 7                                  ✓ substitution             CA
             7x = 21                                            ✓x=3                       CA
             x = 3 and y = 3(3) − 7 = 2                         ✓y=2                       CA
                                                                                           (4)

 4.4         at = v − u                                         ✓ at = v − u                A
                 v−u                                            ✓ answer                   CA
             a=
                    t                                                                      (2)

 4.5         Let the shorter piece be x m.
             x + (x + 2,4) = 12                                 ✓ equation                   A
             2x = 9,6
                                                                ✓ 4,8 m                    CA
             x = 4,8
                                                                ✓ 7,2 m                    CA
             The pieces are 4,8 m and 7,2 m.
                                                                                           (3)

                                                                                           [21]

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### Page 6

Technical Mathematics/P1                          6                            Hlayiso/Mock 2026
                                        NSC – Marking Guidelines

QUESTION 5

 5.1         A = P(1 + in)                                         ✓ formula                  A
             A = 8 500(1 + 0,065 × 3)                              ✓ SF                       A
             A = R10 157,50                                        ✓ answer                  CA
                                                                                             (3)

 5.2.1       A = P(1 + i)n                                         ✓ formula                  A
             A = 8 500(1 + 0,062)3                                 ✓ SF                       A
             A = R10 181,05                                        ✓ answer                  CA
                                                                                             (3)

 5.2.2       Lindiwe’s, because R10 181,05 > R10 157,50            ✓ answer with reason      CA
             (compound interest earns interest on interest).                                 (1)

 5.3.1       0,1 × 14 800 = R1 480                                 ✓ answer                    A
                                                                                              (1)

 5.3.2       Loan = 14 800 − 1 480 = R13 320                       ✓ loan amount             CA
             A = 13 320(1 + 0,16 × 2)                              ✓ SF                       A
             A = R17 582,40                                        ✓ answer                  CA
                                                                                             (3)

                            17 582,40
 5.3.3       Instalment =                                          ✓ dividing by 24          CA
                                24
             = R732,60                                             ✓ answer                  CA
                                                                                             (2)

 5.4         450 × 18,40 = R8 280                                  ✓ multiplication           A
                                                                   ✓ answer                  CA
                                                                                             (2)

                                                                                             [15]

QUESTION 6

                x    −3 −2 −1 0 1 2 3
 6.1          f (x) −5 0      3 4 3 0 −5                           ✓ f : 4 or more correct     A
                                                                   ✓ f : all correct           A
              g(x) −1 0       1 2 3 4 5                            ✓ g: all correct            A
                                                                                              (3)

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### Page 7

Technical Mathematics/P1                                        7                             Hlayiso/Mock 2026
                                                      NSC – Marking Guidelines

 6.2                            y                                                f:
                                                  g
                                                                                 ✓ x-intercepts            CA
                       (0; 4)
                                                                                 ✓ turning point (0; 4)    CA
                                     (1; 3)                                      ✓ shape (concave down) A
                           2                                                     g:
                 −2
                                                                                 ✓ intercepts (−2; 0) and
                                                      x                          (0; 2)                    CA
                                         2
                                                                                 ✓ straight line through (1; 3)
                                                                                 CA
                                                                                                             (5)
                                              f

 6.3.1       (0 ; 4)                                                             ✓ answer                   CA
                                                                                                            (1)

 6.3.2       y ≤ 4 OR               y ∈ (−∞ ; 4]                                 ✓ answer                   CA
                                                                                                            (1)

 6.3.3       x∈R                                                                 ✓ answer                     A
                                                                                                             (1)

 6.4.1       x = −2 or              x=1                                          ✓ x = −2                   CA
                                                                                 ✓x=1                       CA
                                                                                                            (2)

 6.4.2       −2 < x < 1             OR        x ∈ (−2 ; 1)                       ✓ critical values          CA
                                                                                 ✓ notation                 CA
                                                                                                            (2)

                                                                                                            [15]

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### Page 8

Technical Mathematics/P1                         8                            Hlayiso/Mock 2026
                                       NSC – Marking Guidelines

QUESTION 7

               k
 7.1         3=                                                   ✓ substitution of P      A
               2
             k=6                                                  ✓ answer                CA
                                                                                          (2)

 7.2         3 = b2 − 1                                           ✓ substitution of P       A
             b2 = 4
             b = 2 (b > 0)                                        ✓ answer                CA
                                                                                          (2)

 7.3         y = −1                                               ✓ answer                  A
                                                                                           (1)

 7.4         y ∈ R , y ̸= 0                                       ✓ answer                  A
                                                                                           (1)

 7.5         y > −1     OR     y ∈ (−1 ; ∞)                       ✓ answer                CA
                                                                                          (1)

                      6
 7.6         y=         = −2                                      ✓ answer                CA
                     −3
                                                                                          (1)

 7.7         x<0                                                  ✓ answer                  A
                                                                                           (1)

 7.8         x>2                                                  ✓ answer                  A
                                                                                           (1)

                                                                                         [10]

                                                                                   TOTAL: 100

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