# Mathematics Grade 12 Paper 1 Limpopo June 2023 Paper & Memo

> Grade 12 Mathematics Paper 1 question paper and memo from the Limpopo June Exam 2023, 27 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/grade-12-mathematics-2023-limpopo-june-exam-question-paper-and-memorandum-paper-1/01465211-97eb-49dd-86ff-be1d27cadc05
- Updated: 2026-09-28
- Subject: Mathematics
- Grade: Grade 12
- Document type: Question paper and memo
- Year: 2023
- Exam period: Limpopo June Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 27

Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Mathematics](https://www.hlayiso.com/library/grade-12/mathematics) › [Mathematics Grade 12 Paper 1 Limpopo June 2023 Paper & Memo](https://www.hlayiso.com/library/grade-12-mathematics-2023-limpopo-june-exam-question-paper-and-memorandum-paper-1/01465211-97eb-49dd-86ff-be1d27cadc05)

## About this document

Mathematics · Grade 12 · Limpopo June Exam · 2023. Question paper and memorandum, 27 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

Rawnaaded from Stanmanepbysias.com May/June 2023

LIMPOPO

PROVINCIAL GOVERNMENT
REPUBLIC OF SOUTH AFRICA

DEPARTMENT OF

EDUCATION

MOGALAKWENA DISTRICT

MARKS: 150

TIME: 3Hours

Page | 1Copyright reserved Please turn over

### Page 2

Rawnleaded. from Stanmanepbysias.com May/June 2023

INSTRUCTIONS AND INFORMATION:

Read the follo’ instructions carefully before answering the questions.
1. This ques' japer consists of 9 questions.

2. Answer luestions.

3. Clearly show ALL calculations, diagrams, graphs, etc, that you have used in
determining your answers.
4. Answers only will NOT necessarily be awarded full marks.

5. If necessary, round off answers correct to TWO decimal places, unless stated
otherwise.
6. You may use an approved scientific calculator (no programmable and no graphical).

7. Diagrams not necessarily drawn to scale.

8. An information sheet with formulae is included at the end of the question paper.

Page | 2Copyright reserved Please turn over

### Page 3

Rawnleaded. from Stanmenepbysias.com

QUESTION 1
1.1 Solve

1.10Dfx - 3)(5x + 2) = 0
1.1.2 3x?- 10x- 1 = 0 (Correct to two decimal places)
113 x(x- 4) > 21

14 2342

2

1.2. Solve for x and y simultaneously.

x+2y=2
x? + 8y =8
1.2.1
Given: 2**! +2*=3)*2_}, where x and y are integers.
Determine the value of x and y.
1.2.2 10
{x2 43=
eee e-2
1.3. Given:
, -ttye—@

1.3.1 If p = 4, determine the nature of the roots.
1.3.2 Determine the value(s) of p for which the roots are non-real.

1.4 The equations x? +rx+m=0 and x° +mx+r=0 have real and

Solve for the values of r and m if r > 0 and m>0.

Page | 3Copyright reserved

May/June 2023

(2)

(3)

(3)

(4)

(6)
(3)

(5)

(1)

Q)

roots. (6)

[37]

Please turn over

### Page 4

Rawnleaded. from Stanmanepbysias.com May/June 2023

QUESTION 2

The sequence: —67 ; x ; y ; —28 ; —19; ... has a quadratic pattern.

2.1 Dete the value of x and y

2.2 Find pression for the nth term

2.3 Pro\ it the sequence of numbers will never contain a positive term.
QUESTION 3

3.1 The fourth term of a geometric progression is 24 and the ninth term is 768.

Determine the first three terms of the Progression.

3.2

IfsS,, = 207 +3n then

3.2.1 Calculate 7},

3.2.2 Find 7, im its simplest form.
3.3 Caleulate the value ofS (2n _- 1)
3.4 40

Ina converging geometric series S,, = and 7, = > calculate the

z
possible value(s) the first term of the series.
3.5 Given the following geometric sequence:

24
= © 12+ 6x + 3x7+...

3.5.1 Calculate the sum to infinity of the series

3.5.2 Write down the values of x for which this sequence converges.

(6)
(4)
(3)

[13]

(5)

(3)

(4)

(3)

(3)

(4)
(2)
[24]

Page | 4Copyright reserved Please turn over

### Page 5

4.1

4.2

4.3

44

45
4.6

4.7

Rawnleaded. from Stanmanepbysias.com May/June 2023

QUESTION 4

The grap) f(x) and g(x) are shown in the diagram below.

/ \ 7

The turning point of f(x) is A (2 ; 9) and the graphs f and g intersects at B (3; 8). C is a point
on g(x) and is on the axis of symmetry of f.

Show that the function f can be defined by the equation: (4)
fx) = —x?+4xe4+5
Write down the equation of the axis of symmetry of f. (1)
The graph g(x), has the equation y = a*. Determine the value of a . (2)
If it is given that (—1; 0) is one root of f, write down the coordinates of the other root. (1)
For which value(s) of x will f(x) < 0? (2)
Determine the length of AC. (2)
Discuss the nature of the roots of h(x) if, h(x) = f 9. (2)

[14]

Page | SCopyright reserved Please turn over

### Page 6

Rawnladed. from Stanmenepbysjas.com May/June 2023

QUESTION 5

In the diagram below are the sketches of: f(x) = —2x +2 and g(x) = —+ q

x+p

The graph cuts the y-axis at (0; 4).

5.1 Calculate the values of p and q (3)

5.2. Show that a = —2 (2)

5.3 Determine the coordinates of point A the x — intercept of g(x) (2)

5.4 The value of q is increased by 1 unit. What effect will it have on the graph of g? = (1)

5.5 Write the equations of the asymptotes. (2)
[10]

QUESTION 6

1 x
Given: P(x) = (3)
6.1 Write down the equation of p=! inthe form y = -- (2)
62 Sketch the graphs of p and p~! on the given system of axes on the diagram sheet. ()

Show clearly all the intercepts with the axes as well as the line of symmetry.

©3 Determine the value(s) of x for which p71(x) > 1 (2)

Page | 6Copyright reserved Please turn over

### Page 7

Rawnaaded from Stanmanepbysias.com May/June 2023

64 Write down the equation of the graph of h(x), if h(x) is formed by shifting the graph

of p(x) 2 units to the right and then 3 units up.

(2)
[11]
QUESTIO
71 Determine f'(x) from first principles if f(x)=5—2x* . (5)
7.2 Determine 2 if
dx
72 ay ag (3)
7.2.2 xy=5 (2)
7.2.3 so [37 —7x—s
[ ; J 6
[14]

Page | 7Copyright reserved Please turn over

### Page 8

Rawnleaded. from Stanmenepbysias.com

QUESTION 8
Given that = x3 — 6x? + 9x — 4 with x-intercepts at A (1 ; 0) and B, y-intercept at C(0 ;
AandD turning points of f.
yu
f(x)
A(1;0) Be
C (0;-4) D
8.1 Calculate the coordinates of B.
8.2 Calculate the coordinates of the turning point D.
83 Calculate the x-coordinate of the point of inflection of f.
8.4 Write down the values of k for which f(x) = k will have only ONE root.
8.5 Determine the values of x where f '(x) > 0
8.6 Determine the gradient of the line DC.

Page | 8Copyright reserved Please turn over

May/June 2023

-4).

(4)
(5)
(2)
(2)
(3)
(1)

(17

### Page 9

Rawnladed. from Stanmenepbysjas.com May/June 2023

QUESTION 9

Two cyclists start to cycle at the same time. One starts at point B and is heading due north to
point A, whilst the other starts at point D and is heading due west to point B. The cyclist
starti B cycles at 30 km/h while the cyclist starting from D cycles at 40 km/h.
The between B and D is 100 km. After time ¢ (measured in hours), they reach points F

and ively.

A q

Fe

O *
B Cc D
9.1 Determine the distance between F and C in terms of ¢. (4)
9.2 After how long will the two cyclists be closest to each other? (4)
9.3 What will the distance between the cyclists be at the time determined in (2)
QUESTION 9.2?

[10]

Page | 9Copyright reserved Please turn over

### Page 10

Downloaded from Stanmorephysics.com

LIMPOPO

PROVINCIAL GOVERNMENT
REPUBLIC OF SOUTH AFRICA

DEPARTMENT OF

EDUCATION

MOGALAKWENA DISTRICT

MARKS: 150

Page 1 of 18

### Page 11

Downloaded from Stanmorephysics.com

INSTRUCTIONS:
NOTE:

¢ This memorandum consists of 19 pages including the cover page.
e If the’candidate answers a question TWICE, only mark the FIRST

e Consistent accuracy applies in all aspects of the marking memorandum.

Page 2 of 18

### Page 12

Downloaded from Stanmorephysics.com

QUESTION 1
1.1.1 x=3: 2 v¥ ¥each root
. 5
‘al —b+Vb?-4ac
2a
= ~(10)+ ¥C10)?7-4)(-1) ¥ subst
2(3)
_ 104 Viz ¥ Yeach root
= 4 a Pama dil (-1, if incorrect rounding)
‘ , (3)
1.1.3 xv -4x>21 ¥std form
. _— e 0 0 ¥ factors
(x — 7)(x + 3)> a
x<-3andx>7,xeR *eritical pts
¥ ¥ conclusion (5)
114 | 23°45 = 72 (x 2)
p+ gra Sy?
2.2.37 + = >*F
4.F+F=15 ¥ simplify
3° (441) =15 Y factorisation
3=3 ¥ simplify
x=l ¥x=1
(4)

Page 3 of 18

### Page 13

Downloaded from Stanmorephysics.com

1.2
XF DY =D vrrcersee (1) P+8y=8 (2)
x=—2yt2...(3) ¥ substitution
Subst 3 in 2. (-2y + 2)(-2y + 2) + 8y=8
4y’ -8y+4+8y-8=0
4yr-4=0 ¥std form
y-1=0 Y factorisation
v- Dy bs “a vy-value
y=lory=-1
Subst y= 1 in3 Subst y =— 1 in3
x=-2(1)+2 x=-2(-1)+2
x=0 x=4 v¥ vx —values.
Solution (0 ; 1) (43-1)
(6)
1.2.1
2 +2 = pet =
2°(2' +1) =3’G?-1) Y factorise
2*(3) =3'(8)
z=
a)
ees
a
2 3 ¥ simplify equated
2 =3y" bases
“x-3=0 and y-1=0
“x=3 and y=1 ¥ answers
OR OR
ge +2" — 37? ir
2(2' +1) =3°(3?-1) rise
2*(3) =3' (8)
72) — 2993 implified equated
2*(3)=3*(2°) bases
“x=3 and y=1 v answers (3)

Page 4 of 18

### Page 14

Downloaded from Stanmorephysics.com

1.2.2
fe—3 4300
x—-2
Let k=
k+3=
v
k? 43k-10=0 standard form
(k+5)(k-2)=0 v factors
k=-5 or k=2 Y_ both answers for k
Vv ;
Vx-2#-5 or Jx-2=2 selection
2 2
(ve=2) =(2)°
ape Y answer
OR
Je-24+3=—8
x-2
(Vx=2)(vx=2)+3Vx=2 =10
x-2+3V¥x-2 =10
3 v simplified both sid
Ix-2=12-x
2 Y squaring both side
(3Vx=2) =(12-x)
9x-18=144-24x42
x? ~33x+162=0 Y standard form
(x-6)(x-27)=0 F tee
x=6 or x#27 lection (5)
n

Page 5 of 18

### Page 15

Downloaded from Stanmorephysics.com

¥ equal & rational

(1)

ase A= 16-4p = 16-4(4)=0
+: Roots equal rational
—
YA<0 (2)
13.2) a=] <0 non-real vp>4
<-16
>4
1.4

x +rxt+m=0 and x°+mx+r=0
x +rxtm=0

For real and equal roots, A= 0

b? -4ac =0

(r¥ —4(@)(n) = 0

r?>-4m=0

r?>=4m

r

m= -- ()

x°+mx+r=0

b* -4ac =0

nm? —A(1I(r)=0
m’?-4r=0_... (2)
Substitute (1) in (2)

2 2
(= —4r=0
4

4
7 -ar=0
16

r*—64r=0
rr —64)=0

Y substitute into A=0

Y equation for m

Y equation 2

Y substitute for m

Page 6 of 18

### Page 16

Downloaded from Stanmorephysics.com

r(r—4)(r? +4r +16) =0

¥ value of r

“7=4

r
n=—

4

4
=—-

4
m= ¥ value of m

OR

x +rx+m=0 and x*°+mx+r=0
For real and equal roots quadratic must be a perfect
square.

x +rxtem=0 .
(x+4m) =0 (x+ Vm) 5
x +2ymx+m=0

r=2Nm Y equation for m

—=m ra 0 |

x +mx+r=0

(x+F) =0

y :
equation 2

x 4+2Wr+r=0

m= vr

m=4r +2) Y substitute for m

Page 7 of 18

### Page 17

Downloaded from Stanmorephysics.com

Substitute (1) in (2)

4
r*

a |
16

r*-64r=0

r(r’ -64)=0
r(r—4)(r? +4r +16) =0
“rs4

a|%\

=
1
a|*%

3
ll
os

v value of r

¥ value of m

(6)

[37]

Page 8 of 18

### Page 18

Downloaded from Stanmorephysics.com

QUESTION 2
21| —6 Fa
\ \ rll 7 Y first differences
Caw vs
2x—-67 -—2y+x-28 y+3 Y second differences
awe ¥ equating second
mnie differences /
aes vx-value
x=-52
¥ equating second
y-2x-67=-2y+x-28 differences/
3y—-3x=39
y-x=13
i vy-value
y=-39 6)
2.2
-67 -52 -39 -28
15 13011
-2 -2
2a=-2
a=-1 ¥ value of a
3a+b=15
- =1
—— ¥ value of b
b=18
a+b+c=-67
—1+18+c=-67 lue of c
c=-84
T, =—n" +1884 ¥ answer (4)

Page 9 of 18

### Page 19

Downloaded from Stanmorephysics.com

2.3
T,>0 v¥—n’ +18n-84>0
—n+ 84>0
n?-1 4<0 ¥ n° -18n+84<0
no Solution ¥ conclusion
~. the nce will never contain a positive term (3)
[13]
QUESTION 3
3.1
T, =24 T, = 768
vA - rt
T,
768 = r v 18 =r
24 24
32=r° ¥ simplification
2? =r’ vr
r=2
Lop or Ta
7, T,
24 _ ys 768 _
a a
24 768
a=— =
8 256 tz
a=3 a=3
T, =ar""
T, =3-2""
n v T3
3 & so...
OR
(5)

Page 10 of 18

### Page 20

Downloaded from Stanmorephysics.com

T,=24 T, =768
T, =ar""
Ty Or? = Wa ccsscemesccesase’ pation V equations
T, = CS equation2
eq
e 1
ar® =768
ar’? =24
r> =32 vr =32
ao
vr
r=2
ar? =24 or ar’ =768
a-2?=24 a-2° =768
a=3 a=3 va
7.20
T, =3-2""
3 6 1B ¥ T&T;
3.2.1 | Fiz = Siz — Sir “T,=S2—S,
_ a ~ 278 v substitution .
v answer
(3)
322| T,=S,—S,., YT=5.-S,,
=2n? +3n—[(n-1) +3(n-1)] Faibetitetions

=2n? +3n—[2n? —4n+2+3n-3]

=2n? +3n-2n?+n+1

=4n+1

¥ simplification

v answer

(4)

Page 11 of 18

### Page 21

Downloaded from Stanmorephysics.com

3.3 is
Sen 1)
3454+7+...
a=3 S, =F Pa+(n-1a] Yaad
d= =17f.34+(17-12]
2 ¥n=17
n=17 =323 answer (3)
3.4
s, -40 7-3
3 rs
a 40
—= et yt
&, = T, =a
BE)
f=ar=7
a=a-r) r=
= aC - =) “substitution
3 2a
40 100 ¥ standard form
rr ar ¥ factors

¥ answer(both values of a)

(3)

Page 12 of 18

### Page 22

Downloaded from Stanmorephysics.com

aes) a Y value of r
x
- &* 4, 3X 4 x
1 6x 2
S. = Y — substitution in correct
r formula
4
S.= —*—
lz Y correct numerator
48 Y correct denominator
So =
2x —x’? (4)
3.5.2 | It exists when:
ind «3 ie: -—l<r <1 7 Ari
§ ot <2
v answer
ANSWER ONLY: Award full marks. (2)
[24]
QUESTION 4
4.1
f(x) = a(x -2)?+9 YT, =2"
a(3-2)?+9=8 feat
a+9=8 vb 1
» f(x) = -1(x- 2)? +9 v Answer
f(x) = -1(x? -4x +449
f(x) = -x? +4x-449
f(x) = -x? +4x4+5

Page 13 of 18

### Page 23

Downloaded from Stanmorephysics.com

42|x=2 v Answer (1)
43
g(x) = a* ¥ Substitution
a=8
a*=2 ¥ Answer
a=2 (2)
4.4 | (5; 0) ¥ Answer (1)
45 |x<-lorx>5 Mark per answer
v¥x<-1
¥x>5 (2)
4.6 | g(x) = 2%
g(2) =27 =4 ¥g(2)=4
AC =9-4
AC = 5 units ¥ Answer
(2)
4.7 | Roots will be real and equal ¥Real
¥ Equal (2)
[14]
QUESTION 5
5.1
x= intercept of the line: = 2x+2=0 « x=1 ¥x int
p=-l Yp
: . “q
y-intercept of line: y = 2
ge? @)
5.2 - v i
g(x) = +2 ‘Substitute point
x-1 '
a (0:4)
doit 2=4 v
a= -2
(2)

Page 14 of 18

### Page 24

Downloaded from Stanmorephysics.com

5.3
¥Subst y = 0
Sue 2
°A=(;
(2:0) ¥ Answer (2)
5.4 | The graph moves up 1 unit ¥ Answer
(1)
[8]
QUESTION 6
6.1
¥ Interchange
¥ Answer
(2)
6.2 Intercept of p ¥

Intercept of p~*
v

Line: y=x ¥

Shape of p and
pivy

(5)

Page 15 of 18

### Page 25

Downloaded from Stanmorephysics.com

6.3 0 < 1
S*Sq Y critical points
¥ answer
(2)
6.4
x-2 ws —2
h(x) +3 wr aS
(2)
—" [11]
QUESTION 7
7.1
ve) =m Lt W-SLO)
I'(x= a
2 v ituti
5-2(x+h)'-(5-2x°) substitution
= lim ——_—_
a h Y expansion
= lim 2 4A 2h" = 5+ 2x"
= h Y simplification
<lim —4xh-2h°
ho h Y notation and lim (—4x—2h)
._ h(—4x-2h) hoo
= lim ———.
ms h v answer
= lim(-4x-2h)
ho
=-4x
(5)
7.2.1
25"
=7x*+
Ba x aie ;
=7x*+2x? v 2x
i
o. = 28x? +3x? Y 28x° v at (3)

Page 16 of 18

### Page 26

Downloaded from Stanmorephysics.com

722 | *v¥=5
F J
r=
x
» =S5x v y=5x"
v answer
[aTaYey (2)
723 wp [32-18-6 -Tx- -4| V3x-7 ¥ 6x"
_ x
_ 4 ¥ 3and differentiating constant
= D,[3x-7-6x"' | sr
=3+6x° ”)
[14]
QUESTION 8
8.1
x3 —6x?+9x-4=0
(x -1)0?-5x+4) = 0
@-)@-4)A-1) = w= I)
X=1;x=4 (x? -Sx+4)
B(4 ; 0) (x — 4)(x — 1) (4)
.f(@) = -4 ¥ coordinates
D: (3; -4)
8.2 | (x) = x9 -6x?+9x-4
f (Qo) = 3x2 -12x+9=0 3x2 —12x +9
x*-4x+3=0 ¥=0
-(-4) ¥ {C42 - 4D) ¥ subs. into
= 2d) formula
x=3;x=1 = 4
*fG)= -4 « D(3;-4) = (5)
8.3 | f(x) = 3x?-12x+9 % 12 =0
f"(x) = 6x-12=0
wee )
ND
8.4 | k <-4 orlof k>0
wk <-4
(2)
“k>0

Page 17 of 18

### Page 27

Downloaded from Stanmorephysics.com

85 | x <1 or/of x>3 v¥x<1
Yor ¥%x >3
(3)
8.6 |0 v (1)
tra
Jana? [17]
QUESTIO!
9.1
Afters hours: BF = 301 km and CD= 401km -Y BF =30t
-.BC = 100-401 ¥ BC=100—40r
Fc = (Gory + (100 — 407)" ¥ Pythagoras
= V900r? + 10000 — 80007 + 160077
= 425007 — 80007 + 10000 “answer (4)
9.2
FC is a minimum when FC is a minimum. Y BF =30t
¥ BC =100—40r
2 2_
FC* = 2500r° — 80007 + 10000 Py eenisseen
2
are = 50007 — 8000 = 0
dt “answer
t =—— =1,6hrs (96 minutes) (4)
5000
9.3
FC = 4250077 — 80007 + 10000 ¥ subs into equation
= 42500(1.6)° — 1.6)+ 10000
- ¥ 00(1.6)° — 8000(1.6) + Vanswer
They will be 60km apart.
(2)
[10]

Page 18 of 18

## Related documents

- [MATHS P1 GR12 QP JUNE 2023 Afrikaans](https://www.hlayiso.com/library/maths-p1-gr12-qp-june-2023-afrikaans/9cfe815f-db59-438e-bbb0-ad84bef2a887) — Mathematics · QP · 2023 · Afr
- [MATHS P1 GR12 QP JUNE 2023 Afrikaans](https://www.hlayiso.com/library/maths-p1-gr12-qp-june-2023-afrikaans/3f49dc17-3d6a-4adb-aab5-4d32cf7d95c0) — Mathematics · QP · 2023 · Afr
- [MATHS P1 GR12 QP JUNE 2023 English](https://www.hlayiso.com/library/maths-p1-gr12-qp-june-2023-english/245dbccd-9e3b-4add-924a-72f6c39ee078) — Mathematics · QP · 2023 · Eng
- [Mathematics P1 May-June 2023 MG Afr & Eng](https://www.hlayiso.com/library/mathematics-p1-may-june-2023-mg-afr-eng/0ceb885c-ecf6-4f24-ac7b-88b4a828bb7f) — Mathematics · QP · 2023 · Eng
- [Mathematics P1 Nov 2023 MG Afr & Eng](https://www.hlayiso.com/library/mathematics-p1-nov-2023-mg-afr-eng/edb609ea-c44f-4376-983e-124b65725f06) — Mathematics · QP · 2023 · Eng
- [Mathematics P1 Nov 2023 Eng](https://www.hlayiso.com/library/mathematics-p1-nov-2023-eng/9eb0d4a5-b676-4a71-815d-7dc6a6e7c16e) — Mathematics · QP · 2023 · Eng

## More from Grade 12 Mathematics

- [Maths Grade 12 NSC P2 QP September 2025 Eng Northern Cape](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-eng-northern-cape/4ede8207-ca82-4419-bae1-fae734498cc7) — QP · 2025 · Eng
- [MATHS P2 ANSWER BOOK GR12 JUNE 2024 Afr+English](https://www.hlayiso.com/library/maths-p2-answer-book-gr12-june-2024-afr-english/4a71e975-4ac4-4f55-8e23-0b8155a5b0c6) — Addendum · 2024 · Eng
- [MATHS P2 MEMO GR12 JUNE 2024 Afr+English](https://www.hlayiso.com/library/maths-p2-memo-gr12-june-2024-afr-english/4a5f997e-e38d-4abb-b726-8b72e6672832) — Memo · 2024 · Eng
- [MATHS P2 GR12 QP JUNE 2024 English](https://www.hlayiso.com/library/maths-p2-gr12-qp-june-2024-english/f19338aa-ae27-4fb7-9f1b-e60b2b91eb8c) — QP · 2024 · Eng
- [MATHS P1 GR 12 MEMO SEPT 2024 English+Afrikaans](https://www.hlayiso.com/library/maths-p1-gr-12-memo-sept-2024-english-afrikaans/f0eedb8f-ec6f-4f46-9885-3933cd274781) — Memo · 2024 · Afr
- [MATHS P2 QP GR12 SEPT 2024 Afrikaans](https://www.hlayiso.com/library/maths-p2-qp-gr12-sept-2024-afrikaans/8572c34f-22e2-48bc-9d8a-f78c7e29781d) — QP · 2024 · Afr

## Collection

- [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers)

---

Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. When citing, link the page URL above, not a PDF or /api/ URL.