# Mathematics Grade 10 Mpumalanga March Test 2025 Paper & Memo

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

- Page: https://www.hlayiso.com/library/mp-maths-grade-10-march-2025-qp-and-memo/b060da67-d325-4ce7-9913-508da5624cf7
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 10
- Document type: Question paper and memo
- Year: 2025
- Exam period: Mpumalanga March Test
- Curriculum: CAPS
- Pages: 10

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## About this document

Mathematics · Grade 10 · Mpumalanga March Test · 2025. Question paper and memorandum, 10 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

Testpapers.co.za

ce@ty Starynorephysics.com

education

MPUMALANGA PROVINCE
REPUBLIC OF SOUTH AFRICA

. SENIOR CERTIFICATE

MARKS: 50

TIME: 1 HOUR

This question paper consists of 5 pages including the cover page

Please turn over

### Page 2

Testpapers.co.za

wanownloadedifrsm Stanmorephysics.com 24 March 2025

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions:

L

This question paper consists of 3 questi. Answer all questions

An approved calcula c grammable and non-graphical) may be used, unless stated
otherwise.
Number the answers a mbering system used in this question paper

It is in your own interest to write legibly and to present the work neatly.
If necessary, answers should be rounded off to TWO decimal places, unless stated otherwise

Diagrams are not drawn according to scale.

Please turn over

### Page 3

Testpapers.co.za

matownoadediinsm Stanmorephysies.com

QUESTION 1

1.1 Indicate between which two consecutive integers V47 lie
1.2 Write 0,13 as a fraction, clearly show all your steps.

1.3

1.4 Simplify fully:
14.1 (2x-5)(x? -2x+3)

142 x2-x-2
x+l1

143 3, 7.9%
4a

24 March 2025

(2)
(3)

(2)
(2)
(2)
(3)

(2)
(2)

(2)

[20]

Please turn over

### Page 4

Testpapers.co.za

matownoadediinsm Stanmorephysies.com

QUESTION 2

2.2

2.2

23

2.4

Solve for x in each of the following equation:
211 x(5-x)=0

242 48
x

mvs om

Solve the folloy

23,1 -15<1-4x<5 andxeR

2.3.2 Represent your answer above on a number line.

Solve for x and y simultaneously:

x+3y=12
ax+y=9

24 March 2025

(2)
(3)

(3)

(4)

(3)
(2)
(5)

[22]

Please turn over

### Page 5

Testpapers.co.za

matownoadediinsm Stanmorephysics.com 24 March 2025
QUESTION 3
3.1

In APRQ, OR=4cm, PR=5cm andQP =3cm. PQ is perpendicular to
QRandP=0. PR

=
ysics.com

3.1.1 sind (1)
3.1.2 seca GB)
3.13 tan@ ()
3.2 Calculate the size of @ (3)

[8]

TOTAL MARKS: 50

Please turn over

### Page 6

Testpapers.co.za

Downloaded from Stanmorephysics.com

Education

MPUMALANGA PROVINCE
REPUBLIC OF SOUTH AFRICA

MARKS: 50

TIME: 1 HOUR

This marking guidelines consists of 5 pages.

Please turn over

### Page 7

Testpapers.co.za

Bownloadeatafsom Stanmorephysics.com March 2025
QUESTION 1
Il | 36 < 47 < V49 v 36 < V47 < J49 ”
6< 13 <7 Y answer
Between 6 and 7
1.2 ff Letst= 0131313 Y 100x = 13,1313... (3)
100x = 13,1313 Y 100x-x = 13,1313 -0,1313
100x —x = 13,1317 v answer
99x =13
13
99
1.3.1} x?-~3x-10 ¥ ¥ factors (2)
(x+2)(x—5)
1.3.2 | 25x" —4y" ¥ (58° -Qy*l @)
= (Sx’)? -(2y*)? ¥ factors
= (5x? —2y*)(5x? + 2y*)
1.3.3 | 125x3 +27 vv factors (2)
= (5x) +3°
= (5x + 3)(25x* - 15x +9)
OxY —2y +h ¥ 3x(1-2y) Q)
t Y \(-2y+l)
Y answer
Y simplification (2)
v answer
=2x° -9x? +16x-15
14.2| .?-y-2 _ (x= 2x41) ¥ factors (2)
x41 x41 v answer

=x-2

### Page 8

Testpapers.co.za

Bownloadeatafsom Stanmorephysics.com

March 2025
1.4.3 | 36bmpatd 252 350, 750, 950 Y (2.3.7)™
300 = 300 (4)
42 (3:72) 75003500 7500
Se ae 2502-500
= 3500 7500 5500
sila ¥ answer
[20]
Y Y answer (2)
2.12 x-2-8 Y standard form
* ¥ three factors @)
x? -2x=8 Y answer
x? -2x-8=0
(x-4)(x+2)=0
x=4or x=-2
2.1.3 | x4 =27x vx =27
3 5. nf (3)
x= 27 Vx =3
gaat v answer
x=3
2.2 | 4*-3.2"-18 2**-3.2..-18 VOR (4)
2° +3 2* +3 ¥ ¥ factors
_ (2° +3)(2"-6). Y answer
2°43
=?" 6:
2.3.1 | -15<1-4x<5 ¥ -16<-4x<4

### Page 9

Testpapers.co.za

Bownloaded:firom Stanmorephysics.com March 2025
-16<-4x <4 Y end points (3)
4>x>-1 ¥ notation
mae 4

2.3.2 7 a x > vv answer (2)

246 X+3y =12 ¥ equation 3

ax+y=9 ¥ substitution
x=12-3y v simplification
Substitute (3) into vx-value
2(12-3y)+ y=9 % vy-value GS)
24-6y+y=9 fe
~Sy=-15
y=
Subst. y =3 into (3)
x=12-3(3)
x=3

[22]

QUESTION 3

### Page 10

Testpapers.co.za

Bownloadeatafsom Stanmorephysics.com March 2025
3.1.1 | siné ¥ answer qd)
= oR
PR
aye
5.
seca
rae; Jeune B)
__! cosa
- 2 ¥ substitution
= OR Y answer
PR
eo
4
5
>
4
3.1.3 | tand v answer
_QR qd)
“OF
ot
3
3.2.2 ¥ correct ratio (3)
v 7
sin 0 = OR substitution
PR ¥ answer
sin 0 = ha
4
Gm iar
sin ( 3
0 =53,13°
[8]

TOTAL MARKS: 50

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