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SENIOR CERTIFICATE EXAMINATIONS/
NATIONAL SENIOR CERTIFICATE EXAMINATIONS
SENIORSERTIFIKAAT-EKSAMEN
NASIONALE SENIORSERTIFIKAAT-EKSAMEN
MATHEMATICAL LITERACY P2/WISKUNDIGE GELETTERDHEID V2
2023
MARKING GUIDELINES/NASIENRIGLYNE
MARKS/PUNTE: 150
Symbol/Kode Explanation/Verduideliking
M Method/Metode
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
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 17 pages.
Hierdie nasienriglyne bestaan uit 17 bladsye.
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Mathematical Literacy Grade 12 Paper 2 NSC June 2023 Memo
Grade 12 Mathematical Literacy Paper 2 memorandum from the NSC June Exam 2023, 17 pages. Read online or download the free PDF.
Mathematical Literacy · Grade 12 · NSC June Exam · 2023. Memorandum. Read online or download the PDF.
- Subject
- Mathematical Literacy
- Grade
- Grade 12
- Document type
- Memorandum
- Year
- 2023
- Exam period
- NSC June Exam
- Paper
- 2
- Curriculum
- CAPS
- Pages
- 17
- File size
- 897.9 KB
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Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 2 DBE/2023
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 NOT redone 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.
• NOTE: consistent accuracy (CA) does not apply in cases of a 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.
• Rounding is an independent mark.
• In order to award the verification / conclusion mark the candidate must have scored at least one mark in
the calculations preceding the final conclusion.
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 op by die
tweede berekeningsfout.
• Let wel: volgehoue akkuraatheid (CA) geld nie in die geval van 'n afbreuk 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.
• Afronding tel as 'n afsonderlike punt.
• Ten einde die verifikasie/ gevolgtrekking punt toe te ken moes die kandidaat ten minste een punt gekry
het in die berekeninge wat lei tot die finale gevolgtrekking.
Note: Questions marked with * refers to the notes.
Questions where the numbers are encircled are the ones where we have a tolerance range.
QUESTION/VRAAG 1 [28 MARKS/PUNTE] Answer Only AO - full marks
Q/V Solution/Oplossing Explanation/Verduideliking T/L
M
*1.1.1 E A 2A correct option L1
(2) E
M
*1.1.2 C A 2A correct option L1
(2) E
MP
*1.1.3 I A 2A correct option L1
(2) E
MP
*1.1.4 B A 2A correct option L1
(2) E
P
*1.1.5 G A 2A correct option L1
(2) E
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Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 3 DBE/2023
SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T/L
RT RT RT MP
1.2.1 Potatoes, Onions and Cucumber 3RT correct partner L1
Aartappels, Uie en Komkommer (3) E
MP
1.2.2 Six /Ses (6) RT 2RT correct number L1
(2) M
MP
1.2.3 Beans /Bone RT 2RT correct partner L1
M
(2)
MP
*1.2.4 South East OR SE RT 2RT correct direction L1
Suidoos OF SO M
(2)
RT RT MP
*1.2.5 3 and 7 2RT 1st correct number label L1
1RT 2nd correct number label E
(3)
M
C OR/OF π × r2 × h RT
1.3.1 2A correct option L1
(2) E
M
1.3.2 mm3 A 2A correct unit L1
(2) E
M
C
1.3.3 mm to metre = 124 ÷ 1 000 1C correct conversion/dividing by L1
mm tot meter 1 000 E
= 0,124 m A 1A answer in metres
(2)
[28]
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Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 4 DBE/2023
SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
QUESTION/VRAAG 2 [24 MARKS/PUNTE]
Q/V Solution/Oplossing Explanation/Verduideliking T/L
MP
2.1.1 5 A 2A correct number L1
(2) E
MP
2.1.2 Tripod/Driepoot A 2A correct item L1
(2) E
MP
2.1.3 Clockwise/Kloksgewys A 2A correct direction L1
(2) E
MP
2.1.4 H A 2 A correct choice L2
(a) (2) M
MP
2.1.4 G 2 A correct choice L2
(b) A (2) M
MP
2.2.1 65 km × 100 000 L2
1C conversion D
= 6 500 000 cm C
Distance on the map /Afstand op kaart
6 500 000 1MA division by 250 000
= MA
250 000
= 26 cm CA 1CA simplification
OR/OF OR/OF
65 km × 1 000 000
= 65 000 000 mm C 1C conversion
Distance on the map /Afstand op kaart
65 000 000 1MA division by 250 000
= MA
250 000
1CA simplification
= 260 mm CA
OR/OF OR/OF
Map: Reality
Kaart : Werklikheid
1: 250 000
Map Dist/Kaart afstand : 65 km
1MA division by 250 000
Map distance =
Kaart afstand 1C conversion
= 0,00026 km C
= (0,00026 × 100 000) cm 1CA simplification
= 26 cm CA
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Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 5 DBE/2023
SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T/L
OR/OF
1 cm: 250 000 cm
C 1C conversion
Map Dist/Kaart afstand : 65 km
Map distance /Kaart afstand
= MA 1MA division by 2,50
= 26 cm CA
1CA simplification
(3)
MP
2.2.2 Bar scale or line scale or Graphic Scale A 2A correct scale L1
Staafskaal/ Balkskaal of lynskaal of Grafiese skaal (2) E
MP
*2.3.1 Number of reams lengthwise/ Getal rieme in die lengte 1MA dividing lengths L3
MA 1A simplification M
102 cm A
= = 3,65 ≈ 3 R 1R rounding down
27,94 cm
Number of reams widthwise /Getal rieme in die
breedte
44 cm 1MC A reams widthwise
= = 2,04 ≈ 2 MCA
21,59 cm
Number of reams heightwise/Getal rieme in die hoogte
39 cm
= = 6,14 ≈ 6 A 1A reams heightwise
6,35 cm
Total number of reams/Totale getal rieme
= 3 × 2× 6 MCA 1MCA multiplying the
= 36 CA values
1CA total number of reams
(7)
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Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 6 DBE/2023
SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T/L
O MP
2.3.2 To keep them dust free/Om stof af te keer L4
To keep the reams dry/ moisture free E
Om die rieme droog te hou
OR/OF
To keep them safe for later use.
Om hulle veilig te bêre vir latere gebruik
OR/OF
Glass door - For learners to see that the teacher is using
their reams of paper
– Easy to see how many reams are left (record
keeping).
Glasdeure - Sodat leerders kan sien hul onderwyser
gebruik hulle rieme papier
--maklik om te sien hoeveel rieme is oor (hou rekord)
OR/OF 2O reason
Convenient
–Paper is in the class for later usage.
–Keeps the teacher’s table clear/more space on
teacher’s table
–Easily accessible when needed.
–Effective use of space
Gerieflik:
– die papier is in die klas gereed vir later gebruik
– Hou die onderwyser se tafel skoon /meer spasie op
die onderwyser se tafel
– Maklike toegang te hê
– Effektiewe gebruik van spasie
OR/OF
Keeps the classroom neat and in order.
Hou die klaskamer netjies en skep orde.
(2)
[24]
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SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
QUESTION/VRAAG 3 [31 MARKS/PUNTE]
Q/V Solution/Oplossing Explanation/Verduideliking T/L
M
3.1.1 Surface area wall 1 = length × width L2
Oppervlakte muur 1 = lengte × breedte M
= 4,8 m × 2,75 m SF 1SF substitution
= 13,2 m2 CA 1CA simplification
Surface area wall 2 = length × width
Oppervlakte muur 2 = lengte × breedte
= 3,50 m × 2,75 m
= 9,6250 m2 A 1A simplification
Total surface area / Totale oppervlakte
= (13,2 + 9, 625) m2
= 22,8250 m2 CA 1CA simplification
OR/OF
OR/OF
SF 1SF substitution
T SA/ TO = (4,8 m × 2,75 m) + (3,5 m × 2,75 m) 1CA simplification
CA 1A simplification
= 13,2 m2 + 9,6250 m2 A
= 22,8250 m2 CA 1CA simplification
OR/OF OR/OF
A
Surface Area = (3,5 m + 4,8 m) 2,75 m SF 1A adding both wall dimensions
1SF substitution
= 8,3 2,75 CA 1CA simplification
= 22,825 m2 CA 1CA simplification
NPR
(4)
CA from 3.1.1 M
*3.1.2 Volume = Area of wall × thickness of plaster L3
Volume = Opp van muur × dikte van pleister D
C C 2C conversion
12
= ( 22,8250 ×10 000 ) × SF 1SF substitution
10
CA 1CA simplification
= 228 250 cm2 × 1,2 cm
= 273 900 cm3 CA 1CA simplification
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Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 8 DBE/2023
SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T/L
OR/OF
12 mm = 1,2 cm C 1C conversion mm to cm
22,825m2 = (22,825 × 100 × 100) cm2 C 1C conversion m2 to cm2
= 228 250 cm2 CA 1CA simplification
Volume = Area of wall × thickness of plaster
Volume = Opp van muur × dikte van pleister
= 228 250 cm2 × 1,2 cm SF 1SF substitution
= 273 900 cm3 CA 1CA simplification
OR/OF OR/OF
12 mm = (12 ÷ 1 000) 1C conversion m
= 0,012 m C
Volume = 22,825 × 0,012 SF 1SF substitution
= 0,2739 m3 CA 1CA simplification
= (0,2739 × 100 × 100 × 100) C 1C conversion m3 to cm3
= 273 900 cm3 CA 1CA simplification
OR/OF OR/OF
C 1C conversion mm2
Volume = 22 825 000 mm2 12 mm SF 1SF substitution
= 273 900 000 mm3 CA 1CA simplification
C
= 273 900 cm3 CA 1C conversion cm3
1CA simplification
(5)
CA from 3.1.2 M
3.1.3 Number of bags/ Getal sakke L2
273 900 cm 3 MCA 1MCA dividing M
=
15 000 cm 3
= 18,26 CA 1CA simplification
≈ 19 R 1R rounding up
(3)
M
3.1.4 Perimeter / Omtrek L2
RT 1SF substitution E
= 2 × (4,8 + 3,5) m SF 1RT correct values
= 16,6 m CA 1CA simplification
OR/OF OR/OF
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Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 9 DBE/2023
SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T/L
Perimeter/ Omtrek
1MA adding all 4 sides
= 4,8 m + 3,5 m + 4,8 m + 3,5 m MA RT
1RT correct values
= 16,6 m CA 1CA simplification
OR/OF OR/OF
Perimeter/Omtrek MA
= 2(3,5 m) + 2(4,8 m) RT 1MA adding all 4 sides
1RT correct values
= 16,6 m CA 1CA simplification
AO
(3)
P
P = A or/of 0,2 or/of 20%
3.2.1 1 1A numerator L2
5 A 1A denominator E
AO
(2)
P
*3.2.2 P(not appear/ nie verskyn) = 1 – 0,75 MA 1MA subtracting from 1 L2
A M
= 0,25 or/of ¼ or/of 1A simplification
25% AO
(2)
CA from Q3.2.2 P
3.2.3 Less likely /kleiner kans A 2A correct likelihood L2
(2) E
M
*3.3.1 Starting time /Begin tyd L2
MA A 1MA subtract minutes M
= 08:05 – 2 min – 3 min – 4 min 1A all the minutes
= 07:56 CA 1CA simplification
OR/OF OR/OF
Total time to prepare:
= 4 min + 3 min + 2 min
= 9 min A 1A all the minutes
Starting time /Begin tyd
= 08:05 – 9 min MA
1MA subtract minutes
CA
= 07:56 or 4 minutes to eight in the morning 1CA simplification
AO
(3)
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Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 10 DBE/2023
SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T/L
M
3.3.2 Total volume of water /totale volume water L2
MA 1MA multiplying with 5 M
= 7 ℓ × 5 = 35 ℓ
1 gallon/gelling = 3,78541 ℓ
Number of gallons /Getal gellings
35 1C converting
= C
3, 78541
= 9,24602619 ≈ 9,25 R 1R rounded answer
OR/OF OR/OF
1 gallon/gelling = 3,78541 ℓ
n =7ℓ
Number of gallons /Getal gellings
7 1C converting
= C
3, 78541
= 1,849205… ≈ 1,85
For 5 bags/Vir 5 sakke
MA R 1MA multiplying with 5
= 1,85 × 5 = 9,25 gallon / gelling 1R rounded answer
(3)
M
3.3.3 °F – 32° = (1,8 × °C) L2
SF M
73,4 – 32° = (1,8 × °C) 1SF correct substitution
S 1S simplification
41,4° = 1,8 × °C
MCA 1MCA dividing by 1,8
°C = 41,4° ÷ 1,8
1CA simplification
= 23 °C CA (4)
[31]
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SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
QUESTION/VRAAG 4 [30 MARKS/PUNTE]
Q/V Solution/Oplossing Explanation/Verduideliking T/L
MP
4.1.1 12 RT 2RT number of houses L2
(2) E
4.1.2 RT RT 2RT 1st house label or MP
1 , 2 or/of 12 number L2
1RT second M
Any two /Enige twee (3)
M
*4.2.1 The depth 1m or it is shallow/ not too deep. O 2O explanation L4
Die diepte is 1m of dit is vlak/ nie te diep nie. M
(2)
A M
4.2.2 Capacity: the maximum amount of water the pool can L1
hold/contain. M
Kapasiteit is die maksimum hoeveelheid water wat die
swembad kan hou. 2A concept
OR/OF
Capacity: a measure of space covered by pool structure
with water.
Kapasiteit is die mate van spasie wat die swembad met
water vul.
(2)
A 1A radius M
*4.2.3 SF 1SF correct substitutions L3
Volume(cylinder) = 3,142 × ( )2 × 1m
M
= 3,142 × (3,5m)2 × 1m
= 38,4895 m3 CA 1CA simplification
Volume (rectangular) = 6,2 m × 3,25 m × 1,65 m SF 1SF correct values
= 33,2475 m3 CA 1CA rectangular volume
Difference / Verskil = 38,4895 m3 – 33,2475 m3 MCA 1MCA subtracting
= 5, 242 m3 CA 1CA difference
= 5 242 ℓ C 1C conversion
OR/OF OR/OF
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Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 12 DBE/2023
SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T/L
A 2 SF C 1A radius
Volume(cylinder) = 3,142 × (3,5m) × 1m × 1 000 ℓ/ m3 1SF correct substitutions
1C conversion
= 38 489,5 ℓ CA 1CA simplification
SF
Volume (rectangular) = 6,2 m × 3,25 m × 1,65 m × 1 000ℓ/m3 1SF correct values
= 33 247,5 ℓ CA 1CA rectangular volume
Difference / Verskil = 38 489,5 ℓ – 33 247,5 ℓ MCA 1MCA subtracting
= 5 242 ℓ CA 1CA difference
NPR
(8)
O M
4.2.4 To accommodate cutting the tiles or breakages or L4
(a) curved surfaces or keep spares for later usage. 2O reason E
Om voorsiening the maak die sny van teëls of
breekskade of die gekurfde oppervlaktes of om oor te
hou vir latere gebruik. (2)
CA radius form 4.2.3 M
*4.2.4 SA(open cylinder)/BO = 3,142 × radius × (radius + 2 × height) L4
(b) = 3,142 × 3,5 m × (3,5 m + 2 × 1 m) SF 1SF substitution D
= 3,142 × 3,5 m × 5,5 m = 60,4835 m2. CA 1CA area of pool
Area of one tile/ Opp van 1 teël
= m× C 1C conversion
= 0,2 × 0,2 m2 = 0,04 m2. CA 1CA area of a tile
Number of tiles needed /Getal teëls nodig
Area to be tiled
=
Area of single tile
60,4835 MCA 1MCA finding number of
= tiles
0,04 CA
1CA simplification
= 1 512,0875 OR/OF ≈ 1 513
Or rounded up:
Plus 10%
10 MCA
= × 1 512,0875 + 1 512,0875 1 513 × 110% 1MCA calc. 10% and adding
100 it or multiply with 1,10
= 1 664,3
= 1 663,29625 tiles /teëls ≈ 1665
≈ 1 664 tiles /teëls CA 1CA number of tiles
Number of boxes /Getal bokse Boxes
= 1 664 ÷ 16 MCA = 1 665 ÷ 16
1MCA dividing
= 104,06
= 104 CA ≈ 105 1CA number of boxes
VALID/GELDIG O 1O conclusion
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Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 13 DBE/2023
SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T/L
OR/OF
SA(open cylinder)/BO = 3,142 × radius × (radius + 2 × height)
1SF substitution
= 3,142 × 3,5 m × (3,5 m + 2 × 1 m) SF
1CA SA of pool
= 3,142 × 3,5 m × 5,5 m = 60,4835 m . CA2
Area of one tile/ Opp van 1 teël
1C conversion
= m× C
1CA area of a tile
2
= 0,2 × 0,2 m = 0,04 m . 2 CA
Number of tiles needed /Getal teëls nodig
Area to be tiled
= 1MCA finding number of
Area of single tile
tiles
60,4835 MCA
=
0,04 CA 1CA simplification
= 1 512,0875 OR/OF ≈ 1 513
Number of boxes/Getal bokse
= 1 512,0875 ÷ 16 MCA 1MCA dividing
= 94,505… CA 1CA number of boxes
Increased number/ Verhoogde getal
= 94,505… × 110% MCA 1MCA calc. 10% and adding
= 103,95… it or multiply with 1,10
≈ 104 CA 1CA number of boxes
VALID/GELDIG O 1O conclusion
OR/OF OR/OF
SA(open cylinder)/BO = 3,142 × radius × (radius + 2 × height)
= 3,142 × 3,5 m × (3,5 m + 2 × 1 m) SF 1SF substitution
1CA area of pool
= 3,142 × 3,5 m × 5,5 m = 60,4835 m2. CA
Area of one tile/ Opp van 1 teël
= m× C 1C conversion
CA 1CA area of a tile
= 0,2 × 0,2 m2 = 0,04 m2.
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Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 14 DBE/2023
SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T/L
Continue
Area covered by one box/Opp wat een boks bedek 1MCA finding area of box of
= 0,04 m2 × 16 MCA tiles
= 0,64 m2 CA 1CA simplification
Number of boxes/Getal bokse
60,4835 1MCA dividing
= MCA 1CA number of boxes
0, 64
=94,505… CA
1MCA calc. 10% and adding
Increased number/ Verhoogde getal
it or multiply with 1,10
= 94,505… × 110% MCA
= 103,95…
≈ 104 CA 1CA number of boxes
O 1O conclusion
VALID/GELDIG
OR/OF
OR/OF
SA(open cylinder)/BO = 3,142 × radius × (radius + 2 × height)
= 3,142 × 3,5 m × (3,5 m + 2 × 1 m) SF 1SF substitution
= 3,142 × 3,5 m × 5,5 m = 60,4835 m2. CA 1CA area of pool
Increased area/Vergrote opp 1MCA calc. 10% and adding
= 60,4835 × 1,1 it or multiply with 1,10
MCA 1CA simplification
= 66,53185 CA
Area of one tile/ Opp van 1 teël
1C conversion
= m× C
1CA area of a tile
= 0,2 × 0,2 m2 = 0,04 m2. CA
Area covered by one box/Opp wat een boks bedek
1MCA finding area of box of
= 0,04 m2 × 16 MCA
tiles
= 0,64 m2
CA 1CA simplification
Number of boxes/Getal bokse
66,53185 1MCA dividing
= MCA
0, 64
= 103,956… 1CA number of boxes
≈ 104 CA
VALID/GELDIG O 1O conclusion
(11)
[30]
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Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 15 DBE/2023
SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
QUESTION/VRAAG 5 [37 MARKS/PUNTE]
Q/V Solution/Oplossing Explanation/Verduideliking T/L
MP
5.1.1 R617 RT 2RT correct road L1
(2) E
MP
5.1.2 6 km RT 2RT correct distance L1
(2) E
MP
5.1.3 Left /Links A 2A correct direction L1
(2) E
M
5.1.4 Dist. = 16km – dist. Hotel to Mkomazana – dist. L2
Himeville to turn RT 1RT 2 km M
Afstand = 16 km – 4,4 km – 2 km M 1M subtracting of at least one
correct value
= 9,6 km CA 1CA simplification
(3)
MP
5.2.1 The owners need to clean, put fresh linen on the L4
beds and get the cottage ready early for the next M
booking. O
Die eienaars moet die plek skoon maak, skoon
beddegoed opsit en die kothuis vroeg gereed kry vir
die volgende bespreking.
OR/OF
To encourage spending more days at the venue. 2O reason
Om kliënte aan te moedig om meer dae oor te bly.
OR/OF
Breaking a long stay.
To justify renting out the cottage for one night while
someone else might have stayed longer.
Dit onderbreek ‘n lang oorbly.
Om dit te regverdig om die kothuis uit te verhuur vir
een nag terwyl iemand anders langer kon oorbly. (2)
MP
A
5.2.2 Valley Cottage OR Coot Cottage 1A one correct cottage L4
Vallei kothuis OF Bleshoender kothuis E
O
These cottages sleeps 6 persons OR can
accommodate 5 to six OR they are 5 OR Cheaper
option / Price of 6 sleeper accommodation is more 2O reason
reasonable (R2 640 ÷ 6 = R440 pp) compared to the
2 sleeper (R1 150 ÷ 2 = R575 pp) /
Hierdie kothuise is vir 6 persone OF hulle kan 5 tot
6 persone huisves OF die groep is 5 OF Dit is die
goedkoper opsie / Koste per persoon vir die 6-
persoon is goedkoper (R2 640 ÷ 6 = R440 pp)
vergeleke met 2-persoon (R1 150 ÷ 2 = R575 pp)
. (3)
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Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 16 DBE/2023
SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T/L
CA from 5.2.2 M/F
*5.2.3 6 sleeper / slaper: L4
Total cost / totale koste RT RT 1RT correct rate, Thursday M
= R2 640 + R2 730 × 2 MA 1RT correct rate, weekend
1MA multiplying with 2
= R8 100 CA 1CA simplification
VALID/ GELDIG O 1O conclusion
(5)
MP
A
5.3.1 False, the map shows other roads also have toll 1A correct option L4
(a) gates. O 1O reason E
Onwaar, die kaart toon ook ander tolhekke
(2)
MP
5.3.1 True/ Waar A 2A correct option L4
(b) (2) E
O MP
5.3.2 Top view or aerial view or bird's eye view or 2O correct view L1
satellite view E
Bo-aansig of vanuit die lug aansig of voël-aansig (2)
of satelliet aansig.
M
5.3.3 Distance/Afstand = speed/spoed × time/tyd 1SF substitution L2
588 km = speed/spoed × 7 h SF M
Speed/spoed =
588 km A 1A change subject of formula
7h
= 84 km/hCA 1CA simplification
(3)
CA from 5.2.3 M/F
*5.3.4 Expense for tolls / Tol-fooie: RT 1 RT correct 4 tolls L3
= R56,00 + R77,00 + R82,00 + R58,00 D
= R273 CA 1CA simplification
Fuel used/Brandstof verbruik
= 588 km ÷ 100 km × 6,42 ℓ MA 1MA fuel consumption rate
= 37,7496 ℓ A 1A simplification
Fuel cost / Brandstofkoste = 37,7497 × R21,40
= R807,84 CA 1CA fuel cost
Total cost / Totale koste
= R8 100 + R807,84 × 2 + R273,00 × 2 MCA 1MCA return trip
= R10 261,68 CA 1CA total cost for 3 items
Cost per person / Koste per person
= R10 261,68 ÷ 5 MCA 1MCA dividing by 5
= R2 052,34 CA 1CA simplification
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Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 17 DBE/2023
SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T/L
OR/OF
Round trip/ Heer en terug 588 km × 2 = 1176 km
Fuel used/Brandstof verbruik 1MA fuel consumption rate
= 1 176 km ÷ 100 km × 6,42 ℓ MA 1A simplification
= 75,4992 ℓ A
Fuel cost / Brandstofkoste = 75,4992 × R21,40
= R1 615,68 CA 1CA fuel cost
Cost per person/Koste per persoon
= R1 615,68 ÷ 5 MA 1MA dividing by 5
= R323,14
Toll fees / Tol-fooie:
= R56,00 + R77,00 + R82,00 + R58,00 RT 1 RT correct 4 tolls
= R273 CA 1CA simplification
Round trip/ Heen en weer
= R273 × 2 MCA 1MCA return trip
= R546
Cost per person/Koste per persoon
= R546 ÷ 5
=R109,20
Accomodation per person/Verblyf per persoon
=R8100 ÷ 5
=R1 620
Total per person/ Totaal per persoon
= R323,14 + R109,20 + R1 620 MCA 1MCA adding all the values
= R2 052,34 CA 1CA total cost
OR/OF OR/OF
Toll Expenses / Tol-fooie:
MCA RT 1MCA return trip
= 2(R56,00 + R77,00 + R82,00 + R58,00)
1 RT correct 4 tolls
= R546,00 CA 1CA simplification
Fuel Cost /Brandstof koste
Total Distance/Afstand = 588 km × 2 = 1176 km
MA A 1MA fuel consumption rate
1176
Fuel used/Brandstof: 6,42 = 75,4992 ℓ 1A simplification
100
Cost/Koste: 75,4992 × R21,40 = R1 615,68 CA 1CA fuel cost
Total Cost/Totale koste:
CA 1CA total cost for 3 items
R8 100 + R546,00 + R1 615,68 = R10 261,68
MCA CA 1MCA dividing by 5
Cost PP/ Koste PP: R10 261,68 ÷ 5 = R2 052,34 1CA simplification
(9)
[37]
TOTAL/TOTAAL: 150
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