# Mathematics Grade 10 P1 Gauteng November 2022 Memo (Afr)

> Grade 10 Mathematics Paper 1 memorandum from the Gauteng November 2022 (Afrikaans), 10 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/gr-10-mathematics-p1-english-and-afrikaans-november-2022-possible-answers/95c1561e-9f52-43c1-82af-15858ae5bc23
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 10
- Language: Afrikaans
- Document type: Memorandum
- Year: 2022
- Exam period: Gauteng November
- Paper: 1
- Curriculum: CAPS
- Pages: 10

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## About this document

Mathematics · Grade 10 · Gauteng November · 2022 · Afrikaans. Memorandum, 10 pages. Read online or download the PDF.

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

## Matching paper and memo

- [Gr 10 Mathematics P1 (Afrikaans) November 2022 Question Paper](https://www.hlayiso.com/library/gr-10-mathematics-p1-afrikaans-november-2022-question-paper/cb90cd4a-f97b-4561-bdc2-4d79045b30bb) — QP · 2022 · Afr

## Document text

Text extracted from the PDF. Layout, diagrams and tables may not survive extraction.

### Page 1

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             PROVINCIAL EXAMINATION
                   PROVINSIALE EKSAMEN
                            NOVEMBER 2022
                        GRADE 10/GRAAD 10
                  MARKING GUIDELINES/
                            NASIENRIGLYNE
         MATHEMATICS (PAPER 1)/WISKUNDE (VRAESTEL 1)

9 pages/bladsye

 NOTE/LET WEL:

 •   If a candidate answered a QUESTION TWICE, mark only the FIRST attempt.
 •   If a candidate crossed out an answer and did not redo it, mark the crossed-out answer.
 •   Consistent accuracy applies to ALL aspects of the marking guidelines.
 •   Assuming values/answers in order to solve a problem is unacceptable.

 •    As ŉ kandidaat ŉ VRAAG TWEE KEER beantwoord het, merk slegs die EERSTE poging.
 •    As ŉ kandidaat ŉ antwoord deurgehaal het en dit nie oorgedoen het nie, merk die deurgehaalde
      antwoord.
 •    Konsekwente akkuraatheid is van toepassing op ALLE aspekte van die nasienriglyne.
 •    Om waardes/antwoorde te aanvaar om ŉ probleem op te los is onaanvaarbaar.

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                                                                    MATHEMATICS/WISKUNDE
                                                MARKING GUIDELINES/
                                                NASIENRIGLYNE       (PAPER/VRAESTEL 1)
                                                                              GRADE/GRAAD 10

QUESTION/VRAAG 1

1.1 1.1.1   2 x6 − 8
            = 2( x 6 − 4)                                            ✓ factorise/faktoriseer
                                                                     ✓ factorise/faktoriseer
            = 2 ( x3 + 2 )( x3 − 2 )                                                            (2)

    1.1.2   − x + ( x − y )3 + y                                     ✓ −( x − y)
            = ( x − y )3 − ( x − y )                                 ✓ common factor ( x − y)
                                                                       gemeenskaplike faktor
            = ( x − y )[( x − y )2 − 1]
                                                                      ( x − y)
            = ( x − y )( x − y + 1)( x − y − 1)
                                                                     ✓ [( x − y ) 2 − 1]
                                                                     ✓ diff. of two squares/
                                                                       verskil tussen twee
                                                                       vierkante                (4)

                          2
1.2 1.2.1    12   
                                                                         1
                                                                     ✓ x 4 and/en 9
             x − 3
                                                                           1

                 1            1
                                                                     ✓ −6x 2
            = x 4 − 6x 2 + 9                                                                    (2)

    1.2.2   x −3  x −5 + ( x −2 )−1                                 ✓ division x −3+5
            = x −3−( −5) + x 2                                         and brackets/
                                                                       deel x −3+5 en hakies
            = x2 + x2
            = 2 x2                                                   ✓ simplify/vereenvoudig    (2)

    1.2.3   5x +1.(4 x )3
            25 x −2.10 x −1
                 5x +1.(22 x )3
            = 5 x−2                                                  ✓ bases primefactors/
              2 .(5.2) x −1                                            basisse priemfaktore
                  5x +1.26 x
            = 5 x − 2 x −1 x −1
              2 .5 .2
                                                                     ✓ simplify numerator/
            = 5x +1− x +1.26 x −5 x + 2− x +1                          vereenvoudig teller
            = 52.23                                                  ✓ simplify denominator/
            = 200                                                      vereenvoudig noemer

                                                                     ✓ final answer/
                                                                       finale antwoord          (4)

                                                        2

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                                                              MATHEMATICS/WISKUNDE
                                          MARKING GUIDELINES/
                                          NASIENRIGLYNE       (PAPER/VRAESTEL 1)
                                                                        GRADE/GRAAD 10

      1.2.4        − 11            1
                              –                                     ✓ Factorise trinomial/
              2 x − 5 x − 12 4 − x
                   2
                                                                      Faktoriseer drieterm
                      − 11           1
              =                   +
                (2 x + 3)( x − 4) x − 4                             ✓ Change of sign/
                  − 11 + 2 x + 3                                      Verandering van teken
              =
                (2 x + 3)( x − 4)
                     2x − 8                                         ✓ Numerator/Teller
              =                                                     ✓ LCD/KGV
                (2 x + 3)( x − 4)
                    2( x − 4)                                       ✓ Simplified answer/
              =
                (2 x + 3)( x − 4)                                     Vereenvoudigde
                   2                                                  antwoord
              =
                 2x + 3                                                                            (5)

1.3           (3 x + 2)(2 x − 5)                                    ✓ (2 x − 5)
              = 6 x − 11x − 10
                       2
                                                                    ✓ Simplify/Vereenvoudig
               d = −11                                             ✓ Answer/Antwoord              (3)
                                                                                                  [22]

QUESTION/VRAAG 2

2.1 2.1.1     (3x − 1)( x + 2) = 0                                      1
                                                                    ✓
                       1                                                3
              x=         or/of x = –2
                       3                                            ✓ ̶2                           (2)

      2.1.2   3x − 2.3x−1 = 3
              3x (1 − 2.3−1 ) = 3           If answer ONLY          ✓ Factorise/Factorise
                                            NO marks/
                           1
                        3x   = 3          SLEGS antwoord
                           3              GEEN punte nie          ✓ Simplify/Vereenvoudig
                            3x = 32
                                                                    ✓ Answer/Antwoord
                             x=2                                                                   (3)

2.2 2.2.1              2x                                           ✓ Subtract 1 on both sides/
              −1  1 −      5                                        Trek 1 aan albei kante af
                        3
                     2x                                             ✓ 3
              −2  −     4
                     3                                              ✓ (−2) and change of
              −6  −2 x  12
                                                                      inequalities/en
               3  x  −6                                             verandering van
               −6  x  3                                             ongelykhede                  (3)

      2.2.2 9 integers satisfy the equation/9 heelgetalle voldoen   ✓ Answer/Antwoord
            aan die vergelyking                                                                    (1)
                                                    3

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                                                             MATHEMATICS/WISKUNDE
                                         MARKING GUIDELINES/
                                         NASIENRIGLYNE       (PAPER/VRAESTEL 1)
                                                                       GRADE/GRAAD 10

2.3 Solve for x and y simultaneously/Los vir x en y gelyktydig op:

                                        x y                          ✓ Make x the subject of
              x− y =4       and/en       + =5                          the equation (1)/Maak
                                        5 2
     x = y + 4……(1)             2x + 5y = 50……..(2)                    x die onderwerp van die
                                                                       vergelyking (1)
     subst/vervang (1) into/in (2)
                                                                     ✓ LCD/KGV
     2(y+ 4)+ 5y = 50
     2y + 8 + 5y = 50                                                ✓ Substitute/Vervang
     7y = 42
     y=6                                                             ✓ y value/y-waarde

     x=6
     x = 10                                                          ✓ x value/x-waarde

                                     OR/OF
                                                                     ✓ Make y the subject of
      y = x – 4……(1)                                                   the equation (1)/Maak
     subst/vervang (1) into/in (2)                                     y die onderwerp van die
     2x+ 5(x – 4) = 50                                                 vergelyking (1)
     2x + 5x – 20 = 50                                               ✓ Substitute/Vervang
     7x = 70                                                         ✓ Simplify/Vereenvoudig
     x = 10                                                          ✓ x value/x-waarde

     y = 10 – 4
     y=6                                                             ✓ y value/y-waarde           (5)
                                                                                                 [14]

                                                      4

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                                                           MATHEMATICS/WISKUNDE
                                       MARKING GUIDELINES/
                                       NASIENRIGLYNE       (PAPER/VRAESTEL 1)
                                                                     GRADE/GRAAD 10

QUESTION/VRAAG 3

3.1 3.1.1 Zain’s formula/Zain se formule:        Tn = 4(n − 1) + 2          ✓ Method/Metode
                                                    = 4n − 4 + 2
                                                    = 4n − 2                ✓ Answer/Antwoord
            This is the same as Letti’s formula. They are both correct./    ✓ Reason/Rede
            Dit is dieselfde as Lettie se formule. Hulle is albei korrek.                           (3)

     3.1.2 T27 = 4(27) –2 ✓                                                 ✓ Substitution/
                                                                              Vervanging
                = 106 ✓                                                     ✓ Answer/Antwoord       (2)

3.2 3.2.1 Tn = 5n – 8 ✓✓                                                    ✓ 5n
                                                                            ✓–8                     (2)

     3.2.2 The 52th number to end in a 7 will be T105/Die 52ste getal       ✓ n=105
           wat op ŉ 7 einding sal T105 wees.

            T105 = 5(105) – 8                                               ✓ Substitute/Vervang

                 = 517                                                      ✓ 517

            OR/OF

            7 ; 17 ; 27 ; 37

            Tn = 10n –3                                                     ✓ Tn
                                                                            ✓ Substitute/Vervang
            T52 = 10(52) – 3                                                ✓ 517

                = 517

            OR/OF
                                                                            ✓ Pattern/Patroon
            7 ; 17 ; 27 ; 37                                                ✓✓ 517
            By inspection the 52nd term will be 517
            Deur inspeksie is die 52ste term gelyk aan 517                                          (3)
                                                                                                   [10]

                                                     5

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                                                           MATHEMATICS/WISKUNDE
                                       MARKING GUIDELINES/
                                       NASIENRIGLYNE       (PAPER/VRAESTEL 1)
                                                                     GRADE/GRAAD 10

QUESTION/VRAAG 4

4.1 4.1.1     xR                                                    ✓ Answer/Antwoord            (1)

      4.1.2   y  −4 OR/OF y  (−4; )                               ✓ Answer/Antwoord            (1)

      4.1.3   y = −3                                                 ✓ Answer/Antwoord            (1)

4.2 f (x) = ax – 3
    subst/vervang (–2;6)                                             ✓ Substitution/Vervanging
    6 = a–2 – 3
    a–2 = 9
            1
    a–2 =
            9
         1                                                               1
    a = = 3–1                                                        ✓     = 3–1
          3                                                              3                        (2)

4.3 x-intercept exponential graph:/x-afsnit eksponensiële grafiek:
    3–x – 3 = 0
    3–x = 3                                                          ✓ Let/Stel x = 0
    x = –1
      C( ̶ 1 ; 0)                  OR/OF
    x-intercept hyperbola:/x-afsnit hiperbool:
                                                                     ✓ x value/x-waarde
     −4
         –4=0
      x
     −4     4                                                        ✓ co-ordinate/koördinaat
         =                   C( ̶ 1 ; 0)
      x     1
    x = –1                                                                                        (3)

4.4   −1  x  0    OR/OF x  (−1;0)                                 ✓ x  −1
                                                                     ✓ x0                        (2)

4.5 4.5.1 A(–2 ; 6) C(–1 ; 0)
                6−0
          m=
               −2 + 1                                                                 y2 − y1
                                                                     ✓ subst into m =
          m = −6                                                                      x2 − x1
          y = –6x + c                                                                 y −y
          subst/vervang (–1;0) OR/OF subst/vervang (–2;6)             vervang in m = 2 1
                                                                                      x2 − x1
          0 = –6(–1) + c            6 = –6(–2) + c
                                                                     ✓m= ̶6
              c = –6                                                 ✓ subst m and co-ordinate/
                                                                       vervang m en koördinaat
               y = –6x – 6
                                                                     ✓ equation/vergelyking       (4)

      4.5.2   y = −x − 4                                             ✓ –x
                                                                     ✓ –4                         (2)
                                                    6

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                                                              MATHEMATICS/WISKUNDE
                                          MARKING GUIDELINES/
                                          NASIENRIGLYNE       (PAPER/VRAESTEL 1)
                                                                        GRADE/GRAAD 10

4.6              4
      g ( x) = − − 4 and/en B(x ; 6)
                 x
        4                                                    ✓ Subst y = 6/Vervang y = 6
      − −4=6
         x
        4                                                    ✓ x value/
      − = 10
         x                                                     x-waarde
      x = −0, 4
      AB = 2 ̶ 0,4                                           ✓ 1,6
           = 1,6 units/eenhede                                                              (3)
                                                                                           [19]

QUESTION/VRAAG 5

5.1

                                                             ✓ Shape/Vorm
                                                             ✓ x intercepts/
                                                               x-afsnitte
                                                             ✓ Turning point/
                                                                Draaipunt

                                       (0 ;–8)
                                                                                            (3)

5.2 q = –8
    Substitute (–4;0) or (4;0) to calculate a
    Vervang (–4;0) of (4;0) om a te bereken                  ✓Subst coordinate
     y = ax 2 − 8                                             and q = –8/
                                                              Vervang koördinaat
      0 = a (4) 2 − 8                                         en q = –8
      16a = 8
          1                                                  ✓a= 1
      a=                                                             2
          2
             1                                               ✓ Equation/ Vergelyking
       y = x2 − 8
             2                                                                              (3)

5.3 –4  x  4 OR/OF x [–4;4]                               ✓ x  –4
                                                             ✓x4                           (2)

5.4             1                                            ✓ – 1 x2
      k ( x) = − x 2 + 2                                         2
                2                                            ✓ +2                           (2)
                                                                                           [10]

                                                  7

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                                                       MATHEMATICS/WISKUNDE
                                   MARKING GUIDELINES/
                                   NASIENRIGLYNE       (PAPER/VRAESTEL 1)
                                                                 GRADE/GRAAD 10

QUESTION/VRAAG 6

6.1 A = (1 + in)
      = 7 990(1 + (0,15  4)) ✓✓                           ✓ Substitution in correct
                                                              formula/Vervanging
                                                              in korrekte formule
                                                           ✓ Converting 48 months
                                                             to 4 years/
                                                             Omskakeling 48
                                                             maande tot 4 jaar
       = 7 990(1,60)
       = R12 784,00✓                                       ✓ Total amount
                                                             including interest/
                                                             Totale bedrag
                                                             insluitend rente
       = R12 784,00 ÷ 48                                   ✓ Monthly repayment/
       = R 266,33 ✓                                         Maandelikse
                                                            terugbetaling
    Monthly payment including fees/Maandelikse betalings
    insluitend fooie = R266,33 + R13,50
                     = R 279,83 ✓                          ✓ Total monthly
                                                             repayment/
                                                             Totale maandeliks
                                                             terugbetaling              (5)

6.2 A= P(1+ i)n ✓                                          ✓ Correct formula/
                                                             Korrekte formule
      = 5000(1 + 0,2)31 ✓✓✓                                ✓ 20% to/na 0,2
                                                           ✓ 60 – 29, n=31
                                                           ✓ Substitution/
                                                             Vervanging
      = R 1 424 257,88 ✓                                   ✓ Answer/Antwoord            (5)
                                                                                       [10]

                                               8

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                                                               MATHEMATICS/WISKUNDE
                                           MARKING GUIDELINES/
                                           NASIENRIGLYNE       (PAPER/VRAESTEL 1)
                                                                         GRADE/GRAAD 10

QUESTION/VRAAG 7

7.1 7.1.1 P(A or/of B) = 1 – P(A or/of B)'                       ✓ 1 – P(A or/of B)'
                           = 1 – 0,05
                           = 0,95                                ✓ Answer/Antwoord         (2)

     7.1.2 NO/NEE                                                ✓ no/nee
                                                                 ✓ Justification/
             P(A) + P(B)  P(A or/of B)                            Regverdiging            (2)

     7.1.3 P(A or/of B) = P(A) + P(B) – P(A and/en B)                Formula/formule
             P(A and/en B) = 0,25 + 0,89 – 0,95
                          = 0,19                                     Answer/Antwoord       (2)

     7.1.4 P(B) = 0,89 – 0,19                                    ✓ 0,89 – 0,19
                  = 0,6                                          ✓ Answer/Antwoord         (2)

7.2 7.2.1
                          ADD              MULTIPLIER            ✓ 600 – 285 = 315(ADD)
                                                                 ✓ 950 – 285 = 665
                                                                    (MULTIPLIER)
                                                                 ✓ 600 + 950 + 235 –
                 600 – 285 = 315    *285     950 – 285 = 665       1 500 = 285
                                                                 ✓ Correct diagram used/
                                                                   Korrekte diagram
                                                                   gebruik
                                    235
                 600 + 950 + 235 – 1 500 = 285                                             (4)

     7.2.2    285   19                                           ✓ Answer only/
                  =                                                Slegs antwoord
             1 500 100                                                                     (1)

     7.2.3   1 500 − 950    550                                     1 500 − 950
                         =                                       ✓
                1 500      1 500                                       1 500
                           11                                      11
                         =                                       ✓
                           30                                      30

             OR/OF                                               OR/OF

             315 + 235    550   11                               315 + 235
                       =      =
               1 500     1 500 30                                  1 500
                                                                                         (2)
                                                                                       [15]
                                                                         TOTAL/TOTAAL : 100
                                                       9

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                           MATHEMATICS/WISKUNDE
       MARKING GUIDELINES/
       NASIENRIGLYNE       (PAPER/VRAESTEL 1)
                                     GRADE/GRAAD 10

               10

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