# Technical Mathematics Grade 10 P1 2026 Memo (Eng)

> Grade 10 Technical Mathematics Paper 1 memorandum from the November Mock Exam 2 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-2-memorandum-paper-1/f87034e7-288b-47ad-80f1-243b20ef9a2d
- Updated: 2026-10-29
- Subject: Technical Mathematics
- Grade: Grade 10
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: November Mock Exam 2
- Paper: 1
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 8

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## About this document

Technical Mathematics · Grade 10 · November Mock Exam 2 · 2026 · English. Memorandum, 8 pages. Read online or download the PDF.

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## Matching paper and memo

- [Grade 10 Technical Mathematics P1 November 2026 Mock 2](https://www.hlayiso.com/library/grade-10-technical-mathematics-2026-november-mock-exam-2-question-paper-paper-1/282b7846-bdc6-4cc9-ba3e-4df0a8a3bb14) — QP · 2026 · Eng

## Document text

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### Page 1

Mock

                             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

 1.1.1       π                                                      ✓ answer                     A
                                                                                                (1)

             √
 1.1.2           −25                                                ✓ answer                     A
                                                                                                (1)

             √     √      √
 1.2           36 < 40 < 49                                                √
                 √
             6 < 40 < 7                                             ✓ 6 < 40 < 7                 A
                    √                                               ✓ −7 and −6 (correct
             −7 < − 40 < −6
             Between −7 and −6                                      order/sign)                CA
                                                                                               (2)

 1.3            1011012                                             ✓ 101101                 A
             ×       112                                            ✓ 1011010 (shifted row) A
                 101101                                             ✓ answer                CA
                                                                    Answer only: full marks
             + 1011010
                                                                                            (3)
             = 100001112
             (45 × 3 = 135)

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### Page 3

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

             2,5 × 103
 1.4                   = 0,5 × 105                                  ✓ dividing / 0,5 × 105      M
             5 × 10−2
                                                                    ✓ answer in scientific
             = 5 × 104 pieces
                                                                    notation                  CA
                                                                                              (2)

 1.5         (3 − 2i)2 = 9 − 12i + 4i2                              ✓ expansion                A
             = 9 − 12i + 4(−1)                                      ✓ i2 = −1                  A
             = 5 − 12i                                              ✓ answer                  CA
                                                                                              (3)

                                                                                              [12]

QUESTION 2

 2.1         (x − 3)(x2 + 3x + 9) = x3 − 27                         ✓ x3 − 27                  A
             x(x − 2)2 = x(x2 − 4x + 4) = x3 − 4x2 + 4x             ✓ x2 − 4x + 4              A
             x3 − 27 − x3 + 4x2 − 4x                                ✓ −x3 + 4x2 − 4x          CA
             = 4x2 − 4x − 27                                        ✓ answer                  CA
                                                                                              (4)

 2.2.1       2(x4 − 16)                                             ✓ common factor            A
             = 2(x2 − 4)(x2 + 4)                                    ✓ (x2 − 4)(x2 + 4)        CA
             = 2(x − 2)(x + 2)(x2 + 4)                              ✓ answer                  CA
                                                                                              (3)

 2.2.2       (3x − 4)(x + 3)                                        ✓ (3x − 4)                  A
                                                                    ✓ (x + 3)                   A
                                                                                               (2)

 2.2.3       x2 (x − 2) − 9(x − 2)                                  ✓ grouping                 A
             = (x − 2)(x2 − 9)                                      ✓ (x − 2)(x2 − 9)         CA
             = (x − 2)(x − 3)(x + 3)                                ✓ answer                  CA
                                                                                              (3)

 2.2.4       (3 + 2a)(9 − 6a + 4a2 )                                ✓ (3 + 2a)                  A
                                                                    ✓ (9 − 6a + 4a2 )           A
                                                                                               (2)

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### Page 4

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

                 2            8
 2.3         =       −                                              ✓ factorising x2 − 4        A
               x − 2 (x − 2)(x + 2)
                                                                    ✓ LCD                       A
                2(x + 2) − 8                                        ✓ 2x − 4                   CA
             =
               (x − 2)(x + 2)                                       ✓ answer                   CA
                   2x − 4                                                                      (4)
             =
               (x − 2)(x + 2)
                  2(x − 2)      2
             =                =
               (x − 2)(x + 2) x + 2

                                                                                               [18]

QUESTION 3

               2x+1 · 3x+1 · 2x−1
 3.1.1       =                                                      ✓ prime bases (num.)         A
                     22x · 3x
                                                                    ✓ 22x · 3x                   A
             =2 x+1+x−1−2x    · 3x+1−x
                                                                    ✓ adding/subtracting
             = 20 · 31                                              exponents                  CA
             =3                                                     ✓ answer                   CA
                                                                                               (4)

               2x · 22 − 2x
 3.1.2       =                                                      ✓ 2x · 22 and 2x · 2−1      A
               3 · 2x · 2−1
               2x (4 − 1)                                           ✓ common factor 2x          A
             =                                                      ✓ answer                   CA
                3 · 2x · 12                                                                    (3)
               3
             = 3 =2
                 2

 3.2         4a2 · 4a−2                                             ✓ 4a2 and 4a−2              A
             = 16                                                   ✓ 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       x2 − 2x = 15
             x2 − 2x − 15 = 0                                     ✓ standard form             A
             (x − 5)(x + 3) = 0                                   ✓ factors                  CA
             x = 5 or x = −3                                      ✓ both answers             CA
                                                                                             (3)

 4.1.2       2x+1 = 32                                            ✓ 2x+1 = 32              A
             2x+1 = 25                                            ✓ 2 / equating exponents A
                                                                     5

             x+1 = 5
                                                                  ✓ answer                   CA
             x=4
                                                                                             (3)

 4.2         −5 < 3 − 2x ≤ 15                                     ✓ multiplying by 5         A
             −8 < −2x ≤ 12                                        ✓ subtracting 3          CA
             4 > x ≥ −6                                           ✓ −6 ≤ x < 4 (inequality
                                                                  signs reversed)          CA
             −6 ≤ x < 4
                                                                  ✓ number line (closed at −6,
                −7 −6 −5 −4 −3 −2 −1   0   1     2   3   4   5    open at 4)               CA
                                                                                            (4)

 4.3         4x + 3y = 174 …(1)                                   ✓ equation (1)             A
             2x + 5y = 164 …(2)                                   ✓ equation (2)             A
             (2)×2: 4x + 10y = 328 …(3)                           ✓ elimination/substitution M
             (3)−(1): 7y = 154
                                                                  ✓ y = 22 (R22)             CA
             y = 22
             4x + 3(22) = 174 ⇒ 4x = 108 ⇒ x = 27                 ✓ x = 27 (R27)             CA
             Light switch: R27; plug: R22                                                    (5)

 4.4         V 2 = PR                                             ✓ V 2 = PR                  A
                    √                      √
             V = ± PR      (accept V =         PR, V > 0)         ✓ answer                   CA
                                                                                             (2)

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### Page 6

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

 4.5         Let t = time (hours) the motorcyclist travels.
             Van has then travelled for (t + 0,5) hours.                ✓ t + 0,5                  A
             100t = 80(t + 0,5)                                         ✓ equation (equal distances)
                                                                        M
             100t = 80t + 40
             20t = 40
             t = 2 hours                                                ✓t =2                    CA
             07:30 + 2 h = 09:30                                        ✓ 09:30                  CA
                                                                                                 (4)

                                                                                                 [21]

QUESTION 5

 5.1         A = P(1 + i)n                                              ✓ formula                  A
             20 000 = P(1 + 0,075)4                                     ✓ SF                       A
                   20 000
             P=                                                         ✓ answer                 CA
                  (1,075)4
             P = R14 976,01                                                                      (3)

 5.2.1       R5 000                                                     ✓ answer                   A
                                                                                                  (1)

 5.2.2       Simple interest.                                           ✓ simple                   A
             The graph is a straight line (the value increases by the   ✓ reason                   A
             same amount every year).                                                             (2)

 5.2.3       A = P(1 + in)
             6 600 = 5 000(1 + 4i)                                      ✓ SF                       A
             1,32 = 1 + 4i                                              ✓ 1,32 = 1 + 4i /
             i = 0,08                                                   1 600 = 20 000i          CA
             Rate = 8% per annum
                                                                        ✓ answer                 CA
                                                                                                 (3)

 5.3.1       Deposit = 0,15 × 12 600 = R1 890                           ✓ deposit                 A
             Loan = 12 600 − 1 890 = R10 710                            ✓ loan                   CA
                                                                                                 (2)

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### Page 7

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

 5.3.2       A = 10 710(1 + 0,14 × 1,5)                                            ✓ SF (n = 1,5)               A
             A = R12 959,10                                                        ✓ R12 959,10                CA
                          12 959,10                                                ✓ instalment                CA
             Instalment =           = R719,95                                                                  (3)
                              18

 5.3.3       1 890 + 12 959,10 = R14 849,10                                        ✓ answer                    CA
                                                                                                               (1)

                                                                                                               [15]

QUESTION 6

                   x     −4 −2 −1 1                     2    4
 6.1                                                                               ✓ f : 4 or more correct       A
              f (x) 1                  2   4 −4 −2 −1                              ✓ g: 3 or more correct        A
                                                                                   ✓ both tables all correct     A
                   x     −3 −2 −1 0                      1       2                                              (3)
              g(x) 6                   2   0 −1 −1,5 −1,75

 6.2                               y                                               f:
                       g
                                                                                   ✓ shape (2nd and 4th
                                                                                   quadrants)                 A
                                                                                   ✓ asymptotes x = 0, y = 0 /
                                                                                   points plotted           CA
               (−2; 2)
                                                                                   g:
               f                                    x                              ✓ shape (decreasing)       A
                         −1
                              −1               f
                                                   y = −2
                                                                                   ✓ asymptote y = −2         A
                                                                                   ✓ intercepts (−1; 0) and
                                                                                   (0; −1)                  CA
                                                                                                            (5)

 6.3.1       y = −2                                                                ✓ answer                      A
                                                                                                                (1)

 6.3.2       y > −2            OR          y ∈ (−2 ; ∞)                            ✓ answer                    CA
                                                                                                               (1)

 6.3.3       x ∈ R , x ̸= 0                                                        ✓ answer                      A
                                                                                                                (1)

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### Page 8

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

             ( 1 )x
 6.4          2 −2 = 0                                          ✓ 2−x = 21                 A
              −x
             2 = 21
             x = −1 i.e. (−1 ; 0)                               ✓ answer                 CA
                                                                                         (2)

 6.5         −2 < x < 0     OR   x ∈ (−2 ; 0)                   ✓ −2                     CA
                                                                ✓ 0 and notation         CA
                                                                                         (2)

                                                                                        [15]

QUESTION 7

 7.1         q = −8                                             ✓ answer                   A
                                                                                          (1)

 7.2         0 = a(2)2 − 8                                      ✓ substitution of B       A
             4a = 8 ⇒ a = 2                                     ✓ answer                 CA
                                                                                         (2)

                  4−0
 7.3         m=       = −2                                      ✓ m = −2                  A
                  0−2
                                                                ✓ equation (c = 4)       CA
             g(x) = −2x + 4
                                                                                         (2)

 7.4         2x2 − 8 = −2x + 4                                  ✓ equating f and g        M
             2x2 + 2x − 12 = 0 ⇒ x2 + x − 6 = 0
             (x + 3)(x − 2) = 0                                 ✓ x = −3                 CA
                                                                ✓ y = 10                 CA
             x = −3 (x = 2 at B)
                                                                                         (3)
             y = −2(−3) + 4 = 10 E(−3 ; 10)

 7.5         y ≥ −8    OR   y ∈ [−8 ; ∞)                        ✓ answer                 CA
                                                                                         (1)

 7.6         CD = 4 − (−8) = 12 units                           ✓ answer                   A
                                                                                          (1)

                                                                                        [10]

                                                                                   TOTAL: 100

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