Mathematical Literacy Grade 12 P1 2022 Memo (Eng)

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 JUNE 2022 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/RG/RM Reading from a table/graph/map F Choosing the correct formula SF Correct 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 correct rounding off to minimum of two decimal places This marking guideline consists of 8 pages.

Downloaded from hlayiso.com 2 MATHEMATICAL LITERACY P1 (EC/JUNE 2022) 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. • 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. LET WEL: • As ʼn kandidaat ʼn vraag TWEE keer beantwoord merk slegs die EERSTE poging. • As ʼn kandidaat ʼn antwoord van ʼn vraag doodtrek (kanselleer) en nie oordoen nie, merk die doodgetrekte (gekanselleerde) poging. • Volgehoue akkuraatheid (CA) word in ALLE aspekte van die nasienriglyn toegepas, maar dit hou by die tweede berekeningsfout op. • Wanneer ʼn kandidaat aflees van ʼn grafiek, tabel, uitlegplan en kaart en ekstra antwoorde gee, penaliseer vir elke ekstra item. Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/JUNE 2022) MATHEMATICAL LITERACY P1 3 QUESTION 1 [20 MARKS] Que Solution Explanation/Marks AO: FULL MARKS T/L 1.1.1 18,25 1 825 1M fraction F = ✓M 100 10 000 1A answer in a L1 73 reduced form * = ✓A (2) 400 1.1.2 % of price = 100 – 18,25% F ✓ = 81,75% ✓M 1M subtraction L1 81,75 Price = × 380✓M * 100 1M % calculation = R310,65 ✓ CA 1CA answer OR OR 18,25 1M % calculation Reduction = 100 × 380 1M subtraction = R69,35 ✓M 1CA answer Price = R380 – 69,35✓M = R310,65 ✓CA (3) 1.2.1 Difference = R469 – (– R447) ✓CA 1 RT for the two F = R916 million ✓RT correct values L1 1 CA answer (2) 1.2.2 Total = 265+277+326+390+447+458+486 – (469+300) ✓M 1M addition (+) and F = 1880 million ✓CA subtraction (–) of the L1 values 1CA (2) 3 1.3.1 Weekend wage rate = 2 × 25 ✓MA 1MA multiplication F 1A answer L1 = R37,50 ✓A (2) * 1.3.2 ✓M 1M multiplications F Earnings = 6 × 25 + 37,50 × 4 ✓MA L1 1MA addition * = R300 ✓CA 1CA answer (3) 1.4.1 Discrete ✓✓A 2A answer (2) D L1 1.4.2 Game ✓✓RT 2RT answer D (2) L1 1.4.3 Total games = 4 + 6 + 5 + 4 + 1 + 2 + 2 = 24 games ✓M ✓CA 1M adding the games D 1CA answer (2) L1 [20] Copyright reserved Please turn over

Downloaded from hlayiso.com 4 MATHEMATICAL LITERACY P1 (EC/JUNE 2022) QUESTION 2 [18 MARKS] Que Solution Explanation/Marks AO: FULL MARKS T/L 2.1.1 Time 4 hours ✓✓RT 2RT F (2) L2 2.1.2 From graph: F 2 welders complete 1 frame in 4 hours ✓M L3 2:1 1M value from graph 20 : ? frame in 4 hours 1M numerator 20×1 1M denominator Frames = 2 ✓✓ M 1A answer = 10 frames ✓ A OR OR 1SF substitution n×t=8 1S simplification for 20 × t = 8 ✓SF 2,5 frames done in 1 t = 8/20 hour by 20 welders = 0,4 hours to make 1 frame by 20 welders ✓ S 1M multiplication In four hours= 4/0,4 ✓M = 10 frames ✓A 1A answer (4) ✓M 1M correct values for 2.2.1 28−25,81 numerator and F A= × 100%✓MA 25,81 denominator L2 = 8,485% M % calculation = 8,5%✓CA 1CA (3) (NPR) 2.2.2 Cost: Up to 6 kℓ = R0 = R0 ✓M F 1M cost in block 1 L3 6 – 25 kℓ = 19 k × R23,60 = R448,40 ✓M 1M cost in block 2 25 – 30 kℓ = 5 kℓ × R32,20 = R161,00 ✓M ✓M 1M cost in block 3 TOTAL COST = R448,40+R161,00 = R606,40 ✓CA 1M addition all costs 1CA answer (5) 2.3.1 Salary B = R3 192,05+15 761,80 ✓M 1M adding the two F = R18 953,85 ✓CA balances L2 1 CA answer (2) 2.3.2 Bank fees for March = 42,37+17,47+100,88 ✓M 1M adding fees of F = R160,72 ✓CA March L1 1CA answer (2) [18] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/JUNE 2022) MATHEMATICAL LITERACY P1 5 QUESTION 3 [21 MARKS] Quest. Solution Explanation/Marks AO: FULL MARKS T/L 3.1 2020 ✓A Reason: Covid-19 pandemic ✓J 1A year D 1J reason (2) L1 3.2 ✓M 1M subtracting from D C = 25 285,1 – (2093,5+2092,8+2249,4+ 1988,8+1750,5 25 285,1 L2 +1964,7+2067,1+2204,4+2308,0+2267,8+2493,4) 1M addition of all other * = 1804,7 ✓M values ✓CA 1CA answer (3) 3.3 descending order: ✓RT 1RT all values including D 2493,4; 2308,0; 2267,8; 2249,4; 2204,4; 2093,5; 2092,8 value from 3.2 L2 2067,1; 1988,8; 1964,7: 1804,7;1750,5 ✓CA 1CA order with value * from 3.2 (2) 3.4 ✓RT 1RT highest and lowest D Range = 2 262,3 – 33,8 ✓M values L2 = 2 228,5 million ✓CA 1M concept of range 1CA answer (3) 3.5 ✓M 1M concept of mean D 24846,4 1A mean for 2018 L4 Mean income for 2018 = = 2 070,53 million ✓A 12 * 98 18,5 1A mean for 2020 Mean income for 2020 = = 818,21 million ✓A 12 Double mean income for 2020 = 818,21×2 = 1636,42 ✓M Million 1M comparing values of Mean income for 2018 (2 070,53) is greater than double mean 2018 and double mean income for 2020 (1636,42) mean income for 2020 Statement Valid ✓J 1J valid statement. NPR (6) 3.6 From 2018 December income dropped right through up to 1J justification for the D July 2019; then increased from August 2019 to December period Dec 2018 to July L4 2019. It remained high up to March 2020. ✓J 2019 Then it dropped drastically in from April 2020 and remained 1J justification for the low in 2020. ✓J period August 2019 to 2020 (2) 3.7 May ✓A 1A first months D and June ✓A 1A second months. L2 CA from 3.2 (2) [20] Copyright reserved Please turn over

Downloaded from hlayiso.com 6 MATHEMATICAL LITERACY P1 (EC/JUNE 2022) QUESTION 4 [20 MARKS] Que Solution Explanation/Marks AO: FULL MARKS T/L 4.1.1 Values of dependent variable at break-even point 1RT value for income F Income = R300 ✓RT 1RT value for expenses L2 Expenses = R300 ✓RT (2) 4.1.2 Total sales in a week = 37 packets ✓RT 1RT adding sales from F From Graph: Income = R555 ✓RT table L2 Expenses = R385 ✓RT 1RT reading income Profit = R555 – R385 = R170 ✓CA from graph 1RT expenses from graph OR 1CA answer for profit OR Total sales = 37 ✓RT 1RT total sales Income=37 × 15 = R555 ✓SF 1SF for income Expenses = 200+37× 5 = R385 ✓SF 1SF for expenses Profit = R555 – R385 = R170 ✓CA 1CA answer for profit (4) 4.2.1 Year 2009 ✓✓RT 2RT for the year F (2) L2 4.2.2 Fees in 2015 = 1,093 × R12 500 = R13 662,50 ✓M 1M value from F multiplication with L4 Cost of fridge in 2015 = 1,04×R12 500 = R13 000 ✓M education inflation rate. 1M value from Difference = R13 662,50 – R12 500 = R662,50 ✓CA multiplication with general inflation rate 1CA answer (3) 4.2.3 The graph shows education has constantly outstripped 2J justification as from F general inflation. ✓✓J graph. L4 (2) 4.3.1 Arrangement of currencies: £; €; $; P; R; ¥✓RT✓A 1RT all currencies F 1A order according to L3 strength (2) 4.3.2 1¥ = R0,1383 F 3974,85 = R? 1M converting the L2 Cost of 1 in Rands = 3974,85× 0,1383✓M Japanese yens to Rands = R549,72 ✓A 1A cost of one DVD Cost of 500 DVD players = 500 × 549,72 = R274 860,88 ✓CA 1CA answer for cost of 500 DVDs (3) [18] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/JUNE 2022) MATHEMATICAL LITERACY P1 7 QUESTION 5 [25 MARKS] Quest. Solution Explanation/Marks AO: FULL MARKS T/L 5.1.1 Tax bracket = 4 ✓✓RT 2RT bracket (2) F L1 5.1.2 R128 650 ✓✓RT 2RT value of threshold F (2) L2 5.1.3 Monthly income = R35 455 F L4 Annual income = R35 455× 12 = R425 460,00 ✓MA 1MA multiplication by 12 and annual income 7,5 1A the annual pension Pension: 7,5% of R425 460 =100 × R425 460,00 = R31 909,50 ✓A 1CA taxable income Taxable Income = R425 460,00 – R31 909,50 1M use of correct tax = R393 550,50 ✓CA 31 bracket Tax = R67 144 + 100 × (393 550,50 – 321 600) ✓M 31 = R67 144 + 100 × 71 950,50 1CA tax payable before rebates = R67 144 + 22 304,655 = R89 448,655 ✓CA 1RT Total value of ✓RT rebates Tax less the rebates = R89 448,655 – (R14 958+R8199) 1M subtracting rebates Annual tax payable = R66 291,655 ✓M and tax after rebates (7) 5.2.1 ✓RT 1RT all values from D 2,27%; 5,04%; 5,05%; 5,90%; 6,68%; 7,24%; 13,38%; graph L2 16,15%; 38,28%. ✓M 1M arranging in order Median value = 6,68% giving EC ✓CA descending or ascending 1CA median value: EC (3) 5.2.2 5,04+5,05 ✓M 1M concept of getting D Q1 = 2 Quartile 1 L3 = 5,045% ✓A 1A for Q1 13,38 + 16,15 Q3 = 1A for Q3 2 = 14,765% ✓A IQR = Q3–Q1 1M method of = 14,765% – 5,045% ✓M subtracting Q3-Q1 = 9,72% ✓CA 1CA answer (5) 5.2.3 Probability is the chance that an event is likely to 2A explanation P happen. ✓✓A (2) L1 Copyright reserved Please turn over

Downloaded from hlayiso.com 8 MATHEMATICAL LITERACY P1 (EC/JUNE 2022) 5.2.4 Probability for GP = 0,3828 ✓CA 1CA converting 5 to P Probability for EC = 0,0668 ✓CA decimal for QP L3 Probability for a car to be in GP OR EC = 0,3828 + 1CA converting to 0,0668 = 0,4496 ✓A decimal for EC 1A answer (3) [24] TOTAL: 100 Copyright reserved Please turn over