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GRADE 12 MATHEMATICAL LITERACY PAPER 2 MEMORANDUM - NSC PAST PAPERS AND MEMOS NOVEMBER 2017
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GRADE 12
MEMORANDUM
NATIONAL SENIOR CERTIFICATE
NOVEMBER 2017
MARKS: 150
| Symbol | Explanation |
| M | Method |
| MA | Method with accuracy |
| CA | Consistent accuracy |
| A | Accuracy |
| C | Conversion |
| S | Simplification |
| RT | Reading from a table/Reading from a graph/Read from map |
| O | Opinion/Explanation |
| SF | Substitution in a formula |
| P | Penalty, e.g. for no units, incorrect rounding off etc. |
| R | Rounding Off/Reason |
| AO | Answer only |
| NPR | No penalty for rounding |
QUESTION 1 [40 MARKS]
| Ques | Solution | Explanation | T&L |
| 1.1.1 | Decrease amount in thousands = R32 187 × 4,402%✓ ≈ R1 416,87✓ Communication Cost in thousands = R32 187 – R1 416,87 = R30 770,13 = R30 770 Decrease amount = R32 187 000 × 4,402% = R1 416 871,74 ≈ R1 417 000 Comm. Cost = R32 187 000 – R1 417 000 = R30 770 000 OR Communication Cost in thousands = 32 187 – (4,402% × 32 187) = 32 187 – 1 416,87 = 30 770 OR 100% – 4,402% = 95,598 %✓ Communication Cost in thousand = R32 187 × 95,598%✓ = R30 770,12826 ≈ R30 770 OR Communication Cost in thousands = R2 163 571 – R(67 257 + 640 601 + 69 866 + 953 592 + 135 768 + 34 087 + 55 267 + 176 363) = R2 163 571 – R2 132 801 = R30 770 | 1M % calculation 1CA decreased amount 1M subtracting 1R rounding OR 1M subtracting 1M % calculation 1CA decreased amount 1R rounding OR 1M subtracting 1M % calculation 1CA cost 1R rounding OR 1M subtracting 1M adding all other values 1CA total for other values 1CA cost AO (4) | F L2 |
| 1.1.2 | Profits could decrease.✓ OR Imported stock will cost more. ✓ | 2O explanation (2) | F L4 |
| 1.1.3 | For 2015: Percentage profit = 342 534 ×100% ✓ 2 250 041 = 15,22345593% For 2016: Percentage profit = 360 651 ×100% ✓ 1 500 518 617 = 15,00518617 % The profit decreased✓ OR The profit nearly stayed the same. OR NOTE: Calculated profit for 2015 is R343 002 thousand Percentage profit = 343 002 ×100%✓ 2 250 041 ≈ 15,24% For 2016: Percentage profit = 360 651 × 100% ✓ 2 403 509 = 15,00518617 % The profit decreased | 1RT correct values | F L4 (5) |
| 1.2 | Income tax = R147 996 + 39% × R(663 000 – 550 100)✓ = R147 996 + 39% × R112 900✓ = R147 996 + R44 031 = R192 027✓ Total Income Tax (after rebates)✓ = R192 027 – R13 500 – R7 407 OR = R192 027 – R20907✓ = R171 120✓ | 1A correct bracket 1MCA amount above 1S simplification 1CA tax before rebate 1M subtracting both rebates 1CA tax after rebate (6) | F L3 |
| 1.3 | Increase number of donors for 2017 = 110 000 × 9,6% = 10 560✓ Number of donors 2017 = 110 000 + 10 560 = 120 560✓ Increase number of donors for 2018 = 120 560 × 9,6% = 11 573,76✓ Number of donors 2018 = 120 560 + 11 573,76 = 132 133,76 ≈ 132 134✓ OR Number of donors for 2017 =110 000 + (110 000 × 9,6%) = 120 560✓ Number of donors for 2018 =120 560 + (120 560 × 9,6%) = 132 133,76 ≈ 132 134✓ OR Number of donors for 2017 = 110 000 ×109,6% = 120 560✓ Number of donors for 2018 = 120 560 × 109,6% = 132 133,76 ≈ 132 134✓ OR Number of donors for 2018 = 110 000 × 109,6% × 109,6% = 132 133,76 ≈ 132 134✓ | 1M calculating 9,6% | D L3 |
| 1.4.1 | Makes provision for other people who are not Asian, Black, Coloured or White.✓ OR Some donors don't indicate race.✓ OR The percentage of the races do not add up to 100%.✓ OR The other is ‘mixed’ race.✓ OR They are from other countries.✓ | 2O explanation (2) | D L4 |
| 1.4.2 | As the years increase the percentage black donors increase.✓ | 2O increasing trend (2) | D L4 |
| 1.4.3 | The number of donors are different every year.✓ OR The graph represents percentages.✓ OR The percentages are rounded values.✓ OR The graph shows that the bars’ heights are not the same.✓ | 2O explanation (2) | D L4 |
| 1.4.4 (a) | The 2015 donors × 101,02% = 490914✓ Number of donors = 490914 101,02% OR 490 914 4,0102 = 485 957,236… ≈ 485 957✓ | 1MA dividing by 101,02% 1A number of donors NPR (2) | D L2 |
| 1.4.4 (b) | % white = 100% – (8% + 38% + 5% + 2%) = 47%✓ Number of white donors = 485 957 × 47% = 228 399,79…✓ ≈ 228 400 | CA from Q1.4.4 (a) 1MA subtracting from 100% 1CA percentage 1MCA % calculation 1CA rounded number AO (4) | D L3 |
| 1.5.1 | P (Blood Type O ) = (39 + 6)% = 45% OR 9 OR 0,45✓ 20 | 1RT correct two values 1A calculating probability (2) | P L2 |
| 1.5.2 | AB+✓ | 2A correct blood type (2) | P L2 |
| 1.5.3 | No, it is NOT most likely. Can only receive blood from own blood group.✓✓ OR P(O- receiving blood from any donor) = 1 8✓✓ ∴ It is NOT most likely. | 1O verification 2O explanation OR 1A numerator 1A denominator 1O verification (3) | P L4 |
| [40] |
QUESTION 2 [37 MARKS]
| Ques | Solution | Explanation | T&L |
| 2.1.1 | Inland prices have higher costs for transport / storage.✓ OR Coastal storages are close by and transport fees are lower.✓ OR Fuel is imported via harbours.✓ OR Most refineries are along the coast.✓ | 2O reason (2) | F L4 |
| 2.1.2 | S = R2,67 × R616,00✓ R12,32 = R133,50 OR R 2,67 R 2,34 OR Number of litres = R616,00 R12,32 = 50✓ OR R142,50 R2,34 OR R77,00 R1,54 OR R117 R2,34 S = 50ℓ × R2,67/ℓ = R133,50 OR Basic fuel price = R77 × R5,26R1,54 = R263 S = R616 – R142,50 – R77,00 – R263,00 = R133,50✓ | 1M multiplying 1A correct ratio 1CA storage cost 1M dividing 1A litres 1CA storage cost OR 1A basic fuel price 1M subtracting all from total 1CA storage cost AO (3) | F L2 |
| 2.1.3 | Number of litres consumed = 1 250 km × 7,3 ℓ ÷ 100 km = 91,25 ℓ Inland cost = 91,25 ℓ × R12,32/ ℓ = R1 124,20 Coastal cost = 91,25 ℓ × R11,94/ ℓ = R1 089,525 ≈ R1 089,53✓ Statement is NOT valid.✓ OR Litres consumed = 1 250 km ÷ 100 km × 7,3 = 91,25 Difference in fuel price = R12,32 – R11,94 = R0,38 Difference in cost = R0,38/ ℓ × 91,25 ℓ ≈ R34,68✓ Statement is NOT valid.✓ OR Inland Cost / 100 km = 7,3 ℓ × R12,32/ ℓ = R89,94 Number of 100km distances =1250 km ÷ 100 km = 12,5 Cost = 12,5 × R89,94 = R1 124,20 Coastal Cost / 100 km = 7,3 l × R11,94 = R87,16 Number of 100 km distances = 1250 km ÷ 100 km = 12,5 Cost =12,5 × R89,94 = R1 089,53 Difference = R1 124,50 – R1 089,53 = R34,67 Statement is NOT valid.✓ OR Difference = R12,32 – R11,94 = R0,38 Number of 100 km distances = 1 250 km ÷ 100 km = 12,5 Cost = R0,38 × 7,3 × 12,5 = R34,68 Statement is NOT valid.✓ | 1M working with consumption rate | F L4 |
| 2.2.1 | % increase = R70,9billion - R54 billion ×100% R54billion ≈ 31,296 %✓ OR R70,9 Billion× 100% = 131,2962% R54Billion % increase = 131,2962% – 100% ≈ 31,296 %✓ OR Using Trial & Error: R54 billion × 31,3% = R16,9 billion R16,9 billion + R54 billion = R70,9 billion ∴ % increase = 31,3%✓ | 1M % increase 1A correct values 1CA percentage OR 1M % increase 1A correct values 1CA percentage OR 1M % calculation 1A increase amount 1CA percentage NPR (3) | F L2 |
| 2.2.2 | 7 + 118 = 125 7 × Total budgeted income = R70,9 billion ✓ 125 Total budgeted income = R70,9billion ÷ 7 125 = R1 266,07 billion ≈ R1 266 billion✓ OR 7: 118 = R70,9 billion : x 7x = R70,9 billion ×118 x = R70,9 billion × 118 7 ≈ R1 195,17 billion✓ Total budgeted income = R1 195,17 billion + R70,9 billion = R1 266,07 billion ≈ R1 266 billion✓ | 1A adding ratio values 1A using ratio values 1M dividing by ratio 1CA budget value OR 1A using proportion 1S changing subject 1CA other revenues 1CA rounded value in billion (4) | F L3 |
| 2.3.1 | India✓✓ | 2RT country (2) | D L2 |
| 2.3.2 | 0,02 0,52 0,63 0,91 1,12 1,23 2,03 2,17 2,97 3,62 4,11✓ IQR = Q3 – Q1✓ = 2,97 – 0,63✓ = 2,34✓ | 1M use formula of IQR 1A lower quartile 1A upper quartile 1CA IQR AO [Accept 58 – 7 = 51] (4) | D L3 |
| 2.3.3 | Countries with high rankings are developed (rich, 1st world) as well as underdeveloped/developing (poor, 3rd world).✓ OR Countries with low rankings are developed (rich) as well as underdeveloped/ developing (poor).✓ OR Counties listed are from all over the world (different continents).✓ OR Rankings show the sample was chosen randomly.✓ | 2O valid reason (2) | D L4 |
| 2.3.4 | India: Mean Daily wage = 236,51 93,76% ≈ 252,25 Rouble✓ SA: Mean Daily wage = 237,35 26,20% ≈ 905,92 Rouble✓ Difference = (905,92 – 252,25) Russian Rouble = 653,67 Russian Rouble✓ | 1RT reading both values 1MA dividing by % 1A Indian day wage 1A SA day wage 1M subtracting 1CA difference in Rouble (6) | F L3 |
| 2.3.5 | Range = 425,52 – 21,44 = 404,08 Russian Rouble✓ 1 Russian Rouble = 0,016 Euro ∴404,08 Russian Rouble = 404,08 × 0,016 Euro = 6,46528 Euro✓ 1 South African Rand = 0,070 Euro ∴ 6,46528= R92,36 0,07 Learner solution is incorrect✓ OR 1 Russian Rouble = 0,016Rand✓ 0,070 = R 0,2285714286 Range = 425,52 – 21,44 = 404,08 Russian Rouble✓ = 404,08 × 0,2285714286 rand/rouble✓ = R92,36✓ Learner solution is incorrect OR Max. value to rand: 425,52 × 0,016 ÷ 0,07 = R97,26 Min. value to rand: 21,44 × 0,016 ÷ 0,07 = R4,90 Range = R97,26 – R4,90 = R92,36 Learner solution is incorrect.✓ | 1A range 1M multiplication 1C convert to Euro 1C convert to rand 1A rand value 1O verification OR 1C dividing by 0,07 1A conversion factor 1A range 1C conversion 1A rand value 1O verification OR 1C conversion 1CA max value 1CA min value 1M subtracting 1CA rand value 1O verification NPR (6) | D L4 |
| [37] |
QUESTION 3 [40 MARKS]
| Ques | Solution | Explanation | T&L |
| 3.1.1 | 33 Kwela Street ✓✓ | 2A correct number 1A correct street (3) | MP L2 |
| 3.1.2 | Length 22 mm (21 mm to 23 mm)✓ Width 9 mm (8 mm to 10 mm) Scale 25 mm = 30 m (24 mm to 26 mm)✓ ∴ Length = 30× 22 25 = 26,4 m✓ Width = 9 × 30= 10,8 m 25 OR Scale: 25 mm : 30 m (24 mm to 26 mm) 25mm : 30 000 mm✓ 1 : 1 200✓ Length = 22 mm (21 mm to 23 mm) Width = 9 mm (8 mm to 10 mm) Actual length = 22 × 1 200 mm = 26 400 mm = 26,4 m Actual width = 9 × 1 200 mm✓ = 10 800 mm = 10,8 m ✓ | 1A length 1A width 1A measured scale 1M using the scale 1CA length in m 1CA width in m OR 1A measured scale 1M unit scale 1A length 1A width 1CA length in m 1CA width in m (6) | M L3 |
| 3.1.3 | On the enlarged map: Measured length = 62 mm (61mm to 64 mm) Scaled length = 62 mm ÷ 5 = 12,4 mm ≠ 22 mm ∴ NOT valid✓ OR On the enlarged map: The measured width = 24 mm (23 mm to 26 mm) widths: 9 mm × 5 = 45 mm ≠ 24 mm ∴ NOT valid✓ OR On the enlarged map: Measured length = 62 mm (61mm to 64 mm) Measured width = 24 mm (23 mm to 26 mm) Scale factor = 2262 OR width = 924 ≈ 2,82 ≈ 2,67 ∴ Not valid✓ | CA from Q3.1.2 1MCA measured length 1M dividing by 5 1CA simplification 1O verification OR 1A measured length 1M multiplying with 5 1CA simplification 1O verification OR 1A measured 1M dividing 1CA scale factor 1O verification (4) | MP L4 |
| 3.2.1 | Length = 5 240 mm – 2 × 220 mm = 4 800 mm✓ Width = 4 040 mm – 2 × 220 mm = 3 600 mm✓ Floor area = 4 800 mm × 3 600 mm = 17 280 000 mm2 = 17 280 000 ÷ 1 000 000 = 17,28 m2✓ OR Length = 5 240mm = 5,24m Width = 4 040mm = 4,04m Wall thickness = 220mm = 0,22m Interior Length = 5,24m – 2(0,22m) = 4,8m Interior Width = 4,04m – 2(0,22m) = 3,6m Floor Area = 4,8 m × 3,6 m = 17,28m2✓ | 1MA subtracting of thickness 1CA internal length 1CA internal width 1MCA substitution 1C conversion 1CA internal area in m2 OR 1C conversion of all values 1MA subtracting thickness 1CA length 1CA width 1MCA substitution 1CA internal area in m2 (6) | M L3 |
| 3.2.2 | Area of Ceiling board = 2 400 mm × 900 mm = 2 160 000 mm2 Number of boards needed = 17280000 2160000 = 8✓ ∴ Need more than 7 OR Number needed = 4 800 mm ÷ 2 400 mm = 2 for length✓ Number needed = 3 600 mm ÷ 900 mm = 4 for width✓ Total needed = 2 × 4 = 8 ∴ Need more than 7✓ OR Area of one ceiling board = 2,4 m × 0,9 m = 2,16 m2 Total area coved by 7 boards = 2,16 m2 × 7 = 15,12 m2 ∴ Need more than 7✓ | CA from Q3.2.1 1SF substitution 1A area of board 1M dividing 1CA number of boards 1O deduction OR 1M dividing 1CA number length wise 1CA number width wise 1CA number of boards 1O deduction 1SF substitution 1A area of board 1M multiplying 1CA total area 1O deduction (5) | M M |
| 3.2.3 | Length of cornice = 2 × (4 800 mm + 3 600 mm) = 16 800 mm✓ | 1CA values from Q 3.2.1 or RT if reworked 1SF substitution 1CA length (3) | M L2 |
| 3.2.4 | 16 800 ÷ 2 000 = 8,4 Hence 9 lengths cornice needed.✓ Total cost = 8 × R91,44 + 9 × R53,64 = R731,52 + R482,76 = R1 214,28✓ The statement is correct. | CA from Q3.2.3 and Q3.2.2 1CA number of lengths 1A using 2 correct prices 1M multiplying 1CA cost 1O conclusion (5) | F L4 |
| 3.3.1 | Above ground is a higher security risk✓ OR Safety reasons✓ OR Below the ground the cost will be less.✓ OR Above the ground it takes up space.✓ OR Underground, the water stays cooler/fresher than in direct sun/ lessen evaporation.✓ OR Aesthetic reasons.✓ OR Below the ground for water to easily run into it.✓ OR Less maintenance✓ | 2O reason (2) | MP L4 |
| 3.3.2 | 8 000 ℓ = 8 000 000 cm3 = 8 m3✓ Volume of a cylindrical tank = π × radius2 × length 8 m3 = 3,142 × radius2 × 2,9 m (radius)2 = 8m 3 3,142 × 2,9m = 0,87798239…✓ Radius = √0,87798239 ≈ 0,937 m✓ Diameter = 1,874 m✓ OR Volume of a cylindrical tank = π × radius2 × length 8 000 000 cm3 = 3,142 × radius2 × 290 cm✓ (radius)2 = 8 000 000cm3 3,142 × 290cm = 8 779,8239… Radius = √8779,8239✓ ≈ 93,7 cm Diameter = 187,4 cm✓ = 1,874 m✓ | 1C Conversion 1SF substitution 1A change subject of formula 1S simplification 1CA radius 1CA diameter OR 1SF substitution 1A change subject of formula 1S simplification 1CA radius 1CA doubling the radius 1C conversion to m NPR (6) | M L3 |
| [40] |
QUESTION 4 [33 MARKS]
| Ques | Solution | Explanation | T&L |
| 4.1.1 | Dineo's maximum wind speed is 95 (MPH) 95 MPH = 80,4672 × 95km/h 50 = 152,887… km/h = 152,89 km/h✓ OR 50 mile = 80,4672 km 1 mile = 1,609344 km 95 MPH = 95 miles / hour × 1,609344✓ = 152,88768 km/h ≈ 152,89 km/h✓ OR 95 miles – 50 miles = 45 miles 50 miles = 80,4672 km 45 miles = x km x km = 80,4672 km × 45 miles ÷ 50 miles = 72,4205 km ✓ Total distance = 80,4672 km + 72,4205 km = 152,887 km ∴ 95 MPH = 152,89 km/h✓ | 1C conversion 1CA simplification 1R rounding OR 1C conversion 1CA simplification 1R rounding OR 1C conversion 1CA simplification 1R rounding AO (3) | M L2 |
| 4.1.2 | Measured distance between gridlines is 17 mm Measured distance between P and Q is 39 Actual distance = 205,043 km × 39mm 17mm ≈ 470,39 km✓ Distance = Ave. speed × time Ave. speed = 470,39km 24 jours ≈ 19,56 km/h ✓ (Accept 16 mm to 18 mm for gridlines and 38 mm to 42mm for PQ distance) OR App. distance from P to Q is 2 1 gridlines 3 Distance = 2 1 × 205,043 km 3 = 478,4336667 km Distance = Ave. speed × time 478,4336667 km = Ave. speed × 24 hours Ave. speed ≈ 19,93 km/h (Accept 2 1 up to 2 1 ) 6 3 OR 18 mm = 205,043 1 mm = 11,39 Measured distance from the gridline to Q is 3 mm (2 to 4)mm Distance from P to Q = 205,043 + 205,043 + 3 × 11,39 = 444,256 km Ave. speed = 444,256km 24 hours ≈ 18,51 km/h | 1A distance between gridlines 1A distance P to Q 1M using scale 1MCA using correct values 1CA actual distance 1S changing the subject of the formula 1SF substitution 1CA Ave speed NPR (8) 2A distance P to Q 1M multiplying 1A using correct values 1CA actual distance 1SF substitution 1S changing the subject of the formula 1CA ave. speed OR 1A distance between gridlines 1M unit scale 1A distance to Q 1M using scale 1CA actual distance 1SF substitution 1S changing the subject of the formula 1CA Ave speed NPR (8) | M&P L3 (5) Meas L3 (3) |
| 4.2.1 | 10✓✓ | 2RT correct value (2) | D L2 |
| 4.2.2 |
| D L2 | |
| 4.2.3 | North Atlantic✓ | 2RT correct region (2) | D L2 |
| 4.2.4 | Western Pacific: Total storms = 39 + 30 + 52 + 34 + 40 = 195 Damages in million USD = 10 200 + 8 410 + 22 800 + 6 080 + 10 600 = 58 090✓ North Atlantic: Total storms = 12 + 9 + 13 + 19 + 19 = 72 Damages in million USD✓ = 590 + 232 + 1510 + 75 000 +21 000 = 98 332✓ NOT valid statement,✓ Western Pacific had the most storms but North Atlantic had the greatest amount of damages. | 1A number of storms WP 1RT using amounts from table 1MCA adding amounts 1CA number of storms in NA 1RT only using values to 2011 1CA amount of damage 1O not valid 2O reason (9) | D (4) F(4) L4 |
| 4.3 | Growth rate per 1 000 = 38,3 – 11,9 – 1,9 = 24,5 ∴ percentage growth rate =24,5 × 100%✓ 1000 = 2,45% OR Percentage growth rate  | 1MA subtracting rates 1CA growth rate 1MCA calculating percentage (÷1 000 ×100) 1CA simplification OR 1MA subtracting rates 1M calculating percentage 1CA growth rate 1CA simplification AO (4) | D L2 |
| [33] | |||
TOTAL :150