Mathematical Literacy Grade 11 P1 2024 Memo (Eng)

NATIONAL SENIOR CERTIFICATE GRADE 11 NOVEMBER 2024 MATHEMATICAL LITERACY P1 MARKING GUIDELINE MARKS: 100 Symbol Explanation M Method MA Method with accuracy CA Consistent accuracy A Accuracy C Conversion S Simplification RT Reading from a table/a 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 NPR No penalty for correct rounding minimum two decimal places AO Answer only MCA Method with constant accuracy J Justification This marking guideline consist of 6 pages. Hosted by www.ecexams.co.za.

2 MATHEMATICAL LITERACY P1 (EC/NOVEMBER 2024) 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. • If the candidate presents any extra solution when reading from a graph, table, layout plan and map, then penalise for every extra incorrect item presented. QUESTION 1 [26 MARKS] Ques. Solution Explanation T&L 1.1.1 Canva ✓✓A 2 RT correct service F provider L1 Accept Samira Hadid (2) 1.1.2 A = $127 x 4 ✓MA 1 MA multiplying correct F price by 4 L1 = $508 ✓A 1 A answer (2) 1.1.3 Invoice is a list of goods or services rendered as 2O explanation F well as a statement of the amount. ✓✓O (2) L1 1.1.4 B = $369 ÷ $123 ✓MA 1MA dividing the correct F values L1 = 3 ✓A 1A answer (2) 1.1.5 5 March 2024 ✓✓A 2A correct date F (2) L1 1.1.6 Total : R1,00 = $0,053 F ? = $1000 L1 ? = $1000 × R1,00 ✓MA 1MA correct exchange rate $0,053 ? = R18 867,92 ✓CA 1CA answer (2) 1.2.1 Constant/fixed relationship ✓✓A 2A correct answer F (2) L1 1.2.2 R1 200,00 ✓✓RT 2RT F (2) L1 1.2.3 (a) C = 2 ✓✓A 2A correct number F Accept 3 (2) L1 (b) D = R950 ✓✓RT 2RT correct amount F (2) L1 1.3.1 29 + 23 + 3 + 4 = 59 learners ✓✓A 2A correct total D (2) L1 1.3.2 % fail = 29 × 100 ✓MA 1 MA correct percentage D 59 L1 = 49,15 ✓CA 1CA answer (2) 1.3.3 0 – 29 ✓✓A 2A correct level D (2) L1 [26] Copyright reserved Please turn over Hosted by www.ecexams.co.za.

(EC/NOVEMBER 2024) MATHEMATICAL LITERACY P1 3 QUESTION 2 [26 MARKS] Ques. Solution Explanation T&L 2.1.1 Budget is a financial plan outlining income and expenses 2O correct explanation F over a period of time. ✓✓O L1 OR It is a tool to help manage finances, achieve financial goals, and to make informed decisions about spending and saving. ✓✓O (2) 2.1.2 31 029 965: 2A correct F Thirty-one million and twenty-nine thousand, nine L1 hundred and sixty-five. ✓✓A (2) 2.1.3 % increase 1RT correct values F 1MA multiplying by 100 L3 = 44 104 248 - 42 441 422 ✓RT ×100 ✓MA 1MA correct 42 441 422 ✓MA denominator = 3,92 = 4% ✓CA 1CA answer PR (4) 2.2.1 Cost = R50 + R12,00 × (No of km travelled – 3) ✓✓A 2A correct formula F (2) L2 2.2.2 Daniel: R50,00 + R12,00 × (No of km used – 3) F L4 = R50,00 + R12,00 × (80 km ‒3) ✓SF 1SF substitution to a formula = R50,00 + R12,00 × (77 km) ÷ 1 km = R50,00 + R924,00 ✓A 1A simplification = R974,00 ✓CA 1CA answer for Daniel Jerry: R14,00 × R80,00 ✓A 1A multiplying correct amounts = R1 160,00 ✓CA 1CA answer for Jerry His statement is incorrect. ✓J 1J justification (6) 2.2.3 Distance = R50,00 + R12,00 × (No of km travelled ‒3) F L3 R1 214,00 = R50,00 + R12,00 × (D ‒3) ✓SF 1 SF substitution to a formula R1 214,00 – R50,00 = R12,00 × (D – 3) R1 164,00 = R12,00 × (D ‒ 3) ✓S 1S simplification R1 164,00 = D – 3 R12,00 97 = D – 3 ✓S 1S simplification D = 100 ✓CA 1 CA answer (4) Copyright reserved Please turn over Hosted by www.ecexams.co.za.

4 MATHEMATICAL LITERACY P1 (EC/NOVEMBER 2024) 2.3 Discount on school fees = R48 800,00 × 10 ✓MA 1MA multiplying by a F 100 correct percentage L4 = R4 880,00 ✓A 1A simplification Discount on hostel fees = R66 800 × 5 ✓MA 1MA multiplying by 100 correct percentage = R3 340,00 ✓A 1A simplification Total Discount = R4 880,00 + R3 340,00 ✓A 1A adding correct amounts = R8 220,00 ✓CA 1CA answer (6) [26] QUESTION 3 [ 17 MARKS] Ques. Solution Explanation T&L 3.1 India ✓✓A 2A correct country D (2) L2 3.2 103 317 638 ; 15 400 547 ; 15 400 547 ; 12 862 740 ; 2M arranging in correct D 12 615 365 ; 8 485 749 ; 6 692 796 ; 4 684 727 ; order L2 3 740 280 ; 2 713 117 ; 2 544 102 ; 2 505 605 ; 1 854 162 ✓✓M (2) 3.3 Difference = 103 317 638 – 1 854 162 ✓RT 1RT correct values D = 101 463 478 ✓CA 1CA answer L2 (2) 3.4 5 ✓✓RT 2RT D (2) L1 3.5 Thailand ✓✓A CA 3.2 D 2 A answer L2 (2) 3.6 Bar Graph /Line graph ✓✓A 2A Type of graph D (2) L2 3.7 Probability = 9 ✓M ✓M 1M fraction P 13 1M division L2 = 0,69 ✓CA 1 CA answer (3) 3.8 3 countries ✓✓A 2A answer D (2) L2 [17] Copyright reserved Please turn over Hosted by www.ecexams.co.za.

(EC/NOVEMBER 2024) MATHEMATICAL LITERACY P1 5 Question 4 [31marks] Que. Solutions Explanation T&L 4.1.1 Inflation ✓✓A 2A answer (2) F L1 4.1.2 Cost = R2 400,00 + R50,00 × number of learners ✓SF 1 SF formula F = R2 400,00 + R50,00 × 100 ✓MA 1MA multiplying by 100 L3 = R2 400,00 + R5 000,00 ✓M 1M addition = R7 400,00 (3) 4.1.3 16000 15000 14000 13000 12000 11000 10000 Cost in Rands 9000 8000 7000 6000 5000 4000 B 3000 2000 1000 0 0 10 20 30 40 50 60 70 80 90 100 110 Number of Learners 1A starting point (0;0) ✓ F 1A end point (100;15 000) ✓ L2 1A straight line ✓ (3) 4.1.4 24 learners buying 24 t-shirt indicated as B on the graph 2A correct number of F ✓✓A learners indicated on the L2 graph (2) Copyright reserved Please turn over Hosted by www.ecexams.co.za.

6 MATHEMATICAL LITERACY P1 (EC/NOVEMBER 2024) 4.1.5 End of Year One: 10,5 F = R7 500 +  R7 500 ✓M 1 M calculating interest L3 100 = R7 500 + R787,5 = R8 287,50 ✓CA 1CA balance for 1st year End of Year Two: 10,5 = R8 287,50 +  R8 287,50 100 = R8 287,50 + R870,1875 ✓CA 1 CA balance for 2nd year = R9 157, 6875 End of Year Three: 10,5 = R9 157,6875 +  R9 157,6875 100 = R9 157,6875 + R961,5571875 ✓CA 1 CA balance for 3rd year = R10 119, 24 Statement is valid ✓O 1O opinion OR M M M A 3 M multiplication Balance = R7 500 1,105 1,105 1,105 = R10 119,24 1 A answer Statement is valid O 1 O opinion (5) 4.2.1 Discreet ✓✓A 2A answer (2) D L1 4.2.2 ✓A 37 = 1588 + F ✓M 1M adding 44 ✓A 1A substituting 37 D 1588 + F = 1628 ✓A 1A dividing by 44 L3 F = 1628 – 1588 1A multiplying 37 by 44 F = 40 ✓ 1CA answer (5) 4.2.3 12 13 14 15 15 17 18 21 21 22 23 24 24 26 26 26 30 31 1M arranging in order of D 32 33 37 38 39 39 40 40 40 41 41 41 43 43 49 49 50 51 descending or ascending/ L3 53 54 55 60 62 69 64 86 ✓M OR RT correct values Range = 86 – 12 ✓M 1M calculating range = 74 ✓CA 1CA answer Statement is not correct the range is big, so the data is far apart from each other. ✓J 1J correct justification (4) 4.2.4 28 learners ✓✓A 2A correct answer D (2) L1 4.2.5 P = 1 ✓A 1A numerator P 44 ✓A 1A denominator L2 = 0,023 ✓CA 1CA answer (3) [31] TOTAL: 100 Copyright reserved Please turn over Hosted by www.ecexams.co.za.