# Mathematics Grade 10 KZN March Test 2025 Paper & Memo

> Grade 10 Mathematics question paper and memo from the KZN March Test 2025, 14 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/kzn-maths-grade-10-march-2025-qp-and-memo/bc27cbbf-ec6d-4c60-bf7e-25a154d363e4
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 10
- Document type: Question paper and memo
- Year: 2025
- Exam period: KZN March Test
- Curriculum: CAPS
- Pages: 14

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## About this document

Mathematics · Grade 10 · KZN March Test · 2025. Question paper and memorandum, 14 pages. Read online or download the PDF.

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

## Document text

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### Page 1

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KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

a
a
“
"
uo
"
“u
u
“
a
uo

«
| 5

MARKS: 75

TIME: 12 hours

This question paper consists of 5 pages.

Copyright Reserved Please Turn Over
| pyng
L

### Page 2

Math =~ - March 2025 Common Test
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INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.
1. This question paper consists of 3 questions.

yl Answer ALL the questions.

3. Number the answers numbering system used in this question paper.

4. Clearly show ALL cale
answers.

raphs, etc. which you have used in determining your

Ss Answers only will NOT} ded full marks.

6. You may use an approved scientific calculator (non-programmable and non-graphical), unless
stated otherwise.

7. if necessary, round oif answers correci io TWO decimai piaces, uniess siated oitierwise.
& Wiiie nicaiiy aiid iegidiy,
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### Page 3

a , March 2025 Common Test
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QUESTION 1

fala)
Ll arzefo 7 59} , choose the value of x from the given set that will make the

()
qd)
(a)
1.2 Between which twd eonsed joes V170 lie? Show all working details
without the use of a 1 ; (2)
1.3 Simplify the followi
ies (2)
13.2 (5-4x)’ (2)
3 2 1
a ar ape ar &
10773 - 4
Se a «
1.4 Factorise the following expressions fully:
1.4.1 2x7-18 (2)
1.4.2 §+125a° @)
GB)
Q)
(2)
G)
[31]

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### Page 4

Math«
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2.2 the value of b.

23 Given: -9<-2x+3<5
2.3.1 Solve for x .

232. Represent the solution of 2.3.1 on a number line.
2.4 Solve for x and y simultaneously:

2x-y=17
4y+3x=9

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March 2025 Common Test

(2)
(3)
(3)

(4)
(5)

(4)

(2)
(2)

(4)

229]

Please Turn Ove.

### Page 5

Mathematics rf
Teilpapers.caze March 2025 Common Test

QUESTION 3

Su - Ren, “ ri
In the diagram below, APQR is a right-angled triangle, Q=90°, P=0, PQ=5 units

and-PR=13 units.

ut
3.1.1 Determine the numerical value of:

(a) QR (2)
(b) cosé ()
(c) cotd (1)
(d) secQRP (1)
3.1.2 Write down the trigonometric ratio that is equal to ee ¢ (1)
3.2 If A=34° and B=53°, use your calculator to evaluate the following (correct to
TWO decimal places).
3.2.1 cos(B-A) (2)
3.2.2. sin?B Q)
3.3 Without using a calculator, determine the value of:
cosec45° . sin60°
sin 45° . cos30° (5)
TOTAL: ey

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### Page 6

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KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

75

This marking guideline consists of 9 pages.

Copyright Reserved Please turn over

### Page 7

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Mathematics ij March 2025 Common Test
NSC
Marking Guideline
QUESTION 1
1.1.1) 9 vA
5 Q)
1.1.2. | ¢ A
f a)
1.1.3 [0 vA (a)
12 | S769 < V170 < Vi96 YA_ square root of perfect sq.
vA answers
13.1 | (x-2y)(x? +244
Hx +2? y+4ay? —2x° ys VA removing brackets
=x-8 ye
ULL MARKS | Y CA answer
(2)
1.3.2 (5—4x)"
=(5-4x)(5-4x)
=25~20x—20x+16x? vn —
ts
=25-40x+16x° ma
ANSWER ONLY: FULL MARKS
(2)
13.8 3 2 1
YA (x+3)(x-3)
YA numerator
¥YCA_ simplification
YCA answer
(4)

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### Page 8

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Mathematics

NSC
Marking Guideline

March 2025 Common Test

Same genta ge

gate

_ Seats geen at

vA __ prime bases

¥YCA_ removing brackets

YCA simplification

=57'25
i = ¥CA answer
(4)
1.4.1 | 2.7 -18
= 2(x? -9) YA common factor
= 2(x-3)(x+3) ¥CA answer
(2)
VA (2+5a)
YA -10a
VA (Avon 25a")
(3)
YA grouping
vA (x? +5)
YCA (x2)
G3)
TS.1 | 9~
1
>
al
3 vA answer
Q)
135.2. | g*
=9°*
=(2') vA (2')
=(3)'
=27 VA answer
(2)

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### Page 9

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Mathematics
NSC
Marking Guideline

March 2025 Common Test

15.3] (4y
(is)
rest
=i i YA 24
=(2")"
=(3)"
=3" ¥CA _ simplification
1
= 31 YCA answer
(3)
Amorephysies.com [31]
QUESTION 2
2.1.1 | 3x(x-8)=0
vA
x=0 or x=8 VA
2)
2.1.2 | 7x? -5x-2=0
(7x+2)(x-1)=0 YA factors
2
xe or x=1 YCAVCA
G3)
2.1.3. | 3*-729=0
34 =729
35-4 _ 36 vA __ prime base
5x-4=6
5x=10 YCA simplification
y=2 YCA answer
GB)

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### Page 10

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Mathematics March 2025 Common Test
NSC
Marking Guideline

2.1.4 BE |

2x?-=128
a
x7 =64 YA __ division by 2
ieee} 6x (2
xs GI_ 6) YA x reciprocal of exp.
x=Q"
4
L YCA 2
16 ¥CA answer
OR
3
2x? =128
a am YA division by 2
28) gl)
i YA x reciprocal of exp.

“6 ¥CAYCA answer

(4)
YA common factor
¥CA simplification
YCA division by Fy
YCA prime bases
¥CA answer
(5)

i Na
fesart (¢ —2a -a°V fa +2a*)—(a’ -2a% )] =b

el +a ol el -a°+2a™ ‘)= b

vA factorising
YCA _ simplification

YCA _ simplification
YCA answer

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### Page 11

Mathematics

Testpapers.co.za

NSC
Marking Guideline

March 2025 Common Test

OR

(a'+2a*) -(a'-2a°) =6

a°+4+4a° -(a°-4+4a° =6 vA a°+4+4a°
$ 6.6 <6 YA a°-4+44a°
BNE See a al ¥CA _ simplification
8=b YCA answer
(4)
2.3.1 | -9s5=2x+3<5
-9-35-2x<5-3
SIRS eae YA simplification
wol<x $6 ¥CA answer
(2)
2.3.2,
YCA critical values
i— at =e vCA O@
(2)
2.4 2X YAH DTarcrecsereeed equation |
4y+3x=9... equation 2
vA ti
y=2x-17.. i a 6
substitute equation 3 into equation 2
4(2x-17)+3x=9 ¥CA substitution
8x-68+3x=9
11x=77
sgpieel YCA value of x
substitute x — value into equation 3
y=2(7)-17
y=3 YCA value of y
OR

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### Page 12

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Mathematics March 2025 Common Test
NSC
Marking Guideline
DRE YSN oecreecsseeed equation 1
AY PBK EO... ccsssee equation 2
beg 2) equation 3

2
substitute equation 3 into equation 2

4y+3(2417)-9

8y+3y451=18
lly =-33

y=
substitute y—value into equation

4y+3x=9..

substitute x— value into equation 1
2(7)-y=17
y=3

YA equation 3

¥CA_ substitution

YCA value of

YCA value of x

YA 4y=8x-68

YCA equating

YCA value of x

YCA value of y

(4)

[29]

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### Page 13

Mathematics

Testpapers.co.za

NSC
Marking Guideline

March 2025 Common Test

QUESTION 3
3.1.1a | PR?=QR?+QP’ — Pythag
2 if 2 ied
13°= QR’ +5 vA correct substitution
(QR? = Vi3? 3? y
QR=12 CA answer
ANSWER ONLY:FULL MARKS
ude : (2)
3.1.1b 5 vA
cos 0 = —
13
- dd)
ade enlllne St es
12
> da)
311d secre = 13 CA
12
z ad)
212 sing = QR OR cosQrp = 2k #
PR PR
da)
al cos(53°-34°) YA _ correct substitution
YCA
= 0,95
PENALISE HERE FOR ROUNDING
(2)
3.2.2 | sin? B
=(sin 53°) YA correct substitution
=0,64 YCA
(2)
3.3 2 3 v Z
aN A cosec45° =——
2B vA sin ase
wz 2 2
YA cos30°= RE
2
2 2 3 i
ing per YCA . simplification
nD °
22 ¥CA answer

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### Page 14

Mathematics

Testpapers.co.za
NSC
Marking Guideline

March 2025 Common Test

elo v2 YCA simplification
Pp.
1
=2 ¥CA answer

- - (5)
Biafmorephysies.com [15]
TOTAL: | 75

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