# Mathematical Literacy Grade 12 P2 2020 Memo (Eng) > Grade 12 Mathematical Literacy Paper 2 memorandum from the Eastern Cape Mock Exam 2020 (English), 11 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/math-lit-p2-memo-gr12-sept2020-english/e046437f-6884-42d5-944c-ab0ab8e002e1 - Updated: 2026-09-27 - Subject: Mathematical Literacy - Grade: Grade 12 - Language: English - Document type: Memorandum - Year: 2020 - Exam period: Eastern Cape Mock Exam - Paper: 2 - Curriculum: CAPS - Pages: 11 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 P2 2020 Memo (Eng)](https://www.hlayiso.com/library/math-lit-p2-memo-gr12-sept2020-english/e046437f-6884-42d5-944c-ab0ab8e002e1) ## About this document Mathematical Literacy · Grade 12 · Eastern Cape Mock Exam · 2020 · English. Memorandum, 11 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Matching paper and memo - [MATH LIT P2 GR12 ADD SEPT 2020 English](https://www.hlayiso.com/library/math-lit-p2-gr12-add-sept-2020-english/eee4653c-50e8-4339-9537-ac35e93bb04d) — QP · 2020 · Eng - [MATHS LIT P2 QP GR12 SEPT2020 English](https://www.hlayiso.com/library/maths-lit-p2-qp-gr12-sept2020-english/159fc7c8-fb75-42e5-a050-9af674b8bdba) — QP · 2020 · 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 12 SEPTEMBER 2020 MATHEMATICAL LITERACY P2 MARKING GUIDELINE MARKS: 150 Symbol Explanation M Method M/A Method with accuracy MCA Method with consistent accuracy CA Consistent accuracy A Accuracy C Conversion S Simplification RT/RG/RM Reading from a table OR Reading from a graph OR Read from 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 OR Reason AO Answer only NPR No penalty for rounding This marking guideline consists of 11 pages. ### Page 2 Downloaded from hlayiso.com 2 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2020) QUESTION 1 [37] Ques. Solution Explanation Level 1.1.1 Amount used for 3 batches = 3 × 125 MA 1MA Multiplying M = 375 mℓ correct values L2 375 1MA Dividing by Number of cups = 250 MA 250 = 1,5 cups OR 1½ cups A 1A Number of cups (3) 14,99 1.1.2 Price of 3 eggs = M 1M Dividing by 12 F 12 1M Multiply by 3 L3 = R1,249166667 × 3 M 1S Simplification = R3,7475 S 1CA Cost ≈ R3,75 CA (4) 1.1.3 °Fahrenheit = 1,8 × °Celsius + 32° 1SF Substitute M = 1,8 × 180° + 32°SF correct °C L2 = 356°F – 330°F M 1M Subtract correct = 26°F values (2) 1.1.4 Time taken for 9 batches = 25 min + 45 min 1MA Total time for M MA preparation and L4 = 70 min × 9 baking = 630 min CA 1CA Total time for Time in hours and minutes = 10 hours 30 minC 9 batches Time completed = 09 hours 15 min + 10 hours 30 min M 1C Convert to hours = 19:45 CA and minutes Not valid O 1M Adding time 1CA Time completed 1O Not valid (6) Copyright reserved Please turn over ### Page 3 Downloaded from hlayiso.com (EC/SEPTEMBER 2020) MATHEMATICAL LITERACY P2 3 1.1.5 3 1C mℓ to gram F&M Cake flour = × 250 = 150 grams 5 21,99 M 1M Dividing L4 Cost of cake flour in grams = 1C kg to g 2500 C MCA = 0,008796 × 150 × 9 1MCA Multiply by = R11,87 CA 150 and 9 35,99 Cost of oil in millilitres = 1CA Cost of flour 2000 = 0,017995 × 125 × 9 1CA Cost of oil = R20,24 CA 14,99 1CA Cost of eggs Cost of eggs = CA from 1.1.2 12 = 1,249166667 × 3 × 9 1CA Total cost = R33,73CA Total cost = R11,87 + R20,24 + R33,73 1O Invalid = R65,84 CA Invalid O OR 1C mℓ to gram 1M Dividing 3 Cake flour = × 250 = 150 grams C 1C g to kg 5 21,99 M 1M Multiply by Cost for cake flour in kg = 2,5 C 0,15 and 9 = 8,796 × 0,15 × 9 M 1CA Cost of flour = R11,87 CA 35,99 Cost of oil in litre = 1CA Cost of oil 2 = 17,995 × 0,125 × 9 = R20,24 CA 14,99 Cost of eggs = 1CA Cost of eggs 12 = 1,249166667 × 3 × 9 CA from 1.1.2 = R33,73 CA 1CA Total cost Total cost = R11,87 + R20,24 + R33,73 1O Invalid = R65,84 CA Invalid O (9) Copyright reserved Please turn over ### Page 4 Downloaded from hlayiso.com 4 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2020) 1.2.1 7 feet 8 inches = (7 × 0,3048) + (8 × 0,0254) C 1C cm to m M = 2,1336 m + 0,2032 m S 1M Multiplying and L3 = 2,3368 m CA adding 1S Simplification 6 feet 6 inches = (6 × 0,3048) + (6 × 0,0254) 1CA Answer = 1,8288 m + 0,1524 m = 1,9812 m CA 1CA Answer Length = 2,3368 m + 1,9812 m 1CA Length = 4,318 m CA OR 1M Adding Total feet in metres = 7 feet + 6 feet M 1M Multiply = 13 feet × 0,3048 M 1CA Answer = 3,9624 m CA Total inches in metres = 8 inches + 6 inches = 14 inches × 0,0254 C 1C cm to m = 0,3556 m CA 1CA Answer Length = 3,9624 m + 0,3556 m 1CA Length = 4,318 m CA (6) 1.2.2 Top view A 1A View M& All features clearly visible R 2A Reason P (3) L4 1.2.3  People preparing meals in the kitchen. A 2A First reason M&  People helping to prepare meals. A 2A Second reason P  People in and out of the kitchen. A L4  It is frequently visited by all. A Copyright reserved Please turn over ### Page 5 Downloaded from hlayiso.com (EC/SEPTEMBER 2020) MATHEMATICAL LITERACY P2 5 QUESTION 2 [39] Ques. Solution Explanation Level 2.1.1 Simple Interest A 1SF Substitution F Interest = 280 000 × 0,0975 × 2,916666667 SF 1A Number of years L2 = R79 625 CA 1CA Interest (3) 2.1.2 Compound Interest F First year = 280 000 + (280 000 × 0,0825) M 1M Adding and L3(6) = R303 100 CA Multiplying L4(1) Second year = 303 100 + (303100 × 0,0825) 1CA Amt 1st year = R328 105,75 CA A 1CA Amt 2nd year 11 months = R328 105,75 + (R328 105,75 × 0,075625) 1A Interest rate for = R352 918,75 CA 11 months OR 1CA Total amount Amount after 35 months = 280 000 × 1,0825 × 1,0825 × 1,075625 = R352 918,75 1CA Interest Interest = R352 918,75 – R280 000 = R72 918, 75 CA 1O Option 2 She will pay less interest on Option 2 (compound interest) CA from 2.1.1 (7) 2.2.1 Factory 1 M D IQR = 75 000 – 40 000 RD 1RD Q1 and Q3 L3(5) = 35 000 CA 1M Concept of IQR L4(3) Median = 60 000 1CA IQR Factory 2 IQR = 80 000 – 35 000 RD 1RD Q1 and Q3 = 50 000 CA 1CA IQR Median = 50 000 A O Factory 2 performed worse because their IQR is greater 1A Factory 2 and their median is lower O 1O Comparing median 1O Comparing IQR’s (8) 2.2.2 No, factory 2 is still new A CA from 2.2.1 D 2A No with reason L4 (2) 94 MA 2.3.1 Percentage = × 100% M 1MA Dividing M 150 correct values L2 = 62,7% 1M Multiply by 100 (2) Copyright reserved Please turn over ### Page 6 Downloaded from hlayiso.com 6 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2020) 2.3.2 310 𝑚𝑚 M 1A Correct M Over the length = 78 𝑚𝑚 A diameter L4 = 3,97… 1M Divide correct ≈ 3 coffee mugs A 220 𝑚𝑚 values Over the width = 1A Number of 78 𝑚𝑚 = 2,8… mugs (length) ≈ 2 coffee mugs CA 1CA Number of 150 𝑚𝑚 Over the height = mugs (width) 94 𝑚𝑚 = 1,6… 1CA Number of ≈ 1 coffee mug CA mugs (height) Number of mugs in 1 box = 3 × 2 × 1 = 6 coffee mugs CA 1CA Total 66 Number of boxes = 1CA Number of 6 = 11 boxes CA boxes Statement invalid O 1O Invalid (8) 2.4.1 Northeast A 2A Northeast M&P North A 2A North (4) L2 Probability = A 2.4.2 1 1A Numerator P 4 A 1A Denominator L2 (2) 2.4.3 Turn left on the N6 in East London to Queenstown A 1A Left on N6 M&P Turn right on the R56 to Kokstad A 1A Right on R56 L4 Turn left on the N2 A 1A Left on N2 (3) Copyright reserved Please turn over ### Page 7 Downloaded from hlayiso.com (EC/SEPTEMBER 2020) MATHEMATICAL LITERACY P2 7 QUESTION 3 [36] Ques. Solution A Explanation Level 3.1.1 Initial tax = 0,18 × 195A 850 1A Use correct % F = R35 253 1A Use correct L2 amount (2) 3.1.2 7,5 1MA 7,5% of F Pension = × 37 537,75 MA 100 correct value L3 = R2 815,33125 × 12 MA 1MA Multiply by = R33 783,98CA 12 1CA Pension Donation = 575 × 12 amount = R6 900 A 1A Donation amount Total = R33 783,98 + R6 900 1CA Total amount = R40 683,98 CA (5) MA MCA 3.1.3 Taxable income = (37 537,75 × 12) – R40 683,98 CA from 3.1.2 F = R450 453 – R40 683,98 1M Multiply by 12 L2 = R409 769,02 CA 1MCA Subtract pension and donations 1CA Taxable income (3) A 3.1.4 Tax payable = 63 853 + 31% of taxable income above 305 850 CA from 3.1.3 F MCA 1A Correct tax L3&4 = 63 853 + 0,31 × (409 769,02 – 305 850) bracket = 63 853 + 0,31 × 103 919,02 1MCA Amount = 63 853 + 32 214,8962 above S M 1S Simplification = R96 067,8962 – R14 220 1M Subtract rebate 81 847,8962 = M 1M Divide by 12 12 1CA Monthly tax = R6 820,66 CA 1O Invalid or Less Invalid OR less than R6 850 O than (7) 3.1.5 They receive 3 rebates A 2A Explanation F OR L4 Their rebate is higher A (2) Copyright reserved Please turn over ### Page 8 Downloaded from hlayiso.com 8 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2020) 3.1.6 A 37 537,75 1A Correct gross F Gross monthly salary in 2018/2019 = 1,064 MA salary V2 = R35 279,84A 1MA Divide by 1,064 OR 1A Gross salary A CA from 3.1.3 450 453 Gross monthly salary in 2018 = 1A Correct gross 1,064 MA = R35 279,84A salary 1MA Divide by 1,064 1A Gross salary (3) 3.2.1 Value of A = 90 – (8 + 13 + 30 + 15 + 10) 1MA Subtract from D = 90 – 76 MA 90 L2 14 1M Divide by 2 = M 2 1CA Value of A = 7 CA (3) D 3.2.2 16 years A 2A Age (2) L2 3.2.3 Average age MCA CA from 3.2.1 D (13 × 8) + (14 × 7)+ (15 × 13)+(16 × 30)+(17 × 15)+(18 × 7)+(19× 10) 1MCA Adding L3 = 90 correct values 104 + 98 +195 + 480 + 255 + 126 + 190 = 1M Dividing 90 1 448 M 1CA Average age = 90 1O Not valid = 16,088… years CA (4) Statement invalid O 3.2.4 CA from 3.2.1 P Number of boys = 13 + 30 + 15 + 7 1MCA Numerator L2 1CA Denominator 65 MCA 1R 3 decimal Probability = 90 A places = 0,722 R (3) D 3.2.5 The weight of the boys should also be taken into account.A 2A Reason (2) L2 Copyright reserved Please turn over ### Page 9 Downloaded from hlayiso.com (EC/SEPTEMBER 2020) MATHEMATICAL LITERACY P2 9 QUESTION 4 [38] Ques. Solution Explanation Level 4.1.1 14,202957 A R 1A Correct value D The lower the value, the stronger the Rand 1R Reason L2&4 (2) 4.1.2 Amount after exchange fee = 40 830 – (40 830 × 0,045) F = 40 830 – 1837,35 MA 1MA Subtract 4,5% L3&4 = R38 992,65 CA 1CA Value 38 992,65 1MCA Dividing At 14,983385 = MCA 14,983385 correct values = $2 602,39 CA 1CA Dollar value 38 992,65 At 14,398064 = 14,398064 = $2 708,18 CA 1CA Dollar value Difference = $2 708,18 – $2 602,39 = $105,79 CA 1MCA Difference Statement invalid O 1O Invalid OR 40 830 1MA Dividing 14,983385 MA At 14,983385 = CA correct values = $2 725,018412 – (2 725,018412 × 0,045) 1CA Value = 2 725,018412 – 122,6258285 MCA 1MCA Subtract = $2 602,39 MCA 4,5% 40 830 1CA Dollar Value At 14,398064 = 14,398064 = $2 835,797924 – (2 835,797924 × 0,045) = 2 835,797924 – (127,6109066) = $2 708,18 CA 1CA Dollar value Difference = $2 708,18 – $2 602,39 = $105,79MCA 1MCA Difference Statement invalid O 1O Invalid (7) 4.1.3  Political factors A 2A 1st reason D OR 2A 2nd reason L4  Economic factors A OR  Supply and demand of countries A OR  Inflation rate A Copyright reserved Please turn over ### Page 10 Downloaded from hlayiso.com 10 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2020) Accept any other relevant responses (4) 4.2.1 Enlargement: M&P Northern Line (A) = 7,2 cm A (Accept 7,2 cm – 7,4 cm) 1A Measure A L4 Scale is 4,2A cm = 300 km (Accept 4,1 cm – 4,4cm) 1A Measure scale 7,2 𝑐𝑚 1M Ratio Distance = × 300 km M 4,2 𝑐𝑚 1CA Kilometres = 514,286 km CA Map: Northern Line (A) = 1,5 cm A (Accept 1,4 cm – 1,6 cm) A 1A Measure line Scale is 4,2 cm = 1 500 km (Accept 4,1 cm – 4,4 cm) 1,5 𝑐𝑚 1A Measure scale Distance = × 1 500 km 4,2 𝑐𝑚 = 535,714 km CA 1CA Kilometres (7) 4.2.2 Traveller 1 and 2 = $670,36 × 2 1MA Amount for 2 F = $1 340,72 MA travellers L3 Traveller 3 and 4 = 0,239 × 670,36 M 1M 23,9% of 670,36 = $160,22 × 2 1CA Amount = $320,44 CA 1MA Tax amount Tax Amount = $188,64 × 4 1MCA Adding = $754,56 MA amount Total amount = $1 340,72 + $320,44 + $754,56MCA 1CA Total amount = $2 415,72 CA (6) 4.3.1 Range = Highest value – Lowest value 1M Concept of range D 34°F = A - 54°F M 1CA Value of A L2 A = 54°F + 34°F (2) = 88°F CA 5 4.3.2 Probability = A 1A Numerator D 12 A 1A Denominator L2 = 0,416… × 100 = 41,6… 1CA Nearest % ≈ 42% CA (3) 4.3.3 °Celsius = (°F – 32) ÷ 1,8 M = (34°F – 32) ÷ 1,8SF 1SF Substitution L2&4 = 2 ÷ 1,8 S 1S Simplification = 1,1…°C CA 1CA °C Statement not valid O 1O Not valid (4) Copyright reserved Please turn over ### Page 11 Downloaded from hlayiso.com (EC/SEPTEMBER 2020) MATHEMATICAL LITERACY P2 11 4.3.4 1A Jan – Apr D Average Minimum and Maximum 1CA May – L2 temperatures in North Carolina in 2018 Aug from 4.3.1 100 1A Sept – Dec A Temperature in °Fahrentheit 80 A 60 A 40 20 0 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Months (3) TOTAL: 150 Copyright reserved Please turn over ## Related documents - [MATHS LIT P2 GR12 MEMO SEPT2022 English](https://www.hlayiso.com/library/maths-lit-p2-gr12-memo-sept2022-english/dfa5d81c-279c-4516-b124-cfc6103a3517) — Mathematical Literacy · Memo · 2022 · Eng - [MATHS LIT P1 GR12 QP SEPT2020 English (Printready 18 August 2020)](https://www.hlayiso.com/library/maths-lit-p1-gr12-qp-sept2020-english-printready-18-august-2020/f1819888-26f6-4ed6-8382-6909354c4789) — Mathematical Literacy · QP · 2020 · Eng - 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