# Mathematics Grade 9 ANA 2011 Memo (Afrikaans)

> Grade 9 Mathematics memorandum from the ANA 2011 (Afrikaans), 6 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/mathemetics-grade-9-afr-form-b-memo/197b265c-ca88-4ad5-a54b-74ec2523976d
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 9
- Document type: Memorandum
- Year: 2011
- Exam period: ANA
- Curriculum: CAPS
- Pages: 6

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## About this document

Mathematics · Grade 9 · ANA · 2011. Memorandum, 6 pages. Read online or download the PDF.

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

## Document text

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

### Page 1

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basic education

Department:
Basic Education
REPUBLIC OF SOUTH AFRICA

NASIONALE ASSESSERINGS 2010
GRAAD 9 WISKUNDE - AFRIKAANS
VORM B

MEMORANDUM

Belangrike inligting

e — Hierdie is bedoel as ‘n gids om te merk. Leerders kan punte kry in gevalle waar hul ander, maar wiskundig korrekte,

metodes gebruik het om probleme op te los.

e —Leerders kan volpunte kry as hul slegs die regte antwoorde neergeskryf het, buiten as daar verskillende instruksies

gegee is by ‘n vraag.

VRAAG 1
11) A 12 |B 1.3 14 |¢ 15 |¢ Gee 1 punt vir elke regte
1 antwoord.
: 16 |C 17: 1C 18 19 |C 1.10 | B
[10]
VRAAG 2
2.4.4 x 4dr t4—x2 4243 Distributiewe wet: 1 punt
=6x4+7 vy Antwoord: 2 punte (3)
2.1.2 4y2 vv 4:1 punt
x v¥ y? 21 punt
x: 1 punt (3)
2.1.3 vvvv
=8x? +4x—-1—-4x41 Vereenvoudiging van elke breukterm
V (deel): 3 punte
= 8x Hakie uitbreiding:1 punt (5)
Antwoord: 1 punt
2.2.1 Yvv Vir elke korrekte term: 1 punt
=9a? -6a-3 (3)
2.2.2

a> -2a* +4a+2a7-4a+8

=a+8 vv

Uitbreiding: 2 punte

Antwoord: 1 punt

2.3.1 v v
2xy(xy — 2x + 5y)

2xy: 1 punt
(xy — 2x + 5y): 1 punt

(2)

23.2 Yov
(3x + y)Gx-y)

Vir elke korrekte hakie: 1 punt

(2)

2.3.3 v v v
x(x? = 4) = 2x(x + 2)(x - 2)

Gemene faktor: 1 punt
Vir elke korrekte hakie: 1 punt

### Page 2

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                                                                                                       Page 2 of 6

2.4
                        3       1089
                        3       363
                       11       121
                       11       11                                                Regte faktore: 2 punte
                                1

                                                                                      Antwoord: 2 punte
                   1089 =           3 2 . 11 2 = 3 × 11 = 33
                                                                                                                      (4)

2.5.1   3x − 3 − 4 x = 5 − 2 x − 2                                                   Uitbreiding: 1 punt
        ∴ − x − 3 = −2 x + 3                                                     Vereenvoudig x : 1 punt
                                  
        ∴x = 6 
                                                                                    Antwoord = 6: 1 punt              (3)

2.5.2    2 x − 3 = 3x                                                       Vermenigvuldig met 3: 1 punt
                                                                             Korrek vereenvoudig : 1 punt
        ∴ x = −3                                                                      Antwoord: 1 punt              (3)

2.5.3                                                                  Oorkruis vermenigvuldiging: 1 punt
        3( x − 2) − 4( 4 x + 1) = x − 2                        
        ∴ 3x − 6 − 4 x − 4 = x − 2                                            Korrek vereenvoudig: 1 punt
        − x − 10 = x − 2                                                            Vereenvoudig: 2 punte
                                                               
        − 8 = 2x                                               
                                                                                         Antwoord: 1 punt
        ∴ x = −4                                                                                                      (5)

                                                                                                                     [39]

VRAAG 3

               P.n.r
        SI =                    
3.1            100                                                                        Formule: 1 punt

            (5400)(4)6)                                                              Vervanging: 2 punt
        =
               100
                                                                               Vereenvoudiging: 1 punte             (4)
        = R1296
                            n
                 r 
3.2     A = P1 +                   
              100 
                                                                                              Formule: 1 punt
          (8000)(1,05)3             
        =                                                                                 Vervanging: 3 punte
              100
                                                                                            Antwoord: 1 punt
        = R9261                                                                                                      (5)
        Aantal verskuldig = R9261
3.3.1    Omgekeerde eweredigheid
                                                                                            Antwoord: 1 punt
                                                                                                                     (1)
3.3.2   As tyd = t
                                        
                                                                                          Vergelyking: 1 punt
        dan is 16 t = 40
        ∴ t = 2,5                                                                           Antwoord: 1 punt        (2)

                                                                    2

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                                                                                                                  Page 3 of 6

3.4   Koste: 4,5 kg = R36
                                                                                   Bereken koste van 1 kg: 1 punt
      Koste 1 kg: R36 ÷ 4,5 = R8 
      ∴ Koste 2,5 kg = 8 × 2,5 = R20                                                      Koste van 2,5 kg: 2 punte

              4,5 36
                  =
              2,5   x
      of
           ⇒ 4,5 x = 2,5 × 36 = 20                                                                                               (3)

                                                                                                                                [15]

VRAAG 4

             
4.1   17; 21                                                                                Elke term: 1 punt                    (2)

4.2                        
      Tel 4 by om die volgende getal te kry.                                                Antwoord: 1 punt                     (1)

           
4.3   4n + 1                                                                            Antwoord : 2 punte                       (2)

4.4   4n + 1 = 101                                                                 Vergelyking: 2 punte
      4n = 100                                                                    Vereenvoudiging: 1 punt
      n = 25                                                                          Antwoord : 1 punt                        (4)
      daarom T25 = 101

                                                                                                                                 [9]

VRAAG 5

5.1   5.1.1        c                     5.1.2    a                                                  1 punt elk           (2)

5.2    y = −x + 2

       x            -2       -1         0           1
       y             4        3         2           1

      y = 2x − 3          Nota: leerders kan ander x–waardes, as in tabel gegee, gebruik.

       x            -1        0         1
       y            -5       -3         -1

                                                         3

### Page 4

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                                                                                     Page 4 of 6

                                                                                       y = −x + 2
                                              y
                                                                                      4 punte geplot:
                                                                                       1 punt
                                         12
                                                                                      Rigting:
                                         10                                           1 punt

                                                                                      Vorm
                                           8
                                                                                      (Afgebroke
                                                                                      punte):
                                           6                                          1 punt

                       y = −x + 2          4

                                           2

                                                                                                 x
-14   -12   -10   -8   -6   -4      -2                2   4   6   8   10   12   14

                                         -2

                                         -4
                                                                                       y = 2x − 3
                                                                                      Korrekte x en y
                                         -6
                                                                                      afsnitte (twee):
                       y = 2x − 3                                                      2 punte
                                         -8

                                                                                      Reg benoem:
                                         -10                                          1 punt

                                                                                      Vorm (reguit
                                         -12
                                                                                      lyn):
                                                                                      1 punt

                                                                                                     (7)

                                                                                                           [9]

                                                  4

### Page 5

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VRAAG 6
6.1.1 Vv v parallelogram: 1 punt
‘n Parallelogram — beide pare teenoorgestelde sye is gelyk tede:1 punt (2)
6.1.2 Opvl trapesium. APCD = (PC+AD)h Identifisering van trapesium:1 punt
2 Formule: 1 punt
(16+24)12 y Vervanging: 1 punt
~ 2
=240cm2 ¥ Antwoord: 1 punt (4)
v
6.1.3 | Teenoorgestelde (t.o.) sye van reghoek APTC Antwoord: 1 punt (1)
6.1.4 | InA ABP en A CDT
1. AP=TC (tio. sye van reghoek. APTC) ¥ Regte stelling: 1 punt
2. BP=TD=8cem v Regte stelling: 1 punt
mn Regte stelling: 1 punt
3. Pj =T,=90° v
AABP=ACDT  (S,<,S) ¥ Korrekte: 1 punt (4)
6.2 | In AABC A +B=140° (som van <'s in ‘n A) Vv Regte stelling met rede: 2 punte
maar A; =B = 70" (to. hoeke; gelyke sye van A) vv Regte stelling met rede: 2 punte
Cy =A, =70° (Alt <'s; AB || DC
. 2 . ' 3 ( ; IIDC) vy Regte stelling met rede: 2 punte
Ay =C; =40  (Alt<'s; AD || BC) Regte stelling met rede: 2 punte
“BAD =110° v Antwoord: 1 punt (9)

[6.3.1 | A ABE | Antwoord: 1 punt (1)
AB_BE_ AE i , Proporsionele sye : 2 punte
i’ === =" (proporsionele sye van gelykvormige Ae) vv
6.3.2 aC CD Ep PO ye van gely| ige Ae)
9 3 v Vervanging: 1 punt
AC 8
AC = 24cm v AC = 24:1 punt
“. BC = 24-9 .
cleem Vv Antwoord: 2 punte 6)
6.4.1 | Volume = area A ABC xh vy Area = opvl Formule: 4 punt
- 2 «20 m3 v Vervanging: 1 punt
= 600 m? v Antwoord: 1 punt (3)
64.2 | InAABC: BC2=25+ 144m? (Pyth) yy
= 169 m2
DaaromBC=13m Vv Gebruik Pythagoras vir BC: 4 punte
Opvl = 2 x area A ABC + (AB+AC+BC)xAD  v Formule: 4 punt
=2 5(12) +6020) vy Vervanging: 2 punte
2 (8)
= 60 + 600
= 660 m2 v Antwoord : 1 punt
[38]

### Page 6

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                                                                                         Page 6 of 6

VRAAG 7

7.1         Stam   Blaar                    Stam   Blaar        2 punte for elke groep
        3          81                          3   18
        4          579                         4   579
        5          2 258                       6   2 258
        6          84                          6   48
        7          42                          7   24
        8          14                          8   14                                                   (4)

7.2.1   Verspreiding = 84 – 31 = 53                      Korrekte antwoord: 2 punte                   (2)
7.2.2   Modus = 52                                        Korrekte antwoord: 2 punte                   (1)
7.2.3   Mediaan = 55                                     Korrekte antwoord: 2 punte                   (2)
7.2.4               870                                    Korrekte antwoord: 3 punte                   (3)
        Gemiddelde=     = 58          
                    15
7.3     7 leerders                                       Korrekte antwoord: 2 punte                   (2)

                                                                                                       [14]

VRAAG 8

8.1                2 1
        P(> 6) =    =      
                                                                   Antwoord: 2 punte
                   8 4                                                                                  (2)
8.2                         4 1
        P (priem getal) = =                                        Antwoord: 2 punte
                           8 2                                                                        (2)
8.3                          4 1
        P (‘n faktor van 8) = =                                    Antwoord: 2 punte
                             8 2                                                                      (2)

                                                                                                        [6]

                                                                               Totaal              [140]

                                                   6

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