# 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. - Page: https://www.hlayiso.com/library/mathematical-literacy-p2-may-june-2023-mg-afr-eng/ebcf5077-e428-4eb2-8a9f-77b84c95ab3d - Updated: 2026-10-29 - Subject: Mathematical Literacy - Grade: Grade 12 - Document type: Memorandum - Year: 2023 - Exam period: NSC June Exam - Paper: 2 - Curriculum: CAPS - Pages: 17 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Mathematical Literacy](https://www.hlayiso.com/library/grade-12/mathematical-literacy) › [Mathematical Literacy Grade 12 Paper 2 NSC June 2023 Memo](https://www.hlayiso.com/library/mathematical-literacy-p2-may-june-2023-mg-afr-eng/ebcf5077-e428-4eb2-8a9f-77b84c95ab3d) ## About this document Mathematical Literacy · Grade 12 · NSC June Exam · 2023. 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Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com 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. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 2 Downloaded from hlayiso.com 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 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Downloaded from hlayiso.com 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] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Downloaded from hlayiso.com 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 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Downloaded from hlayiso.com 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) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com 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] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 7 DBE/2023 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 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Downloaded from hlayiso.com 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 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com 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) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com 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] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 11 DBE/2023 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 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com 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 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Downloaded from hlayiso.com 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. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Downloaded from hlayiso.com 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] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Downloaded from hlayiso.com 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) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 Downloaded from hlayiso.com 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/hCA 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 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 17 Downloaded from hlayiso.com 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 Copyright reserved/Kopiereg voorbehou ## Related documents - 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