# Mathematics Grade 10 Paper 2 KZN November 2025 Paper & Memo

> Grade 10 Mathematics Paper 2 question paper and memo from the KZN November Exam 2025, 22 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/kzn-maths-grade-10-november-2025-p2-and-memo/417e9b18-c39d-4fb2-9e32-c089e8fbfd22
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 10
- Document type: Question paper and memo
- Year: 2025
- Exam period: KZN November Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 22

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## About this document

Mathematics · Grade 10 · KZN November Exam · 2025. Question paper and memorandum, 22 pages. Read online or download the PDF.

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## Document text

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### Page 1

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KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

MARKS: 100

TIME: 2 hours

This question paper consists of 11 pages, including an information sheet.

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INSTRUCTIONS AND INFORMATION
Read the following instructions carefully before answering the questions.
ls This question paper consists of 9 questions.

2 Answer ALL the questions.

3) Number the answers correctly according to the numbering system used in this question paper.
4. Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining your
answers.
3. Answers only will NOT necessarily be awarded full marks.
—

6. You may use an approve
stated otherwise.

or (non-programmable and non-graphical), unless

a If necessary, round off answ O decimal places, unless stated otherwise.

8. Write neatly and legibly.

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QUESTION 1

DBE/November 2025

The table below shows the number of public schools from the NINE provinces in South Africa.

Provinces

Number of public schools

Eastern Cape

5 468

Free State

1214

Ll ()
1.2 (2)
1.3 Write down the media qd)
1.4 Determine the interqua (3)
1.5 Due to the population: as uth Africa, 0,68% more schools will be required in
the next 15 years. Determine the number of schools needed in the next 15 years. (2)
19]
QUESTION 2
The following table represents the grade 10 mathematics test results at Excel Secondary School.
MARKS (in %) FREQUENCY
20<x<30 6
30<x<40 ll
40<x<50 9
50<x<60 7
60<x<70 8
70<x<80 13
80<x<90 6
2.1 How many learners wrote the test? (1)
22) Calculate the estimated mean. (4)
2.3 Calculate the percentage of learners who obtained less than SIXTY PERCENT. (2)

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(7]

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QUESTION 3

3.1 In the diagram below, trapezium QHFS is drawn, such that QH//SF.
ry

v

x
3.1.1 Determine the gradient of SF. (2)
3.1.2 Determine the equation of QH. (3)
3.1.3 Calculate the value of k. (2)
3.1.4 Determine the length of HF. (2)
3.1.5 Calculate the coordinates of M, the midpoint of HS. (3)

3.2 Line segment AB is perpendicular to the segment CD. A(-5 ; 2) and B(p ;-1) are
points on AB. If the gradient of segment CD is 2, calculate the value of p. (3)

[15]

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QUESTION 4

a
4.1 if Yay: 7,7° and y=52,3°. Calculate the following correct to TWO decimal places.
(2)
(2)
(2)
42 ( ms (correct to ONE decimal place).
(3)
4.2.2 4tan0= (3)
4.3 In which quadrant d¢ e | n of the angle @ lie if:
43.1 sin@<0 and (1)
43.2 tand<0a (1)
[14]

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QUESTION 5

artesian plane such that XOP = 0.

Siahimorephysics.com

S-Lel Calculate the length of OP. (2)

$112. Hence, WITHOUT the use of a calculator, determine the value of:

a) tanO ()

b)  (sind+cos0)" (3)

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5.2 In the diagram below WMT = 90°, MS_L WT, W =70° and MW =40 units.

T
5.2.1 (3)
52:2 (5)
[14]
QUESTION 6
Given: g(x)=cosx+1 and f(x)=2sinx for x <[0° ; 360°].
6.1 On the grid provided in your SPECIAL ANSWER BOOK, sketch the graphs of f and
gon the same set of axes for x e[0° ; 360°]. (4)
6.2 Write down the period of /. qd)
6.3 Determine the range of g. (2)
6.4 For which value of x will al itis =1? (2)
19]

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In the next TWO questions, ensure you give reasons for each statement you make.
QUESTION 7

7.1 In the diagram below, ABCD is a parallelogram with AB//EG//DC.

B
v
G
Zz
Cc
Talal Determine the es of and w and x. (4)
71.2 Calculate the values of y and z. (4)
7.2 Given AABC, where AM = MB and AN=NC.
A
N
B. C

5
Prove the theorem which states that MN =5BC 6)

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73 In the diagram below KR//CP, PX//RC , MP =PR =9cm and CK =15cm.

K
yr
=
x
M 99cm jf
U s -_ U

73.1 Prove that M (3)
73.2 Calculate tahmore chess Co: (2)
7.3.3. Hence, prove that MRK = 90° (2)

[20]

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QUESTION 8

In the diagram below, OB bisects ABC. O is the centre of the circle and ABC = 60°.

8.1 Determine the size of (1)
8.2 Calculate the size of 0, ue (3)
8.3 Hence, prove that AOAB= AOCB. (3)

[7]

QUESTION 9

A globe of the world, spherical in shape, is placed in a geography classroom.
The globe has a radius of 21 cm.

91 Calculate the volume of the globe, correct to ONE decimal place. (2)
9.2 Determine the area of the globe, correct to ONE decimal place. (2)
9.3 If the radius is doubled, what would be the ratio of the volume of
the original sphere to the volume of the new sphere? (1)
[5]
TOTAL MARKS: 100
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DBE/November 2025

INFORMATION SHEET: MATHEMATICS

D=bEND’ —4ac

2a
A= P(1+ni) A=P(1-ni) A=P(1-i)" A=P(1+i)"
T, =a+(n-l)d S, =Sea+(n—nd]
T,=ar™" Ss _ alr" -1) ;r#l S,=—¢; -ler<l
. r-1 1-r
pa+i' = past-0+9"]
j ;
7 xt+h)—f(x
P(x) = im LEADS)
h>0 h
y=mxt+e 2a Ns m=tan@

In AABC:

snA sinB sinC

a =b° +c? —2be.cosA

area AABC = 5absinC
sin(@ + #) =sina cos B+ cosa@sin #
cos(a + #) =cosacos # —-sinasin B

cos’ a-sin’ @
cos2a@ =41—2sin> @
2cos* a1

Copyright Reserved

sin(a@ — #) =sin acos B —cosasin B

cos(a — #) =cosa cos B+ sina sin B

sin2@ =2sina.cosa

9 2
De 3)
of =
n

P(A ot B) = P(A) + P(B) — P(A and B)

De-*O-H)

= *"Se-a

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KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

TO: DIRECTOR: CURRICULUM GRADES 10 - 12
DISTRICT DIRECTORS
DISTRICT HEADS OF CURRICULUM GRADES: 10 - 12
DISTRICT HEADS OF EXAMINATIONS AND ASSESSMENT
DISTRICT HEADS OF CIRCUIT MANAGEMENT
CIRCUIT MANAGEMENT
MATHEMATICS SUBJECT ADVISORS
PRINCIPALS OF SECONDARY SCHOOLS

ADDENDUM TO THE PROVINCIAL COMMON EXAMINATION FOR GRADE 10
MATHEMATICS PAPER 2 NOVEMBER 2025

Please note that question 7.2 from this question paper has to be removed and replaced as shown below:
Remove question 7.2:
Given AABC, where

M, y (5)

Prove the theorem which states that MN=>BC B c

Replace question 7.2 with:

Given quadrilateral ABCD as shown below, with AD=BC and AD || BC. Prove that ABCD is a
parallelogram.

PLEASE NOTE: THIS QUESTION MUST BE ANSWERED ON PAGE 16 UNDER ADDITIONAL
SPACE

Your, eration in this regard is highly appreciated.
(oe 04% 41.2025
DRC PENNISTON DATE
CHIEF DIRECTOR EXAMINATIONS, ASSESSMENT
AND QUALITY ASSURANCE

Scanned with
SJ CamScanner’

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KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

MARKS: 100

This marking guideline consists of 10 pages.

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GRADE10
Marking Guideline
QUESTION 1
1. x =2083 AY answer ()
1.2 _ 23718
9
mean =2635,33 | ANSWER ONLY: FULL MARKS AV’ answer
PENALISE HERE FOR ROUNDING (2)
1.3 Mpumalanga ALSO ACCEPT 1725 CAY answer (1)
14 IQR = Q,-Q, CAY Q, and Q,
IQR = 4667,5—1332 CAY correct subst
= 3335,50 CAY answer (B)
1.5 Number of schools = 23718(0,68%)e= AY
= 161,28 ‘
= 162 schools AY
ALSO ACCEPT
Number of schools = 2 5 AY
= 23879,28
= 23880 schools Av Q)
19]
QUESTION 2
21 A ¥ answer (1)
.7) + (65.8) +(75.13)+(85.6) | A ¥ correct midpts
CAY 60
CAY 3330
CAY
R ONLY: FULL MARKS answer
—_ 3330 BS" B80
60 CAY 60
¥=55,50 CAY answer
(4)

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GRADE10
Marking Guideline
v
2.3 naa. «100 AVY 33
| 605) CAY answer
[55% (2)
uif 17]
QUESTION 3 |
3.1.1 = 5 =
fig y27)1
x, —%
O+1 AY - subst into
Msp = 3-0 correct formula
Mp =; ANSWER ONLY: FULL MARKS __ | CAY answer
(2)
CAY may = Me
CAY correct subst
fafimorephysies.com
CAV answer
CAY Moy = Msp
CAV correct subst
CAV answer
(3)

Piet

3
k=-2
OR

CAY correct subst

CAV answer
k<0

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GRADE10
Marking Guideline
AY correct subst
CAY answer
k<0 (2)
3.1.4
AY — correct subst
CAY answer
(2)
3:15
AY correct subst
CAV 2
cay 3
2:
(3)
3.2 1
AY my = “3
CAY equate toM,,
CAY answer
(3)
[15]
QUESTION 4
4.1.1 sin x
2
_ sin 67,7° AY correct subst
2 CAY answer
=0,46 (2)
4.1.2 cos(3y)
_ cos(3(52,3°)) AY correct subst
v
=-0,92 CAY answer
(2)

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GRADE10

Marking Guideline

November 2025

AY correct subst

ad CAY answer (2)
3+cos40 = 3,706 AV
cos 40 = 0,706 cos 40 = 0,706
40 = 45,08961075° CAY
@=T11,3° 40 = 45,08961075°
CAY answer (3)
4.2.2 4tan0=7-—4tan
8tand=7 AY 8tan0=7
tand=7 CAY tano=
6 8
O=41,2° CAY answer (3)
4.3.1 Third quadrant AY (l)
4.3.2 Fourth quadrant AY (l)
[14]
QUESTION 5
5.1.1 a? +y =r Pythagoras
v
(-12)’ +(5)° = OP? AY — correct subst
OP = 13units ANSWER ONLY: FULL MARKS
CAY answer
(2)
vo
512 2) wd=— is A
12
a)

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GRADEI

0
Marking Guideline

November 2025

b)-(sind +cos0)°
isla)
=| —+| —
13 13
(UB

=o
169

OR

=sin’ 0+2sin@cos@+ cos’ @

CAY Y correct subst

CAY answer

AY simplification
CAY correct subst

_ 25, =120 144
169 169 169 a
_ 49 CAY answer
169 (3)
5.2.1 §, = 90° angles on str line AY S,=90°
70° = WSs AY correct subst
costae = “40. into trig ratio
WS = 13,68 units CAY answer (3)
P Y tri i
Ded cos W = WM AY trig ratio
WT
cos.70° = 40° AY correct subst
WT CAY
BNE opeekie WT=116,952176
ST=116,952176 -13,68 CAV difference
ST =103,27 units CAY answer
OR
T=20° sumofA AY SIR
in20°= 40 CAY correct subst
sa WT into trig ratio
WT =116,952176 cay
ST=116,952176-13,68 WT=116,952176
ST =103,27 units CAY difference
CAY answer
[14]

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GRADE1L

0
Marking Guideline

November 2025

QUESTION 6

6.1

g(x)
Av shape
AV Tpt

wae)

Av shape
AY x int

(4)

6.2

360°

(1)

6.3

O<y<2 or ye[0;2]

(2)

6.4

a(x)— SQ) =3
x=270°

ANSWER ONLY: FULL MARKS

(2)

19]

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GRADE10
Marking Guideline
QUESTION 7
711 | 2y4210=70° corr Zs AB//EG AY S/JR
w=30° AV answer
5x+45°+70°=180° sum Zs in A AVY S/R
x=DBF AY answer (4)
7.1.2 | y+70°+45°=180° sum Zs inA AV SJR
y=65° AY answer
z=70° opp Zs of parm YSYR (4)
72
A i D
B
In AABC and AADC
AC is common vS
Ce Ay alt Zs AD|IBC vS/IR
BC=AD given vs
. AABC=AADC SAS vS/R
A,=C, congruent As vS
-. AB|| DC alt Zs equal
«. ABCD is a parm
2 pairs of opposite sides parallel
ALTERNATIVELY
AD=BC given vS
AD || BC given vs
-. ABCD is a parm vs
One pair of opposite sides equal and parallel vYR
(5)

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GRADE10
Marking Guideline
7.3.1 |MC=KC line through midpt // to 2nd side AY S/R
MG =15 Av 15
MX=CX Line through midpt // 2nd side AY SIR
MC=+=MX+CxX
2MxX=15
MX = os.
2 (3)
7.3.2 KR= 16
5
KR= 16 x 15 AY correct subst
5 2
RB- 24 AY answer (2)
7.3.3 | MK* =30°
MK’ =900 AY
MR? +KR? = 187 +24?
MR? +KR? =900 CAY
MRK=90° _convers (2)
[20]
QUESTION 8
8.1 ABO = 30° OB bisects ABC AVS (1)
8.2 A=ABO Zs opp equal sides AY S/R
A=30° AY A=30°
0, =ABO+A Ext Zs of A
AY answer
O, = 60°
OR
A=ABO Zs opp equal sides AV SIR
AOB = 120° sum of A AY SIR
Bs 0, = 60° Zs onastr line AY SIR
(3)

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GRADE1

0
Marking Guideline
8.3 In AABO and ACBO
1) BO is common AVY S/R
2) ABO =CBO=30° —OB bisects ABC AY SIR
3) A=ABO=30° Zs opp = sides
C=CBO=30° Zs opp = sides
A=C
:. AOB = COB sum of A
AABO =ACBO SAA Av SAA @)
[7]
QUESTION 9
9.1
V= 4 ar
AY correct subst
4 3
V=—2(21
7 (21)
V =38792,4 cm’® CAPanswer
Provided z and r used (2)
in the formula
9.2 A=4ar?
A=42(21) AY correct subst
. CAY answer
A=5541,8 em Provided and r used
in the formula
Q)
9.3 1 CAY answer
1:8 or —
8 qd)
[5]

TOTAL MARKS: 100

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