Tuesday, 14 September 2021 08:28

## MATHEMATICAL LITERACY PAPER 1 GRADE 12 MEMORANDUM - 2018 SEPTEMBER PREPARATORY EXAM PAPERS AND MEMOS

MATHEMATICAL LITERACY PAPER 1
NATIONAL SENIOR CERTIFICATE
MEMORANDUM

SEPTEMBER 2018

 Symbol Explanation M Method M/A Method with Accuracy MCA Method with Consistent Accuracy CA Consistent Accuracy A Accuracy C Conversion S Simplification RT/RG/RM Reading from a table OR Reading from a graph OR Read from map F Choosing the correct formula SF Substitution in a formula J Justification P Penalty, e.g. for no units, incorrect rounding off etc. R Rounding Off OR Reason AO Answer only NPR No penalty for rounding

QUESTION 1 [30 MARKS] INTEGRATED QUESTION

 Question Solution Explanation/MarksAO: FULL MARKS Lev. 1.11.1.1 Inflation is the increase in the price of goodsand services over time.✓✓AORThe decrease in the purchasing value of themoney over time. ✓✓A 2A Explanation(2) FL1 1.1.2 Difference = R125 – R120 = R5 ✓M= 5 × 100 = 500 cents ✓A 1M Difference1CA In cents (2) FL1 1.1.3 Price = 112.5 x 125             100✓M= R140,60 ✓CAORProfit = 12,5 × 125 = R15,625✓M OR             100Price = 125 × 1,12Price = 125 + 15,625 = R140,625 = 140, 625= R140,60 ✓CA= R140,60 1M Multiply112,5 by 125 anddividing by 1001CA PriceOR1M For profit1CA PriceNPR(Concept of money)(2) FL1 1.21.2.1 Cost of a dozen = 90 x 12 ✓ MA                             60= R18,00 ✓ CAORDozens = 60                12= 5 ✓Cost of a dozen =90                             5= R18 ✓ 1MA Divide by 60 andmultiply by 121CA Cost1MA Number of dozens1CA Cost (2) FL1 1.2.2 % Profit = (( 60 + 75 ) 90 ) x 100                           90= 45 x 100% ✓M   90= 50% ✓CA 1M Subtraction for profit1M Fraction 45/90 multiplyby 1001CA Percentage(3) FL1 1.3 Time = 24:00 – 16:25 = 7h 35 min ✓MFrom midnight to 3:30 am = 3h30 min ✓M∴ Total time = 7 hours 35 min + 3h30min= 11h 05 min ✓CA 1M For 7h 35min1M For 3h 30min1CA Time(3) ML1 1.4 Plastic cups = 2 x 1000 ✓M                          275= 7,2727= 7 ✓A 1M Correct numerator anddenominator1A Rounding down(2) ML1 1.5 1.5.1 Length = 2𝑐𝑚 × 75 𝑚 ✓✓2M                 100 𝑐𝑚= 1,5 m ✓ 1M for Numerator1M for Denominator1CA With correct unit(3) ML1 1.5.2 Scale = 100 : 7 500 ✓✓2M= 1 : 75 ✓ 1CA 1M Ratio form1C Conversion1CA Answer(3) MPL1 1.6 1.6.1 72,67% = Nelson Mandela District ✓✓ A 2A District(2) DL1 1.6.2 Districts are: Chris Hani East, OR TamboCoastal, ✓AAmathole East and Amathole West ✓A 1A First 2 districts1A Second 2 districts(2) DL 1 1.6.3 Nelson Mandela, Sarah Baartman OR TamboInland, Chris Hani West, Alfred Nzo West, JoeGqabi, Buffalo City ✓M∴ 𝑀𝑖𝑑𝑑𝑙𝑒 𝑑𝑖𝑠𝑡𝑟𝑖𝑐𝑡 = CHRIS HANI WEST ✓CA 1M Identifying 7 bestperforming districts inorder1CA for middle district(2) DL1 1.6.4 Candidates failed = 26.1 × 313 030 ✓M                                100= 81 701 ✓CAORPassed = 73,9 × 313 030                 100= 231 329 ✓MNumber failed = 313 030 – 231 329= 81 701✓CA 1M Correct values1CA Candidates1MA Candidates passed1CA Candidates failedAccept 81 700 (2) DL1 

QUESTION 2 [46 MARKS] FINANCE

 Question Solution Explanation/MarksAO: FULL MARKS Lev. 2.1 2.1.1 Value of A = R3 250 + R4 500 + R1 200 ✓ M= R8 950 ✓ AORValue of A = R7 950 + R1 000 ✓ M= R8 950✓ A 1M Addition1A Value of A1M Addition1A Value of A (2) L2 2.1.2 Value of B = R1 440 – R1 500 ✓M= -R60 ✓ AValue of B = R1 440 – R1 500 ✓M= (R60) ✓ A 1M Subtracting correctvalues1A Negative answer1M Subtracting correctvalues1A Answer in brackets(2) 2.1.3 Value of C = R2 600 – R200 ✓ M= R2 400 ✓ CAORValue of C = R5 780 – (R880 + R1 000 + R1 500) ✓M= R2 400 ✓CA 1M Subtraction1CA Value of C1M Subtraction1CA Value of C (2) L1 2.1.4 Delivery ✓✓A 2A Correct answer (2) L1 2.2 2.2.1 Annual gross salary = R65 000 ×12 ✓M= R780 000 ✓CA 1M Multiply by 121CA Annual salary(2) L1 2.2.2 Mrs John’s annual pension fund contribution:=    7.5    (100 × 65 000) ×12 ✓M= R58 500 ✓CA 1M Multiply by 12CA Pension contr.(2) L1 2.2.3 Medical Aid contribution = R1 050 ×12 ✓M= R12 600 ✓A 1M Multiply by 121A Contribution (2) L1 2.2.4 Mrs John’s performance bonus = 75                                                     100 ×65 000 ✓M= R48 750 ✓A 1M Correct values1M Multiply by 75%(2) L1 2.2.5 Annual taxable income= R780 000 + R48 750 – (58 500 + R12 600) ✓✓M= R828 750 – R71 100 ✓M= R757 650 ✓CACA from 2.2.1, 2.2.2 and 2.2.3 1M Total income1M Total contributions1M Subtraction1CA Taxable income (4) L2 2.3 2.3.1 R701 301 and above ✓✓RT 2RT Correct group(2) L1 2.3.2 Rebate Mrs John will receive = R13 500 + R7 407 ✓M= R20 907 ✓A1M Adding correct rebates 1A Total rebate (2) L1 2.3.3 Actual tax = Income tax calculated on taxableincome Rebate= 206 964 + 41 % of the amount above 701 300 –R20 907= 206 964 + 0,41 × (757 650 – 701 300) –20 907 ✓SF= 206 964 + 0,41 × 56 350 – 20 907 ✓S= 206 964 + 23103,50 – 20 907 ✓S= R209 160,50 ✓CA CA from 2.2.5, 2.3.1 and 2.3.21SF Substitution1S Simplification1S Simplification1CA Nearest rand(4) L3 2.3.4 Net annual salary = Annual taxable income –Actual annual tax= R757 650 – R209 160,50 ✓SF✓M= R548 489,50 ✓CA CA from 2.2.5 and 2.3.31SF Substitution1M Subtraction1CA Net salary (3) L1 2.4 2.4.1 Interest 1st year = 15.5                              100 × 400 000= R62 000 ✓A2nd year amount = R400 000 + R62 000= R462 000 S✓Interest 2nd year = 15.5                                100 × R462 000= R71 610✓STotal interest = R62 000 + R71 610= R133 610 ✓A 1A Interest1S New amount 2nd year1S Interest in 2nd year1A Total interest(4) L2 2.4.2 Phone D:Cost excluding VAT = 100                                   115 × 1 750= R1 521,74 ✓MVAT = R1 750 – R1 521, 74 ✓M= R228,26 ✓AORVAT = 15  ✓M × 1 750 ✓M         115= R228,26 ✓A 1M Cost without VAT1M Subtraction1A VAT value1M Fraction1M Multiplication1A VAT value (3) L2 2.4.3 Phone D : Phone E3 : 260 : E ✓ME = 60 x 2          3= 40 ✓CA 1M Ratio form1CA Number of phones(2) 2.4.4 Total cost = 60 × 1 750 + 40 × 2 000= 105 000 ✓M + 80 000 ✓M= R185 000 ✓CA CA from 2.4.31M Cost for phones D1M Cost for phones E1CA Total cost (3) L1 2.4.5 1: 0,52709185 000 : Total cost in CYNTotal cost = 185 000 × 0,52709 ✓M= CYN 97511,65 ✓A✓A 1M Conversion1A Total cost1A Answer in Yuan(3) L2 

QUESTION 3 [25 MARKS] MEASUREMENT

 Question Solution Explanation/MarksAO: FULL MARKS Lev. 3.1.1 Length = 4 880 mm ÷ 1 000 ✓C= 4,88 m ✓A 1C Divide by 10001A Metres(2) L1 3.1.2 Distance A = 3 (150 mm) ✓M= 450 mm ✓A 1M Multiplication by 31A Distance(2) L1 3.1.3 Height of the wall = 2,1 m + 450 mm ÷ 1 000 ✓C= 2,1 m + 0,45 m ✓S= 2,55 m ✓CA CA from 3.1.21C Conversion1S Simplification1CA Height (3) L1 3.1.4 Area = Length × Width= 4,88 ✓ × 2,1 m ✓M      2 = 5,124 m2 ✓CA ✓Unit CA from 3.1.11M Dividing by 21M Multiplication1CA Area1A Unit (4) L2 3.1.5 Area covered by bricks= Area of garage – Area of double door= (2,55 m × 5,18 m) ✓ – (4,88 m × 2,1 m) ✓M✓M= 13,209 m2 – 10,248 m2 ✓S= 2,961 m2 ✓CA CA from 3.1.1 and 3.1.31M Area of garage1M Area of double door1M Subtraction1S Simplification1CA Area covered by bricks (5) L2 3.2.1 Height of the bricks and cement = (12 x 2) + 76 mm ✓M= 100 mm✓CA 1M Multiplication by 2 and addition1CA Height(2) L1 3.2.2 Number of rows of bricks = 2 500 ✓M ✓M                                             100 = 25 ✓CA CA from 3.2.11M Conversion1M Division1CA Number of rows of bricks (3) L1 3.2.3 Volume = 23 cm ×11 cm × 7,6 cm ✓SF ✓C= 1 922,8 ✓ A cm3 ✓A 1SF Substitution1C Conversion1A Volume1A Units(4) L2 

QUESTION 4 [31 MARKS] DATA HANDLING

 Question Solution Explanation/MarksAO: FULL MARKS Lev. 4.1. Range = 3,316 kg – 0,182 kg ✓RT ✓M= 3,134 kg ✓CA 1RT Correct values1M Subtraction1CA Range(3) L2 4.2 1,668 kg ✓✓A 2A Median(2) L2 4.3 Average =1,26 ×2 +1,371 ×9 +1,668 ×8 + 1,746 ×4 + 1,849 ×8 + 2,163 +2,333 + 3,128 ×2 ✓ M= 60,731 ✓M      35= 1,735= 2 kg ✓R 1M Adding1M Concept of mean1R Average to nearest kg(3) L2 4.4 22 ✓✓ A 2A Explanation(2) L1 4.5 Probability = 8  ✓M × 100%✓M                   35= 22,9 % ✓CA 1M Fraction value1M Multiply by 1001CA Percentage(3) L1 4.6 4M For the first 4 bars plotted correctly1M For the next 2 bars plotted correctly1M For the last 2 bars plotted correctly (6) L2 4.7 0,182; 0,182; 0,182; 0,309; 0,729; 0,729; 0,729; 0,856; 0,856; 0,856; 0,8560,936; 2,448; 2,448; 2,449; 3,038; 3,316; 3,316; 3,316; 3,316; 3,316; 3,316Q2Q2=Median= 0,856+0,936 = 0,896 ✓✓ MM                               2Q3=3,316 ✓ CA 1M Correct values1M Value or position of median1CA Q3(3) L1 4.8 3,316 kg ✓✓ 2A RT Modal value(2) L1 4.9 % of total fishes =  8  ✓× 100 ✓M                              57= 14,04 % ✓ 1M Fraction1M Multiply by 1001CA Percentage(3) L1 4.10 3rd hour ✓✓ 2RT Correct hour(2) L1 4.11 1,1,1,1,2,3,3,4,6 ✓✓ 2M Arrangement(2) L1 

QUESTION 5 [18 MARKS] MAPS, PLANS and OTHER REPRESENTATIONS

 Question Solution Explanation/MarksAO: FULL MARKS Lev. 5.1 There is no seat for Lundi here ✓✓A 2A No seat(2) MPL1 5.2 South ✓✓A 2A Direction(2) L1 5.3 A8 ✓ RPA11 ✓ RPA15 ✓ RP 1RP First seat1RP Second seat1RP Third seat (3) L1 5.4 35 ✓✓RP 2RP Available seats(2) L1 5.5 B14 ✓✓ RP 2RP Asi’s seat no.(2) L1 5.6 Row J ✓✓ RP 2RP Furthest row(2) L1 5.7 Side BB ✓✓RP 2RP Correct side(2) L1 5.8 P(seat from side AA) = 20  ✓A ✓A                                   19 4= 0,103 ✓CA 1A Numerator1A Denominator1CA Answer(3) PL2  TOTAL: 150