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GRADE 12 MATHEMATICAL LITERACY PAPER 1 MEMORANDUM - AMENDED SENIOR CERTIFICATE PAST PAPERS AND MEMOS MAY/JUNE 2017
Share via Whatsapp Join our WhatsApp Group Join our Telegram GroupMATHEMATICAL LITERACY P1
GRADE 12
SENIOR CERTIFICATE EXAMINATIONS
MAY/JUNE 2017
| Symbol | Explanation |
| M | Method |
| MA | Method with accuracy |
| CA | Consistent accuracy |
| A | Accuracy |
| C | Conversion |
| S | Simplification |
| RT/RG/RD | Reading from a table/graph/diagram |
| SF | Correct substitution in a formula |
| O | Opinion/Example/Definition/Explanation |
| R | Rounding off |
| NPR | No penalty rounding or omitting units |
| AO | Answer only, full marks |
QUESTION 1 [30 Marks]
| Ques | Solution | Explanation | T/L |
| 1.1.1 | R8,70 × 40 = R348✓✓ | 1MA multiplying with 40 1A box price AO (2) | F L1 |
| 1.1.2 | A profit is made when the selling price is more than the cost price.✓✓ OR A profit is the amount added to the cost price✓✓ OR Making more money than the cost price.✓✓ OR Positive difference between income and expenditure.✓✓ OR Income is more than cost or expenses.✓✓ OR Gained/extra money from the sale of a product ✓✓ | 2O explanation OR 2O explanation OR 2O explanation OR 2O explanation OR 2O explanation OR 2O explanation (2) (Except a correct example as an explanation) | F L1 |
| 1.1.3 | Amount = 40% × R435,04✓ = R174,016 ≈ R174,02✓ | 1MA calculate 40% of R435,04 1A VAT amount AO NPR (2) | F L1 |
| 1.1.4 (a) | Total cost = R10,04 + R8,70 + R20,66 + R6,73 + R29,99✓✓ = R76,12 ✓ | 1RT all correct values 1M adding at least 3 correct amounts 1CA total AO (3) | F L1 |
| 1.1.4 (b) | Selling price = R22 770✓ 230 = R99,00 ✓ | 1MA dividing correctly 1A selling price AO (2) | F L1 |
| 1.2.1 | South African Revenue Services✓✓ | 2A full name (2) | F L1 |
| 1.2.2 | R61 296✓✓ | 2RT correct amount (2) | F L1 |
| 1.2.3 | R542 096.76✓ 12 = R45 174,73 ✓ | 1MA dividing correctly 1A monthly salary AO (2) | F L1 |
| 1.2.4 | Tax bracket 4✓✓ OR 406 401 – 550 100✓✓ OR 96 264 + 36% of taxable income above 406 400✓✓ | 2RT correct tax bracket (2) | F L1 |
| 1.3.1 | 1 unit on the map is 200 units in reality✓✓ OR The real one is 200 times bigger✓✓ OR The drawing is 200 times smaller✓✓ | 2A explanation (2) | Maps L1 |
| 1.3.2 | Perimeter = 4 cm + 2 cm + 4,25 cm + 2,55 cm✓✓ = 12,8 cm✓ | 1C converting 1M adding 4 sides 1CA perimeter AO (3) | Meas L1 |
| 1.4.1 | January 2015✓ OR 01/2015✓ | 1A correct month 1A correct year (2) | D L1 |
| 1.4.2 | The price of cake went down/decreased/ dropped/ declined / less✓✓ | 2A description (2) | D L1 |
| 1.4.3 | 100%✓✓ | 2A correct index No penalty if % omitted Penalise if the index is given as R100 (2) | D L1 |
| [30] |
QUESTION 2 [35 MARKS] Topic Finance
| Ques | Solution | Explanation | T/L |
| 2.1.1 | R8 060,27 + R600 = R8 660,27 ✓✓ | 1RT reading both correct amounts 1CA balance AO (2) | L1 |
| 2.1.2 | R4 050,98 – R4 034,77 = R16,21 ✓✓ | 1M subtracting 1CA interest AO (2) | L1 |
| 2.1.3 | Accept any account number from 143260000 to 143269999 ✓✓ OR Writing only the FOUR missing digits | 2A possible number (2) | L1 |
| 2.1.4 | Mdiso Khaile ✓✓ | 2A correct person (2) | L1 |
| 2.1.5 | 0 OR none ✓✓ | 2A correct number (2) | L1 |
| 2.1.6 | 0 OR 0% OR impossible✓✓ | 2A correct probability (2) | P L2 |
| 2.1.7 (a) | R1,50 × 4 + R0,40 × 6 + R1,20 + R5,00 × 2 = R19,60 ✓✓ | 1M adding values 1A correct values (2) | L1 |
| 2.1.7 (b) | Amount without VAT = R19,60 = R17,19 ✓ 114% VAT amount = R19,60 – R17,19 ✓ = R2,41✓ OR VAT amount = 14%×R19,60 ✓ 114% ✓ = R2,41✓ | 1M dividing by 114% 1M subtracting 1A VAT amount OR 1M dividing by 114% 1M working with ratio 1A VAT amount AO (3) | L2 |
| 2.2.1 | Service charges ✓✓ | 2A correct item (2) | L1 |
| 2.2.2 | R4 253 219 thousand – R4 165 225 thousand ✓ = R87 994 thousand ✓ | 1M subtracting correct values from table 1A difference in thousands (2) | L1 |
| 2.2.3 | R2 878 830 thousand = R2 878 830 000 ✓✓ ≈ R2,9 billion | 1RT correct expected income 1A expanding the amount 1CA income in billions AO (3) | L1 |
| 2.2.4 | B = 4 253 219 – (794 866 + 2 694 542 + 34 044 + 211 526) ✓✓ = 518 241 ✓ | 1M subtracting 1MA adding correct values 1CA value AO (3) | L2 |
| 2.2.5 | Total income = 716 603 + 2 227 636 + 51 027 + 519 604 + 312 290 ✓ = 3 827 160 ✓ Total expenditure = 886 355 +34 657 + 481 980 + 71 180 + 1 780 120 + 238 + 875 072 = 4 129 602✓ A = R3 827 160 – R4 129 602 = – R302 442✓ or (R302 442) It is a DEFICIT✓ | 1MA adding correct values 1A income 1A expenditure 1CA amount 1CA deficit (5) | L3 |
| 2.2.6 | Percentage increase = Difference in remuneration × 100% Original budget remunearaion = 43 033 000 - 42 350 000 × 100% ✓ 42 350 000 ≈ 1,613 % ✓ | 1RT reading correct values 1SF substitution 1CA % increase AO (3) | L2 |
| [35] |
Related Items
QUESTION 3 [28 MARKS] Topic Measurement
| Ques | Solution | Explanation | T/L |
| 3.1.1 | B ✓✓ OR 325 × 325 × 325 ✓✓ | 2A correct letter OF 2A dimensions (2) | L1 |
| 3.1.2 | Area = 1 200 mm × 325 mm ✓✓ = 120 cm × 32,5 cm ✓ = 3 900 cm2 ✓ OR Area = 1 200 mm × 325 mm ✓✓ = 390 000 mm2✓ = 3 900 cm2 ✓ | 1RT correct dimensions 1SF substitution 1C converting 1CA area OR 1RT correct dimensions 1SF substitution 1A area 1C converting AO (4) | L2 |
| 3.1.3 | Number of boxes on ground = 24 = 12 ✓ 2 Total area needed = 12 × 1 056,25 cm2 ✓ = 12 675 cm2 ✓ OR Total area = 1 056,25 cm2 × 24 = 25 350 cm2 ✓ Total needed = 25350cm2 ✓ 2 = 12 675 cm2 ✓ | 1MA dividing number of boxes by 2 1M multiplying area of 1 box by number of boxes in one layer 1CA area OR 1MA multiplying area by 24 1M dividing total area by 2 1CA area AO (3) | L1 |
| 3.1.4 | 600 : 325 ✓✓ = 24 : 13 ✓ | 1RT correct two values 1A ratio correct order 1S simplification AO (3) | L1 |
| 3.1.5 (a) | Volume = 1 500 mm × 475 mm × 462,5 mm ✓ = 1,5 m × 0,475 m × 0,4625 m✓ = 0,32953125 m3✓ Inside volume = 0,32953125 m3 – (0,32953125 m3 × 9,36%)✓✓ = 0,298687125 m3 ≈ 0,299 m3 OR Volume = 1 500mm × 475mm × 462,5mm✓ = 1,5 m × 0,475 m × 0,4625 m✓ = 0,32953125 m3✓ 100% – 9,36% = 90,64% ✓ Inside volume = 0,32953125 m3 × 90,64% ≈ 0,299 m3 ✓ | 1SF substitution 1C conversion 1CA volume 1CA subtracting 1M multiplying by 9,36% OR 1SF substitution 1C conversion 1CA volume 1A subtraction 1M multiply with 90,64% (5) | L3
L3 |
| 3.1.5 (b) | Number of boxes = 6m 3 0.299m3 ✓ ≈ 20,066 ✓ ≈ 20 ✓ | 1MA dividing 1A simplification 1R rounding down AO (3) | L1 |
| 3.1.5 (c) | Volume needed = 148 × 0,299 = 44,252 ✓ Truck loads = 44.252m3 ✓ 6m3 = 7,375333... ≈ 8 ✓ OR Truck loads = 148✓ 20 = 7,4 ✓ ≈ 8 ✓ | 1A total volume 1M dividing by 6 m3 1R rounding up OR 1M working with ratio from Q3.1.5(b) 1A total volume 1R rounding up AO (3) | L2 |
| 3.2.1 | 5¼inches OR 5,25 inches ✓✓ | 1A radius 1A inches (2) | L1 |
| 3.2.2 | h = Volume (in cm3) ¼×π×(diameter in cm)2 h = 20 000 cm3) ✓ ¼×3.142×(10½×2.54cm)2 ✓ = 20 000 cm3) 558.717431cm2 ≈ 35,8 cm ✓ | 1SF correct substitution (20 000 and 2110) 1C convert inch to cm 1CA height NPR AO (3) | L2 |
| [28] |
QUESTION 4 [23 MARKS] Topic Maps, Plans and other
| Ques | Solution | Explanation | T/L |
| 4.1.1 | North West or NW ✓✓ | 2A direction (2) | L2 |
| 4.1.2 | It indicates the BORDER between South Africa and Botswana ✓✓ | 2O explanation Accept: border /fence/ boundary (2) | L1 |
| 4.1.3 | Travel from Johannesburg to Zeerust via Koster, then then from Zeerust to Abjaterskop Gate✓✓✓ OR Take the N14, N4, then turn on to the R49 ✓✓✓ | 1A Koster or N14 1A Zeerust or N4 and 1A Abjaterskop Gate or R49 (3) | L1 |
| 4.1.4 | Distance = 221,2 km – (62,4 km + 88,1 km)✓✓ = 70,7 km ✓ | 1M subtracting 1RT correct distances 1CA distance AO (3) | L1 |
| 4.1.5 | Via Koster: 70 km + 71,9 km + 35,2 km = 177,1 km ✓✓✓ | 1A correct distances 1M adding 1CA shortest route distance CA from 4.1.3 (3) | L2 |
| 4.2.1 | Left-hand side ✓✓ | 2A correct side (2) | L1 |
| 4.2.2 | 3 × 31 = 93 ✓✓✓ | 1RT 31 cottages 1MA multiply 3 1CA number of guests AO (3) | L2 |
| 4.2.3 | Walk towards reception and pass between reception and cottage number 17.✓ Continue pass the ablusion block✓ Cross the road to the swimming pool✓ OR Turn right into the road passing the petrol station, reception and shop✓ Turn left into the road✓ Continue straight, the swimming pool is on your right-hand side✓ | 1A passing reception 1A passing ablusion 1A crossing road OR 1A passing petrol station, reception and shop 1A turn left into road 1A swimming pool on your right hand side (3) | L2 |
| 4.2.4 | P(not a night drive) = 2 or 66,67% or 0,67✓✓ 3 | 1A numerator 1A denominator (2) | P L2 |
| [23] |
QUESTION 5 [34 MARKS] Topic Data
| Ques | Solution | Explanation | T/L |
| 5.1.1 | Free State ✓✓ | 2A correct province (2) | L1 |
| 5.1.2 | 66 007 + 24 475 + 74 823 + 96 057 + 57 108 +34 936 + 8 972 + 26 194 + 36 451 = 425 023 ✓✓✓ | 1RT all correct values 1M adding (min 8 prov.) 1CA total teachers AO (3) | L1 |
| 5.1.3 | 6156 ×100% ✓ 25720 ≈ 23,93% ✓ | 1RT correct values 1MA % calculation 1CA % schools AO NPR (3) | L2 |
| 5.1.4 | LSR = Total number of learners Total number of schools = 2129526✓✓ 2649 ≈ 803,898 ≈ 804 ✓ | 1RT correct values 1SF substitution 1CA ratio AO NPR (3) | L2 |
| 5.1.5 (a) | 30,1 ✓✓ | 2A mode (2) | L1 |
| 5.1.5 (b) | 31,5 30,1 30,1 30,0 29,8 29,4 28,9 28,5 27,2 ✓✓ | 1A all the values 1A correct order (2) | L1 |
| 5.1.5 (c) | 29,8 ✓✓ | 2A median CA from Q5.1.5 (b) (2) | L2 |
| 5.1.6 |  | ||
6 × 1A for each correct bar
| (6) | L2 | |
| 5.2.1 | 0,1 = 10% ✓✓ | 1A identifying the correct value 1CA writing it as a percentage (2) | L1 |
| 5.2.2 |
| 2A outcome at (a) 2A outcome at (b) (4) | L1 |
| 5.2.3 | 0,05 = 5 = 1 ✓✓ 100 20 | 1RT correct probability 1CA simplified fraction AO (2) | P L2 |
| 5.2.4 | 1 562 × 0,8 = 1 249,6 ✓✓ ≈ 1 249 or 1250 ✓ | 1RT correct values 1CA simplification 1R rounding AO (3) | L1 |
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