# Mathematical Literacy Grade 11 P2 2017 Memo (Eng) > Grade 11 Mathematical Literacy Paper 2 memorandum from the Eastern Cape November 2017 (English), 8 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/maths-lit-p2-gr11-memo-nov2017-english/d1e2ed82-2941-4a4c-8a81-6976dd707a71 - Updated: 2026-09-27 - Subject: Mathematical Literacy - Grade: Grade 11 - Language: English - Document type: Memorandum - Year: 2017 - Exam period: Eastern Cape November - Paper: 2 - Curriculum: CAPS - Pages: 8 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 11](https://www.hlayiso.com/library/grade-11) › [Mathematical Literacy](https://www.hlayiso.com/library/grade-11/mathematical-literacy) › [Mathematical Literacy Grade 11 P2 2017 Memo (Eng)](https://www.hlayiso.com/library/maths-lit-p2-gr11-memo-nov2017-english/d1e2ed82-2941-4a4c-8a81-6976dd707a71) ## About this document Mathematical Literacy · Grade 11 · Eastern Cape November · 2017 · English. Memorandum, 8 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Matching paper and memo - [MATHS LIT P2 GR11 QP NOV2017 English](https://www.hlayiso.com/library/maths-lit-p2-gr11-qp-nov2017-english/5e367a82-a186-4920-b623-cb6113e80d94) — QP · 2017 · Eng ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 11 NOVEMBER 2017 MATHEMATICAL LITERACY P2 MARKING GUIDELINE MARKS: 100 Symbol Explanation M Method MA Method with accuracy CA Consistent accuracy A Accuracy C Conversion S Simplification RT/RG/RM Reading from a table/Reading from a graph/Reading from a map F Choosing the correct formula SF Substitution in a formula J Justification P Penalty, e.g. for no units, incorrect rounding off etc. R Rounding Off/Reason AO Answer only NPR No penalty for rounding This marking guideline consists of 8 pages. ### Page 2 Downloaded from hlayiso.com 2 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2017) QUESTION 1 [29] Question Solution Explanation Topic and Level 1.1 Total number of cars and microbuses = (10 × 5) + (5 × 5) 1 M Calculated total P = 50 + 25 number of cars and L2 = 75  microbuses 50  1 A Numerator Probability = 75  1A Denominator Accept 0,667 OR 66,7% (3) 1.2 Amount for cars washed Monday – Friday: 1M Multiplied by 10 F 10 × 30 = 300 × 5  and 5 L4 = R1 500  1CA Amount for cars 160 1M Increase by 60% Amount for cars Saturday and Sunday =  × 10 100 1CA Number of cars = 16  1CA Amount for cars = 16 × 30 × 2 on Saturday and = R960  Sunday Amount for microbuses (Monday – Friday) = 5 × 40 1CA Number of = R200 microbuses 200 × 5 = R1 000  1CA Number of Amount for microbuses (Saturday and Sunday): microbuses 130 × 5 = 6,5 =7  1CA Amount for 100 microbuses 7 × 40 × 2 = R560  Total Amount = 1 500 + 1 000 + 960 + 560 1CA Total amount = R4 020  1 O Claim valid Claim is valid  (10) Copyright reserved Please turn over ### Page 3 Downloaded from hlayiso.com (EC/NOVEMBER 2017) MATHEMATICAL LITERACY P2 3 1.3 Water per car = 25 × 3 CA from 1.1 = 75 litres M L3 Water for cars in 5 days = 10 × 5 × 75 1M Water for cars = 3 750 litres  Mondays to Fridays Water for cars on Saturdays and Sundays = 16 × 2 × 75 1CA Water for cars on = 2 400 litres  Saturday and Sunday Water per microbus = 25 × 4 = 100 litres 1CA Water for Water per microbus in 5 days = 5 × 5 × 100 microbuses Mondays to = 2 500 litres  Fridays Water per microbus on Saturdays and Sundays 1CA Microbuses on = 7 × 2 × 100 Saturdays and Sundays = 1 400 litres  Total number of litres = 3 750 + 2 400 + 2 500 + 1400  1M Addition = 10 050 litres  1CA Answer OR Amount of water = amount for cars + amount for microbuses = (3 × 25 × 82) + (4 × 25 × 39)  = 6 150 litres  + 3 900 litres  = 10 050 litres  (6) 1.4 Water for 7 days = 10 050 litres CA from 1.3 M&F Water for a month = 10 050 × 4 L3 = 40 200 litres  1MA Amount of litres 40 200 l = 40,2 kilolitres  1C Amount of kilolitres Cost of water from table: 1M First tariff 6 × 8.66 = R51,96  1M Second tariff 9 × 10.02 = R90,18  1M Third tariff 15 × 12,28 = R184,20  1M Fourth tariff 10,2 × 15.25 = R155,55  Total amount = R481,89 1M Calculated VAT = R481,89 x 1,14  1CA Amount with = R549,35  VAT (8) 1.5 Most people do not work on Saturdays and Sundays.  2O Opinion DH It is the only time they have available to wash their L4 cars.  Accept any other relevant answer. (2) Copyright reserved Please turn over ### Page 4 Downloaded from hlayiso.com 4 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2017) QUESTION 2 [24] 2.1 2.1.1 R212  1RT Correct value from DH table (2) L2 2.1.2 188−76  1M Used correct values F × 100 76  1M Divided by correct L4 = 147,37% OR 147,4% value 1CA % Statement is valid  1J Valid % increase is more than 100% NPR (4) 2.1.3 Cost of burying a 5-year-old is less than that of F burying a 15-year old.  L4 Accept any other relevant answers. 2J Opinion (2) 2.2 2.2.1 Diameter = 150 cm 1M Subtraction M Length of figure = 100 cm 1M Used 60 L4 A = 100  – 60  1CA Value of A = 40 cm  B = 150 – 100 = 50 cm ÷ 2  1 Divided by 2 = 25 cm  1CA Answer 25 : 40 1M Ratio =5:8 No penalty if not simplified Penalise 1 mark if ratio is written with units (6) Copyright reserved Please turn over ### Page 5 Downloaded from hlayiso.com (EC/NOVEMBER 2017) MATHEMATICAL LITERACY P2 5 2.2.2 Area of cross shaped glass = Area of longer rectangle + 2 × area of smaller rectangles M&F = l × b + 2(l × b) 1M Substituted in L4 = 100 × 40 + 2(30 ×40)  formula = 4 000 + 2 400 = 6 400 cm2  1CA Simplification OR 4(40 x 30) + (40 x 40)  = 6 400 cm2  OR 2(100 x 40) – 402 = 8 000 –1 600  = 6 400 cm2  6 400 1C Converted to square = 0,64 m2  10 000 metres (m²) Area wasted = Area of circle – Area of shape 1M Used formula = πr2 – 0,64 1SF Substitution = 3,142 × 0,75 × 0,75  – 0,64 = 1,767375 – 0,64  1 CA Unused area = 1,127375m2  Amount lost = 1,127375 × 15 1CA Total amount = R16,91  1M Added VAT 114 With VAT = × 16,91  100 1CA Answer = R19,28  1O Opinion (10) Statement is valid.  Copyright reserved Please turn over ### Page 6 Downloaded from hlayiso.com 6 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2017) QUESTION 3 [16] 3.1 Company A: DH L3&4 29 000+25 000+24 000+15 000+15 000+8 000 1M Addition Mean =  6 1M Divided by 6 116 000 = 6 = R19 333,33 1CA Mean Company B: 31 000+29 000+17 0000+14 000+13 000+7 000 Mean = 6 111 000 = 6 =R 18 500  1CA Mean Difference = 19 333,33 – 18 500 = R833,33  1CA Difference Statement is not valid 1O Not valid (6) 3.2 No modal salary for company B.  2A DH (2) L2 3.3 Modal value = 15 000  1RT Correct modal value 15 1M Calculated 15% of × 15 000  100 modal value DH&F = R2 250  1CA Value L3 105 Year 1 = 100 × 2 250  = R2 362,50  1M Increased by 105 5% Year 2 = 100 × 2 362,50 1CA First year = R2 480,63  1CA Final amount (6) 3.4 0 OR 0% OR Impossible  DH 2A Answer (2) L2 Copyright reserved Please turn over ### Page 7 Downloaded from hlayiso.com (EC/NOVEMBER 2017) MATHEMATICAL LITERACY P2 7 QUESTION 4 [31] 4.1 4.1.1 A = 21 168 700 + 2 305 800 + 677 000 + 2 214 400 1 MA Added correct DH = 26 365 900  values L2 1CA Total 22 165 000 B = 27 635 900  × 100 1 Correct values = 80,2%  1CA % (4) 4.1.2 Indian / Asian  2A Answer DH (2) L2 4.1.3 Two million two hundred and fourteen thousand four 2A Answer DH hundred  (2) L2 4.2 4.2.1 57  – 38  = 19  1M Correct number of M&P seats on the left L2 1M Subtraction 1M Correct number of seats on the right 1CA Answer (4) 4.2.2 4 1RM Correct values M&P Percentage of seats for the handicapped 38 x 100  numerator and L3 denominator 10,526315789 1M Multiplied by 100 1CA Answer (3) 10,53%  4.3 4.3.1 Total distance = 35 + 57 + 11 + 21 + 59 + 41  1RM Correct values M, = 224 km  𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 M&P Speed = 𝑡𝑖𝑚𝑒 1CA Distance L3&4 224 105 = 𝑡𝑖𝑚𝑒  1M Substituting in 224 formula Time = 105 = 2,133 hours  1 S Change subject of Minutes = 0,133 × 60  formula = 7,98 1CA Time in hours 2 hours 8 minutes  1C Time in minutes Time of arrival = 7:00 + 2 hours 8 minutes 1CA Arrival time = 9:08 am  1O Claim valid (8) Claim is valid  Copyright reserved Please turn over ### Page 8 Downloaded from hlayiso.com 8 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2017) 4.3.2 CA from 4.3.1 F 2,82 1M Divided by 1,068 L3 Amount before increase =1,068  1A Amount = R2,64  I CA Distance Total distance travelled = 224 x 2 = 448 km  1M Multiplied by amount calculated Amount = 448 x 2,64  1CA Amount = R1 182,72  1R Nearest rand (6) = R1 183  4.3.3 R61  1A R61 M&P R349  1A R349 (2) L2 TOTAL: 100 Copyright reserved Please turn over ## Related documents - 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