MATHS LIT GRADE 12 2025 P1 MEMORANDUM

NATIONAL SENIOR CERTIFICATE GRADE 12 SEPTEMBER 2025 MATHEMATICAL LITERACY P1 MARKING GUIDELINE MARKS: 150 Symbol Explanation M Method MA Method with accuracy CA Consistent accuracy A Accuracy C Conversion S Simplification RT Reading from a table/graph/document/diagram SF Correct substitution in a formula O Opinion/Explanation P Penalty, e.g., for no units, incorrect rounding off, etc. R Rounding off/Reason NPR No penalty for correct rounding minimum two decimal places AO Answer only MCA Method with constant accuracy This marking guideline consists of 9 pages.

2 MATHEMATICAL LITERACY P1 (EC/SEPTEMBER 2025) MARKING GUIDELINE NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • If a candidate has crossed out an attempt of a question and not redone the question, mark the crossed-out version. • Consistent accuracy applies in ALL aspects of the marking guideline. Stop marking 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 and table, 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. Topics: F – Finance, DH – Data Handling, P – Probability QUESTION 1 [34 MARKS] Que. Solution Explanation T&L 1.1.1 December ✓✓A 2A reading F correct month L1 (2) 1.1.2 Mr Ravenswood ✓✓A 2A correct name F (2) L1 1.1.3 Costs = R631,70 + R1 399,00 + R243,50 + R243,50 + R64,37 1MA Adding F + R68,02 + R21,28 + R94,43 ✓MA correct amounts L1 = R2 765,80 ✓CA 1CA answer (2) 1.2.1 Cost = R325,50 × 12 ✓MA 1MA multiplying F = R3 906,00 ✓CA by 12 L1 (2) 1.2.2 Cost for one tyre in cents = R899,00 × 100 = 89 900 cents ✓✓A 2A correct amount F (2) L1 1.2.3 R225 ∶ R398 ✓RT 1RT correct F amounts L1 225 398 ∶ ✓M 1 M division 225 225 1CA simplification 1 ∶ 1,77 ✓CA (3) 1.3.1 Rent ✓✓A 2A correct fixed F expense L1 OR Cell phone contract ✓✓A OR transport ✓✓A OR savings ✓✓A (2) Copyright reserved Please turn over

(EC/SEPTEMBER 2025) MATHEMATICAL LITERACY P1 3 1.3.2 ✓MA F Total Expenses = R6 000,00 + R3 500,00 + R2 000,00 + R300,00 + 1MA for adding L1 R2 000 correct values = R15 800,00 ✓CA 1CA simplification (2) 1.3.3 Deficit ✓✓A 2A correct F identification L1 (2) 1.4.1 IsiZulu ✓✓A 2A answer D (2) L1 1.4.2 1 A for English D Response Tally Frequency 1 A for Afrikaans L1 English |||| |||| |||| |||| |||| |||| |||| |||| |||| ✓A 45 Afrikaans |||| || ✓A 7 1 A for IsiXhosa IsiXhosa |||| |||| |||| |||| ||| 23 and IsiZulu IsiZulu |||| |||| |||| ✓A 14 (3) 1.4.3 Probability (IsiXhosa) = 25,8% ✓✓RT 2RT correct P OR percentage L1 (2) 23 ✓RT = 0,26 𝐎𝐑 26% ✓A 89 1.5.1 D ✓✓A 2 A answer F (2) L1 1.5.2 C ✓✓A 2 A answer F (2) L1 1.5.3 B ✓✓A 2 A answer P (2) L1 1.5.4 A ✓✓A 2 A answer D (2) L1 [34] Copyright reserved Please turn over

4 MATHEMATICAL LITERACY P1 (EC/SEPTEMBER 2025) QUESTION 2 [31 MARKS] Que. Solution Explanation T&L 2.1.1 Max monthly earnings = R13 766 ✓✓RT 2RT Correct F monthly earnings L1 (2) 2.1.2 Annual earnings = R165 192 ✓✓RT 2RT correct annual F earnings L1 (2) 2.1.3 Annual Tax = R165 192 × 18 ✓A CA from 2.1.2 F 100 1A correct bracket L3 = R29 734,56 ✓A 1A simplification = R29 734,56 – (R17 235,00 + R9 444,00 + R3 145) ✓MA 1MA subtracting all = -R89,44 ✓A rebate The employee does not pay any tax because the rebate is 1A answer larger than the amount of tax. SARS must pay the employee 2A justification a refund of R89,44 ✓✓J (6) 2.2 Monthly salary = R4 118,40 × 100 ✓MA 1MA multiplying by F = R411 840,00 ✓M 100 L2 = R411 840,00 ÷ 12 ✓MA 1M simplification = R34 320 ✓CA 1MA dividing by 100 1CA answer (4) 2.3.1 Cost for 20 laptops = R5 000,00 × 20 ✓MA 1MA multiplying by F = R100 000,00 ✓A 20 L4 1A simplification Selling price for 10 laptops = R6 000,00 × 10 = R60 000,00 ✓MA 1MA simplification 10 Discount on remaining = R60 000 × 100 = R6 000 ✓CA 1CA simplification = R60 000,00 – R6 000 = R54 000,00 ✓CA 1CA discount Income = R60 000,00 + R54 000,00 ✓M 1M adding correct = R114 000,00 amounts 1MCA subtracting Profit = Income - Expenses correct values = R114 000,00 – R100 000,00 ✓MCA 1CA simplification = R14 000,00 ✓CA Copyright reserved Please turn over

(EC/SEPTEMBER 2025) MATHEMATICAL LITERACY P1 5 OR OR Cost for 20 laptops = R5 000,00 × 20 ✓MA 1MA multiplying by = R100 000,00 ✓A 20 1A simplification Selling price for 10 laptops = R6 000,00 × 10 = R60 000,00 ✓MA 1MA simplification 90 Discount on remaining = R60 000 × 100 ✓M = R54 000,00 ✓CA 1M multiplication 1CA discount Income = R60 000,00 + R54 000,00 ✓M = R114 000,00 1M adding correct amounts Profit = Income - Expenses 1MCA subtracting = R114 000,00 – R100 000,00 ✓MCA correct values = R14 000,00 ✓CA 1CA simplification (8) 2.3.2 % profit = Profit × 100 CA from 2.3.1 F Cost 1 RT correct value L3 1MA multiplying by = 𝑅14 000,00 × 100 ✓MA 100 100 000 ✓RT 1CA simplification (3) = 14% ✓CA 2.4 1 Euro(€) = R19,92 F ? = R10 000,00 L2 = R10 000,00 ✓C 1C conversion R19,92 = € 502,00 ✓CA 1CA answer = €502 ‒ € 500 ✓MA 1MA subtracting = € 2 ✓CA currencies 1CA simplification Remaining = R19,92 × € 2 ✓C 1C conversion = R39,84 = R40 ✓R 1R rounding OR OR 1 Euro(€) = R19,92 500(€) = R? 1RT correct value 1C conversion = 500 ✓RT × 19,92 ✓C 1CA answer = 9 960 ✓CA 1RT correct value Remaining = R10 000 ✓RT – R9 960 ✓M 1M subtraction = R40 ✓CA 1CA answer (6) [31] Copyright reserved Please turn over

6 MATHEMATICAL LITERACY P1 (EC/SEPTEMBER 2025) QUESTION 3 [28 MARKS] Que. Solution Explanation T&L 3.1.1 2022 ✓✓A 2A correct year D (2) L1 3.1.2 2 ✓✓A 2A answer D (2) L1 3.1.3 Difference = 6,883 – 5,424 ✓RT ✓MA 1RT reading correct values D = 1,459 thousands OR 1459 ✓CA from the table L2 1MA subtracting correct OR values 1CA answer Difference = 6 883 – 5 424 ✓RT ✓MA = 1 459 ✓CA (3) 3.1.4 Total number (2023): D = 8,860 + 6,765 + 5,821 + 5,689 + 4,742 + 3,871 + 3,338 L2 + 3,274 ✓MA 1MA correct values = 42,36 thousand ✓CA 1CA total ✓RT 1RT correct value P (United Kingdom) = 5,821 + 5,689 + 4,742 42,36 16,252 = 42,36 × 100 ✓M 1M multiplying by 100 1CA correct percentage = 38,37% ✓CA (5) 3.1.5 Discreet ✓✓A 2A answer (2) D L1 3.2.1 FET % = 15% +10% ✓RT ✓MA 1RT correct % D = 25% ✓A 1MA adding correct % L2 1A correct % (3) 3.2.2 ✓ RT 1RT correct total D 20 1MA multiplying by 20% L2 No of learners = 780 × 100 ✓MA 1CA correct answer = 156 learners ✓CA (3) 3.2.3 Grade 8 = 780 × 30 ✓MA 1MA multiplying by 30% D 100 1A correct number of L3 = 234 learners ✓A grade 8 learners 25 Grade 9 = 780 × 100 ✓MA 1MA multiplying by 25% = 195 ✓A 1A correct number of grade 9 learners Difference = 234 – 195 ✓MA = 39 learners ✓CA 1MA subtracting correct numbers 1CA answer (6) [26] Copyright reserved Please turn over

(EC/SEPTEMBER 2025) MATHEMATICAL LITERACY P1 7 QUESTION 4 [26 MARKS] Que. Solution Explanation T&L 4.1.1 Developing questions; collecting data; ✓A 1A 1st two stages D Organising data; Summarising data; ✓A 1A for 2nd two stages L1 Representing data graphically; Analyse data ✓A 1A for 3rd two stages (3) 4.1.2 Bar graph, Double bar graph, Histogram, Stacked bar 3A any three correct graphs D graph, Scatter Plot, Line graph ✓✓✓A (3) L2 4.2.1 2022/23 fare: R14,50 F 2023/24 fare: R15,00 L2 ✓ RT 1RT correct values % increase = R15,00 – R14,50 × 100 ✓MA 1MA multiplying by 100 R14,50 ✓M 1M dividing by R14,50 0,5 = 14,5 × 100 1CA answer = 3,45% ✓CA (4) 4.2.2 ✓A ✓MA 1A for max distances F 5 km × R10,00 = R50,00 ✓A 1MA multiplying by tariff L3 5 km × R12,50 = R62,00 1A 4 amounts 5 km × R15,00 = R75,00 1CA total before VAT 9,1 km × R17,00 = R154,70 1M VAT = R341,70 ✓CA × 1,15 ✓M 1CA VAT inclusive = R393,53 ✓CA 1J justification Nancy’s statement is correct ✓J (7) 4.3.1 36,5 ✓✓A 2A correct value D (2) L2 4.3.2 20 ✓✓A 2A correct age D (2) L2 4.4. BMI = mass(kg) D (Height(m))2 1 SF correct substituting in a L3 = 75 kg ✓SF formula (1,71m)2 ✓C 1 C conversion = 75 kg 2,9241 m2 ✓S 1S squaring 1,71 = 25,6489 kg/m2 ✓CA 1CA simplification = 25,6 kg/m2 ✓R 1R rounding (5) [26] Copyright reserved Please turn over

8 MATHEMATICAL LITERACY P1 (EC/SEPTEMBER 2025) QUESTION 5 [31 MARKS] Que. Solution Explanation T&L 5.1 25 Deposit = R12 000 × 100 ✓MA 1MA multiplying by F 25% L4 1A simplification = R3 000,00 ✓A Loan = R12 000,00 – R3 000,00 1A loan = R9 000 ✓A 15 Interest charged = 100 × R9 000 × 3 years ✓MA 1MA multiplying by = R1 350 × 3 3 years = R4 050,00 ✓CA 1CA interest Total repayment amount = R9 000,00 + R4 050 = R13 050 ✓A 1A total amount Monthly instalment = R13 050,00 ÷ 36 ✓MA 1MA dividing by 36 = R362,25 ✓CA 1CA simplification OR OR Loan = 100% – 25% = 75% ✓MA 1MA subtraction 1M multiplication 75 = 100 × R12 000 ✓M = R9 000 ✓A 1A loan 15 Interest charged = 100 × R9 000 × 3 years ✓MA 1MA multiplying by 3 = R1 350 × 3 years = R4 050,00 ✓CA 1CA interest Total repayment amount = R9 000,00 + R4 050 = R13 050 ✓A 1A total amount Monthly instalment = R13 050,00 ÷ 36 ✓MA 1MA dividing by 36 = R362,25 ✓CA 1CA simplification (8) 5.2.1 1A title D South African Team Netball Results A 3A for game 1,2 and 4 L4 80 75 bars 70 A A 58 61 60 A 55 Points scored 50 40 30 20 10 0 Game 1 Game 2 Game 3 Game 4 Number of games (4) Copyright reserved Please turn over

(EC/SEPTEMBER 2025) MATHEMATICAL LITERACY P1 9 5.2.2 Mean = 55 + 58 + 61 + 75 ✓RT 1RT correct values D 4 ✓M 1M dividing by 4 L4 = 249 4 = 62,25 ✓CA 1CA simplification Median: 55 58 61 75 ✓M 1M arranging in ascending or = 58 +61 ✓M descending order 2 = 59,5 ✓A 1M correct values 1A simplification Difference = 62,25 – 59,5 ✓M = 2,75 1M subtracting 2 values The coach is correct ✓J 1J explanation (8) 5.3.1 Credit card ✓✓A 2A answer F (2) L2 5.3.2 Total cost of comfort luxury = R72,99 × 2 ✓M = 1 M multiply by 2 F R145,98 ✓A 1 A total amount L2 1RT correct discount Discount R45,98 ✓RT 1 M multiply by 100 1 A answer 45,98 Percentage discount: 145,98 × 100 ✓M = 31,50% ✓A (5) 5.3.3 15% = R54,97 1M multiplication F 115% = ? 1 CA answer L1 115 VAT inclusive = 15 × 54,97 ✓M = 421,44 ✓CA 1MA subtracting Amount paid = R421,44 – R63,50 ✓MA correct amounts = R357,94 ✓CA 1CA answer OR OR Total amount = R1,48 + R72,99 + R72,99 + R25,99 + R24,99 + R57,98 + R82,99 + R34,99 + R105,00 – (R11,98 1MA adding correct + R45,98) ✓MA values = R421,44 ✓CA 1 CA answer 1MA subtracting Amount paid = R421,44 – R63,50 ✓MA correct amounts = R357,94 ✓CA 1CA answer (4) 5.4 Real increment = R5,5% – 3,0% ✓MA 1MA subtracting F = 2,5% ✓A correct % L4 1A answer (2) [33] TOTAL: 150 Copyright reserved Please turn over