# Mathematics Grade 10 Paper 1 KZN November 2025 Paper & Memo

> Grade 10 Mathematics Paper 1 question paper and memo from the KZN November Exam 2025, 18 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/kzn-maths-grade-10-november-2025-p1-and-memo/ece9abee-2009-4e22-8de3-34ccb229e5cb
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 10
- Document type: Question paper and memo
- Year: 2025
- Exam period: KZN November Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 18

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## About this document

Mathematics · Grade 10 · KZN November Exam · 2025. Question paper and memorandum, 18 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

tanmorephysics. com

MARKS: 100

TIME: 2 hours

This question paper consists of 8 pages, including an information sheet.

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CAPS — Grade 10

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

Ll. This question paper consists of 7 questions.

2: Answer ALL the questions.

3s 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.

—

$; Answers only Wi be awarded full marks.

6. You may use an apj
unless stated othe

calculator (non-programmable and non-graphical),

Ti If necessary, round off answers correct to TWO decimal places, unless stated otherwise.

8. Write neatly and legibly.

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Cc

S — Grade 10

12.3 593-4 0

1.3 Calculate the value of the following, WITHOUT the use of a calculator:

2010 (2013? —9)x 2012
2016x (2011? —1)

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(2)
(3)
(3)

(2)
(3)
(3)

(4)
[20]

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Cc

'S — Grade 10
QUESTION 2

2.1 Solve for x without using a calculator:

2.2 Ataschool event, tickets are sold at two different prices: R20 for adults and R12 for
learners. A total of 20 tickets were sold, collecting R320 in total. How many adult
tickets and learner tickets were sold?

Let the number of adult tickets be x and the number of learner tickets be y.

QUESTION 3

3.1 The n" term of an
3.1.1 Determin ri of the sequence.
3.1.2 Which term is equal to 683?

3.2 A grade 10 learner wrote the word WHEAT over and over again on a piece of paper as

follows:

d position, the second H is in the seventh position, the
osition, and so forth. Determine in which position the

1;-7x4;-11x7;-15x10.

Determine the 62"! term of the above pattern.

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(3)

(4)

(6)
[13]

(1)
(3)

(1)
(3)

(3)

(4)
[15]

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CAPS — Grade 10

QUESTION 4

Amanda is planning a trip to England in 2 years. She considers two different ways to save her

money.
meena Option 2
She invests RS 000 at 6% She invest R8 000 at 5,5%
per annum simple interest per annum compound
—— interest for 2 years.
4.1 Calculate the total man ill have after 2 years if she chooses the 2" option. (2)

a

4.2 Which option gives aa com return after 2 years? Show calculations to support

your answer. (3)
4.3 If the exchange rate is £1 = R22,50. How many British pounds (£) will she have after 2
years if she chooses the better option? (2)
4.4 Suppose instead she wants to use option 1, but save monthly. If the bank agrees to let her
save a fixed amount every month for 2 years, and she wants a total of R9 000, calculate
the fixed monthly payment.
(Assume the interest is only calculated on the total of all deposits at the end of 2 years) (3)
[10]

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CAPS — Grade 10

QUESTION 5
5.1 Sketched below are the graphs of f(x) = 2s and g(x)=—2x—-1. Points A and B are
x

x-intercepts of f and g respectively. E is the y-intercept of g. C and D are the points of

intersection of f and g.

5
g&
x
any
S74
| ee
f
5.1.1 Write down:
a) the y-coordinate of E.
b) the equation of the vertical asymptote of

5.1.2 Determine the coordinates of B.
5.1.3. Calculate the length of AB.
5.1.4 Determine the coordinates of C and D.

5.1.5 Determine the value(s) of x for which:

a) f is increasing.
g(x)
b 2—<0
£0)

5.1.6 If h(x) = f(—x)+3. Describe in words the transformation from f to h.

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(1)
Q)
(2)
(4)
(5)

(1)

(3)

(2)

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'S — Grade 10

5.2 Use the following properties to sketch the graph of p if p(x)=axb'+q:
ea>0
°0<b<l
+q<0 (3)
[22]
QUESTION 6
Given: m(x) =—2x’ +8
6.1 Determine the x — intercepts of m. (2)
6.2 Sketch the graph of m. (3)
6.3 Determine the value(s) of k for which m(x)=k will have no solution. (1)
6.4 Determine the maximum value of sn (2)
[8]
QUESTION 7
7.1 If P(D)=0,25 ; P( P(D or K) = 0,625
7.1.1 Calculate | (2)
7.1.2 Are eve nts D entary? Give a reason. (2)
7.2 A survey was conducted among 100 learners to determine their participation in two
subjects: Maths (M) and Life Orientation (LO). The following results were obtained:
e 42 learners participate in Maths.
e 35 learners participate in Life Orientation.
x learners participate in both subjects.
re
nt the above information. (3)
s who participated in both subjects. (2)
learner selected at random participates in:
(1)
(1)
()
[12]

TOTAL MARKS: 100

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'S — Grade 10

INFORMATION SHEET: MATHEMATICS

~b+tAb° —4ac
xa
2a
A=P(1+ni) A= P(1-ni) A=PQ-i" A=P(1+i)”
n
T, =a+(n-Nd 8, = 5 a+ (-Dd]
I, =r" Ss _ alr -1) 3;r#l S,=—";-l<r<l
. r-1 l-r
a (+1) 1 p= sls)"
t I
xth x
f= im a F(x)
m=tand
In AABC: gg On
snA sinB sinC
a =b? +c? —2be.cosA
area AABC = 5 absin c
sin(a + B)=sin acos 8 +cosa@sin B sin(a@ — 8) =sin acos B-cosasin B
cos(@ + 8) = cosacos f-sinasin B cos(a — 8) = cosa cos #+sinasin B
cos? a—sin? @
cos2@ =41-2sin’?a sin2@ = 2sina.cos a
2cos*a-1
n 2
2s >, -¥)
x= oe o =a
n
nA) _
P(A)= “AS) P(A orB)=P(A) + P(B) — P(A and B)
. p_ 80-7)
p=atbx

 LG-3"

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INAL

KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

MARKS: 100

This marking guideline consists of 9 pages.

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CAPS — Grade 10
Marking Guideline
QUESTION 1
1.1.1 2 -6x— v
SEI Ps - 62-9 Answer only; Award Full Marks 1 products
* A answer
4x°-9
2)
1.1.20) §5(4x? 4x41) +43 YA 4x°-4x41
—20x° +20x° —5x+3 YCA ~20x° + 20x?
¥CA 5x43
@)
11.3 _ 23% ¥ A prime bases
pgs wa ad
=29x3° ¥CA simplification
=] ¥ CA answer
GB)
1.2.1 | 2p+5)\(p-3) VA(2p+5)¥ A(p-3)
(2)
1.2.2 | =y—-1+by by-b-1 v A rearranging

=(y-)+b-D +b)—(b+1) V A factors
=(y-l(i+d) b\(y-1) v A answer
: GB)
¥ A common factor
VA (m-2)
VA (m>+2m+4)
G3)

Answer Only; 0/4

OR

v A numerator (factors)
vA denominator

¥ A simplification
v A answer

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'S — Grade 10
Marking Guideline

DBE/November 2025

let 2010 =x
3[(+3)'-9 (+2)
(e+) =a]

v A numerator (factors)

bal oA 6)(x° + 2x) ¥ A denominator
_ xx(x+2)(x+6)
~x(x+6)(x+2) vA simplification
6ie
= 2010 v A answer
(4)
[20]
QUESTION 2
2.1.1 | 234-99 YA2°
3x-4=-9 ¥CA equating exp.
5
x= “3 ¥CA answer
GB)
212; 4 10 2 VA x(x-5)
x-5 x x(x-5) ¥ A restrictions
LCD = x(x-5) x#00rx#5
restrictions : x #0 or x #5
4x-10(x-5) =2 ¥ A simplification
6x =-48
x=8 ¥CA answer
(4)
2.2 X+Y=20 ore (1) vA Equation 1
20x +12y =320......(2) YA Equation 2
y=20-x aes (3) ¥CA_ Equation 3
20x +12(20-x) =320 YCA substitution
20x + 240 -12x =320
8x =80
z= 10 ¥CA_ x-value
pail ¥CA_ y-value

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CAPS — Grade 10
Marking Guideline
OR vA Equation |
a w th, Wapato
20x+12y =320 oe) YCA_ Equation 3
x=20-y sseene(3) ¥CA substitution
20(20- y)+12y =320
400-20) +12y =320
8y=80 YCA_ x-value
y=10 ¥CA_ y-value
on=i0. (6)
-. 10 adult tickets and 10 learner tickets were sold.
[13]
QUESTION 3
SiLel T, =4 VA 4
dQ)
3.1.2 | 683=7n— VA equating
686=7n YCA simplification
n=98 YCA answer (néN)
GB)
S21, |'T vA answer qd)
3.2.2 | 508 VA dividing by 5
5
=101,6
0,6x5=3
93 ibrar v A remainder times 5
v A answer
OR OR
510 —e
7 — (510 is divible by 5) VA any number nearest
=102 to 508 & divisible by 5
510 or 505
510 to 508 =3 steps backwards 7x method
=> 3" letter (backwards or forward)
iE v A answer
GB)

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Grade 10
Marking Guideline
ar n-3 YA 5Sn-3
hia 5(103)—3 YCA subst. n = 103
Ls 12 ¥CA answer
Ir md (3)
3.3 lah +1)3d —2) Vv AV A(-4k +1)(3d -2)
iz ; (62) + 1][3(62) —2] ¥CA substitution
|= +247x184 ¥CA answer
Accept: —45448 as the final answer (4)
[15]
QUESTION 4
4.1 A=P(1+i)"
= 8000(1 40,0 vA subs. into correct
formula
=R8904
¥CA answer
OR
A= eft + 1 mai imorephysies.com
r vA subs. into correct
= s000( 33 } formula
00 ¥ CA answer
=R8904,20 (2)
4.2 A=P(1+ni)
7 ae 8000(1 +2 0,06) v A subs. into correct
| teal ' formula
¥CA future value
¥CA answer
OR

vA subs. into correct
formula

¥CA future value
| ¥ CA answer
LL @)
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CAPS — Grade 10
Marking Guideline
43 R8960x £1 ¥CA dividing by 22,5
R22,50
=£398,22 ¥ CA answer
(2)
44 A=P(i+niy
9000 = P(1+ 20,06) VA if wrong formula used
P=R8035,71 Y A value of P
8035,71 Y A dividing P by 24
Fixed monthly payment= 24 Y CA answer
=R334,82 (3)
[10]
QUESTION 5
5.1.1a)] y=-l * VA y=-l (1)
5.1.1b) | x=0 VA x=0 (1)
5.1.2 0=-2x- v A equating to 0
x=-t
7 wer only; Award Full marks Y CA answer (in
a(-5:0] coordinate form)
(2)
513 [5 2, VA f(x)=0
x
BeAr
x Y CA x,
x=-2 A(-2;0)
1 Y CA difference
ee 1 Sunt
AB 53 +2=1,5units Y CA answer
OR OR
rynor‘en YA f()=0
ee
=
x Y CA x,
x=-2 A(-2;0)
1 2 5 Y CA using the
AB= [-3+2] +(0-0) distance formula
v CA answer (4)
=1,5units

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CAPS — Grade 10
Marking Guideline
Slt 2 = 2x-] VA equating
x
ud =-2x
x
x -1=0 V A std. form
Y CA factors
+1)(x-1)=0
rea) Y CA both values of
x=-l or x=1 %
y=l or y=-3
C11) Y CA both values of
D(1;-3) y
(5)
5.1.5a)| xeR;x¥0 VA
OR
x>0 or x<0 VA
OR 4
xe(-20 ; 0) or VA
Q)
5150) | oan VCAYCAVA
GB)
OR
x€(-0;-2) or xe|-3:0) ¥CAYCAVYA
2
G3)
5.1.6 | - Reflected about the y — axis or line x = 0. vA
- Shifted or translated 3 units up. YA
Q)
5:2
¥ vA shape (decreasing
function)
¥ A both intercepts
ax x with the axes < 0
|
Rec, VA horizontal
al asymptote < 0
GB)

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CAPS

— Grade 10
Marking Guideline

hysics.com

DBE/November 2025

OR

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¥

vA shape
(decreasing
function)

YA point of
origin

¥ A horizontal
asymptote < 0

A shape
(decreasing
function)

YA both
intercepts with
the axes

¥ A horizontal
asymptote < 0

[22]

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CAPS — Grade 10
Marking Guideline

QUESTION 6
6.1 0==2x7 +8
0=x’-4 vA factors or
square root on
(&+2)(x—2)=0 both sides
x= -2 or x=2
v A (both
values of x)
Q)
6.2
Y A turning
point
5
" | ¥ CA both x -
intercepts
v A shape
(concave down)
GB)
6.3 k>8& or keE(8;0) ¥CA (from the
graph in 6.2)
Q)
6.4 Max value of m=8
8-5=3 YA3
3
«. The maximum value is 27. v A answer
Q)
[8]

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APS — Grade 10
Marking Guideline

QUESTION 7

DBE/November 2025

Tedel

P(Dand K) = P(D) + P(K)—P(D or K)

=0,25+0,5-0,625

vA substitution

= 0,125 v CA answer (2)
7.1.2-~1.No, vA
- Events are not mutually exclusive / P(Aand B) #0
_OR vA reason
- events are not exhaustive / P(A)+P(B) =1 (2)
Toul

¥ A x on the intersection

vA both 42 —x& 35 —x

YA 38 (3)
7.2.2 vA equating
¥ CA answer
2)
7.2.3 VA
dd)
vA
dd)
vA
Q)
[12]

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TOTAL: 100 MARKS

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