# Mathematics Grade 10 April Test 2021 Paper & Memo

> Grade 10 Mathematics question paper and memo from the April Test 2021, 10 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/math-april-qp-memo-2021-gr10/a4907300-7572-4895-bfcc-4000378a73ba
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 10
- Document type: Question paper and memo
- Year: 2021
- Exam period: April Test
- Curriculum: CAPS
- Pages: 10

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## About this document

Mathematics · Grade 10 · April Test · 2021. 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

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NSC

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

il This question paper consists of 6 questions.
2. Answer ALL the questions.
ch Clearly show ALL calculations, diagrams, graphs, etc. which you have used in

determining your answers.
4 Answers only will NOT necessarily be awarded full marks.

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

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

c Diagrams are NOT necessarily drawn to scale.

8. Write neatly and legibly.

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Mathematics 3 KZN/March 2021
: NSC

“QUESTION 1

Ll Determine the product of the following and simplify fully:

11 ab? (20 +40) (2)

1.1.2 (x-2)(x? +2x+8) (3)
1.2 Factorise the following expressions fully:

1210 2x? +7x-4 (2)

122 ax-ay-b’x+by (3)

1.2.3. 125-27m5 (2)

ile} Simplify the following expressions fully:

13.1 25xty*x (5x*y°) G)
fs
132 +7xt+10 x+2 (4)
x+5 5
9.3"? 45,3"
1330 4
a @)
[23]
QUESTION 2
2.1 Solve for x in each of the following equations:
2.1.1 1-Sx=6 (2)
212 (x+2)(x-4)=0 (2)
14+2
213 g=—— @)
x
214  3.7°'4+7=154 (3)
2.2 Solve for x : 3(2x+4)-3(x-3)<0. Represent your answer in interval notation. (3)
oe Solve for x and y simultaneously:
4x-2y=6
2y+3x=8 (5)
[18]

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QUESTION 3
3 1 Simplify the following expression:
x-2 x+4
ti |
3 6
-2
3.1.2 | Hence, determine the value of x if: ae = =.

32 Without using a calculator, simplify the following expression fully:

2021
1-2020°

Give reasons for your statements in the answers to QUESTIONS 4, 5 and 6.
QUESTION 4

4.1 In the diagram below, F = 2x, €,= 75°, €,=2x—15° and B, =135°.

A
E af, F

iz
2x 135°

Tee
f 2x - 15°

Cc 3

4.1.1 Determine the value of x.

4.1.2 Determine if EF //GH. Motivate your answer.

(3)

(2)

QB)
[8]

(3)
(3)

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Mathematics 5 KZN/March 2021

NSC

42 In the diagram below, square ACDE has an area of 256 cm? and AB = BC.

B

A * C
4 ;

E , D

4.2.1 Determine the length of AC. (Let AC =x)

4.2.2 Hence, determine the length of AB. (Correct to TWO decimal places).

QUESTION 5
5.1 The angles of a certain quadrilateral are in the ratio 2:3:4: 6.
Calculate the size of the largest angle.

5.2 Complete the statements below for the properties of a rhombus:
(write the question number and missing word / term only)

\21 Opposite anglesare.
5.2.2 Allsidesare........ and, opposite sides are parallel to each other.

5.2.3. Diagonals bisect each other at ........ degrees.

(3)

(3)
[12]

(2)

qd)
qd)
dd)

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By) The diagonals of rhombus PQRS intersect at T.

| $, = 2x and BP, = 3x +20°

Pd
S ‘ R
5.3.1 State the value of Q, , in terms of x. qd)
5.3.2 Hence, calculate the value of x. (3)

19]
QUESTION 6

In the diagram below, ABCD and EFCD are parallelograms.

A B

D Cc

Prove that AFDE isa parallelogram. [5]

TOTAL: 75

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Education

REPUBLIC OF SOUTH AFRICA

SENIOR CERTIFICATE

GRADE 10

SRR See ceescoEaas

¢
fepeseseuee See eeeeeseeeeseesesss

MARKS: = 75

TIME: 1% hours

This memorandum consists of 7 pages.

Copyright Reserved

KwaZulu-Natal Department of Education

RESESESR ESSE.

: MATHEMATICS .
: COMMON TEST :
: MARCH 2021 i
: MARKING GUIDELINE : cf :

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2

KZN/March 2021

Mathematics
NSC - Markimg Guidelines
QUESTION !
eet at? (-2a2 + 4b) yee
=—2a0°b? + 4ab* vo 4ab* Q)
V2
(x-2)(x? +2x+8)
Dy? + 8x 237 4-16 ¥ = .
=x +4y—16 oe B)
ee ee ae yarn
=(2x-1Nx+4) ae )) (2)
122 | @x-ay—b'x+by
ae Y common factors
¥(a-8) V
=x(a~bYa+b)-y\a-b) (a~ a+b)
=(a-b)[x(a+b)-y]
=(a-bYax+bx-y) ce G)
123) Fi
125-27 oe
=(5~3m\25+15m+ 9m") Yo (25415m49m2) Q)
[ee eta
vo sty
¥ simplification of
indices
1
ee
xy 3)
Es2eee
SET tlO x42
xts OS ¥¥% both factors
_ (+5) 042) 9 a
x45 x+2 x+2
7 4
= 7 (4)
133 | 9.37453"
each
IS H5.3" as
3-33 oa
3 (8145)
31-3) ¥ _ factorisation of 3°
28
2
v 4)
=-43 Ge (4)
te {23}

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Mathematics 4 KZN/March 2021
NSC — Marking Guidelines
dx—-2y=6  Eql
2y+3x=8  Eq2 Flea a
‘ te stitution of (ig
ae 2 into (/:q 2)
ys2x-3
Sub Eq3 into Eq2: | 2(2x~-3)+3x=8 x=2
4x—643x=8
Tx=l4
x=2 substitution of x =2
x=2:  4(2)-2y=6
Subx=2 (2)-2y (5)
i) |
= vel i
[18]
QUESTION 3
F aE =
BT [Sal | xn2_ x44 a
a 6
Ux-2)_ (x44) ico
-x-4
answer @)
312
substitution of answer
from 3.1.1
ca (2)
32 2021
1-2020°
= 2021
(1=2020)(1+ 2020) factorisation
____2021
*2019)(2021) simplification
--—_L
2019 answer
(3)
[8]

Mathematics 3 KZN/March 2021
NSC - Marking Guidelines
QUESTION 2
ae Pan
Y simplification
Answer only: 2/2 Q
xe-l ¥__answer
21.2 | (x4.2)(x-4)=0
w=72 or x=4 vv _answers (2)
2.13 1+2x
g=
x
gralt2x Vv gronLHs
gx-2y=1
xg-2)=1 ve factorisation
1
) Me answer Q)
214 13.7475 154
3.77 =147
a ell
malt 3
3
T1349
ei Y prime base
x-1=2
= ¥ answer
: G)
22 [ Hax+4)-3x-30<0
6x+12—3x+9<0 - fo
simplification
3r421<0 :
3x<-21
eT Yo x<-7
s.xe(—01-7) Y__answer (3)
23 | 4x-2y=6 Egil
2y+3x=8  Eq2
Egl+£q2:  7x=14 v Egl+ Eq2
(elimination)
x=2 Yo Ix=l4
Subx=2:  4(2)-2y=6 Vx?
-2y=-2 y substitution of x=2
ye v (5)
OR
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Mathematics: = KZN/March 2021
NSC ~ Marking Guidelines
QUESTION 4
at
eee (ext Zof :
135° = 2x-+75°
2x =135°-75°
2x = 60°
= 30° answer (3)
44.2 | G =2(30°)-15° ‘
C, =45
By + C, = 135° + 45° = 180° statement
2 EFO (co-int Zs supplementary) reason
OR
2(30°)- 15°
45° é, 45°
EBC = 45° (Zsonstraight line)
oC, = EBC statement
2 &FO (alternate Zs equal) Teason 6)
ae
< YY trea of 2 ACKAE :
B :
x7 =256cm"
A Cc
answer
E D
(3)

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6
NSC - Marking Guidelines

KZN/March 2021

4.2.2 | Inf AB= BC (given)
and AB +BCh= AC? (pythag) |) gp
2AB =16° v
AB =128
AB=8 Yom
AB=11,3lom 4 ca (3)
L— (12)
QUESTION 5
5.1 2x + 3x+4x+ 6x = 360° (sum of Zs in quad)
15x = 360°
— v — xvalue
«. largest angle =6x24° = 144° v_ca Q)
5.2_[ 5.2.1 |equal v answer (|
322 [equal 7 answer @
v answer, aw)

(523 [90

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Mathematics 7 KZN/March 2021
NSC ~ Marking Guidelines

53 Pp Q
s LO R
33.1 |G =2x (angles opp=sides) “4 y «ty
532 7 SIR
F=90° ) (diags of rhombus) a sir
ind : Q
3x+20° +90°42x= 180°
: me ¥ answer | 3)
OR \
B= 3x 20° (diags of rhombus) ¥ SR
Ind oS Oo [
65-4404 2420 = 180°
lox = 140°
i:
¥__answer (3)
[9
QUESTION 6
A B
E.
D fe
ED= FC (opp sides of parm EFCD =) vs
AF = FC (diags of parm ABCD) vy SIR
ED= AF ada
but EDD vy oSR
AFDEis a parallelogram
i _ Rk 15]
(2 opposite sides equal and parallel)
TOTAL: 75

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