# Mathematics Grade 12 Paper 2 North West Mock 2018 Memo

> Grade 12 Mathematics Paper 2 memorandum from the North West Mock Exam 2018, 19 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/2018-grade-12-math-trial-exam-paper-2-nw-memo/2584de7a-c322-4b38-a3a8-4ce30dd2fb3f
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2018
- Exam period: North West Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 19

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## About this document

Mathematics · Grade 12 · North West Mock Exam · 2018. Memorandum, 19 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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Education and Sport Development

Department of Education and Sport Development
Departement van Onderwys en Sportontwikkeling
Lefapha la Thuto le Tlhabololo ya Metshameko

NORTH WEST PROVINCE

NATIONAL
SENIOR CERTIFICATE/
NASIONALE
SENIOR SERTIFIKAAT

" MATHEMATICS P2/WISKUNDE V2 "
nm . u
1 SEPTEMBER 2018 1
iy n
: MEMORANDUM *
+

SESS SSSR SSS SEU

MARKS/PUNTE: 150

This memorandum consists of 22 pages./
Hierdie memorandum bestaan uit 22 bladsye.

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Mathematics P2/Wiskunde V2

NOTE:

2

NSC/NSS — Memorandum

NW/September 2018

Ifa candidate answers a question TWICE, only mark the FIRST attempt.

* Ifa candidate has crossed out an attempt of a question and not redone the question, mark

the crossed out version.

* Consistent accuracy applies in ALL aspects of the marking memorandum. Stop marking at

the second calculation error.

¢ Assuming answers/values in order to solve a problem is NOT acceptable.

NOTA:

¢ As ‘n kandidaat ‘n vraag TWEE KEER beantwoord, merk slegs die EERSTE poging.
¢ As ‘n kandidaat ‘n antwoord van ‘n vraag doodtrek en nie oordoen nie, merk die

doodgetrekte poging.

¢ Volgehoue akkuraatheid word in ALLE aspekte van die nasienmemorandum toegepas.
Hou op nasien by die tweede berekeningsfout.
¢ Aanvaar van antwoorde of waardes om ‘n probleem op te los, word NIE toegelaat nie.

QUESTION/VRAAG 1
11 _ 5478 v 5478
11 ll
¥ = 498 NOTE: ANSWER ONLY, FULL MARKS v 498 @
12° | o, =119,47 v Vanswer/antwoord
@)
1.3 ( -o,;X+0,)
(498-119,47 ; 498 +119,47) Y 378,53
(378,53 ; 617,47) Y 617,47
«5 distances /afstande v 5
Q)
1.4.1 | new average/nuwe gemiddeld : Y k=505
- 5555
b=—= lly = 5555-5478) y 4
> 77
k = 505 OR/ OF = ir OR/OF
the value of y/ die waardevany : Y Ily=5555 — 5478
=k —¥ = 505-498 =7cm =7om
y y Y y=7
; (2)
1.4.2 | 119,47 vv correct answer /
korrekte antwoord
(2)
[11]

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Mathematics P2/Wiskunde V2

3 NW/September 2018
NSC/NSS - Memorandum

QUESTION/VRAAG 2
2.1 puathx Vv a=57,87
a = 57,87 and/en b = 0,05 ¥ b=0,05
a Y equation/
“. D = 57,87 +0,05x vergelyking
(3)
2.2 r= 0,93 vv 0,93
(2)
2.3 Strong positive correlation / Sterk positiewe korrelasie ¥ Strong / sterk
¥ Positive / positief
(2)
2.4 Jy = 57,87+0,05(465) Y — substitute 465
Dp =81,12% into eq. / vervang 465
in vgl.
VY 81,12 (2)
OR/ OF
J =80,81% (calculator / sakrekenaar) VV 80,81% / 81%
= 81%
[9]

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Mathematics P2/Wiskunde V2 4 NW/September 2018
NSC/NSS — Memorandum
QUESTION/VRAAG 3 y
B
x
E(-11 ; 0) F AC3;0) 9
3.1 Length of AD / Lengtevan AD: v
—_——— 2
dy = y(-3+8) +(0-5) dip = y(-3+8) +(0-5)°
dp = V'50 Y dyy =5V2 or/ of 7,07
diy = 52 or / of 7,07 units / eenhede
Length of DE / Lengte vanDE:
doy = \(-8 +11) +(5-0)°
v =
dy, = 34 or / of 5,83 units / eenhede dog = V34 ot / of 5,83
VY AE =8
Length of AE / Lengte van AE = 8units / eenhede
Perimeter / Omtrek A ADE :
= 5/2 +3448 =20,9units / eenhede v 2 aVi48 4g
or / of 7,07+5,83+8= 20,90 ovlof 7,07 + 5, 6)
3.2.1 | Line from mdpt 1* side of A, parallel to 2™ side, bisects | “reason/rede
3" side / Lyn uit midpt van een sy van A, ewewydig aan
2” sy, halveer 3® sy, qd)
3.2.2 —11+(-8) 0+5 ¥  method/metode
oD Yom 19
and/en Yo xg=-—
9 oy 25 2
“XG 3 “Ng 3 v ye 5
2 2
(3)
3.2.3 5/2 ¥  answer/antwoord

FG =—AD= — orf of 3,54 units / eenhede

L
2

(1)

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Mathematics P2/Wiskunde V2

5
NSC/NSS — Memorandum

NW/September 2018

3.2.4 0-5

My =>
—3-(-8)
a>

5

=-1

Meg = Mgp [AD| FG]

Through the point / Deur die punt (-4 |

Y Me, = Myy [AD| FG]
Y Mg = ~1

¥ Subst./Vervang

(-2:3)
2°2

=~, 19,5
» 2-2

ys-x-7 Vy=-x-7

: (4)
33 | tanDAO = m,,

tan DAO = ~1 ¥ tanDAO = -1
. DAO = - 45°+180°
DAO = 135° Y DAO =135°

2)

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Mathematics P2/Wiskunde V2 6 NW/September 2018
NSC/NSS — Memorandum
3.4 DAB = 75° [opp.Zs eyelic quad./ teenoorst.Zekvh| ¥ DAB = 75°
OAB = 135°-75° yok
= 60° OAB = 60
tan OAB = May
tan 60° = m,, = V3 Y tty =v3
equation / vergelyking AB through / deur (-3;0):
y-0=43 (x+3)
y = v3x43V3 vy = V3x4+3V3
at / by B: v
SAS, E58 = V3x43V3 ABSA, Mes3 = Vie 3V3
“Bah, ~ Jax =3V3 — esV8
(4 8) _ sp 228
Vez
x=2 x=2
and/en
oy =¥3(2)+3V3 OR/OF y= SEAS (HSB
ny = 53 V y= 53
. (7)
£B(2:5V5) 3]
QUESTION/VRAAG 4
4.1 v4yer?
¥ subst. into eq./vervang
vty? =(23) in vgl.
x+y? =12 ¥ answer/antwoord
Q)
421 | 3? —6x4(-3P +y? -2y4(—1)? = 6+(-3" +(-1) Y (x-3P +(y-1)
: = Y 16
.(x-3P +(y-1? = 16
. (2)
4.2.2 |M@G;1) are
v =]
(2)
4.2.3 | 4 v answer/antwoord
(1)

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Mathematics P2/Wiskunde V2 7 NW/September 2018
NSC/NSS — Memorandum
4.2.4
Q&3y) YMQLPQ
dyp = (6-3) +(-2-1)° v
dyp = V18 dyp = V18 or / of 3N2
dup = 3/2
Pythagoras :
PQ? = 3V2)?-4? ¥ PQ? = (3V2)?~4?
PQ’ =18-16

“PQ= V2 units / eenhede

¥PQ = V2
(4)

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Mathematics P2/Wiskunde V2 8 NW/September 2018
NSC/NSS — Memorandum
43 y
B(2k; 10)
20 95
C} —;—
13°13
. .
ACK; 5)
\ x,
43.1 | AB =6+7=13 V AB=6+7=13

dy, = «f(2k-(-K)) +(10-5)

13° = (3k) +(5)’

169 = 9k? +25

144 = 9k?

16 =k?

k=2+4

k= 4 (seediagram / sien diagram)
“.A(~4;5) and / en B(8;10)
Gradient of AB /

Gradiént van AB

¥ 13? = Bk) +(5)P

Vi? =16
Vk=4
v .
10-5 5 . :
aa tary Correct subst. into gradient /

vervang korrek in gradiént

(5)

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Mathematics P2/Wiskunde V2 NW/September 2018
NSC/NSS — Memorandum
5 5
= Y¥M,,=—
Mas = 35 @ 3k
95
B™ 95
Mc 20 B 5
+k YM ys = 3
13 4k
13
_ 95-65
20+13k
5 30 v = - 95 —65
3k 204+13k 3k = 20+13k
100 + 65k = 90k
100 = 25k
k=4 Vk=a
v
iM. => a> Correct subst. into gradient /
@3(4) 12 vervang korrek in gradiént

G)

4.3.2

radius | tangent / radius Lraaklyn
o Meangent/raaklyn --2
5
Equation of common tangent /
Vergelyking van gemeenskaplike raaklyn

95 12 *)
Pa t4{ , 48
13. °=«5 13

12
“ysz-sxtll
y 5

12

Y Menges /rcakiyn = 7

Y Subst. / vervang (32)

G)
[19]

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Mathematics P2/Wiskunde V2 10 NW/September 2018
NSC/NSS — Memorandum
QUESTION/VRAAG 5
5.1.1. | sin106°
= sin (180°-74°)
= sin 74°
= cosl6° ¥ cosl6°
_ vk vik (2)
=o 7
S12 | 5 \e-k make use of diagram/
sinl6° = ; maak van diagram gebruik
y Pk vv ye ~k
OR/OF 6 | |

sin?16°+cos? 16° = 1

v Vek

t

2
OR/OF
sorrers[) =1
k ¥ sin? 16°+cos?16° =1
in? 16° = 1-— k
si a v 5
: ek
216° = a
sin* 16 fa v ‘ 6)
2
sinl6° = rok
t
5.1.3. | cos16° = cos 2(8°) = 2.cos?(8°)—1 ¥ cos2(8°)
vk = 2c0s?(8°)—1 v
cos16° = 2cos?(8°)~1
wea = 20s? (8°) v
VE 1 vk = 2cos?(8°)~1
op ty = 8B) t
cos8? = vk 1 = Vie +t ve 1 or/of
22° V2 V2 2

Vk +t
2t

(4)

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Mathematics P2/Wiskunde V2 11 NW/September 2018
NSC/NSS — Memorandum
5.2 LHS/ LK
yA(L= cos @)(1+ cos) (1—cos6)(1+ cos 6)
sin20 v ;
=1~-cos’é
_ /4(1—cos? 6)
~~ Qsin 8 cosO
¥  1-cos* =sin? 6
+2
= 2ysin’ VY sin26 =2sin@cosd
2sin@cosO
2sin@cos@ 2sin 0 cosO
4
=| _pns/rx 4)
cos@
5.3.1 | sin p+V3cosp =1
1. 3B 1
1 gin +33. cos =i (+2) Yo sin p +508 p “2
ger PET 2
cos 60°sin p+ sin 60° cos p = 1 (special angles / spesiale hoeke) ¥ cos 60°sin p
1 2 ¥ sin 60° cos p
sin (60°+ p) = 3 (compound angles / saamgestelde hoeke) (3)
532; 1 v
sin (60°+ p) = > ref.Z/verw.Z = 30°
ref.Z/verw.Z = 30° v 30°4m360°/
60°+ p =30°+n,360° or/of 60°+p = (180°-30°)+n.360° | 2 = 330 oe eo
p= ern,
= - oO hea = Oo Oo
p= —30°+n,360° or/of p=90°+n360° (neZ) Y p= 90°+n360°
Y (neZ)if used in correct
context / as dit in korrekte
konteks gebruik is
(4)
5.4 cos 0° + cos1°+cos 2° +... + cos(180 — 2)° + cos(180~1)° + cos 180° + 2 ¥ cos(180—2)° + cos(180~1)°

1+cos1°+cos2°...+ (—cos2°) +(—cos1°)+ (1) +2
1+(-)+2
=2

v

...(— cos 2°) +(~cos1°)
v2 6)
[23]

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12

Mathematics P2/Wiskunde V2

QUESTION/VRAAG 6

NSC/NSS ~ Memorandum

NW/September 2018

VV p =30°
6 p = 30° and/en g =1 pP *
2 V¥qa-4
2
(4)
6.2 x =—150° or/of x = 30° v¥~150°
v¥ 30°
(2)
6.3 The graph of f must be translated 60° to the left / Die grafiek | YV60° to the lef/na
van f moet 60° na links getransleer word. links (2)

The graph of f must be translated 120° to the right and then be
reflected about the x-axis/ Die grafiek van f moet 120° na regs
getransleer word en dan in die x-as gereflekteer word.

OR/OF

v¥ 120° right/regs
v — reflect about x-
axis/reflekteer in die x-as.

{8}

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Mathematics P2/Wiskunde V2

13 NW/September 2018

NSC/NSS— Memorandum

‘D
QUESTION/VRAAG 7 E
1
1
1
H
1
1
'
tT
i
1
A ,
71 | B=90°
BC x
cosa = — = —
AC AC
x
“ AC =
COS a @)
72 | ECA =90°-6 YECA = 90°-6
AEC = 180° - 6-(90°—6) ¥ AEC = 180°~8-(90°-6)
AEC = 90°- 8 +0 or/of 90°-(B-0) y AEC = 90°-8+6 or/ of 90°-(B-8)
or/ of 90°+(6-f) or/ of 90°+(8—-f)
(3)
73 InA ABC
AC =— (see / sien 7.1)
cos@
InAACE
. ares Y Correct application of sine-rule/
sing = SMAEC korrekte toepassing van sinus-reél
EC x
OS a
. * _xsin B
xsin B VEC = OS ____
= COS sin[90° -(2-6)]
sin[90°-(f-6)]
xsin B
= COS a
cos(—A) ¥ sin[90°—(2 —0)] = cos(8—-@)
_ xsin B
cosa.cos(f—6) G3)

[8]

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Mathematics P2/Wiskunde V2 14 NW/September 2018
NSC/NSS — Memorandum
GEOMETRY/MEETKUNDE
Ss A mark for a correct statement

(A statement mark is independent of a reason.)

‘n Punt vir ‘n korrekte bewering
(‘n Punt vir ‘n bewering is onafhanklik van die rede. )

R A mark for a correct reason
(A reason mark may only be awarded if the statement is correct.)

‘n Punt vir ‘n korrekte rede

(‘n Punt word slegs vir die rede toegeken as die bewering korrek is.)

S/R Award a mark if the statement AND reason are both correct.
Ken ‘n punt toe as beide die bewering EN rede korrek is.
QUESTION/VRAAG 8
8.1
8.1 6, = G +A [ext.Z of A/buite Z van Al YSIR
But /maar G =A [Zs opp. equal radii / Ze teenoor gelykeradiusse] | VS/R
“0, = 2C, VS
In the same manner / op dieselde wyse :
6, =2¢, vs
6, +6, = 26, + 2,
.. AOB = 2¢, +2,
= 2( C+ ¢,) vS
= 2ACB (5)

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Mathematics P2/Wiskunde V2 15 NW/September 2018
NSC/NSS — Memorandum

8.2

8.2.1 | D, = 65° [Zat centre = 2x Zat circumference / midpt.Z = 2x omtreksZ] | ¥SYR (2)
8.2.2 B, = 65° [tangent chord theorem /raaklyn — koordstelling | YSYR Q)
8.2.3 | BAD = 65° [alt Zs/verwiss Ze; AD / /EF] yYSR (1)
8.2.4 | 6 =115° loppZs of cyclic quad / teenoorst.Ze van kvh] YSYR (2)
8.2.5 | GBF = 90° [rad 1 tangent / rad  raaklyn| VSIR
DBF = 65° [alts / verwiss Ze; AD | /EF] YSIR
B, = 90°-65° = 25° vs
OR/OF
_ B, + B, =90° [rad 1 tangent /rad L raaklyn] VSR
A v
G, = 90° [alts /verwiss Ze; AD //EF | eR
B, = 180°— 65°-90° = 25°[ Zs of A/binne Ze van A) @)
8.2.6 | G, = 90° [co-interior Zs / ko - binneZe; AD | /EF]/ [Proved in 8.2.5] ¥ en
AG = GD [line from centre | chord; Loodlyn vanuit midpt sirkel na koord ]
.op=¥7 vi
4 4
3)
[18]

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Mathematics P2/Wiskunde V2 16
NSC/NSS — Memorandum

QUESTION/VRAAG 9

Let/Laat é, =X,

NW/September 2018

911 G = A, =x [tangent chord theorem /raaklyn — koordstelling | VSvR
@uhe equal circles, equal chords, equal “s |
/gelyke sirkels, gelyke koorde, gelykeZe YSYR
, =C, (4)
92 1) C40, =A, +A, [ext. Zs of cyclic quad / buiteZe van kvh] vSYR
But/Maar C, =A, = x [in9.1] “Ss
AGA,
OR/OF
E, =180°-2x [sum of sof A /somvanbinneZevan A] “SIR
B =180°-(1 80°—2x) =2x [opps of cyclicquad / teenoorst.Ze vankvh] | YS/R
6,+6,4+6, = B+A, [ext.Zof A / buiteZ van A]
But / Maar 6, +¢, =x+x = 2x=B YS
u, =A, (3)

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Mathematics P2/Wiskunde V2 “417 NW/September 2018
NSC/NSS —- Memorandum

9316 =A, [see/ sien 9.1]
“CE = AE [sides opp. = Zs/sye teenoor = Ze| VSR
A, =D [tan-chord theorem / raaklyn — koord — stelling] VSR
A, =¢, [from / van 9.2]
ude,
“.CE=DE [sides opp. = Zs/sye teenoor = Ze| YSIR
CE = AE =DE “8
. Eis the centre of the circle/is die middelpunt van die sirkel (4)
9.4 | AE = EC = ED [radii / radiusse] vs
But / Maar CD = AE [Given / gegee]
“. CD = DE = CE vs
~ ..AECD is equilateral / gelyksydig (13)

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QUESTION/VRAAG 10

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Mathematics P2/Wiskunde V2 18
NSC/NSS — Memorandum

NW/September 2018

R

10.1 Parallel to the third side of the A/ Ewewydig aan die derde sy van | YR qd)

die A
10.2.1 | R = 90° [tangent radius /raaklyn Lradius| YSVYR

Q, =90° [tangent _L radius /raaklyn Lradius] YS/R

Q, =R =90°

. PQ//SR [corresponding /s are equal / ooreenk. Ze is gelyk] YR

OR/ OF ‘

R =90° [tangent 1 radius /raaklyn Lradius| VYSYR

Q, = 90° [tangent L radius /raaklyn Lradius| YSIR

“R+Q, =180°

“.PQ{|SR [co-interior Zs = 180°; ko —binneZe saam180°] vR (4)
10.2.2

= = oR [prop.theorem; PQ |[SR /eweredigheidst;PQ||SR] | vg vp
Tp = FQPS
QR

PC = BP and/en SC = SR [radii / radiusse] vS/R

PS = PC+CS = BP+SR vs

TP = LQBP+SR) (4)

QR

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Mathematics P2/Wiskunde V2 19

NW/September 2018
NSC/NSS — Memorandum

10.2.3 | InATQP and/en ATRS
A on vw
T =T [common/ gemeenskaplik] s
Q, =R [corresp.2s / coreenk.Ze;PQ || SR] vs
B =§ [sum of Zs A/som van Ze ny vs
- ATQP ||| ATRS
OR/ OF
T=T [common/ gemeenskaplik | vs
Q, =R or /of 8, =§, [corresp.Zs / ooreenk.Ze;PS || QR] vS
- ATQPIATRS [2,4,2] VR

(3)
10.2.4 | TS’-SR? = TR? [Pythagoras] vs
~ CS= SR [radii / radiusse]
.. TS? -CS? = TR?
TQ_ QP LT Ihhas] VSIR
TR RS_ TS
rR = QRS vs
QP
But / Maar QP = BP and/enRS=CS [radii/radiusse]
“TR= TQ.CS
BP

TQ? = TP? +PQ? -2TP.PQ.cosP,
But/Maar PQ = BPand/en §, = P,
TQ = (TP? +BP? —2TP.BP cosS)
_ TACs?

TR? -
BP

TO’Cs?
BP?

. ee - (te? +BP? —2TP.BP cosS) CS

BP

“. TS? -CS? =

Vuse cosine-rule
to find TQ/
Gebruik cosinus-
reél om TQ te vind

2 2
/ TQS
BP?
v
Ts cs? = 1YCs"
BP*
(6)
[18]

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- [2018 GRADE 12 TRIAL EXAM P2 FS .pdf](https://www.hlayiso.com/library/2018-grade-12-trial-exam-p2-fs-pdf/f06a5592-1e1b-494c-b6ce-40234267eaea) — Mathematics · QP · 2018
- [2018 GRADE 12 MATH TRIAL EXAM PAPER 2 MP](https://www.hlayiso.com/library/2018-grade-12-math-trial-exam-paper-2-mp/75be9d76-1568-4e33-96ae-c09ea34accee) — Mathematics · QP · 2018
- [Mathematics P2 May June 2018 Eng](https://www.hlayiso.com/library/mathematics-p2-may-june-2018-eng/4f3d38e9-a707-40ca-beee-fea324344ead) — Mathematics · QP · 2018 · Eng

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## Collection

- [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers)

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