# Mathematics Grade 8 Paper 2 November Mock 2026 Memo

> Grade 8 Mathematics Paper 2 memorandum from the November Mock Exam 2026, 7 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/grade-8-mathematics-2026-november-mock-exam-memorandum-paper-2/2effe792-c42c-4f34-87fa-def729d8bc70
- Updated: 2026-10-26
- Subject: Mathematics
- Grade: Grade 8
- Document type: Memorandum
- Year: 2026
- Exam period: November Mock Exam
- Paper: 2
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 7

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## About this document

Mathematics · Grade 8 · November Mock Exam · 2026. Memorandum, 7 pages. Read online or download the PDF.

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## Document text

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### Page 1

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Mock
hlayiso mock exams
Practice Series:
Grade 8 Exam Preparation
WWW.HLAYISO.COM
SENIOR PHASE
END-OF-YEAR EXAMINATION
GRADE 8
MATHEMATICS P2
NOVEMBER 2026 (MOCK)
MARKING GUIDELINES
MARKS: 75
These marking guidelines consist of 7 pages.

### Page 2

Mathematics/P2
2
Senior Phase – Marking Guidelines
Hlayiso/Mock 2026
NOTE:
• If a learner answers a question TWICE, only mark the FIRST attempt.
• Consistent accuracy (CA) applies in ALL aspects of the marking guidelines, unless stated
otherwise. Stop marking at the second calculation error.
• Accept answers rounded correctly to TWO decimal places. Penalise units only where the question
asks for them.
• Accept any other correct method and award marks accordingly.
GEOMETRY
S A mark for a correct statement (a statement mark is independent of a reason).
R A mark for the correct reason (a reason mark may only be awarded if the statement is correct).
S/R Award a mark if the statement AND reason are both correct.
Accept the usual abbreviations, e.g. ∠s on a str line; vert opp ∠s; corresp ∠s, AB∥CD; alt ∠s; co-int
∠s; sum of ∠s in △; ext ∠of △; ∠s opp = sides; opp ∠s of ∥m; sides of rhombus.
QUESTION 1
1.1
C
(reflex: between 180◦and 360◦)
✓C
(1)
1.2
C
(x = 180◦−115◦= 65◦; co-int ∠s, KL ∥MN)
✓C
(1)
1.3
A
(360◦−100◦−80◦−95◦= 85◦)
✓A
(1)
1.4
D
(rhombus)
✓D
(1)
1.5
A
(SAS / s∠s)
✓A
(1)
1.6
B
(52 +122 = 169 = 132)
✓B
(1)
1.7
C
(3,14×102 = 314 cm2)
✓C
(1)
1.8
D
(100×100 = 10000 cm2)
✓D
(1)
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### Page 3

Mathematics/P2
3
Senior Phase – Marking Guidelines
Hlayiso/Mock 2026
1.9
A
( 5
10 = 1
2)
✓A
(1)
1.10
B
(x; y) →(−x; y) gives (−4; −3)
✓B
(1)
[10]
QUESTION 2
2.1.1
3x+10◦+2x+20◦= 180◦
∠s on a str line
5x = 150◦
∴x = 30◦
✓180◦
✓reason
✓x = 30◦
(3)
2.1.2
DˆEB = AˆEC = 3(30◦)+10◦= 100◦
[vert opp ∠s]
OR
CˆEB = 80◦, so DˆEB = 180◦−80◦= 100◦[∠s on a str
line]
✓100◦(reason required; CA
from 2.1.1)
(1)
2.2
a = V ˆWX = 64◦
[corresp ∠s; PQ ∥RS]
b = V ˆXW = 47◦
[alt ∠s; PQ ∥RS]
c = W ˆVX = 180◦−64◦−47◦= 69◦[sum of ∠s in △]
OR
Q ˆVW = 180◦−64◦= 116◦[∠s on a str line];
c = 116◦−47◦= 69◦
✓S ✓R (a)
✓S ✓R (b)
✓S ✓R (c)
(6)
2.3.1
2x+15◦+3x−5◦= 180◦
[co-int ∠s; AB ∥CD]
5x+10◦= 180◦
5x = 170◦
x = 34◦
✓S
✓R
✓x = 34◦
(3)
2.3.2
B ˆGH = 2(34◦)+15◦= 83◦
A ˆGH = 180◦−83◦= 97◦
[∠s on a str line]
OR
A ˆGH = G ˆHD = 3(34◦)−5◦= 97◦[alt ∠s]
✓97◦(CA from 2.3.1)
(1)
[14]
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### Page 4

Mathematics/P2
4
Senior Phase – Marking Guidelines
Hlayiso/Mock 2026
QUESTION 3
3.1.1
Rectangle
✓answer
(1)
3.1.2
Kite
✓answer
(1)
3.1.3
Parallelogram
✓answer
(1)
3.2
ˆQ = ˆR1
[∠s opp = sides]
ˆR1 = 180◦−(4x+10◦) = 170◦−4x [∠s on a str line]
4x+10◦= 2x−10◦+(170◦−4x)
[ext ∠of △]
6x = 150◦
x = 25◦
OR
2x−10◦+2(170◦−4x) = 180◦[sum of ∠s in △]
⇒6x = 150◦, x = 25◦
(Check: ˆP = 40◦, ˆQ = ˆR1 = 70◦, P ˆRS = 110◦)
✓S/R ˆQ = ˆR1
✓S/R ˆR1 = 170◦−4x
✓S/R ext ∠/ sum of ∠s
✓x = 25◦
(4)
3.3
AB = BC
[sides of rhombus]
∴ˆA1 = B ˆCA
[∠s opp = sides]
ˆA1 = 180◦−116◦
2
= 32◦
[sum of ∠s in △]
✓S/R ˆA1 = B ˆCA (AB =
BC)
✓R sum of ∠s in △
✓32◦
(3)
3.4
AB = DE (= 5 cm)
∴△ABC ≡△DEF
[SAS / s∠s]
✓DE
✓SAS
(2)
3.5
YZ
LM = XY
KL
[||| △s]
YZ
6 = 10
4
(scale factor 2,5)
YZ = 15 cm
✓correct proportion / scale
factor 2,5
✓YZ = 15 cm
(2)
[14]
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### Page 5

Mathematics/P2
5
Senior Phase – Marking Guidelines
Hlayiso/Mock 2026
QUESTION 4
4.1
BC2 = AC2−AB2
[Pythagoras]
BC2 = 262 −102 = 676−100 = 576
BC = 24 cm
✓correct use of Pythagoras
✓576
✓BC = 24 cm
(3)
4.2
PQ2+ QR2 = 92 +402 = 81+1600 = 1681
PR2 = 412 = 1681
PQ2+ QR2 = PR2, so △PQR is right-angled (at Q)
[converse of Pythagoras]
✓1681 (sum of squares)
✓412 = 1681
✓conclusion
(3)
4.3
brace2 = 1,62 +1,22 = 2,56+1,44 = 4
brace = 2 m
✓substitution
✓2 m
(2)
[8]
QUESTION 5
5.1.1
CE2 = BC2+ BE2 = 62 +82 = 100
[Pythagoras]
CE = 10 cm
✓substitution
✓10 cm
(2)
5.1.2
P = AE + EC + CD + DA
= 22+10+14+6 = 52 cm
✓52 cm (CA from 5.1.1) (1)
5.1.3
A = 14×6+ 1
2 ×8×6
= 84+24 = 108 cm2
✓84 and 24
✓108 cm2
(2)
5.2.1
C = πd = 3,14×14 = 43,96 m
✓43,96 m
(1)
5.2.2
r = 7 m
Pond = πr2 = 3,14×72 = 153,86 m2
Lawn = 20×20 = 400 m2
Grass = 400−153,86 = 246,14 m2
✓3,14×72 (153,86)
✓400
✓246,14 m2
(3)
[9]
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### Page 6

Mathematics/P2
6
Senior Phase – Marking Guidelines
Hlayiso/Mock 2026
QUESTION 6
6.1
V = l ×b×h = 30×20×15
= 9000 cm3
✓substitution
✓9 000 cm3
(2)
6.2
TSA = 2(30×20)+2(30×15)+2(20×15)
= 1200+900+600
= 2700 cm2
✓any two pairs of faces
correct
✓all three pairs
✓2 700 cm2
(3)
6.3
9000 cm3 = 9000 mℓ
= 9 ℓ
✓9 000 mℓ(CA from 6.1)
✓9 ℓ
(2)
6.4
5 faces (2 triangles and 3 rectangles)
✓5
(1)
[8]
QUESTION 7
7.1
A′(1; −1), B′(4; −1), C′(1; −3)
[(x; y) →(x; −y)]
✓any two correct
✓all three correct
(2)
7.2
(x; y) →(x−5; y+2)
✓answer
(1)
7.3
B′′(−4; −1)
[(x; y) →(−x; −y)]
✓answer
(1)
[4]
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### Page 7

Mathematics/P2
7
Senior Phase – Marking Guidelines
Hlayiso/Mock 2026
QUESTION 8
8.1
7; 7; 11; 12; 13; 14; 15; 15; 15; 16; 18
✓correct order
(1)
8.2
Range = 18−7 = 11
✓11
(1)
8.3
Mean =
7+7+11+12+13+14+15+15+15+16+18
11
=
143
11 = 13
✓143
✓13
(2)
8.4
Median = 6th value = 14
✓14 (CA from 8.1)
(1)
8.5
Mode = 15
✓15
(1)
8.6
Marks more than 14: 15; 15; 15; 16; 18 (5 learners)
P(more than 14) = 5
11 (≈0,45 or 45,45%)
✓5 favourable outcomes
✓5
11
(2)
[8]
TOTAL: 75
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