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GRADE 12 MATHEMATICAL LITERACY PAPER 1 MEMORANDUM - NSC PAST PAPERS AND MEMOS NOVEMBER 2017
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GRADE 12
NATIONAL SENIOR CERTIFICATE
MEMORANDUM
NOVEMBER 2017
MARKS: 150
| Symbol | Explanation |
| M | Method |
| MA | Method with accuracy |
| CA | Consistent accuracy |
| A | Accuracy |
| C | Conversion |
| S | Simplification |
| RT/RG | Reading from a table/Reading from a graph |
| 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 [30 MARKS]
| Ques | Solution | Explanation | Topic/L |
| 1.1.1 | D ✓ | 2RT correct letter (2) | F L1 |
| 1.1.2 | G ✓ | 2 RT correct letter (2) | D L1 |
| 1.1.3 | C ✓ | 2 RT correct letter (2) | M L1 |
| 1.2.1 | Profit = R18 700 – R 14 960 = R 3 740✓ | 1M subtracting correct values 1A calculating profit AO (2) | F L1 |
| 1.2.2 | 10:15 + 5h50 = 16:05 16:05 OR 4:05 pm✓ OR 5 past 4 in the afternoon | 1M adding 1A correct time of sale AO (2) | M L1 |
| 1.2.3 (a) | Radius = 32,8 mm ÷ 2 = 16,4 mm | 1MA dividing diameter by 2 1CA radius AO (2) | M L1 |
| 1.2.3 (b) | Distance = (71,8 mm – 32,8 mm) ÷ 2 = 19,5 mm✓ OR 71,8 mm ÷ 2 = 35,9 mm✓ Distance = 35,9 mm – 16,4 mm = 19,4 mm✓ | 1MA subtracting and dividing 1CA distance OR 1MA subtracting and dividing 1CA distance AO (2) | M L1 |
| 1.3.1 | Cost of diluted juice per litre = R 44,95 ÷ 14 ℓ = R 3, 210714286 ≈ R 3,21✓ | 1MA dividing 1CA cost per litre NPR AO (2) | M L1 |
| 1.3.2 | 2 ℓ : 12 ℓ 1 : 6✓ | 1A correct volume of water and order 1CA simplification Accept 1 6 AO (2) | M L1 |
| 1.3.3 | Number of glasses of juice = 14 0,175 = 80✓ | 1MA dividing the correct values 1CA simplification to a whole number AO (2) | M L1 |
| 1.4.1 | 35 39 39 60 63 84 93 107 117 120 126 142✓✓ | 1RT all values 1MA ascending order (2) | D L1 |
| 1.4.2 | July OR 7th month✓✓ | 2A correct month (2) | D L1 |
| 1.4.3 | 9✓✓ | 2A correct mode (2) | D L1 |
| 1.4.4 | April OR 4th month✓✓ | 2A correct month (2) | D L1 |
| 1.4.5 | May and July OR 5th month and 7th month✓✓ | 1A May 1A July (2) | D L1 |
| [30] |
QUESTION 2 [46 MARKS]
| Ques | Solution | Explanation | Topic/L |
| 2.1.1 | R465,00✓✓ | 2RT correct bus fare (2) | F L1 |
| 2.1.2 | Queenstown and King William's Town✓✓ | 2RT correct cities (2) | F L1 |
| 2.1.3 (a) | Port Elizabeth to Bloemfontein = R435,00 | 1RT R435 1CA cost Accept trial and error method AO (2) | F L1 |
| 2.1.3(b) | King William's Town✓✓ | CA from Q2.1.3(a) 2RT correct city (2) | F L2 |
| 2.1.4 | Cost excluding VAT = R365,00 × 100 114 = R320,175… ≈ R320,18✓ OR Cost excluding VAT = R365 ≈ R320,18✓ 1,14 OR 114 : 365 = 100 : x x = price excl. VAT x = R365,00 × 100 114 = R320,175… ≈ R320,18✓ OR VAT = R365 × 14 114 = R44,82 Cost excluding VAT = R365 – R44,82 ≈ R320,18✓ | 1M × 100 1M ÷ 114 1CA simplification OR 1M dividing 1MA 1,14 1CA simplification OR 1M proportion 1M x as subject of formula 1CA simplification OR 1M multiplying with ratio 1M subtracting VAT 1CA simplification NPR AO (3) | F L2 |
| 2.1.5 | From Queenstown to Bloemfontein return trip = R410 × 2 = R820 Total travelling cost = 12 × R820 = R9 840✓ OR Number of trips = 2 × 12 = 24 Total travelling cost = 24 × R410 = R9 840✓ OR One way cost for a year = R410 × 12 = R4 920✓ Total traveling cost = R4 920 × 2 = R9 840✓ OR Traveling cost = R410 × 2 × 12 =R9 840✓ | 1RT correct fare 1CA for calculating the return trip 1M multiplying by 12 1CA total cost OR 1M multiplying by 12 1CA total trips 1RT correct fare 1CA total cost OR 1RT correct fare 1M multiplying with 12 1M multiplying with 2 1CA total cost OR 1RT correct fare 1M multiplying with 2 1M multiplying with 12 1CA cost AO (4) | F L2 |
| 2.2.1 | July 2013 OR 07/2013 OR 07/13✓✓ | 1RT month 1RT year (2) | F L1 |
| 2.2.2 | Water and Sewerage Refuse Removal✓✓ | 1RT water and/or sewerage 1RT refuse Penalty for including property rates (2) | F L1 |
| 2.2.3 | November = 3 days, December = 20 days end date 2016/12/20 OR 20 December 2016✓✓ | 1M adding 1A end date 20 Dec Accept 19 Dec AO (2) | F L1 |
| 2.2.4 | Daily average consumption = 12,00 kℓ ÷ 23 days ≈ 0,522 kℓ✓ OR Verifying the consumption rate per day: = 12,00 kℓ ÷ 0,522 kℓ/day ≈ 23 days✓ OR 0,522 kℓ/day × 23 days ≈ 12,00kℓ✓ | 1RT correct value 1M dividing in correct order OR 1RT correct value 1M dividing in correct order OR 1M multiplying 1A volume (2) | F L1 |
| 2.2.5 | Water The amount of water consumption is not the same every month.✓✓✓ | 1R variable expense 2O explanation clearly showing change (3) | F L1 |
| 2.2.6 (a) | A = R690 000 × R0,0069160 ÷ 12 = R397,67✓✓ | 1RT all values from bill 1CA simplification Note value for B can be used to calculate A AO (2) | F L1 |
| 2.2.6 (b) | B = R397,67 – R115,27 = R282, 40✓✓ OR B = R880,10 – R167,58 – R430,12 = R282,40✓✓ | 1M subtracting correct values 1CA simplification OR 1M subtracting correct values 1CA simplification AO (2) | F L1 |
| 2.2.7 | Sewerage rate per m2 = R 298,36 463 = R0,6444060475 OR 463m2 : R 298,36 1m2 : R0, 6444…✓ | 1RT correct values 1A simplification OR 1RT Correct values 1A simplification NPR AO (2) | F L1 |
| 2.2.8 | R919,33 | 2RT unpaid amount (2) | F L1 |
| 2.2.9 | Rounding up OR Rounding (off) to the nearest R10,00✓ OR Rounding (off) to the nearest R100,00✓ | 2A Rounding up OR 1A rounding 1A nearest 10 rand OR 1A rounding 1A nearest 100 rand (2) | F L1 |
| 2.3.1 | Commission = 1,95% × £360,00 = £7,02✓ | 1MA calculating % 1A commission in pound AO (2) | F L1 |
| 2.3.2 | £360,00 = 360 0,05773 = R6 235,9258.. ≈ R6 235,93 or R6 235 or R6 236✓ OR £1 = R1,00 0,057773 = R17,32201628 £360 = R17,32201628 × 360 = R62 35,925862 ≈ R6 235,93✓ OR R1,00 = £0,05773 x = £360,00✓ x =R 1 ×360 0,05773 = R6 235,93✓ | 1MA conversion 1A simplification 1CA rounding OR 1MA conversion 1A simplification 1CA rounding OR 1A multiplying with 360 1MA conversion 1CA rounding NPR AO (3) | F L2 |
| 2.3.3 | Interest after 1 year = R5 000 × 6,3% | 1M calculate interest for first year 1A simplification 1CA 2nd year amount 1M half year interest 1CA simplification OR 1M calculate interest for first year 1A simplification 1M 2nd year rate 1CA half year interest 1CA simplification OR 1M calculate amount for first year 1A simplification 1CA 2nd year amount 1M half year 1CA simplification (5) | F L2 |
| [46] |
QUESTION 3 [21 MARKS]
| Ques | Solution | Explanation | Topic/L |
| 3.1.1 | Number of tables = 240 ÷ 8 = 30 Number of balloons = 4 × 30 = 120✓✓ | 1A correct number of tables 1CA minimum number of balloons AO (2) | M L1 |
| 3.1.2 | Length of decorative ribbon in cm = 2 × (length + width) + 1 = 2 × (10 + 6) + 1 = 33✓✓ | 2SF substituting correct values into the formula 1A minimum length AO (3) | M L2 |
| 3.1.3 | Volume = π × (radius)2 × height = 3,142 × (6 cm)2 × 28 cm = 3 167,136 cm3✓✓✓ | 1A radius 1SF correct height and 3,142 1CA simplification NPR (3) | M L2 |
| 3.1.4 | Volume = 1 680 cm3 × 45% = 756 cm3 Mass of sand = 756 cm3 × 1,53g/cm3 = 1 156,68 g ÷ 1 000 ≈ 1,16 kg✓✓✓✓ OR 1,53 g/cm3 = 0,00153 kg/cm3 Volume = 1 680 cm3 × 45% = 756 cm3 Mass of the sand = 0,00153 kg/cm3 × 756 cm3 = 1,15668 kg ≈ 1,16 kg✓✓✓✓ OR Mass of sand in a full vase = 1 680 cm3 × 1,53g/cm3 = 2 570,4 g = 2,5704 kg✓✓ Mass of sand if filled to 45% = 2,5704 kg × 45% = 1,16 kg✓✓ | 1A calculating 45% 1M multiply by rate 1CA mass in grams 1C converting to kg to 2decimal places OR 1C converting to kg 1A calculating 45% 1M multiplying with the rate 1 CA mass in kg to 2 dec. places OR 1M multiplying with the rate 1A mass 1C conversion 1CA mass of sand to two decimal places (4) | M L2 M L2 |
| 3.2.1 | Area of triangle = ½ × 4 cm × 3,464 cm = 6,928 cm2✓✓✓ | 1A substituting correct values in formula 1RT height 1CA simplification NPR AO (3) | |
| 3.2.2 | Total surface Area of a triangular prism = 2 × 6,928 + 3 × 6 cm × 4cm = 13,856 cm2 + 72 cm2 = 85,856 cm2✓✓✓✓ | CA from Q3.2.1 1CA substituting area of triangle 1SF substituting correct values in formula 1CA simplification 1CA total surface area (4) | M L3 |
| 3.2.3 | 30 minutes = 1 800 seconds Average time to cover 1 box = 1800seconds 20 = 90 seconds✓ OR Average time to cover 1 box = 30 min = 1,5 min 20 = 1,5 min × 60 sec/min = 90 seconds✓ | 1 C conversion to seconds 1CA simplification OR 1M time per box 1C conversion AO (2) | M L1 |
| [21] |
QUESTION 4 [27 MARKS]
NOTE :MPU & NC maximum [23 MARKS] to be scaled to 27 MARKS
| Ques | Solution | Explanation | Topic/L |
| 4.1.1 | Bar scale OR Scaled bar OR Linear scale OR Graphical scale ✓✓ | 2A identifying type of scale (2) | M&P L1 |
| 4.1.2 | Top view OR Aerial view OR Bird’s eye view OR Satelite view✓✓ | 2A correct view of the map (2) | M&P L1 |
| 4.1.3 | South East OR SE OR East of South✓✓ | 2A identifying correct direction (2) | M&P L1 |
| 4.1.4 | 5✓✓ | 2A exact number of medical points Accept 4 (2) | M&P L2 |
| 4.1.5 | Mowbray and Observatory✓✓ | 2A identifying correct suburbs Accept Maitland and Saltriver (2) | M&P L1 |
| 4.1.6 | Castle De Goede Hoop, Old Biscuit Mill , Planetarium OR 4, 5 and 6✓✓✓ | 3A identifying correct tourist attractions (3) | M&P L2 |
| 4.2.1 | D; B; E; A; C✓✓ | NOTE: [MPU & NC not to be marked] 1A order BEA 1A end with C (2) | M&P L2 |
| 4.2.2 | E OR B✓✓ | NOTE: [MPU & NC not to be marked] 2A correct letter (2) | M&P L1 |
| 4.2.3 (a) | 0 % OR Impossible OR 0 OR 0 OR None✓✓ 130 | 2A probability (2) | P L2 |
| 4.2.3(b) | Total blocks = 20 + 25 + 28 + 30 + 27 = 130 Probability of taking out a blue block = 25 130 OR 5 OR 19,23% OR 0,19✓✓✓ 26 | 1A total 130 1A numerator 1A denominator AO (3) | P L2 |
| 4.2.4 (a) | Number of layers = 35 cm ÷ 16, = 2,12… ≈ 2✓✓ | 1MA dividing correct values 1CA exact number of layers AO (2) | M&P L1 |
| 4.2.4 (b) | Number of cans which can be packed lengthwise = 56 cm ÷ 12,6 cm = 4,444… ≈ 4 Number of cans which can be packed width-wise = 41 cm ÷ 12,6 cm = 3,253… ≈ 3 Maximum number of cans = 4×3×2 = 24✓✓✓ | 1MA dividing the width or length by 2,6 1A rounding both down to whole numbers 1CA for max number of cans AO (3) | M&P L3 |
| [27] |
QUESTION 5 [26 MARKS]
| Ques | Solution | Explanation | T/L |
| 5.1.1 | Broken line graph OR line graph ✓✓ | 2A correct type of graph (2) | D L1 |
| 5.1.2 | Number of candidates = 287 453 + 389 615 = 677 068✓✓ | 1M adding Math and Math Lit 1CA max number of candidates AO (2) | D L2 |
| 5.1.3 | 100% OR 1 OR certain OR definite✓✓ | 2A correct probability (2) | P L2 |
| 5.1.4 | Accounting, Business Studies, Economics and Mathematical Literacy✓✓✓ | 1RT 1st subject 1RT 2nd subject 1RT last two subjects (3) | D L1 |
| 5.1.5 | Mathematics✓✓ | 2RT correct subject (2) | D L1 |
| 5.1.6 | The data of one variable is grouped into subjects OR The data of one variable is not numerical✓✓ | 2A explanation (2) | D L1 |
| 5.1.7 | Business Studies✓✓ | 2RT correct subject (2) | D L1 |
| 5.2.1 | Copyright payments, advertising costs, bursary, grants etc. (OR any other valid expenditure)✓✓ | 2O an example of other type of expenditure (2) | D L1 |
| 5.2.2 | Donations | 1M subtracting from R63 billion 1CA simplification in billions 1CA donations as a % OR 1M percentage income shown 1M subtracting from 100% 1CA simplification OR 1M percentage calculation 1M subtracting from 100% 1CA simplification NPR AO (3) | D L2 |
| 5.2.3 | Interest in Rand = 54 100 000 000 × 0,7% = 378 700 000 OR 378,7 million✓✓✓ OR Interest in rand = 54,1 billion × 0,7% = 0,3787 billion = 378 700 000 OR 378,7 million✓✓✓ | 1RT correct amount 1M multiplying with 0,7% 1CA interest amount OR 1RT correct amount 1M multiplying with 0,7% 1CA interest amount AO (3) | F L1 |
| 5.2.4 | Difference = income – expenditure = R63 billion – R54,1 billion = R8,9 billion = R8 900 million OR R8 900 000 000✓✓✓ OR Difference = income – expenditure = R63 000 million – R54 100 million = R8 900 million OR R8 900 000 000✓✓✓ | 1M subtracting 1CA simplification in billions 1C for difference in millions OR 1M subtracting 1C converting to millions 1CA difference in millions (3) | D L2 |
| [26] |
TOTAL: 150