SENIOR CERTIFICATE EXAMINATIONS/
NATIONAL SENIOR CERTIFICATE EXAMINATIONS
SENIORSERTIFIKAAT-EKSAMEN/
NASIONALE SENIORSERTIFIKAAT-EKSAMEN
MATHEMATICAL LITERACY P2/
WISKUNDIGE GELETTERDHEID V2
MAY/JUNE/MEI/JUNIE 2024
MARKING GUIDELINES/NASIENRIGLYNE
MARKS/PUNTE: 150
Symbol/Kode Explanation/Verduideliking
MA Method with accuracy/Metode met akkuraatheid
CA Consistent accuracy/Volgehoue akkuraatheid
A Accuracy/Akkuraatheid
C Conversion/Herleiding
S Simplification/Vereenvoudiging
RT Reading from a table/a graph/document/diagram/Lees vanaf tabel/grafiek/diagram
SF Correct substitution in a formula/Korrekte vervanging in formule
O Opinion/Explanation/Reasoning /Opinie/Verduideliking/Redenasie
P Penalty, e.g. for no units, incorrect rounding off, etc./Penalisasie, bv. vir geen
eenhede/verkeerde afronding, ens.
R Rounding off/Afronding
NPR No penalty for correct rounding/Geen penalisasie vir korrekte afronding nie
NPU No penalty for omitting correct unit/Geen penalisasie vir die uitlos van die korrekte
eenheid nie
AO Answer only/Slegs antwoord
MCA Method with constant accuracy/Metode met volgehoue akkuraatheid
RCA Rounding consistent with accuracy/Afronding met volgehoue akkuraatheid
These marking guidelines consist of 20 pages.
Hierdie nasienriglyne bestaan uit 20 bladsye..
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Mathematical Literacy Grade 12 P2 NSC June 2024 Memo (Afr)
Grade 12 Mathematical Literacy Paper 2 memorandum from the NSC June Exam 2024 (Afrikaans), 20 pages. Read online or download the free PDF.
Mathematical Literacy · Grade 12 · NSC June Exam · 2024 · Afrikaans. Memorandum, 20 pages. Read online or download the PDF.
- Subject
- Mathematical Literacy
- Grade
- Grade 12
- Language
- Afrikaans
- Document type
- Memorandum
- Year
- 2024
- Exam period
- NSC June Exam
- Paper
- 2
- Curriculum
- CAPS
- Pages
- 20
- File size
- 515.4 KB
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NSC June Exam 2024
Document textSearch extracted text and jump to a page.
Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 2 DBE/May/June/Mei/Junie 2024
SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
NOTE:
• If a candidate answers a question TWICE, only mark the FIRST attempt.
• If a candidate has crossed out (cancelled) an attempt to a question and did NOT redo the solution,
mark the crossed out (cancelled) version.
• Consistent accuracy (CA) applies in ALL aspects of the marking guidelines; however, it stops at the
second calculation error or breakdown.
• If the candidate presents any extra solution when reading from a graph, table, layout plan and map,
then penalise for every extra item presented.
• As a general marking principle, if a candidate has incurred one mistake and there is evidence of sound
mathematics thereafter, then that candidate should lose one mark only.
• A conclusion mark can only be given if relevant calculations precede it (at least 1 mark before
conclusion).
• Rounding is an independent mark.
• No penalty for rounding (NPR) if the first decimal is correct, except questions involving money.
LET WEL:
• As 'n kandidaat 'n vraag TWEE KEER beantwoord, sien slegs die EERSTE poging na.
• As 'n kandidaat 'n antwoord van 'n vraag doodtrek (kanselleer) en nie oordoen nie, sien die
doodgetrekte (gekanselleerde) poging na.
• Volgehoue akkuraatheid (CA) word in ALLE aspekte van die nasienriglyne toegepas, dit hou egter op
by die tweede berekeningsfout of afbreuk ’break down’ nie
• Wanneer 'n kandidaat aflesings vanaf 'n grafiek, tabel, uitlegplan en kaart geneem en ekstra antwoorde
gee, penaliseer vir elke ekstra item.
• 'n Algemene nasienbeginsel is dat indien 'n kandidaat een fout maak en daarna voortgaan met korrekte
wiskunde, dat die kandidaat slegs een punt verloor
• 'n Gevolgtrekkingspunt kan slegs gegee word indien relevante berekeninge dit voorgaan (ten minste een
punt voor die gevolgtrekking).
• Afronding tel as 'n onafhanklike punt.
• Geen penalisering vir ronding (NPR) as die eerste desimaal korrek is nie, behalwe as vrae geld insluit.
QUESTION/VRAAG 1 [26 MARKS/PUNTE] Answer Only AO - full marks
Q/V Solution/Oplossing Explanation/Verduideliking T/L
1.1* 1.1.1 E ✓✓A 2A correct option M
L1
1.1.2 G ✓✓A 2A correct option P
L1
1.1.3 F ✓✓A 2A correct option M
L1
1.1.4 B ✓✓A 2A correct option M
(8) L1
✓✓ A MP
1.2.1 Numerical /Number/ ratio scale. 2A type of scale L1
Numeriese- / Nommer- /Getalle- /syfer-/verhouding- (2) E
skaal.
✓✓A MP
1.2.2 1 unit on the map is equivalent to 50 000 units in real life. 2A relationship L1
1 eenheid op die kaart is gelykstaande aan 50 000 M
eenhede in werklikheid
OR/OF
The map is 50 000 times smaller than real life.
Die kaart is 50 000 keer kleiner as werklikheid (2)
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Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 3 DBE/May/June/Mei/Junie 2024
SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T/L
MP
1.2.3* 1: 25 000 ✓✓ A 2A correct scale L1
(Accept B) E
(2)
✓A ✓A M
1.3.1* Rectangle and a circle. 1A rectangle L1
Reghoek en ‘n sirkel 1A circle E
(2)
✓✓A MP
1.3.2 144 km 2A correct answer L1
Accept 144 E
(2)
MP
1.3.3* It is the maximum speed a motorist can travel on the L1
road. ✓✓A M
Dit is die maksimum spoed wat ‘n motoris mag ry op die
pad. 2A correct explanation.
OR/OF
The motorist can cover a distance of 120 km in 1 hour.
Die motoris kan 120 km aflê in 1 uur (2)
M
1.3.4* Distance/Afstand (Jhb – Trompsburg) L1
M
= 534 – 27 ✓ RT 1RT both correct values
✓A
= 507 km 1A distance
NPU
(2)
MP
1.3.5 North /N/Noord / N ✓✓A 2A correct direction L1
(2) E
M
1.3.6 90 𝑐𝑚 1MA dividing by 100 L1
✓MA
100
E
1A simplification
= 0,9 m ✓A
(2)
[26]
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Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 4 DBE/May/June/Mei/Junie 2024
SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
QUESTION/VRAAG 2 [29 MARKS/PUNTE]
Q/V Solution/Oplossing Explanation/Verduideliking T/L
MP
2.1.1 2 ✓✓A 2A correct number L1
(2) E
MP
2.1.2 6 ✓✓A 2A correct road L1
(2) M
MP
2.1.3* 7 ✓✓A 2A correct number L1
(2) E
✓✓A P
2.1.4* C 2A correct choice L2
(2) M
✓✓A MP
2.1.5 South East (SE) / Suidoos (SO) 2A correct direction L2
(2) M
MP
2.1.6 Length / Lengte = 65 m L3
= 65 000 mm ✓C 1C conversion M
Scale/ Skaal: 1 : 8 000
n : 65 000
65 000 ✓ MA 1MA dividing
n=
8 000
✓CA 1CA simplification
= 8,125 mm
✓R 1R rounding
≈ 8 mm
OR/OF OR/OF
Scale/Skaal: 1 : 8 000
n : 65
65 ✓ MA 1MA dividing
n=
8 000
= 0,008125 m ✓CA 1CA simplification
= 8,125 mm ✓C 1C conversion
≈ 8 mm ✓R 1R rounding
(4)
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SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T/L
MP
2.2* 2.2.1 C ✓A L2
2.2.2 E ✓A M
2.2.3 D ✓A 5A correct order
2.2.4 B ✓A
2.2.5 A ✓A (5)
✓✓O MP
2.3* Only use the go cart on level ground. / smooth, flat, L4
hard, tarred, road surface E
Gebruik die knortjor slegs op gelyke grond/ gladde , 2O Explanation for 1st picture
plat, harde, pad, geteerde oppervlakte or for 2nd picture
OR/OF
Do not use the vehicle on a long grassy surface.
Moet nie in lang gras ry nie.
(2)
✓MA P
2.4.1 X = 2 840 – 1 476 – 1 024 = 340 ✓CA L1
1MA subtracting from total E
OR/OF 1CA simplification
✓MA
X = 565 – 163 – 62 = 340 ✓CA AO
(2)
P
2.4.2 2 840 + 796 ✓RT 1RT numerator L2
P(not a horse) = 1RT denominator M
4 996 ✓RT
3 636
=
4 996
909
= ✓A 1A simplification
1 249
OR/OF OR/OF
1 360
P(horse) = 1RT both values
4 996 ✓RT
1 360
P(not a horse) = 1 – ✓MCA 1MCA subtracting from 1
4 996
909 ✓A
= 1A simplification
1 249
(3)
✓RT P
st
2.4.3 363 1RT 1 value L3
P= × 100%
796 1RT 2nd value correctly place M
✓RT
= 45,60301508 % ✓CA 1CA simplification
NPR
(3)
[29]
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SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
QUESTION/VRAAG 3 [33 MARKS/PUNTE]]
Q/V Solution/Oplossing Explanation/Verduideliking T/L
M
3.1.1 2 000 kW ✓✓RT 2RT correct kW L1
NPU E
(2)
M
✓✓O
3.1.2 To allow rotor blade to produce more energy. L4
Om met die rotorlem meer energie op te wek. E
OR/OF
Advanced technology to have material that can
allow a big structure to stand firm on the ground.
Verbeterde tegnologie om materiaal te hê wat
toelaat dat so ‘n groot struktuur standvastig kan
staan
OR/OF 2O reason (more electricity)
Larger rotor diameters allow wind turbines to
sweep more area, capture more wind and produce
more electricity
Groter rotormiddellyne laat die windturbines ‘n
groter area dek, meer wind vang en so meer krag
opwek
OR/OF
Demand for electricity increased/ demand for
cleaner electricity
Verhoogde aanvraag vir elektrisiteit/ aanvraag vir
skoner elektisiteit (2)
M
3.1.3 Max. height (in m) = Poles height + radius of L2
rotor M
Maks. Hoogte (in m) = Paal hoogte + radius van
rotor ✓ RT
124 1RT both correct values
= 114 +
2 ✓MA 1MA divide by 2 to determine
the radius
= 114 + 62
= 176 ✓CA 1CA simplification
OR/OF OR/OF
Pole + rotor/ Paal + rotor
=124 + 114 ✓ RT 1RT both correct values
= 238
Maximum height /Maksimum hoogte in m
= 238 – (124÷2) ✓MA 1MA divide by 2 to determine
= 238 – 62 the radius
= 176 ✓CA 1CA simplification
AO
(3)
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SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T/L
M
124
3.1.4* Radius = 2 = 62 ✓ A 1A radius L4
M
Area /Oppervlakte
= 3,142 × (62)2 ✓ SF 1SF substitution squared
= 12 077,848 m2 ✓ CA 1CA simplification
✓O
Not valid. / Nie geldig nie 1O invalid
(4)
✓ RT
3.1.5 5 000 − 800 1RT 1st correct value M
% increase/verhoging = × 100% ✓ MA 1A denominator
800 L2
✓A 1MA percentage M
= 525 % ✓CA 1CA simplification
OR/OF OR/OF
Current percentage / Huidige persentasie
✓RT 1RT 1st correct value
5 000
= 800 × 100% 1A denominator
✓A
= 625%
% increase/verhoging
625% – 100% ✓ MA 1MA percentage difference
= 525 % ✓CA 1CA simplification
(4)
✓✓ A M
3.1.6* Generators OR solar power OR hydro-power OR L1
nuclear power 2A source E
Kragopwekkers OF sonkrag OF hidro-elektrisiteit OF
kernkrag (2)
✓ RT 1RT correct values M
✓ SF
3.2.1 Perimeter/Omtrek = 2 × (2,3 + 2,3 + 2,3 + 3) m 1SF substitution L2
✓ MA E
= 2 × (6,9 + 3) m 1MA 6,9
= 19,8 m ✓ CA 1CA answer
OR/OF OR/OF
Perimeter/Omtrek 1RT correct values
✓ RT ✓ SF 1SF substitution
= 3 + 2,3 + 2,3 + 2,3 + 3 + 2,3 + 2,3 + 2,3 m ✓MA 1MA 6,9
= 19,8 m ✓CA 1CA answer
(4)
✓RT ✓ A M
1RT correct numbers 5 and 7
3.2.2* 5 + 7 + 4 = 16 boards /planke ✓CA L3
1A on 4
M
1CA simplification adding
AO
(3)
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SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T/L
M
5
3.2.3 Height /hoogte B= × 0,5 m L3
7 D
= 0,3571428571 m✓ A 1A height box B
Vol B = ( 2,3 × 3 × 0,3571428571)𝑚3
= 2,464285714𝑚3 ✓ CA 1CA volume of B box
Vol A = (2,3 × 3 × 0,5) 𝑚3 ✓ SF 1SF volume of A box
= 3,45𝑚3 ✓A 1A simplification 3,45
CA total volume
Total/Totaal = 2,4642514𝑚3 + 3,45𝑚3
= 5,914285714𝑚3 ✓ CA
Capacity / Kapasiteit = 1 000 × 5,914285714 1CA answer in litres
= 5 914,285714 litres. ✓ CA
OR/OF
OR/OF
Height of section B/ Hoogte van boks B
5
= × 0,5 m 1A height box B
7
= 0,3571428571 m ✓ A
Vol = (length×width×height) + (length×width×height)
✓ SF 1SF volume of A box
= (2,3 m × 3 m × 0,357… m) + (2,3 × 3 m × 0,5 m)
✓ CA ✓A 1CA volume of B box
= (2,464285714 + 3,45) m3. 1A 3,45m3
✓ CA
= 5,914285714 m3 1CA total volume
Capacity / Kapasiteit = 1 000 × 5,914285714
= 5 914,285714 litres. ✓ CA 1CA answer in litres
OR/OF OR/OF
Volume A = length × width × height/
lengte × breedte ×hoogte
= (3 m × 2,3 m× 0,5 m) ✓SF 1SF volume of A box
= 3,45 m3 ✓A 1A simplification 3,45
∴ 3 450 litres ✓C 1C conversion
3 450 ℓ 1A ratio
Volume B = ×5 ✓A
7
= 2 464,285714 litres ✓CA 1CA volume box B
∴ 𝑇𝑜𝑡𝑎𝑙 = 3 450 + 2 464,285714
1CA answer in litres
= 5 914,285714ℓ ✓ CA
NPR
(6)
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Q/V Solution/Oplossing Explanation/Verduideliking T/L
M
3.3*
o
C = 5 × (oF – 32o) L2
9
M
✓ SF
70° = 5 × (oF – 32o) 1SF substituting in formula
9
9
70° × = °F – 32 ✓ MA
5 1MA changing subject
126° = °F – 32
°F = 158 ✓ CA 1CA answer
(3)
[33]
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QUESTION/VRAAG 4 [33 MARKS/PUNTE]
Q/V Solution/Oplossing Explanation/Verduideliking T/L
✓RT 1RT 30 MP
4.1.1 30 : 6 ✓RT 1RT 6 L2
= 5 : 1 ✓A 1A simplification E
(3)
✓✓A ✓A 2A 1st room MP
4.1.2 Reading room and computer lab 1A second room L1
Leeskamer en rekenaarlokaal (3) E
✓✓A MP
4.1.3 Stairs / trappe 2A stairs L2
(2) E
✓✓A ✓A MP
4.1.4 Multi-media room 1 / Multi-mediakamer 1 2A correct room L3
1A correct number M
(3)
M
4.1.5 Area/ Opp A = length × width /lengte × breedte L4
= 11 m × 3,5 m ✓SF 1SF substitution D
= 38,5 m2 ✓ MCA 1MCA simplification
Area/ Opp B = length × width /lengte × breedte
= 14 m × 3,5 m
= 49 m2 ✓ MA 1MA simplification
Floor area/Vloer opp.
= 38,5 m2 + 49 m2
= 87,5 m2 ✓ MCA 1MCA simplification total
area
Area of tile = length × width
Opp van teël = lengte × breedte
= 600 mm × 600 mm
= 360 000 mm2 ✓ A
1A area tile
360 000
∴ = 0,36 m2 ✓C 1C conversion
1000 000
87,5
Number of tiles/Getal teëls = ✓ MCA
0,36 1MCA dividing
≈ 243,056 tiles✓ CA 1CA number of tiles
244
Number of boxes/ Getal bokse =
5
= 48,8
✓ CA
= 49 1CA number of boxes
INVALID/ ONGELDIG. ✓ O
1O opinion
OR/OF OR/OF
1SF substitution
✓ SF ✓ MA ✓A 1MA adding areas
Floor Area/vloeropp = 11m ×7 m + 3,5 m × 3 m
= 77 m2 + 10,5 m2 1A 3
= 87,5 m 2
✓ CA 1CA area
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SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T/L
600 mm
Tiles /Teëls = = 0,6 m ✓ C 1C conversion
1000
Area of a tile / Opp van teël 1MCA area of tile
= 0,6 m × 0,6 m = 0,36 m2 ✓ MCA
87,5 1MCA dividing areas
Number of tiles/Getal teëls = ✓ MCA
0,36
1CA number of tiles
≈ 243,056 tiles✓ CA
244
Number of boxes/ Getal bokse =
5
= 48,8
= 49 ✓ CA 1CA number of boxes
INVALID/ ONGELDIG. ✓ O 1O conclusion
OR/OF OR/OF
✓ SF ✓ A 1A 3
Floor Area/ Vloer opp = 11m ×7 m + 3,5 m × 3 m 1SF substitution
= 77 m2 + 10,5 m2 ✓ MA 1MA adding areas
= 87,5 m2 ✓ CA 1CA area
600 mm
Tiles / Teëls = = 0,6 m ✓ C 1C conversion
1000
Area of a tile / Opp van ‘n teël = 0,6 m × 0,6 m
= 0,36 m2 ✓ MCA 1MCA area of tile
87,5
Number of tiles / Getal teëls = ✓ MCA 1MCA dividing areas
0,36 ✓ CA
≈ 243,056 tiles 1CA number of tiles
tiles in 40 boxes / teels in 40 bokse = 40 × 5 = 200
40 boxes is not enough or 200 < 244 ✓ CA 1CA less than
40 bokse is nie genoeg nie of 200 < 244
INVALID./ ONGELDIG ✓O 1O conclusion
OR/OF OR/OF
✓ SF 1A 3
✓A
Floor Area/vloeropp = 14m ×7 m – 3,5 m × 3 m 1SF substitution
= 98 m2 – 10,5 m2 ✓ MA 1MA subtracting areas
= 87,5 m2 ✓ CA 1CA area
600 mm
Tiles /Teëls = 1 000 = 0,6 m ✓C 1C conversion
Area of a tile / Opp van teël
= 0,6 m × 0,6 m = 0,36 m2 ✓ MCA 1MCA area of tile
87,5
Number of tiles/Getal teëls = ✓ MCA 1MCA dividing areas
0,36
≈ 243,056 tiles ✓ CA 1CA number of tiles
244
Number of boxes/ Getal bokse =
5
= 48,8
✓ CA 1CA number of boxes
= 49
INVALID/ ONGELDIG. ✓ O 1O conclusion
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SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T/L
OR/OF OR/OF
Area A = length × width/ lengte × breedte
= 11 m × 3,5 m ✓SF 1SF substitution
= 38,5 m2 ✓ MCA 1MCA simplification
Area B = length × width/ lengte × breedte
= 14 m × 3,5 m
= 49 m2 ✓ MCA 1MCA simplification
Floor area/Vloer opp.
= 38,5 m2 + 49 m2 1MCA simplification total
= 87,5 m2 ✓ MCA area
Area of tile = length × width / Opp van teël= l × b
= 600 mm × 600 mm
= 360 000 mm2 ✓A 1A area tile
360 000 1C conversion
∴ 1000 000 = 0,36 m2 ✓C
87,5
Number of tiles/Getal teëls = ✓ MCA 1MCA dividing
0,36 1CA number of tiles
≈ 243,056 tiles ✓ CA
244
Number of boxes/ Getal bokse =
5
= 48,8
✓ CA 1CA number of boxes
= 49
INVALID/ ONGELDIG. ✓ O 1O opinion
OR/OF OR/OF
Area of tile / Opp van teël
= 600 mm × 600 mm
= 360 000 mm2 ✓A 1A area tile
L = 14 m × 1 000
= 14 000 mm ✓C 1C conversion
B = 7 m × 1 000
= 7 000 mm
∴ 𝐴𝑟𝑒𝑎/𝑂𝑝𝑝 = 14 000 𝑚𝑚 × 7 000 𝑚𝑚 ✓SF 1SF substitution
= 98 000 000 mm2 ✓ MCA 1MCA simplification
∴ 𝐴𝑟𝑒𝑎/𝑂𝑝𝑝 = 3 500 𝑚𝑚 × 3 000 𝑚𝑚
= 10 500 000 mm2 ✓ MCA 1MCA simplification
Total area/ Totale opp
= 98 000 000 mm2 – 10 500 000 mm2 1MCA simplification total
= 87 500 000 mm2 ✓MCA area
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Q/V Solution/Oplossing Explanation/Verduideliking T/L
Number of tiles/Aantal teëls
87 500 000 𝑚𝑚2 1MCA dividing
= ✓MCA = 243,0555556 tiles ✓CA
360 000 𝑚𝑚2 1CA number of tiles
Number of boxes/Getal bokse
243,0555556
= 5
= 48,61 ≈ 49 ✓CA 1CA number of boxes
∴ INVALID/𝑂𝑁𝐺𝐸𝐿𝐷𝐼𝐺 ✓O 1O opinion
OR/OF OR/OF
✓SF ✓ A 1A 3
Area (Lab) = (7 × 14 – 3 × 3.5) 𝑚2 1SF substitution
= (98 – 10,5) 𝑚2 ✓MA 1MA subtracting
= 87,5 𝑚2 ✓ MCA 1MCA simplification total
area
Tile side / Teël sy = 600 ÷ 1 000 = 0,6 m ✓ C 1C conversion
Area covered by a box of tiles
Oppervlakte bedek deur ‘n boks teëls
= (0,6 × 0,6) × 5 ✓MCA 1MCA area of 1 tile
= 1,8 m2 ✓ CA 1CA area box of tiles
Number of boxes / Getal bokse
87,5 1MCA dividing
= 1,8 ✓ MCA
= 48,6 ≈ 49 ✓ CA 1CA number of boxes of
tiles
INVALID / ONGELDIG ✓ O 1O opinion
OR/OF OR/OF
Calculating 3 areas/Berekening 3 opp.
A1 = 3,5 ×11 ✓SF
= 38,5𝑚2 1SF substitution
A2 = 3 × 3,5 ✓ A 1A 3
=10.5𝑚2
A3 = 3,5 × 11
= 38.5𝑚2 1MA adding
TOTAL =38,5𝑚2 +10,5𝑚2 +38,5𝑚2 ✓MA
=8 7,5𝑚2 ✓MCA 1MCA simplification total
area
87,5
Number of tiles/Getal teëls = ✓ MCA 1MCA dividing
0,36
≈ 243,056 tiles✓ CA 1CA number of tiles
244
Number of boxes/ Getal bokse =
5
= 48,8
= 49 ✓ CA 1CA number of boxes
INVALID/ ONGELDIG ✓O 1O opinion
(10)
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Q/V Solution/Oplossing Explanation/Verduideliking T/L
MP
4.2.1* 4 ✓✓ A 2A number of countries L1
(2) E
MP
4.2.2 Harare ✓✓ A 2A correct town L1
(2) E
M
4.2.3* ✓ SF ✓A 1A 179 L3
(a) 97 km/h = 179 km ÷ tyd 1SF substitution 97 D
Time =distance ÷ speed
Tyd = afstand ÷ spoed
179 1MCA change formula
= ✓ MCA
97
= 1,845 hours ✓ CA 1CA time in hours
Time duration / tydsduur = 1,845 hours./uur
= 1 hour/uur + 0,845 × 60 min
= 1 hour/uur 51 min
Arrival time / Aankomstyd:
= 09:55 + 1 h 51 min
= 11: 46 ✓ CA 1CA answer
(5)
✓MA M
4.2.3 Distance/Afstand = (713 – 263) + 2(18) 1MA subtracting L2
(b) correct values M
= 450 + 36 ✓MA 1MA getting 36
= 486 km ✓ CA 1CA total distance
OR/OF OR/OF
Distance /Afstand 1MA subtracting
✓ MA ✓ MA correct values
= (713 – 552) + 18 + 18 + (552 – 263) 1MA adding distances
= 161 + 18 + 18 +289 = 486 km ✓ CA 1CA answer
(3)
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Q/V Solution/Oplossing Explanation/Verduideliking T/L
OR/OF OR/OF
✓MA
Distance/Afstand = (455 – 5) + 2(18) 1MA subtracting
correct values
= 450 + 36 ✓MA 1MA getting 36
= 486 km ✓ CA 1CA total distance
OR/OF OR/OF
Distance /Afstand
✓ MA ✓ MA 1MA subtracting correct values
= (166 – 5) + 18 + 18 + (455 – 166) 1MA adding values
= 161 + 18 + 18 +289 = 486 km ✓ CA 1CA answer
OR/OF OR/OF
Distance/Afstand
✓MA ✓MA 1MA subtracting correct values
= 179 + 18 + (552 – 263) km 1MA adding values
= 486 km ✓CA 1CA answer
(3)
[33]
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QUESTION/VRAAG 5 [29 MARKS/PUNTE]
Q/V Solution/Oplossing Explanation/Verduideliking
T/L
MP
5.1.1* 8 ✓✓A 2A correct number L1
(2) E
MP
5.1.2 Front entrance portal /Voorste ingangsportaal 1MA subtracting from 58 L3
✓MA ✓RT ✓RT 1RT room dimensions M
= 58 – (11 × 4 + 2 × 4) ✓A 1RT wall thickness
1A multiplying with 4
= 6 feet/ voet ✓CA 1CA simplification
(5)
MP
5.1.3 There are no walls separating the kitchen, dining room L4
and living room. ✓✓O 2O reason E
Daar is geen mure wat die kombuis, eetkamer en (2)
woonvertrek skei nie
MP
✓✓A
5.1.4* Toilet OR bath OR basin or sink 2A correct feature L1
Toilet OF bad OF wasbak (2) E
✓O ✓O MP
rd
5.1.5 3 floor and B that it is the second apartment L4
3de vloer en B is die tweede woonstel M
OR/OF
✓O
Block B, Number 3 ✓O
Blok B, nommer 3
OR/OF
1O numbering of the floors
✓O 1O numbering of the
3 Floor, unit on the left/right ✓O
rd apartments
de
3 vloer, die eenheid links/ regs
OR/OF
✓O
3rd Floor, B-wing ✓O
3de vloer, B -vleuel
(2)
M
5.1.6 17,6784 m = 58 feet/voet L2
(a) Conversion factor/ Herleidings faktor: M
✓RT 1RT 58
= 3,28083989… ✓MA
58
1m= 1MA simplification
17, 6784
≈ 3,281 feet ✓R 1R rounded answer
(3)
CA from 5.1.6 (a) M
✓RT
5.1.6 40 1RT correct width L2
Width / Breedte =
(b) 3, 281 ✓MCA 1MCA dividing M
= 12,191405 m ✓CA 1CA simplification
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Q/V Solution/Oplossing Explanation/Verduideliking T/L
OR/OF OR/OF
58 feet/voet = 17,6784 m
40 feet/voet = n
✓RT 1RT correct width
40 1MA working with ratio
n = 58 × 17,6784 ✓MA
1CA simplification
= 12,191405 m ✓CA
NPR
(3)
M
5.2.1 Area /Oppervlakte = length × width / lengte × breedte L2
= 0,614 m × 0,474 m ✓SF 1SF substitution E
= 0,291036 m2
= 0,3 m2 ✓R 1R simplification
NPU
(2)
CA from Q 5.2.1 M
5.2.2* Area for 6 panels /Opp van 6 panele = 0,3 m2 × 6 L4
= 1,8 m2 ✓MCA 1MCA simplification D
Cost for 6 panels /Koste van 6 panele
= 1,8 m2 × R490/m2 = R882 ✓MCA 1MCA simplification cost
Mass of the 6 panels / Massa van 6 panele
= 1,8 m2 × 15 kg/m2 = 27 kg ✓MCA 1MCA simplification: mass
Delivery mass / Aflewerings massa= 20 kg + 7 kg
Cost of delivery / Afleweringskoste
✓MA 1MA cost of 1st 20kg
= R820 + R53,50 × 7 kg ✓MCA 1MCA add and multiply
= R1 194,50 ✓CA 1CA simplification
Total cost / Totale koste = R882,00 + R1 194,50
= R2 076,50 ✓CA 1CA simplification
INVALID/ ONGELDIG ✓O 1O verification
OR/OF OR/OF
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Q/V Solution/Oplossing Explanation/Verduideliking T/L
Using unrounded area
Area for 6 panels /Opp van 6 panele
= 0,291036 m2 × 6
= 1,746216 m2 ✓CA 1CA simplification
Cost for 6 panels /Koste van 6 panele
= 1,746216 m2 × R490/m2 = R855,65 ✓CA 1CA simplification cost
Mass of the 6 panels / Massa van 6 panele
= 1,746216 m2 × 15 kg/m2 = 26,19324 kg ✓CA 1CA simplification: mass
Delivery mass / Aflewerings massa= 20 kg + 7 kg
Cost of delivery / Afleweringskoste
✓MA 1MA cost for 1st 20 kg
= R820 + R53,50 × 7 kg ✓MCA 1MCA add and multiply
= R1 194,50 ✓CA 1CA simplification
Total cost / Totale koste = R855,65 + R1 194,50
= R2 050,15 ✓CA 1CA simplification
INVALID/ ONGELDIG ✓O 1O verification
(8)
[29]
TOTAL/TOTAAL: 150
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NOTE/LET WEL:
1.1 1.1.1 Circumference E The boundary that surrounds the circular shape.
1.1.2 Probability G The likelihood that something may happen. Full marks
1.1.3 One hour F A time measurement equivalent to three thousand for written
six hundred seconds. explanations
1.1.4 Temperature B The measure of hotness or coldness.
1.2.3 B 2 out of 2
1.3.1 Accept round (for circle) 2 out of 2
1.3.3 A motorist can only travel up to 120 km/h on the road.
‘n Motoris mag net tot 120km/h ry op die pad, 2 out of 2
120 km/h is the speed limit./ Do not exceed 120 km/h on this road
120km/h is die spoedbeperking/ Jy mag nie 120km/h oorskry op die
pad nie
1.3.4 For candidates writing 534 – 144 = 390 1 out of 2
Listing all seven correct:
2.1.3 1, 5, 8, 9, 10, 11, 12 1 out of 2
Vehicle entrance, cattle vehicle, etc.
2.1.4 Accept Certain /Beslis 2 out of 2
2.2 5A correct order
C E D B A
(5)
2.3 Do not drive off the road/ Moenie van die pad af gaan nie. 2 out of 2
3.1.4
Using 124 m as radius, but correct calculation 48 311,392 m2
2 out of 4
and conclusion
The following words can be used:
3.1.6 2 out of 2
Water, coal, sun, inverters
3.2.2 12 3 out of 3
3.2.2 15 2 out of 3
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Using this formula correctly – no part marks
3.3 9
℉ = (℃ × 5 ) + 320
3 out of 3
0 9 0
= (70 × 5 ) + 32
= 158
4.2.1 Zambia, Zimbabwe, South Africa, Botswana 1 out of 2
4.2.3 Accept 11:45 5 out of 5
(a)
5.1.1 6 or 2 1 out of 2
5.1.4 Accept door 2 out of 2
Area for 6 panels/Oppervlakte van 6 panele
5.2.2 = 0,3 m2 × 6
= 1,746216 m2
= 2 m2 7 out of 8
Cost for 6 panels/Koste van 6 panele
= 2 m2 × R490/m2 = R980,00
Mass of the 6 panels/Massa van die 6 panele
= 2 m2 × 15 kg/m2 = 30 kg
Delivery mass = 20 kg + 10 kg
Cost of delivery/Afleweringskoste
= R820 + (R53,50 × 10)
= R1 355,00
Total cost/Totale koste
= R980,00 + R1 355
= R2 335,00
INVALID/ONGELDIG
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