MATHEMATICAL LITERACY PAPER 2
GRADE 12
NATIONAL SENIOR CERTIFICATE EXAMINATIONS
MEMORANDUM
MAY/JUNE 2019
| 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 rounding |
| AO | Answer only |
| MCA | Method with constant accuracy |
NOTE:
QUESTION 1 [28 MARKS]
| Q | Solution | Explanation | T&L |
| 1.1.1 | A = 750 50 = 15 occupants/bewoners B = 2 500 × 30 = 75 000 OR Using ratios A : 22 500 8 : 12 000 A = 822500× = 15 or A = 22 500= 15 12000 1 500 50 : B 1 : 1 500 B = 1 500 × 50 = 75 000 | 1MA dividing by 50 1A occupants 1MA multiplying 1A litres OR 1M dividing and multiplying 1A occupants 1M multiplying 1A litres AO (4) | D L2 |
| 1.1.2 | Number of days = 1500 50 = 30 But July has 31 days His statement is NOT valid. OR Number of days in July 31 1 500 × 31 = 46 500 or 50 ×31 = 1 550 Not valid since it is not the same values OR/ July has 31 days 1500 ÷ 31 = 48,39 ℓ /pp 48,39 < 50 Not valid [using any of maximum litres/month or maximum litres/day] | 1MA dividing correct pair 1A number of days 1O verification OR 1A number of days 1M multiply 1O verification OR 1 A days in July 1M dividing 1O verification (3) | M L4 |
| 1.1.3 (a) | Total occupants = 2 + 4 + 2 + 2 = 10 Volume of water allowed/ = (10 × 50) × 31 = 15 500 ℓ Extra = 20% × 15 500 = 3 100 ℓ Total volume = 15 500 + 3 100 = 18 600 ℓ = 18,6 kℓ OR Total occupants = 2 + 4 + 2 + 2 = 10 Volume of water allowed per day = 50 × 10 = 500 ℓ Extra/Ekstra = 20% × 500 = 100 ℓ Total volume per day = 500 + 100 = 600 ℓ Total volume for May = 600 × 31 = 18 600 ℓ = 18,6 kℓ OR Total occupants = 2 + 4 + 2 + 2 = 10 Increased quota per day = 50 + 20% × 50 = 60 Maximum consumption = 60 ×10 × 31 = 18 600 ℓ =18,6 k ℓ | 1MA no. of occupants 1MCA 500 1M multiplying by 31 1CA no. of litres 1CA calculating 20% 1CA Adding litres 1C Converting to kilolitres (7) 1MA no. of occupants 1MCA 500 1CA calculating 20% 1CA Adding litres 1M multiplying by 31 1CA no. of litres 1C Converting to kilolitres OR 1MA no. of occupants 1CA calculating 20% 1CA Adding litres 1CA 600 1M multiplying by 31 1CA no. of litres 1C Converting to kilolitres (7) | M L3 |
| 1.1.3 (b) | Amount payable First 6 kℓ = 6 × R29,93 = R179,58 Next 4,5 kℓ= 4,5 × R52,44 = R235,98 Next 4,8 kℓ = 4,8 × R114,00 = R547,20 Total amount = R179,58 + R235,98 + R547,20 = R962,76 | 1MA multiplying by rate 1CA correct answer 1M same correct column calculation 1M same correct column calculation 1M adding 1CA total (6) | F L3 |
| 1.1.3 (c) | Accept one of the following applicable reasons
OR | 2O relevant answer (2) | M L2 |
| 1.2 | Labour day 1 = 6 hours × R129,99/h = R779,94 Day 2 = 2 hours × R129,99/h = R259,98 Total = R779,94+ R259,98 = R1 039,92 Cost of installing the tank = R12 958,00 + R1 943,70 + R1 039,92 = R15 941,62 Mr Vellem's budget is NOT enough OR Total labour = 6 hours + 2 hours = 8 hours Labour cost = 8 × R129,99 = R1 039,92 Cost of installing the tank = R12 958,00 + R1 943,70 + R1 039,92 = R15 941,62 Mr Vellem's budget is NOT enough OR Total labour = 6 hours + 2 hours = 8 hours Budget = R15 900 – R12 958,00 – R1 943,70 – R129,99 × 8 = – R41,62 Mr Vellem's budget is NOT enough | 1R rounding 1M 1st day labour calculation 1CA 2nd day labour calculation 1CA Adding 2 day values 1CA total cost 1O verification OR 1R rounding 1CA total hours labour 1M labour calculation 1CA labour cost 1CA total cost 1O verification OR 1R rounding 1CA total hours labour 1M subtracting from budget 1CA labour cost 1CA simplification 1O verification (6) | F L4 |
| [28] |
QUESTION 2 [32 MARKS]
| Q | Solution | Explanation | T&L |
| 2.1.1 (a) | NOTE: 2.1.1 IS NOT TO BE MARKED, MARKS WILL BE SCALED Total surface area = 2(L × W) + 2(L × H) + 2(W × H) = 2(6,5 × 6,5) + 2(6,5 × 12,5) + 2(6,5 × 12,5) cm2 = 2(42,5) cm2 + 2(81,25) cm2 + 2(81,25) cm2 = 84,5 cm2+ 162,5 cm2 + 162,5 cm2 = 409,5 cm2 | 1C conversion 1SF substitution 1S simplification 1CA total area (4) | M L2 |
| 2.1.1 (b) | To appeal to young children. OR To advertise the medicine OR To show it is for children OR Accept any valid reason | 2O reason (2) | M L4 |
| 2.1.2 | radius = 2cm ,522 = 1,26 cm 10 mℓ = 3,142 × (1,26 cm)2×ℎP 10cm3 4,9882392cm2 = h 2,0047…m = h | 1A radius 1SF substituting volume 1M changing the subject of the formula 1CA height NPR (4) | M L3 |
| 2.2 | Number of boxes in one carton = 6 × 8 × 4 = 192 Total number of boxes = 192 × 5 = 960 OR Total number of boxes = 6 × 8 × 4 × 5 = 960 In each carton 1 layer has 6 × 8 = 48 boxes Each carton has 4 layers 48 × 4 = 192 boxes They ordered 5 cartons 192 × 5 = 960 boxes | 1M multiplying 1A number per box 1M multiply 1A total OR 1M multiplying 2 values 1M multiplying with 3rd value 1M multiplying with 4th value 1A total (4) | MP L2 |
| 2.3.1 | Range = N – Lowest value 4 527 = N – 612 4 527 + 612 = N 5 139 = N | 1M writing formula 1SF substitution 1M change the subject of the formula 1CA when 76 for unknown age is used (4) | D L2 |
| 2.3.2 | 612 , 1 280 , 2 221 , 3 051 , 3 429 , 5 139 Interquartile Range = 3 429 – 1 280 = 2 149 | CA from 2.3.1 1M arranging 1A Q1 1A Q3 1M subtraction 1CA Simplify (5) | D L3 |
| 2.3.3 | Total = 1 280 + 612 + 3 051 + 2 221 + 5 139 + 3 429 + 76 = 15 808 Percentage = 76 ×100% 15808 = 0,48 ≈ 0,5% It is correct, due to rounding. Dit is korrek as gevolg van afronding. | CA from 2.3.1 1MCA adding all values 1RT unknown age value 1M % calculation 1CA simplification 1O explanation (5) | D L4 |
| 2.3.4 | Number hospitalised < 6 months = 1 280 × 44,2% = 565,76 Number hospitalised 20+ = 3 429 × 7,6% = 260,6 Difference = 565,76 – 260,60 = 305,1 ≈ 305 | 1MA % calculation 1A simplification 1A simplification 1CA difference NPR (4) | D L3 |
| [32] |
QUESTION 3 [26 MARKS]
| Q | Solution | Explanation | T&L |
| 3.1.1 | 3 200 [Accept values from 3 100 to 3 250 | 2RT number of copies (2) | F L2 |
| 3.1.2 | Contract 2 | 2RT correct contract (2) | F L2 |
| 3.1.3 | Total cost = fixed cost + cost per page = R625 per month for the first 600 pages + (R1 475 – R625) ÷ (4 000 – 600) per page more than 600 = R625 for the first 600 pages + R0,25 per page extra OR Total cost = R625 + R0,25 (n – 600) where n is the number of pages more than 600. | 1A setting up the equation 1RT constant cost 1RT values from graph 1M calculating the increment per page 1CA cost per page extra OR 1RT constant cost 1M calculating the increment per page 1CA cost per page extra 1A setting up the equation 1A explaining the unknown in the equation (5) | F L4 |
| 3.1.4 | 1A Starting point (0 copies ; R0,00 charge) 1A end point (4 000 ; 2 800) 1A connecting points with a straight line. (3) | F L2 | |
| 3.2 | The electrical lead is crossing the floor. This can be dangerous since persons can step on it and perhaps unplug the copier which might damage the machine or A person can trip over the lead and fall causing injury. OR Copier in the middle of the room takes up space, if it is against the wall the room is not so crowded OR Not suitably placed. Directly facing the window, it can attract criminals OR The copier is suitably placed since it can now be accessed from all sides. | 1A justification 2O reason (3) | MP L4 |
| 3.3.1 | 100% – 58,5 % = 41,5 % Length of truck on original picture = 76mm 41,5% ≈ 183 mm Length of the real truck = 183 mm ×50 = 9156 mm = 9,156 m OR Length /Lengte = 76 mm ×50 = 3 800 mm = 3,8 m 100% – 58,5 % = 41,5 % Length of real truck = 3,8 m 41,5 % = 9,157 m | 1M subtraction from 100% 1M dividing the 76 mm with the percentage 1M Multiplying by 50 1S simplifying 1C conversion NPR (5) 1M Multiplying by 50 1S simplifying 1C conversion 1M dividing 3,8 m by the percentage 1CA real length (5) | M L3 |
| 3.3.2 | A$ 45 × 300 × R9,41564/A$ A$45<A$47,26 | 1M multiplying by 300 1C conversion 1CA when 15% is used 1CA simplification 1CA adding all costs 1O verification OR 1M multiplying by 300 1C conversion 1A total tax rate 1CA simplification 1CA adding all costs 1O verification OR 1M dividing by 300 1A total tax rate 1M dividing by 119,7% 1S simplification 1C conversion 1O verification NPR (6) OR 1M multiplying by 300 1C conversion 1A using total tax rate 1M multiplying with total rate 1CA simplification 1O verification | F L4 |
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QUESTION 4 [34 MARKS]
| Q | Solution | Explanation | T&L |
| 4.1.1 | SW (South west/Suidwes) | 2A reading direction (2) | MP L2 |
| 4.1.2 | Part or sections of the railway line are not seen from above. OR The road stays continuous (whole) while the railway line is in sections. | 2A description (2) | MP L2 |
| 4.1.3 | Toyota or 11 | 2A correct circle (2) | MP L2 |
| 4.1.4 | Proceed straight on Stateway Street until you turn right at the City Council into Arrarat Street. Then proceed straight until Alma. Destination is on the left-hand side.
Bingo (10), Alfa(8) and Engen(4) (from each circle taking 2nd exit to NE). | 1A straight on Stateway 1A turn right 1A Arrarat 1A straight until Alma 1A destination on left-hand OR 1A exit point 1A correct street 1A exit point 1A description 1A naming the circles (5) | MP L3 |
| 4.1.5 | Distance between Alfa and Engen circles = 5mm ∴5mm = 500m 1mm = 100m or 1: 100 000 Distance between circles 13 and 14 is 28 mm = 1,4 km ∴28 mm = 1 400 m 1mm = 50m or 1 : 50 000 This map is NOT drawn to scale. 4 – 7 for distance between Afla and Engen 24 – 29 for distance between 13 and 14 | 1A measuring given distance 1CA simplification or scale 1A measure distance 1CA simplification or scale 1O explanation (5) | MP L3 |
| 4.1.6 | 5 minutes = 5 ≈ 0,083 hour 60 Distance = Speed × time 4 = speed × 5 min Speed = Distance = 4 Time 0,083 = 48 km/h The car's speed was within the speed limit. OR Speed = 4 km ÷ 5 min = 0,8 km/min × 60 min/hour = 48 km/h The car's speed was less than the limit. | 1C minutes to hours 1MA substituting 1CA Speed value 1O conclusion NPR OR 1MA substituting 1C converting 1CA Speed value 1O conclusion (4) | M L4 |
| 4.1.7 | P = 3 20 = 0,15 Valid | 1RT numerator 1RT denominator 1S simplification 1O conclusion (4) | P L4 |
| 4.2.1 | 104 : 88 = 13 : 11 | 1RT correct values 1A correct order 1S simplification (3) | D L2 |
| 4.2.2 | 203 1 724 = 0,11774942 ≈ 0,12 | 1RT numerator 1RT denominator 1CA simplification NPR (3) | P L2 |
| 4.2.3 | Total NOT electrical repairs = 1 + 206 + 103 = 310 P(NOT)= 310× 100% 368 ≈ 84% OR P(electr) = 58 368 P(NOT) = 1 – 58 368 = 310 × 100% 368 ≈ 84% OR P(electr) = 58 × 100% 368 ≈ 16% P(NOT) = 100% – 16% = 84 % | 1A numerator 1RT denominator 1M multiplying with 100% 1CA rounded simplification OR 1RT denominator 1A numerator 1M multiplying with 100% 1CA rounded simplification OR 1RT denominator 1M multiplying with 100% 1A subtracting from 100% 1CA simplification (4) | P L3 |
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QUESTION 5 [30 MARKS]
| Q | Solution | Explanation | T&L |
| 5.1.1 | Motor claims = R100 712 182 – (R18 513 071 + R15 498 565 + R7 339 724 + R6 463 292) =R100 712 182 – R47 814 652 = R52 897 530 ≈ R53 million OR Motor claims = 53% × R100 712 182 = R53 377 456 ≈ R53 million | 1RT correct values 1M subtracting from the total 1CA rounded value OR 1RT correct values 1M percentage calculation 1CA rounded value (3) | F L2 |
| 5.1.2 | Total 2016 × 60% = R59 438 533 Total 2016 = R59 438 533 ÷ 60% = R99 064 221,67 Difference = R99 064 221,67 – R87 101 354 = R11 962 867,67 OR 2016 60% - R59 438 533 10% - R9 906 422,17 16% - R15 850 275,47 7% - R6 934 495,52 × 2 Total/Total : R99 064 221,67 Difference= R99 064 221,67 – R87 101 354 M = R11 962 867,67CA | 1M relating values 1M dividing 1A simplification 1M subtracting from 2017 value 1CA difference OR 1M finding rand values 1M double the 7% value 1A simplification 1M subtracting from 2017 value 1CA difference (5) | F L3 |
| 5.1.3 | Percentage difference = New value - Old value × 100 % Old value = R11829111 - R15 498 565 × 100 % R15 498 565 = – 23,676…% ≈ – 24% | 1A concept of percentage difference 1M difference 1RT correct values 1CA percentage (4) | D L3 |
| 5.1.4 (a) | Percentage Household/Persentasie Huishoudelik = 7 339724 × 100 % 100712182 = 7,28782… % Percentage Other/Persentasie Ander = 6 463 292 × 100 % = 6,41758… 100712182 Her statement is valid; the percentage should be 6% if rounded down. OR Motor claims 2015 = 53% × R100 712 182 = R53 377 456 Total of the claims = R18 513 071 + R15 498 565 + R7 339 724 + R53 377 456 + R6 463 292 = R101 192 108 Other % = 6 463 292 × 100 % = 6,41758… 100712182 = 6,38% Household % = 7 339724 × 100 % 100712182 = 7,25% Her statement is valid; the percentage should be 6% if rounded down. OR Rand value of the sectors /Randwaarde van die sektore = 7% × R100 712 182 = R7 049 852,74 Both household and other were supposed to be R7 049 852, but it is not. ⸫ Her statement is valid | 1RT correct values 1M multiplying with 100% 1A simplification 1A simplification 1O verification OR 1RT correct values 1M multiplying with 100% 1A simplification 1A simplification 1O verification OR 1RT correct values 1MA percentage calculation 1A simplification 1O explanation 1O verification (5) | D L4 |
| 5.1.4 (b) | When subtracting the percentages of Commercial, Home owner, Household and motor from 100% Other will be 7% due to %values in a circle diagram. OR Percentages were rounded. | 2O reflecting (2) | D L4 |
| 5.1.5 | Number of successful claims = 14,0858% × 2 144 ≈ 302 Average paid out = R 11 829 111 302 = R39 169,24 | 1MA % calculation 1A simplification 1M dividing 1CA simplification (4) | F L3 |
| 5.1.6 | The percentage of commercial claims went down from 2015 to 2016 but then again went up from 2016 to 2017. OR From 2015 to 2017 the trend is it increased | 1A down 2016 1A up 2017 (2) | D L4 |
| 5.2 | Number of days/Aantal dae = 21 (July/Julie) + 31 + 30 + 31 + 3 = 116 It is not valid./Dit is nie geldig nie. OR 131 days is more than 4 months It is not valid | 1MA adding correct days 1A simplification 1O verification (3) | D L4 |
| 5.3 | Accept one of the following The insurance company believes the claim is not valid. They suspect it is a fraudulent claim. They don't believe the item was specified. Under insured / Unpaid premiums Too many claims to date Negligence on the side of the client OR Any other valid reason/Enige ander geldige rede | 2O reason (2) | F L4 |
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