# Mathematics Grade 12 Paper 2 NSC November 2026 Memo (English)

> Grade 12 Mathematics Paper 2 memorandum from the NSC November Exam 2026 (English), 13 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/grade-12-mathematics-2026-nsc-november-exam-unofficial-memorandum-paper-2/378d19a9-7237-4fc4-9e8c-829acae368a9
- Updated: 2026-10-27
- Subject: Mathematics
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: NSC November Exam
- Paper: 2
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 13

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## About this document

Mathematics · Grade 12 · NSC November Exam · 2026. Unofficial memorandum, Paper 2, 13 pages. Unofficial marking guidelines with full worked answers by Hlayiso for the Grade 12 Mathematics Paper 2 NSC November 2026 exam, plus how our prediction compared. Not the official DBE memorandum. Read online or download the PDF.

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## Document text

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### Page 1

UNOFFICIAL
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Unofficial memo
hlayiso mock exams
Practice Series:
Grade 12 NSC Preparation
WWW.HLAYISO.COM
NATIONAL
SENIOR CERTIFICATE
GRADE 12
MATHEMATICS P2
NOVEMBER 2026 — UNOFFICIAL MARKING GUIDELINES
MARKS: 150
These marking guidelines consist of 13 pages.

### Page 2

UNOFFICIAL
Mathematics/P2
2
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
IMPORTANT: These are unofficial marking guidelines prepared by Hlayiso for revision. They are
not the DBE’s official memorandum; the official marking guidelines may differ in mark allocation
and accepted alternatives.
NOTE:
• If a candidate answers a question TWICE, only mark the FIRST attempt.
• If a candidate has crossed out an attempt of a question and not redone the question, mark the
crossed-out version.
• Consistent accuracy (CA) applies in ALL aspects of the marking guidelines. Stop marking at the
second calculation error.
• Assuming answers/values in order to solve a problem is NOT acceptable.
• Answers only will NOT necessarily be awarded full marks; where an answer-only allocation
applies it is stated in the row.
• Rounding: penalise incorrect rounding (or no rounding where TWO decimal places are required)
ONCE per question, not per item.
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.
READINGS OF THE QUESTION PAPER
• Q1: the ogive points are read as (20;0), (40;8), (60;20), (80;44) and (100;50). The ANSWER
BOOK table is not visible; the classes are taken as 20 ≤x < 40, 40 ≤x < 60, 60 ≤x < 80,
80 ≤x < 100.
• Q9.2: ˆO2 = 4x is read as the angle AOC on the side of D (the arc is drawn towards D); ˆO1 is the
reflex angle AOC.
QUESTION 1
1.1
50 vehicles
✓50
(1)
1.2
Median = 25th value: ≈64 km/h
✓answer
Note: Accept 62–66 km/h read
from the curve.
(1)
1.3
50−20 = 30 motorists
✓30
(1)
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UNOFFICIAL
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NSC – Unofficial Marking Guidelines
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1.4
Speed (km/h)
Frequency
Cum. freq.
20 ≤x < 40
8
8
40 ≤x < 60
12
20
60 ≤x < 80
24
44
80 ≤x < 100
6
50
✓8 and 12
✓24 and 6
(2)
1.5
¯x = 30(8)+50(12)+70(24)+90(6)
50
= 3060
50
= 61,2 km/h
✓midpoints
✓∑fx = 3060
✓61,2 km/h
Note: CA from 1.4.
(3)
[8]
QUESTION 2
2.1.1
20
40
60
80
100
hours
%
0
2
4
6
8
10
(vertical axis starts at 20 for clarity)
✓1–4 points correct
✓5–8 points correct
✓all 10 points correct
(3)
2.1.2
a = 79,94
b = −4,11
ˆy = 79,94−4,11x
✓a = 79,94
✓b = −4,11
✓equation
Note: Exact: a = 79,936...,
b = −4,109...; r = −0,79 (not
asked).
(3)
2.1.3
ˆy = 79,94−4,11(7) = 51,17%
✓substitution of x = 7
✓51,17%
Note: Accept 51,17–51,18%.
CA from 2.1.2.
(2)
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### Page 4

UNOFFICIAL
Mathematics/P2
4
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Hlayiso/November 2026
2.1.4
σ = 16,20%
(mean = 59,8%)
✓16,20
Note: Accept sample SD
s = 17,07.
(1)
2.2
T1 = 100, T2 = 98 = 100(0,98), so the hours form a
geometric sequence with a = 100, r = 0,98
100(0,98)n−1 = 54,55
(0,98)n−1 = 0,5455
n−1 = log0,98 0,5455 = 29,998... ≈30
n = 31, i.e. in the 31st month
OR: 98(0,98)n−2 = 54,55 ⇒n−2 = 29 ⇒n = 31
✓r = 0,98
✓100(0,98)n−1 = 54,55
✓use of logs
✓n = 31
Note: Check: 100(0,98)30 =
54,548... ≈54,55. Answer
only: 1/4.
(4)
[13]
QUESTION 3
3.1
mCD = 1−(−9)
4−2
= 10
2 = 5
✓substitution
✓5
(2)
3.2
F
(4+2
2
; 1+(−9)
2
)
= F(3; −4)
✓x = 3
✓y = −4
(2)
3.3
mEF = −1
5
(EF ⊥CD)
y−(−4) = −1
5(x−3)
At E, y = 0: 4 = −1
5(x−3) ⇒x−3 = −20
x = −17 ∴E(−17; 0)
✓mEF = −1
5
✓substitution of F
✓y = 0; x = −17
(3)
3.4
EF =
√
(3+17)2 +(−4−0)2 =
√
416 = 4
√
26
CD =
√
(4−2)2 +(1+9)2 =
√
104 = 2
√
26
Area △CDE = 1
2 CD.EF (EF ⊥CD)
= 1
2
√
104
√
416 = 1
2(208) = 104 square units
✓EF
✓CD
✓1
2 CD.EF
✓104
Note: Other valid methods (e.g.
area rule with CˆED) score full
marks.
(4)
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UNOFFICIAL
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3.5
mAB = mDC = 5
(opp sides of parm)
Inclination of AB: tan−1 5 = 78,69◦
mBC = 1−4
4+1 = −3
5; inclination of BC
= 180◦−30,96◦= 149,04◦
A ˆBC = 149,04◦−78,69◦= 70,35◦(ext ∠of △)
OR: B ˆCD from inclinations of CB and CD (149,04◦
and 78,69◦): B ˆCD = 109,65◦;
A ˆBC = 180◦−109,65◦= 70,35◦(co-int ∠s)
✓mAB = 5 and 78,69◦
✓mBC = −3
5
✓149,04◦
✓70,35◦
Note: Accept 70,34◦–70,35◦; a
cosine-rule method in △ABC is
acceptable.
(4)
3.6
Diagonals of a parm bisect each other:
(x+4
2
; y+1
2
)
=
(−1+2
2
; 4−9
2
)
⇒A(−3; −6)
OR: B→A is the same translation as C→D:
(x;y) →(x−2;y−10)
AB ∥CD and EF ⊥CD ⇒E ˆHA = 90◦
∴EA is a diameter of the circle (chord subtends 90◦)
Centre = midpoint of EA
=
(−17−3
2
; 0−6
2
)
= (−10; −3)
r2 = (−17+10)2 +(0+3)2 = 58
(x+10)2 +(y+3)2 = 58
✓x = −3 ✓y = −6
✓E ˆHA = 90◦⇒EA
diameter
✓centre (−10;−3)
✓r2 = 58
✓equation
Note: Finding H
(
−31
13;−38
13
)
and solving with three points is
valid but long; full marks if
correct. CA from A.
(6)
[21]
QUESTION 4
4.1
r2 = AK2 = (−7+4)2 +(0+4)2 = 25
∴(x+4)2 +(y+4)2 = 25
✓substitution of K
(1)
4.2
mAK = 0+4
−7+4 = −4
3
✓−4
3
(1)
4.3
mtan = 3
4
(tan ⊥radius)
y−0 = 3
4(x+7)
y = 3
4x+ 21
4
✓m = 3
4
✓substitution of K
✓equation
Note: CA from 4.2.
(3)
4.4
x2 −2x+1+y2 −2y+1 = 3+1+1
(x−1)2 +(y−1)2 = 5
M(1; 1)
✓completing the square (x)
✓completing the square (y)
✓M(1;1)
(3)
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4.5
r =
√
5 units
✓
√
5
(1)
4.6
B: y = 0: (x+4)2 +16 = 25 ⇒x+4 = ±3
x = −1 or x = −7 (K) ∴B(−1; 0)
Reflection in y = −4: (x;y) →(x;−8−y) ∴
T(−1; −8)
✓y = 0 substituted
✓B(−1;0)
✓xT = −1 ✓yT = −8
(4)
4.7
mKA = −4
3 (4.2); mAT = −8+4
−1+4 = −4
3
mKA = mAT and A is a common point ⇒K, A, T
collinear
OR: midpoint of KT =
(−7−1
2
; 0−8
2
)
= (−4;−4) = A
✓mAT = −4
3
✓equal gradients and
common point
OR: ✓midpoint ✓= A (2)
4.8
AM =
√
(1+4)2 +(1+4)2 =
√
50 = 5
√
2 ≈7,07
Touching externally: AM = R+rnew
rnew = 5
√
2−5 ≈2,07
% reduction =
√
5−(5
√
2−5)
√
5
×100%
= 7,38%
✓substitution
✓AM = 5
√
2
✓rnew = 5
√
2−5
✓% formula
✓7,38%
Note: Rounding rnew to 2,07
gives 7,43%: accept. CA from
4.4/4.5.
(5)
[20]
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UNOFFICIAL
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QUESTION 5
5.1.1
sin12◦=
√
1−cos2 12◦=
√
1−k2
(or from a sketch: x = k, r = 1, y =
√
1−k2)
✓identity/sketch
✓
√
1−k2
(2)
5.1.2
tan(−12◦) = −tan12◦= −
√
1−k2
k
✓−tan12◦
✓answer
(2)
5.1.3
1−2sin2 39◦= cos78◦= sin12◦=
√
1−k2
✓cos78◦
✓
√
1−k2
(2)
5.2
LHS = cosθ ·cos2 θ +sin2 θ(−sinθ)
cosθ −sinθ
= cos3 θ −sin3 θ
cosθ −sinθ
= (cosθ −sinθ)(cos2 θ +sinθ cosθ +sin2 θ)
cosθ −sinθ
= 1+sinθ cosθ = RHS
✓cosθ ✓cos2 θ
✓−sinθ (numerator)
✓−sinθ (denominator)
✓difference of cubes
✓sin2 θ +cos2 θ = 1
(6)
5.3
Undefined when sinθ −cosθ = 0
tanθ = 1
θ = 225◦
(θ ∈[180◦;360◦])
✓sinθ −cosθ = 0
✓tanθ = 1
✓225◦
Note: θ = 45◦included: max
2/3.
(3)
[15]
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QUESTION 6
6.1.1
cos25◦= cos(35◦−10◦)
= cos35◦cos10◦+sin35◦sin10◦
✓cos(35◦−10◦)
✓expansion
(2)
6.1.2
cos35◦cos10◦+sin35◦sin10◦−2sin10◦sin35◦
= cos35◦cos10◦−sin35◦sin10◦
= cos(35◦+10◦) = cos45◦=
√
2
2
✓substitution of 6.1.1
✓cos45◦
✓
√
2
2 or
1
√
2
(3)
6.2
LHS = 4sinθ cosθ(cos2 θ −sin2 θ)
= 2sin2θ cos2θ
= sin4θ = RHS
✓common factor
✓2sin2θ and cos2θ
✓sin4θ
(3)
6.3.1
A = 53,13◦
✓53,13◦
(1)
6.3.2
sinA = 4
5, cosA = 3
5
Divide by 5: 4
5 cosx−3
5 sinx = 1
5
sinAcosx−cosAsinx = 1
5
sin(A−x) = 1
5
(ref ∠= 11,54◦)
A−x = 11,54◦+k·360◦or A−x = 168,46◦+k·360◦
x = 41,59◦+k ·360◦or x = −115,33◦+k ·360◦,
k ∈Z
OR: cos(90◦−A) = 4
5: 5cos(x+36,87◦) = 1
x+36,87◦= ±78,46◦+k ·360◦(same solutions)
✓cosA = 3
5
✓dividing by 5
✓sin(A−x) = 1
5
✓both equations with
k ·360◦
✓41,59◦✓−115,33◦
Note: Accept 244,67◦+k ·360◦
for the second family. k ∈Z
must be stated once. Check:
4cos41,59◦−3sin41,59◦= 1.
(6)
[15]
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QUESTION 7
7.1
y ∈[0; 2]
or
0 ≤y ≤2
✓answer
(1)
7.2
180◦
✓180◦
(1)
7.3
x
y
−180◦−150◦−120◦
−90◦
−60◦
−30◦
-2
-1
1
2
f
g
(−180◦;0)
(−90◦;−1)
(0◦;−2)
✓shape (increasing to −2 as
x →0◦)
✓endpoints (−180◦;0) and
(0◦;−2)
✓(−90◦;−1)
(3)
7.4.1
AB = f(x)−g(x) = cos2x+1+cosx+1
= 2cos2 x−1+cosx+2 = 2cos2 x+cosx+1
✓f(x)−g(x)
✓cos2x = 2cos2 x−1
(2)
7.4.2
AB = 2
(
cosx+ 1
4
)2 +1−1
8 = 2
(
cosx+ 1
4
)2 + 7
8
Minimum AB = 7
8 = 0,875 units, when cosx = −1
4
(x = −104,48◦, which lies in [−180◦;0◦])
OR:
d
d(cosx): 4cosx+1 = 0 ⇒cosx = −1
4; AB
= 2
( 1
16
)
−1
4 +1 = 7
8
✓completing the square /
cosx = −1
4
✓7
8
(2)
[9]
QUESTION 8
8.1
x
y
C(0;0)
B(a;0)
A(bcosC;bsinC)
b
c
A(bcosC; bsinC) and B(a; 0)
c2 = AB2 = (bcosC −a)2 +(bsinC −0)2
= b2 cos2C −2abcosC +a2 +b2 sin2C
= a2 +b2(sin2C +cos2C)−2abcosC
= a2 +b2 −2abcosC
✓coordinates of A and B
✓distance formula
✓expansion
✓sin2C +cos2C = 1
(4)
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UNOFFICIAL
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8.2.1
(4p)2 = (2p)2 +42 −2(2p)(4)cosθ
16p2 = 4p2 +16−16pcosθ
cosθ = 16−12p2
16p
= 4−3p2
4p
✓cosine rule substitution
✓making cosθ the subject
(2)
8.2.2
AˆEB = 180◦−θ
(∠s on a str line)
In △ABE:
(
√
5p)2 = (2p)2 +22 −2(2p)(2)cos(180◦−θ)
5p2 = 4p2 +4+8pcosθ
p2 = 4+8p
(4−3p2
4p
)
= 4+8−6p2
7p2 = 12 ⇒p =
√
12
7 = 1,31
Height of the tree = AD = p = 1,31 m
✓180◦−θ
✓cosine rule in △ABE
✓cos(180◦−θ) = −cosθ
✓substitution of 8.2.1
✓1,31 m
Note: θ = 102,60◦. Otherwise:
DB = 2p, DE =
√
3p, DC
=
√
15p (Pythagoras, AD ⊥
plane), then the cosine rule at E
in △DBE and △DEC with
supplementary angles gives
28p2 = 48, the same p.
(5)
[11]
QUESTION 9
9.1.1
ˆP1 = ˆT2 = 40◦
[∠s in same seg]
✓S ✓R
(2)
9.1.2
P ˆRS = 180◦−(35◦+40◦) = 105◦
[opp ∠s of cyclic quad]
✓S ✓R
(2)
9.2
ˆO1 = 360◦−4x
[∠s round a pt]
ˆD = 1
2 ˆO1 = 180◦−2x
[∠at centre = 2∠at
circumference]
ˆD = C ˆBE = 2x+10◦
[ext ∠of cyclic quad]
180◦−2x = 2x+10◦⇒x = 42,5◦
ˆD = 2(42,5◦)+10◦= 95◦
OR: A ˆBC = 1
2 ˆO2 = 2x
[∠at centre = 2∠at
circumference]
2x+2x+10◦= 180◦
[∠s on a str line]
x = 42,5◦; ˆD = 180◦−A ˆBC = 180◦−85◦= 95◦
[opp ∠s of cyclic quad]
✓S/R
✓S ✓R
✓S/R
✓x = 42,5◦
✓95◦
OR:
✓S ✓R
✓S/R
✓x
✓S ✓R (95◦)
Note: Check: ˆO2 = 170◦,
A ˆBC = 85◦, C ˆBE = 95◦.
(6)
[10]
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QUESTION 10
10.1
HK
KC = EL
LC = 1
3
[prop theorem; KL ∥HE]
HK = 1
3(12) = 4 cm
✓S ✓R
✓4 cm
(3)
10.2
AH
HC = BE
EC = 2
4
[prop theorem; HE ∥AB]
HC = 16 ⇒AH = 8, AC = 24 cm
△BDE ||| △BAC
[∠∠∠; DE ∥AC]
BD
BA = BE
BC = 2
6 = 1
3
(DE = 8 cm)
Area △BDE
= 1
2 BD.BE.sin ˆB = 1
9
(1
2 BA.BC.sin ˆB
)
= 1
9 area
△ABC
Area DECA = 8
9 area △ABC
Area △BDE
Area DECA = 1
8
✓BD
BA = 1
3 (S/R; via ||| or
prop th)
✓area rule / ratio
(1
3
)2
✓area DECA = 8
9△ABC
✓1
8
Note: A candidate who uses DE
= 8 and AC = 24 with heights
in the ratio 1 : 3 scores full
marks.
(4)
[7]
QUESTION 11
11.1
R
S
T
G
H
P
K
L
Constr: Mark G on RS and H on RT so that RG = PK
and RH = PL. Join GH.
In △RGH and △PKL:
RG = PK; RH = PL
[constr]; ˆR = ˆP
[given]
∴△RGH ≡△PKL
[SAS]
∴R ˆGH = ˆK = ˆS
∴GH ∥ST
[corresp ∠s =]
∴RG
RS = RH
RT
[line ∥one side of △/ prop theorem;
GH ∥ST]
∴PK
RS = PL
RT, i.e. RS
PK = RT
PL
✓construction
✓SAS (S/R)
✓R ˆGH = ˆS
✓S/R (GH ∥ST)
✓R (prop theorem)
✓conclusion with RG = PK,
RH = PL
Note: Construction mark is also
given if G, H are placed on PK,
PL (in the smaller triangle)
correspondingly.
(6)
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11.2.1
Line from centre ⊥to chord (OL ⊥PT)
✓R
(1)
11.2.2
In △PML and △OMP:
ˆM = ˆM
[common]
PˆLM = 90◦
[given PT ⊥KM]
OˆPM = 90◦
[tan ⊥radius]
∴△PML ||| △OMP
[∠∠∠]
✓S
✓S
✓S/R
✓R
(4)
11.2.3
PL
OP = ML
MP
[||| △s]
PL = 1
2PT
[11.2.1]; OP = 1
2PS
[radii; POS
diameter]
1
2PT
1
2PS = ML
MP ⇒PT
PS = ML
MP
∴PT
ML = PS
MP
✓S ✓R
✓PL = 1
2PT
✓OP = 1
2PS
✓substitution and
conclusion
(5)
11.2.4
From △PML ||| △OMP:
MP
OM = ML
MP ⇒MP2 = OM.ML
In △OLP and △PLM: OˆLP = PˆLM = 90◦;
L ˆOP = LˆPM
[= 90◦−ˆM, sum of ∠s in △]
∴△OLP ||| △PLM
[∠∠∠]
⇒PL
ML = OL
PL ⇒PL2 = OL.ML
MP2
PL2 = OM.ML
OL.ML = OM
OL
MP2
1
4PT2 = OM
OL ⇒MP
PT = 1
2
√
OM
OL
✓MP2 = OM.ML
✓||| (S/R)
✓PL2 = OL.ML
✓MP2
PL2 = OM
OL
✓PL = 1
2PT and conclusion
Note: Alternative: OP2 =
OL.OM (△OLP ||| △OPM)
combined with 11.2.3; mark
similarly.
(5)
[21]
TOTAL: 150
Copyright reserved
Please turn over

### Page 13

UNOFFICIAL
Mathematics/P2
13
NSC – Unofficial Marking Guidelines
Hlayiso/November 2026
HOW DID OUR PREDICTION DO?
The real November 2026 P2 compared with the Hlayiso 2026 Maths P2 Prediction paper
Q
Topic
Real paper asked
Our closest prediction item(s)
Verdict Covered
1
Ogive
Read n, median, “more than 60”;
frequencies from ogive; grouped mean
Q2: cumulative freq., ogive, “more than
R220”, estimated mean

8/8
2
Regression;
sequence
Scatter plot, regression line, predict, SD (9);
geometric decrease of battery hours (4)
Q1.1, 1.3, 1.5 (regression, prediction, SD).
No sequence (a P1 skill) in our P2
/
9/13
3
Analytical
geometry (lines)
Gradient, midpoint, perpendicular line to E,
area △CDE, A ˆBC via inclinations, 4th
vertex of parm, circle through 3 points with
a 90◦angle
Q3.2–3.6: gradient/line, inclination, A ˆBC,
area, circle through A, B, C with a right
angle (diameter)

19/21
4
Circles
Circle eq., tangent at K, centre/radius from
general form, reflection in y = −4,
collinearity, % reduction to touch externally
Q4.1–4.4: centre/radius, tangent, touching
externally (distance between centres).
Reflections were listed as “unlikely”
/
16/20
5
Trig: reduction
cos12◦= k; reduction identity using a
difference of cubes; when undefined
Q5.2 (sin34◦= m), Q5.3 (reduction),
Q6.1.2 (undefined)

13/15
6
Trig: compound
angles
cos(35◦−10◦) evaluation; sin4θ identity;
4cosx−3sinx = 1 general solution
Q5.1.3, Q6.1 (compound/double angles),
Q6.2 (general solution of a different form)
≈
9/15
7
Trig graphs
Range, period, sketch −cosx−1;
minimum vertical distance AB
Q7.1–7.2 (period, sketch). Vertical distance
was listed as “unlikely”
≈
5/9
8
Cosine rule / 3D
Proof of the cosine rule (4); cosine rule
twice with 180◦−θ to find a height
Q8 (3D tower with the sine rule). Rule
proofs were listed as “unlikely”
/≈
2/11
9
Circle geometry
Same seg, opp ∠s cyclic quad; centre ∠+
ext ∠cyclic quad with x
Q9.2 (centre ∠, opp ∠s, ext ∠of cyclic
quad)

10/10
10 Proportionality
Prop theorem with ∥lines; area ratio △BDE
: DECA
Q10.2 (two ∥lines, prop theorem, area ratio)

7/7
11
Similarity
Equiangular ⇒sides in proportion proof;
similar △s with tangent/diameter; ratio
proofs
Q10.1 (same theorem proof, same wording);
Q11.3–11.4 (similar △s and product/ratio)

16/21
Marks of the real paper covered by our prediction
114/150
A learner who mastered our prediction could have answered about 114/150 (76%) of this paper with confidence.
Much of that comes from topics that appear every year, so the extra value of the prediction lies in the specific
calls (the proof, the right-angle circle, the area ratio).
What we got right
• The theorem proof: equiangular triangles, sides in
proportion (6 marks), our “most overdue” pick, set
almost word for word.
• The overall structure and weighting: ogive +
regression, about 40 marks of analytical geometry,
50 of trigonometry and 38 of Euclidean geometry.
• Circle through three points using a 90◦angle
(diameter), angle via inclinations, area of a triangle.
• Circle-geometry and proportionality riders: ext ∠
of cyclic quad, centre angle, area ratio with ∥lines,
similar triangles with a ratio proof.
What we missed
• Three items we explicitly listed as “unlikely” all
came up: the proof of the cosine rule, reflection of a
point, and the minimum vertical distance between
two graphs (about 12 marks).
• 4cosx−3sinx = 1 (reducing acosx+bsinx to a
single ratio), a rare and hard type.
• A P1 geometric sequence inside the statistics
question (4 marks).
• The hardest geometry: 11.2.4 (two similarities,
PL2 = OL.ML), and the 3D cosine-rule system in
8.2.2.
Rating: 7/10. The core skills and the theorem proof were right, but the “unlikely” list cost about 12
marks and the hardest 15–20 marks went beyond our paper.
Copyright reserved

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