Mathematical Literacy Grade 12 P1 NW June 2025 Memo (Eng)

PROVINCIAL ASSESSMENT GRADE 12 MATHEMATICAL LITERACY P1 JUNE 2025 MARKING GUIDELINES MARKS: 100 Symbol Explanation M Method MA Method with accuracy MCA Method with consistent 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/Reasoning P Penalty, e.g. for no units, incorrect rounding off, etc. R Rounding off NPR No penalty for correct rounding AO Answer only These marking guidelines consist of 7 pages. Copyright reserved Please turn over

Mathematical Literacy/P1 2 NW/June 2025 Grade 12 – Marking Guidelines 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.  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 penalize for every extra item presented.  General principle of marking, 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 opinion type questions marks will only be awarded if relevant calculations are shown. QUESTION 1 [20 MARKS] Answer only AO FULL MARKS Q Solution Explanation T/L 1.1.1 Bar graph/horizontal bar graphA 2A correct graph type D (2) L1 E 1.1.2 Ratlou A 2A correct LEO D (2) L1 E 1.1.3 = JB Marks + Matlosana + Maquassi 1RT correct values D MA 1MA adding L1 = 48 329 + 85 016 + 22 653 RT 1CA simplification E = 155 998 CA (3) 1.1.4 Naledi: Rustenburg 1A correct values D 51 701: 122 035A A 1A correct order L1 1: 2,36CA 1CA simplification M NPR (3) 1.2.1 Total amount F L1 RT MA 1RT correct values E =R250,44 + R99 + R89+ R135,25+ R686,64+R237,78 1MA addition = R1 498,11CA 1CA simplification (3) 1.2.2 RT 1RT correct values F R51,92 1MA dividing L1 = ×100 MA R1 287,62 M 1A simplification = 4,03%A NPR (3) Copyright reserved Please turn over

Mathematical Literacy/P1 3 NW/June 2025 Grade 12 – Marking Guidelines 1.2.3 = Front brake pads + Rear brake pads 1RT correct values F RT L1 = R328,57 + R619,38MA 1MA addition M = R947,95A 1A simplification (3) 1.2.4 VW Citi Golf CA CA from Question 1.2.1 F 2CA correct vehicle L1 (2) E [20] QUESTION 2 [43 MARKS] 2.1.1 A charge in rand per measuring unit of F electricity.O 2O explanation L1 (2) E 2.1.2 F (a) No. of units purchased =142,4 kWh – 5,3kWh MA 1MA subtracting L3 = 137,1 kWh A 1A difference in units M Cost for first 100 units = 100 × R2,775755 MCA 1MCA multiplying with tariff = R277,5755 CA 1 CA simplifying Next 37,1 units = 37,1 × R3,248405MCA MCA multiplying with tariff = R120,5158255CA 1 CA simplifying Total cost = R277,5755 + R120,5158255 CA 1CA adding = R398,09 ≈R398 OR OR 100 kWh × R2,775755 MA 1MA multiplying with tariff = R277,5755A 1A simplification R398– R277,5755 = R120,5755 CA 1CA cost of R120,5145 R120,5755 Amount of kWh = MCA 1MCA dividing with tariff R3,248405 = 37,0707188605 kWhCA 1CA simplification Total kWh purchased = 100 kWh + 37,0707188605 kWh = 137,0707188605 kWhCA New reading on meter 1CA kWh purchased = 137,0707188605 kWh + 5,3 kWh CA = 142,371886 kWh 1CA adding new = 142,4 kWh (7) 2.1.2 𝑅398 F = 1,15 MA 1MA dividing by 1,15 (b) L2 = R346,0869565…MA 1MA simplification M = R398 – R346,0869565 … = R51,91A 1A VAT Copyright reserved Please turn over

Mathematical Literacy/P1 4 NW/June 2025 Grade 12 – Marking Guidelines OR OR 15 15 R398 × 115MA 1MA multiply with 115 = R51,91304348MA 1MA simplification = R51,91A 1A VAT (3) (c) = 142,4 kWh – 82 kWh MA 1MA subtracting F = 60,4 kWh CA 1CA simplification L4 E Not valid O 1O opinion. (3) 2.2.1 Annual taxable income F R36 780 × 12MA 1MA multiplying by 12 L2 = R 441 360,00A 1A simplification E 1MCA tax bracket Tax bracket 3  MCA AO (3) 2.2.2 Use tax bracket 3 O 1O choose to use tax bracket F 3 above L3 Highest amount at tax bracket 3 = 512 800RT 1RT maximum value for tax D bracket 3 used. Formula for tax bracket 3: 77 362 + 31% of taxable income above 370 500 = R77 362 + 31% of (R512 800− R370 500)SF 1SF substitution 1MCA simplification = R77 362 + 0,31 × (R142 300)MCA 1CA add the values = R77 362 +R 44 113CA = R121 475 (5) 2.2.3 15 1SF Substitution F Gratuity = R42 270 × 26 × 29 SF 1A simplification L2 = R 707 209,6154 A M 16 = R 707 209,6154 × MCA 1MCA multiplying by 16% 100 = R113 153,54CA 1CA simplification (4) 2.2.4 Interest after 1 year = R 25 600 × 8,60% 1MA calculating interest for F = R2 201,60  MA 1st year L3 D Amount after 1 year = R25 600 + R2 201,60 1A simplification = R27 801,60A Interest for full 2nd year = R27 801,60 × 8,60% = R2 390,9376CA 1CA for 2nd year 1 𝑅 2 390,9376 1MCA half year  interest for year = MCA 2 2 = R1 195,4688 CA 1 CA simplification Amount of the fixed deposit = R27 801,60 + R1 195,4688 = R28 997,07CA 1CA simplification Copyright reserved Please turn over

Mathematical Literacy/P1 5 NW/June 2025 Grade 12 – Marking Guidelines OR OR Interest after 1 year = R 25 600, × 8,60% 1MA calculating interest for = R2 201,60MA 1st year Amount after 1 year = R25 600 + R2 201,60 1A simplification = R27 801,60 A 8,60% 2nd year interest rate = 2 = 4,3% CA 1CA for 2nd year 1 Interest for year = R27 801,60 × 4,3% 2 = R1 195,44MCA 1MCA half year Amount of the fixed deposit = R27 801,60 + R1 195,4688MCA 1MCA adding values = R28 997,07 CA 1CA simplification OR OR Amount after 1 year = R 25 600,00 × 108,6%MA 1MA calculating interest for = R27 801,60A 1st year 1A simplification 8,06% 2nd year interest rate = 1CA for 2nd year 2 = 4,3% CA 1MCA half year MA 1 Amount for year = R27 801,60 × 104,3%MCA 1MA multiply with 104,3% 2 = R28 997,07 CA 1CA simplification (6) 2.3.1 Public Primary A 2A correct item F (2) L1 D 2.3.2 New price −old price 1RT correct values F Percentage Change = × 100% old price 1SF substitution L2 RT 1A answer M 37 840 000 −39 236 000 = × 100%SF 39 236 000SF NPR = - 3,56% A [No penalty for omitting million] (3) 2.3.3 RT 1RT correct value F 0,512 1MA multiplying L4 = 597 063 000 × 100 MA 597 063 000 by 0,512% M = R 3056 962,56 CA 1CA simplification Actual Expenditure 1MCA subtracting values = R597 063 000 – R 3 056 962,56MCA 1CA simplification = R594 006 037,40CA 1O opinion (6) Valid O [44] Copyright reserved Please turn over

Mathematical Literacy/P1 6 NW/June 2025 Grade 12 – Marking Guidelines QUESTION 3 [23 MARKS] 3.1.1 65 cm A 2RT reading from chart D (2) L2 E 3.1.2 RT RT 2RT correct percentiles D Between 10 and 25th Percentile th (2) L2 E 3.2.1 DA 2A correct option D (2) L1 E 3.2.2 Range = Maximum value – Minimum value 1RT correct values D RT L2 = 179 968 – 14 152MA 1MA concept of range E = 165 816 A 1A simplification (3) 3.2.3 The number of grade 10 learners in the Northern Cape D is more less than O the second lowest number of 2O conclusion L4 grade 10 from Free state with 65 970 in the data (2) M set.O 3.2.4 Mean of grade 11= MA 1MA concept of mean D 120 993+45156+166 069+240 146+125357+84902+16189+53 708+75494 L4 9 928 050 M = 9 = 103 116, 6667  CA 1CA simplification Difference 1MCA difference = 103 116,6667 – 81122,78 MCA 1CA answer = 21 993,89 CA 1O conclusion Not valid O (5) NPR 3.2.5 CA the total grade 11 P RT leaners from Question 3.2.4 L2 125 375 M = ×100%MCA 928 050 1RT 125 375 1MCA concept of percentage = 13,51%CA 1CA answer NPR (3) 3.2.6 O 2O decrease D Number of learners decrease from grade 10 to 12 in all L4 Provinces D Drop out from school/ Failure rate.O 2O conclusion (4) [23] Copyright reserved Please turn over

Mathematical Literacy/P1 7 NW/June 2025 Grade 12 – Marking Guidelines QUESTION 4 [13 MARKS] Q Solution Explanation T/L 4.1.1 The exchange rate where one unit of a foreign currency is 2A explanation F expressed in terms of South African rand.A L1 OR E It is what one foreign currency is worth in terms of rands.A (2) 4.1.2 Indian RupeeRT 2RT correct currency F (2) L1 E 4.1.3 Amount in Canadian dollar F L2 1R = 0,077572 RT 1RT identifying correct E exchange rate 0,077572 CAD 1MA multiplying by the = × R1670,00 MA exchange rate R1 = 129,54524 CAD A 1A simplification OR OR 1RT Identifying correct 1CAD = R12, 891276  RT exchange rate R1670,00 1MA multiplying by the = ×1CAD MA R12,891276 exchange rate = 129,54524 CADA 1A simplification (3) 4.2.1 D 196,95 million A 2A correct value L2 (2) E 4.2.2 D 52,58 = Q3 – Q1 MA 1MA concept of IQR L3 RT 1RT correct value (Q1) E 52,58 = Q3 – 180,16 1MA changing the subject Q3 = 180,16 + 52,58MA of the formula Q3 = 232,74 million CA 1CA simplification (4) [13] TOTAL: 100 Copyright reserved