Thursday, 15 December 2022 07:13

## MATHEMATICS LITERACY PAPER 1 GRADE 12 MEMORANDUM - NSC PAST PAPERS AND MEMOS JUNE 2022

NATIONAL
SENIOR CERTIFICATE
JUNE 2022
MATHEMATICAL LITERACY P1
MARKING GUIDELINE
MARKS: 100

 Symbol Explanation M Method MA Method with accuracy CA Consistent accuarcy 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 Answers only NPR No penalty for correct rounding off to minimum of two decimal places

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.

 QUESTION 1 (20 Marks) Que Solution Explanation/Marks  AO: FULL MARKS T/L 1.1.1 18,25  =     1825      ✓M   100        10000 = 73         ✓A   400 1M fraction1A answer in a reduced form (2) F L1 * 1.1.2 % of price = 100 – 18,25%                 = 81,75%Price =  81,75  × 380  ✓M               100 = R310,65 ✓ CA                ORReduction = 18,25 × 380                      100= R69,35  ✓MPrice = R380 – 69,35  ✓M = R310,65  ✓CA 1M subtraction1M % calculation1CA answer    OR1M % calculation 1M subtraction1CA answer                                       (3) F L1 * 1.2.1 Difference = R469 – (– R447)  ✓CA  = R916 million  ✓RT 1 RT for the two correct values 1 CA answer (2) F L1 1.2.2 Total = 265+277+326+390+447+458+486 – (469+300) = 1880 million  ✓CA 1M addition (+) and subtraction (–) of the values1CA             (2) F L1 1.3.1 Weekend wage rate = 3/2 × 25   ✓MA = R37,50   ✓A 1MA multiplication 1A answer F L1 * 1.3.2 ✓MEarnings = 6 × 25 + 37,50 × 4   ✓MA        = R300  ✓CA 1M multiplications1MA addition1CA answer (3) F L1 * 1.4.1 Discrete  ✓✓A DL1 1.4.2 Game ✓✓RT DL1 1.4.3 Total games = 4 + 6 + 5 + 4 + 1 + 2 + 2 = 24 games ✓M ✓CA 1M adding the games 1CA answer (2) DL1 

### Related Items

 QUESTION 2 [18 MARKS] Que Solution Explanation/Marks AO: FULL MARKS T/L 2.1.1 Time 4 hours   ✓✓RT 2RT    (2) F L2 2.1.2 From graph:2 welders complete 1 frame in 4 hours2 : 120 : ? frame in 4 hoursFrames =  20×1 ✓✓M                   2              = 10 frames ✓                        ORn × t = 820 × t = 8 ✓t = 8/20 = 0,4 hours to make 1 frame by 20 welders ✓SIn four hours= 4/0,4 ✓M = 10 frames ✓A 1M value from graph 1M numerator1M denominator1A answer              OR1SF substitution1S simplification for 2,5 frames done in 1 hour by 20 welders1M multiplication1A answer (4) F L3 2.2.1 ✓MA = 28−25,81  × 100%  ✓MA           25,81 = 8,485% = 8,5% ✓CA 1M correct values for numerator and denominatorM % calculation1CA                    (3)(NPR) F L2 2.2.2 Cost: Up to 6 kℓ = R0 = R0 ✓M 6 – 25 kℓ = 19 k × R23,60 = R448,40  ✓M 25 – 30 kℓ = 5 kℓ × R32,20 = R161,00   ✓M                                  ✓MTOTAL COST = R448,40+R161,00 = R606,40   ✓CA 1M cost in block 11M cost in block 21M cost in block 31M addition all costs1CA answer           (5) F L3 2.3.1 Salary B = R3 192,05+15 761,80   ✓M = R18 953,85   ✓CA 1M adding the two balances1 CA answer (2) F L2 2.3.2 Bank fees for March = 42,37+17,47+100,88  ✓M = R160,72   ✓CA 1M adding fees of March1CA answer     (2) F L1 

 QUESTION 3 [21 MARKS] Que Solution Explanation/Marks AO: FULL MARKS T/L 3.1 2020  ✓ A Reason: Covid-19 pandemic    ✓J 1A year1J reason (2) DL1 3.2 ✓MC = 25 285,1 – (2093,5+2092,8+2249,4+ 1988,8+1750,5+1964,7+2067,1+2204,4+2308,0+2267,8+2493,4)= 1804,7  ✓M     ✓CA 1M subtracting from 25 285,11M addition of all other values1CA answer (3) DL2* 3.3 descending order:            ✓RT2493,4; 2308,0; 2267,8; 2249,4; 2204,4; 2093,5; 2092,82067,1; 1988,8; 1964,7: 1804,7;1750,5 1RT all values including value from 3.21CA order with value from 3.2 (2) DL2* 3.4 ✓RT Range = 2 262,3 – 33,8   ✓M = 2 228,5 million    ✓CA 1RT highest and lowest values1M concept of range 1CA answer (3) DL2 3.5 ✓MMean income for 2018 =  24846,4  = 2070,53 million  ✓AMean income for 2020 = 9818,5 = 818,21 million  ✓ADouble mean income for 2020 = 818,21×2 = 1636,42 Million  Mean income for 2018 (2 070,53) is greater than double mean income for 2020 (1636,42)Statement Valid  ✓J 1M concept of mean1A mean for 20181A mean for 20201M comparing values of mean 2018 and double mean income for 20201J valid statement. NPR DL4* 3.6 From 2018 December income dropped right through up to July 2019; then increased from August 2019 to December 2019. It remained high up to March 2020.   ✓JThen it dropped drastically in from April 2020 and remained low in 2020.  ✓J 1J justification for the period Dec 2018 to July 20191J justification for the period August 2019 to 2020 (2) DL4 3.7 May and June 1A first months1A second months.CA from 3.2 (2) DL2 

 QUESTION 4 [20 MARKS] Que Solution Explanation/Marks AO: FULL MARKS T/L 4.1.1 Values of dependent variable at break-even pointIncome = R300   ✓RTExpenses = R300   ✓RT 1RT value for income1RT value for expenses FL2 4.1.2 Total sales in a week = 37 packets  ✓RTFrom Graph: Income = R555  ✓RT Expenses = R385   ✓RT Profit = R555 – R385 = R170   ✓CA                      ORTotal sales = 37  ✓RTIncome=37 × 15 = R555  ✓SF   Expenses = 200+37× 5 = R385   ✓SFProfit = R555 – R385 = R170   ✓CA 1RT adding sales from table1RT reading income from graph1RT expenses from graph1CA answer for profit                OR1RT total sales1SF for income1SF for expenses1CA answer for profit (4) FL2 4.2.1 Year 2009 ✓✓RT 2RT for the year    (2) FL2 4.2.2 Fees in 2015 = 1,093 × R12 500 = R13 662,50  ✓MCost of fridge in 2015 = 1,04×R12 500 = R13 000   ✓MDifference = R13 662,50 – R12 500 = R662,50   ✓CA 1M value from multiplication with education inflation rate.1M value from multiplication with general inflation rate1CA answer (3) FL4 4.2.3 The graph shows education has constantly outstripped general inflation.  ✓✓J FL4 4.3.1 Arrangement of currencies: £; €; \$; P; R; ¥ ✓RT ✓A 1RT all currencies1A order according to strength (2) FL3 4.3.2 1¥ = R0,13833974,85 = R?Cost of 1 in Rands = 3974,85× 0,1383  ✓M = R549,72  ✓ACost of 500 DVD players = 500 × 549,72 = R274 860,88  ✓CA 1M converting the Japanese yens to Rands1A cost of one DVD1CA answer for cost of 500 DVDs (3) FL2 

 QUESTION 5 [25 MARKS] Que Solution Explanation/Marks AO: FULL MARKS T/L 5.1.1 Tax bracket = 4 ✓✓ RT 2RT bracket   (2) FL1 5.1.2 R128 650 ✓✓ RT 2RT value of threshold  (2) FL2 5.1.3 Monthly income = R35 455Annual income = R35 455× 12 = R425 460,00 ✓MAPension: 7,5% of R425 460 =  7,5  × R425 460,00                                                100                                          = R31 909,50  ✓ATaxable Income = R425 460,00 – R31 909,50                           = R393 550,50  ✓CATax = R67 144 +  31  × (393 550,50 – 321 600)  ✓M                           100 = R67 144 +  31  × 71 950,50                     100 = R67 144 + 22 304,655 = R89 448,655    ✓CA                                                                         ✓RTTax less the rebates = R89 448,655 – (R14 958+R8199)Annual tax payable = R66 291,655   ✓M 1MA multiplication by 12 and annual income1A the annual pension1CA taxable income1M use of correct taxbracket1CA tax payable before rebates1RT Total value of rebates 1M subtracting rebates and tax after rebates     (7) FL4 5.2.1 ✓RT 2,27%; 5,04%; 5,05%; 5,90%; 6,68%; 7,24%; 13,38%; 16,15%; 38,28%.   ✓MMedian value = 6,68% giving EC  ✓CA 1RT all values from graph1M arranging in order descending or ascending1CA median value: EC DL2 5.2.2 Q1 = 5,04+5,05  ✓M                   2        = 5,045%   ✓AQ3 = 13,38 + 16,15                  2        = 14,765%   ✓AIQR = Q3–Q1       = 14,765% – 5,045%    ✓M         = 9,72%   ✓CA 1M concept of getting Quartile 11A for Q11A for Q31M method of subtracting Q3-Q11CA answer (5) DL3 5.2.3 Probability is the chance that an event is likely to happen.  ✓✓A 2A explanation   (2) PL1 5.2.4 Probability for GP = 0,3828  ✓CAProbability for EC = 0,0668   ✓CAProbability for a car to be in GP OR EC = 0,3828 +0,0668 = 0,4496   ✓A 1CA converting 5 to decimal for QP1CA converting to decimal for EC1A answer                         (3)  TOTAL: 100