# Mathematics Grade 12 Paper 2 North West Mock 2016 Memo

> Grade 12 Mathematics Paper 2 memorandum from the North West Mock Exam 2016, 20 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/2016-grade-12-math-trial-exam-paper-2-nw-memo/27b1ef31-88b2-4ae5-989b-3803275ad4dc
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2016
- Exam period: North West Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 20

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## About this document

Mathematics · Grade 12 · North West Mock Exam · 2016. Memorandum, 20 pages. Read online or download the PDF.

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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

MARKS: 150
PUNTE: 150

This memorandum consists of 17 pages.
Hierdie memorandum bestaan uit 17 bladsye.

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Mathematics/P2/Wiskunde/V2 2 NW/September 2016
NSC/NSS — Memorandum

QUESTION 1/ VRAAG 1

11 y= 12,01 + 0,88x Y value of a/waarde

van a

v value of b/waarde
van b

Y equation /
vergelyking (3)

1.2 y= 12,01 +0,88(46) v sub. 46 into the
= 52% equation / vervang 46
in die vergelyking
v answer / antwoord
(2)
13 No, the preparatory exam mark is the independent v answer / antwoord
variable. Hence we cannot determine the prep. exam v reason /rede
marks using the final exam./Nee,die voorbereidende (2)
eksamenpunt is die onafhanklike veranderlike. Dus kan
ons nie die voorbereidende eksamenpunt met behulp
van die finale eksmenpunt bepaal nie.
14 x =60,58 Y x = 60,58
y = 65,33 Vv y =65,33
LHS/ ILK = y = 65,33 ¥ sub. into RHS /
RHS/RK = 12,01 + 0,88(60,58) = 65,32 vervang in RK
LHS/ILK = RHS / RK
(x ; y) lies on the regression line v LHS = RHS and
conclusion / LK = RK
en gevolgtrekking
(4)
1.5 r = 0,98 Y value of r/ waarde
van r (1)
1.6 There is a very strong positive correlation between ¥ very strong / baie

prep. marks and final marks./
Daar is’n sterk positiewe korrelasie tussen die
voorbereidende punte en die finale punte.

sterk
Y positive / positief
(2)
[14]

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Mathematics/P2/Wiskunde/V2 3 NW/September 2016
NSC/NSS — Memorandum
QUESTION 2 / VRAAG 2
21|/a=6 Y answer /
antwoord
()
22
Expenditure (in rand) Frequency Cumulative
' frequency
50 < x <100 24 24
100 < x <150 ae 76
150 < x <200 14 90
v 76; 90
200 < 25
x <250 6 96 ¥96:100
250 < x <300 4 100 2)
231] 120
3
sg
2 100
8
&
g
= 804
3
3
3
S69
SI
Fs
x 40
&
£
3 20
|
=|
0 it
0 50 100 150 200 250 300 350
Expenditure/Uitgawes
¥ Correct points Y Shape V Grounding /¥ Korrekte punte ~Vorm “grondvlak @)
2.4 | 100 < x <150 ¥ answer / antwoord
()
(7]

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Mathematics/P2/Wiskunde/V2 4
NSC/NSS — Memorandum

QUESTION 3 / VRAAG 3

NW/September 2016

y 7
A(455)
WA, —>
x
BC-3 ; -2

D

C(6 ; -5)
31) Bc = Yl—x) +O.-9)

J(6+3)? + (-5+2)

90 or/of 3V10 or /of 9,49

I

¥ sub. into the distance
formula / Vervang in
die afstandformule

¥ answer / antwoord

(2)

32 _
Msc peTerae

—5+2
64+3
1

3
+5 = eG) or/ of yr2= Le 43)
2 3 3

y= -$x-3 or 3y=-x-9

Y sub. into gradient
formula / vervang in

die gradiént formule
gos
3

v equation / vergelyking
@)

ee

¥ Map = 3

¥ sub. of the point /
vervanging van die punt

¥ equation / vergelyking
GB)

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Mathematics/P2/Wiskunde/V2 5

NSC/NSS —- Memorandum

NW/September 2016

3.4 1
a . Y equating the two
equations / stel die
_ twee vergelykings gelyk
ee aan mekaar
-10x = -12
6
x=
5
6 v x- value / x-waarde
y = 3(2)-7
5
ul
a :
6 17 y-value / y-waarde
DG : -=) @)
3.5 _ 5+2
AB = ———
4
+3 V map = 1
=1
¥ tana = 1
tana = 1
Ya = 45°
a = 45°
tanB = 3
B =71,57° ¥B =71,57°
BAD =71,57°—45°
= 26,57° ¥ answer / antwoord
GS)
3.6 | Equation of a line AE || BC/ vergelyking van AE || BC

1
—~5=-—(x-4
y 3% )

3y-15 = -x+4
AE: 3y +x =19
x-intercept/x-afsnit is 3(0) +x =19

x =19
E (19: 0)

1
Y sub. of 2 and point
into the equation /
7 1
vervanging van — 3 en

die punt in die
vergelyking

Y equation of AE/
vergelyking van AE

v x-intercepts / x-afsnit
Yanswer / antwoord

(4)
[20]

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Mathematics/P2/Wiskunde/V2

6
NSC/NSS — Memorandum

NW/September 2016

QUESTION 4 / VRAAG 4 ys
NQ;
R
»M(4 ; 2)
oO
41 («-4" +(y-2" =P ¥ sub. of M into equation
of a circle / vervang M
(0-47 +(0-2P =r in die vergelyking van
die sirkel
0~=r vsub. of O(0; 0) /
vervang (0;0)
2 2 Y value of r’/ waarde
(x-4)°+(y-2)° = 20 ae 3)
4.2 (x-4) + (y—-2)" = 20
Subst/Verv(2 ; y) ¥ sub of N(2; y)/
2 _4y +(y- 2 = 20 vervang N(2 ; y)
4+y-4y+4=20
vy —4y-12 =0
y- 4y-12 =0
Y(- 6) +2) = 0
(y-6)(y +2) = 0
vy =6 4)
y =6 or/of y =-2 N/A
OR / OF
OR/OF
(x-4) +(y-2)° = 20
¥ sub of N(2; y)/
Subst/Verv(2 ; y) vervang N(2 ; y)
(2-4) +(y-2) = 20 2
Vv (y-2) =16
(y-2) =16
Voy, 2 = 4
y-2 =+4
y = 6 ot/of y =—2 N/A/INVT vy =6 (4)

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Mathematics/P2/Wiskunde/V2 u

NSC/NSS — Memorandum

NW/September 2016

43 2 tl 1
Moy “42 ¥ Mou a5
Mog =—2 Y Mop = —2
Equation of OR is /Vergelyking van OR
ya-2 Vy =-2x @)
44 -2 Ym, = —-2
m, —— =_ 2) ‘MN
MN 2-4
Myr =; Y Myx =>
7 =+(-2)
ee 2 Vv y-6= 1-2)
2y-12 = x-2 2

2(-2x) - x- 10 =0

-5x = 10
x=-2
y=2(2)=4

RC 2; 4)

OR

RO? = NR?

vty = (x-2P + (W-6"
ty =x -4x4+44 y? —12y +36
4x +12y =40

x+3y= 10 and y =-2x

x +3(-2x)= 10
x-6x= 10
-5x= 10
x=-2
y=-2(-2) =4 RC 2;4)

v 2(-2x) - x- 10 =0

(6)

Ve +y = (x29

+ (0-67
v4x +12y =40
Y x+3y= 10

¥ x +3(-2x)= 10

v
x=-2
vy=4 ©

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Mathematics/P2/Wiskunde/V2 8 NW/September 2016
NSC/NSS — Memorandum
4.5 MNRO is a kite because/vlieér omdat OR =RN and /en ¥ Kite/Vlieér
MN = OM ¥ adjacent sides equal
/aangr.sye gelyk
2)
[18]
QUESTION 5 / VRAAG 5
5.1 ‘o Oy) ein 2(Q(0 _
cos(180°+ x) TELL x).sin“ (90° — x) Aghbe:
sin(180° —x)
(—cos x).(—tan x).cos? x Aer Y¥—cos x
sin x v— tanx
¥cos*x
sinx 4 vsin x
cos x. COS” X
= snk +sin’ x
sinx
= cos’ x + sin? x ¥ cos’ x + sin? x
=1 Yanswer/antwoord (6)
5.2.1 | cos(A — B) — cos(A + B) ¥ expansion of
cos(A-B) /
= cos A cos B+ sin A sin B — [cos A cos B — sin A sin B] | uitbreiding van
cos(A-B)
= cos A cos B + sin A sin B—cos A cos B + sin A sin B | Yexpansion of
cos(A+B) /
= 2sin A sinB uitbreiding van
cos(A+B)
¥ simplification /
vereenvoudig (3)
22 cos15° — cos 75° = cos(45° — 30°) — cos(45° + 30°) v 45° — 30° and /en
: 45° + 30°
= 2sin 45°.sin 30° v 2sin 45°.sin 30
_ axr2yt or/of 2x-Lyb yy Lyd
272 v2 2 yO ap Db
V2 1 ¥ answer / antwoord
=— oof —~=
2 v2
(4)

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Mathematics/P2/Wiskunde/V2 9

NSC/NSS — Memorandum

NW/September 2016

OR

cos15° — cos 75°
= cos(45° — 30°) — cos(45° + 30°)
= cos 45°cos30°+sin45°sin 30°— [cos 45°cos 30°

sin45°sin30°]
= 2sin45° sin 30°
= axd2 yh or/of 2x-Lxt
D V2 2
v2 1
=— orof —
2 oT

v 45° — 30° and /en
45° + 30°

v 2sin 45°.sin 30

ya2 1b 1d

Me yx

22 2

¥ answer / antwoord

4)

5.3

AB? = (cos 0-6)" (sin0-7)
86 = cos’0 — 12cos 0+ 36 +sin’0 — 14sin 6 +49
86 = 1+ 36+ 49 - 12cos 0 -14sin 6
0 = - 12cos0 -14sin 0

14sin8 = — 12cos 0

sin® _ -12

cosO 14

tan® =-©/.0,86
7

Y sub. into the
distance formula /

vervang in die

afstandformule

¥ simplification /

vereenvoudig

v14sin0 = —12cos 0

¥ tand =-$/0,86 (4)
a7)

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NW/September 2016

10
NSC/NSS — Memorandum

Mathematics/P2/Wiskunde/V2

QUESTION 6 / VRAAG 6

= Tae ae Cat
Q eS @ ial
» o
g
’ AR fh ~7
ToincccpessqeceneeesPooe —————— . leis §
\ 4 H ' A H a] S|] s8 ~
1 1 ' ' 1 =o ° lo
1 4 t q d ' ' ww S I~ |~].- ¢
' 7 t 1 ' ' 7 ' S isa wala fala” WO &
H 0 ' 0 5 0 4 ' of SCT O/DlOonm gs
1 1 1 1 ' 1 ' " 4 B/Blo-
H 4 H H i H ' " i ~..8
Suge - eee ba------- La------ de------ phe ee eee ee Fee — " Qala se
SD H H H fe) t H ! t a RR x X88] 1 oO 8
'
H H : i H H ! H . 8 oS > op hs fh >
: D ' p ' ' : 1 2 a
' ; : ‘ 1 ' ! ' 5s 5
! H ‘ i t H 3 1 a ¢
ee ee ee Sey ae ee een ee ee Deere eres || on ek
r iH g : ' ' sag
H ' ' ' ' i Bo
H 4 H H t H ' H Sox
' : \ ' ' ' ! ' 32s
4 ' ' a) eS
wn-deneee--- b--------f------- N.----- ha------ Je------- L------- Ae fee Ege
' H H 1 ao
' 1 1 1 mn > 2°
1 ' 1 ete ha}
H ! ’ : Ea |
' i t H sas 5 7
a a eos Kay
= 4 : S wa — R
4 : 2 ot ah v7
' j sas °
1 1 3 oO 2.
eas : > 2
a a DD» COUR, CS eee as S al
1 ' a on a
a= a iS} ay
1 q £8 e r {| j= 3S
' ; 2 7.
f oS isa} no {Ml Ss 8
ee Aco = R bal
1 H £8 " i] °
' H go R > Isl Io &
' i iS) iS} +
' ‘ Sw — = > wm |
' ' So fs} ke 6 +
ce
i q ans) 5 iS) | Vv
2 ° 5°
prc te cence en penne ne | ie hel ~ Yay onl roy R
! ! ee t Shel ie
' 4 ion I par 7 can) Vv
i H r °
' ' 8 " jes 2
US le x |uj_|y as
SoooosS beee-f os a R O77
gl NN jes a
ee N N |Qais
Ne} Neo} So jololo

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Mathematics/P2/Wiskunde/V2 ll

NW/September 2016

NSC/NSS — Memorandum

QUESTION 7 / VRAAG 7

A AC’ = 10? + 6? — 2(10)(6) cos100° ¥ sub. into cosine rule /
vervang in die kosinus-

= 156,8377813 reél

AC = 12,5 units/eenhede Y 156,8377813
B = 80°( opp angles of a cyclic ¥ 12,5 units / eenhede
quad/teenoorst<e van koordev ) A
vB = 80°
In AABC, ue = ae
sin40° sinB
BC 12,5 ¥ sub. into sine rule /
, ea vervang in die sinus-
sin40  sin80 reél
pow l2oxsin 4d
sin 80°
Apne Y answer / antwoord
= 8,2 units (6)
1 1 re)
72 Area of AABC = Fey Y 60°
1 : ¥ sub. into the area

= 298.2) sin 60° tule / vervang in die

= 44,4 square units

opperviakte-reél

v answer / antwoord
G3)
[9]

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Mathematics/P2/Wiskunde/V2 12

NSC/NSS — Memorandum

NW/September 2016

QUESTION 8 / VRAAG 8
8.1.1 | 90° v ans /antwoord
qd)
8.1.2 | Angle in the alternate segment ¥ ans / antwoord
Die hoek in die teenoorstaande segment dd)
8.2 5 A
o
D
oe
x . VS VR
8.2.1 | Bs =E=x (tan chord theorem / hoek tussen raaklyn en
koord) YS YR
B, =A=x ( corresponding angles / vS VR (6)
ooreenkomstige hoeke)
B, =E=x (radii / radiusse OE = OB)
A A VS YR
8.2.2 | B2 +B3 =90° (subtended by a diameter / onderspan
deur middellyn) | V ans /antwoord
CBE = 90°+x (3)
8.2.3 | In ADBE,
An = = (line | to one side of a A/ lyn | aaneen |VS VR
syvanA )
But/maar ee) =1 (radii / radiusse) vs
OD
EF = FB v EF = FB (4)

F is the midpoint of EB / F is die middelpunt van EB

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Mathematics/P2/Wiskunde/V2 13 NW/September 2016
NSC/NSS — Memorandum
OR
In A EOF and ABOF
E =B2 (Proven in 8.2.1/Bewys in 8.2.1)) A OK
VE =B2
EO = OB (radii/radiusse)
Di = B3 (Zs opp = sides/ Z e teenoor gelyke sye) A i
A A ¥D1 =B3
Di = 03 (corresp Z s/ooreenkomstige Ze BD | AO) A a
A A ¥Di = 03
-. B3 = O03
“. B3 =O2 (alt Z s/verwisselende Ze BD | AO)
“03 =O2
AEOF = ABOF (AAS/HHS) YAEOF = ABOF
(AAS/HHS)
EF = FB (4)
8.2.4 OF 1 EB (line from centre to a midpoint / lyn uit v SIR
middelpunt van sirkel na middelpunt van koord) g
EF = 4 (Fis the midpoint / F is die middelpunt) | ¥ EF = 4
OE’ = OF +EF
= 3 442
= 25
VYOE =5
OE =5
Y ans / antwoord
ED = 10cm
OR/OF OR/ OF
F 3 = 90° (corresponding angles / v S/R
ooreenkomstige hoeke)
EF =4 (Fis the mid point / F is die VEF =4
middelpunt)
OE’ = OF +EF
= 3° +4?
= 25
VYOE =5
OE =5
Vans/antwoord (4)
ED = 10cm

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Mathematics/P2/Wiskunde/V2 14 NW/September 2016
NSC/NSS — Memorandum

OR/ OF OR/ OF
OF = spp (midpoint theorem/middelpunt 1
2 ¥Y OF = —DB
formule ) 2
v DB = 6
DB = 6cm
In AEDB, Y Applicati f
ED’ = 6° +8” (Pythagoras thm/ Pythagoras 6 Ppiication 0
ythagoras
formule) thm/Toepassing van
formule
= 100
Y ans/antwoord (4)
ED =10
[19]
QUESTION 9 /VRAAG 9

A A

91 A =C (tan CD and chord CE / raaklyn CD YA = Co
en koord CE) | “reason / rede

= 3 (tan AEH and chord ED / raaklyn AEH | ¥ C = E3

en koord ED) | V reason / rede
But they are corresponding angles / Maar hulle is
ooreenkomstige hoeke
AB ||ED

¥ corresponding angles /
ooreenkomstige hoeke

GS)

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Mathematics/P2/Wiskunde/V2 1S

NSC/NSS — Memorandum

NW/September 2016

OR/OF

A+ + Ei =180° (sum of angles in a A/som van

die binne Z° vanA )

Cc stE2 +Di =180° (Sum of angles ina A/ som van

OR/OF
VY A+Ci +E: = 180° and

C +E +Di = 180°

die binne Z° vanA ) VA =C
v reason / rede
But A=C2 (tan CD and chord CE/ raaklyn cD | ¥ Ei = Ds
en koord CE)
E.=Di (tan AEH and chord CE/ Y altangles /
raaklyn AEH en koord CE) | verwisselende hoeke
but they are alt.angles/ Maar hulle is verwisselende
hoeke
AB||ED
9.2 ACDE is a parallelogram because one pair of opposite | V answer / antwoord
sides (AC and ED) are equal and parallel / ACDE is ‘n_ | Vreason / rede
parallelogram omdat een paar teenoorstaande sye (AC (2)
en ED) gelyk en ewewydig is
9.3 In AABH,
bce ( proportionality thm or AH || CD/ | ¥ AC _ HD
CB DB CB DB
eweredigheidstelling of AH | CD ) | “reason / rede
HE = HD (proportionality thm or AB | ED/ HE = BD
EA DB EA DB
eweredigheidstelling AB | ED) | “reason / rede
AC _ HE “
CB EA [11]

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Mathematics/P2/Wiskunde/V2 16 NW/September 2016
NSC/NSS — Memorandum
QUESTION 10
E F
B 2
10.1 Const: On AB ,mark off AP = DE and on AC, mark ¥ Construction/

off AQ = DF/ Op AB, merk op AP = DE en
op AC, mark of AQ = DF.
Proof/Bewys: In AAPQ and A DEF:

AP = DE (const/konstr)
AQ = DF (const/ konstr)
A=D ( given/ gegee)
AAPQ = ADEF (SAS/SHS )
Pi =B ( = B)
PQ || BC (corresp. angles =/
ooreenkomstige hoeke)
AB = AC (line // oneside of a A/lyn// aan een sy van A
AP AQ
AB _ AC (AP = DE and AQ = DF)
DE DF

konstruksie

v
AAPQ = ADEF (SAS)
(SHS)

VP =E

v PL =B

¥ PQ || BC

y AB_AC
AP AQ

Vine // to one side of a
triangle / lyn // aan een
sy van ‘n driehoek

(7)

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Mathematics/P2/Wiskunde/V2 17

NSC/NSS — Memorandum

NW/September 2016

10.2.1 Q =R2 (tan-chord theorem/ tan-koord) VYSYR
=S2 (alt angles QS//RE/verwisselende) vS
QR = RS (side opp. of equal angles/ sye VR (4)
teenoor gelyke hoeke)
10.2.2 | In ARST and APQT
=T (common/gemeenskaplik) ¥T=T
R2+R3=P (ext. angle of a c.q PQRS/buite ¥ R2+R3=P
hoek van kdvh PQRS ) | VReason/Rede
$3 =Q,+Q; (ext. angle of c.q or 3“ angle inA/ v3" angle or
buite hoek van kdvh PQORS ) Reason /3° hoek of rede
(4)

ARST || APQT (AAA/ HHH)

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Mathematics/P2/Wiskunde/V2

18 NW/September 2016

NSC/NSS — Memorandum

v
10.2.3 | RS _ ST _ RT (ARST || APOT) R(ARST ||| A PQT)
PQ QT PT
RS _ RT eee
PQ” PP wrest (1) PQ PT
In AQST, QS || RE
_SE_ QR
“ET RT (line drawn parallel to one side of a A/ SE_ QR
lyn parallel sy van ‘nA) | ¥ 2. —= = ———
lyn pi lel aan een sy ) Er RT
SE RS v Reason/rede
“Er "Rr (QR =RS proved/bewys in 10.2.1)
_SE _ RS
“ET RT ER
_ PQ ( from equation/uit vergelyking (1)) ve. Be = = (5)
PT
[20]

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TOTAL / TOTAAL: 150

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Mathematics/P2/Wiskunde/V2 19 NW/September 2016
NSC/NSS — Memorandum
ANALYSIS GRID

Q.no Concepts Level 1 | Level2 | Level3 | Level4 | Total
11 Equation of Regression line 3
TZ Prediction 2
1.3 Estimation and Analysis 2
14 @; y) 4
1.5 Correlation coefficient i
1.6 Relationship between the 2

two variables
Total 5 ul 2 14
21 Finding the value of 1

unknown
2.2 Cumulative frequency table | 2
2.3 Ogive 3
24 Modal class 1
Total 4 3 7
3.1 Distance 2
| Equation of st. line 3
3.3 Altitude 3
3.4 Points of intersection 3
3.5 Angle between two lines 5)
3.6 Analysis 4
Total 5 6 5 4 20
4.1 Equation of a circle 3
4.2 Distance 4
43 Equation of a tangent 3
44 Point of intersection 6
4.5 Analysis 2
Total 5 4 9 18
5.1 Reduction and identity 6
5.2.1 Compound angle 3
5.2.2 Deduction and special 4

angle
Bis) Trig and distance 4
Total 9 4 4 17
6.1 Sketching of graphs 6
6.2.1 Solutions 2
6.2.2 Asympote 2
6.2.3 Range 1
6.2.4 Solutions 1
6.2.4 Inequality 3
Total 4 2 9 15
qi Application of cosine rule 3 3

and sine rule
ie Area rule 3
Total 6 3 9

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Please turn over/Blaai om asseblief

### Page 20

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Mathematics/P2/Wiskunde/V2 20 NW/September 2016
NSC/NSS — Memorandum

8.1.1 Diameter 1

8.1.2 Tan-chord 1

8.2.1 Geo. Reasoning 6

8.2.2 App.tan-chord 3

8.2.3 Proving mid point 4

8.2.4 App. of thm 4
| Total 2 7 10 19
9.1 Tan- chord 3)

9.2 Parallellogram 2

9.3 Proportinality 2 2

Total 5 4 2 11
10.1 Similarity theorem 7 :
10.2.1 Side opp. of equal angles 4

7 Cyclic quad. 4

10.2.3 Application of Similarity 5

Total 4 11 5 20
Gr.Total 29 60 44 17 150
Expected 30 52,5 45 22,5 150
marks

Actual 19,3 % 40% 29,3 % 11.3% 100
%

Expected 20% 35% 30% 15% 100
%

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