MATHEMATICAL LITERACY PAPER 2
GRADE 12
NATIONAL SENIOR CERTIFICATE
JUNE 2018
MEMORANDUM
| Symbol | Explanation |
| M | Method |
| MA | Method with accuracy |
| CA | Consistent accuracy |
| A | Accuracy |
| C | Conversion |
| S | Simplification |
| RT/RG/RM | Reading from a table/Reading from a graph/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/Reason |
| AO | Answer only |
| NPR | No penalty for rounding |
QUESTION 1 [36]
| Ques. | Solution | Explanation | Level |
| 1.1.1 | Total actual expenditure value for 2017 = R62 459,75 + R125 000,05 + R63 241,20 +R200 541,65 = R451 243,10 ✓✓ OR Total actual expenditure value for 2017 = R461 864,70 – R10 621,60✓✓ = R451 243,10 | 1RT Correct values 1M Addition 1RT Correct values 1M Subtraction (2) | L2 F |
| 1.1.2 | Actual value is the amount of money that was either received or spent.✓✓ Budgeted value is the amount of money that is predicted to be either received or spent✓✓ OR Amount of money planned to cover all expenses. Accept any logical explanation. | 2A Explain actual value 2A Explain budgeted value (4) | L4 F |
| 1.1.3 | Teaching resources✓ School bought most of the teaching resources the previous year✓✓ OR They received teaching resources from donors✓✓ Accept any other valid reason. | 1RT 2R Reason (3) | L2 & L4 F |
| 1.1.4 | Disagree, because the schools budget for 2018 shows a negative balance✓✓ OR From 2016, the balance decreased.✓✓ Accept any other explanation. | 2A Explanation (2) | L4 F |
| 1.1.5 | Percentage increase for 2017 = R164 535,70 − R149 567,00 x 100%✓✓ R149 567,00 = 10%✓ Percentage increase for 2018 = R180 976,00 − R164 535,70 x 100%✓ R164 535,70 = 9,99% ≈ 10%✓ Statement is valid✓ | 1F Correct formula 1SF Correct values 1CA Percentage 1SF Correct values 1CA Percentage 1O Valid (6) | L4 F |
| 1.1.6 | School fee amount in 2015 | 1MA 2016 value divided by 1,1 1A 2015 School fee 1MA 2016 value divided by 1,1 1A 2015 School fee (2) | L2 F |
| 1.2.1 | Mean of Excelsior = 15+ 50 + 43 + 34 +19 + 67+ 29 + 87 + 94 + 79 + 96 + 99 + 43✓ 13 ✓ = 755 13 = 58,08% OR 58,1% OR 58%✓ Mean of Whittlesea = 25 + 27 + 32 + 38 + 40 + 45 + 53 + 59 + 60 + 67 + 75 + 78 + 84 + 89 + 91 + 97 16 ✓ = 960 16 = 60%✓ Statement is valid ✓ | 1M Concept of mean 1M Divide by 13 1CA Mean 1MA Add and divide by 16 1CA Mean 1O Valid NPR (6) | L3 & L4 DH |
| 1.2.2 | IQR for Excelsior 15; 19; 29; 34; 43; 43; 50; 67; 79; 87; 94; 96; 99 Quartile 2 (Median) = 50%✓ Quartile 1 (Lower) = 29+34✓ 2 = 31,5% Quartile 3 (Upper)= 87 + 94✓ 2 = 90,5% IQR = 90,5% – 31,5%✓ = 59%✓ Learner’s solution is incorrect✓ | 1M Arrange 1A Concept of median 1MA Correct values divided by 2 1CA Q1 1CA Q3 1M Concept of IQR 1CA IQR 1O Incorrect (8) | L2, L3 & L4 DH |
| 1.2.3 | P( at least 75%) = 14 29 ✓✓ = 0,35897… = 0,359✓ | 1A Numerator 1A Denominator 1CA Rounding (3) | L2 P |
QUESTION 2 [19]
| Ques. | Solution | Explanation | Level |
| 2.1 2.1.1 | To see whether they have a market for the super-sized tuna tin. Accept any other logical explanation | 2A Reason (2) | L4 D |
| 2.1.2 | Volume of the original tin = 𝜋 x radius2 x height ✓ = 3,142 x 6 cm x 6 cm x 7 cm✓ = 791,784 cm3✓ Volume of the super-sized tin = 𝜋 x radius2 x height✓ = 3,142 x 12 cm x 12 cm x 7 cm = 3 167,136 cm3✓ Not valid✓ The volume of the super-sized tin is not double the volume of the original tin.✓ | 1A Radius 1SF Substitution 1CA Volume 1A Radius 1CA Volume 1O Not valid 1O Explanation (7) | L3 & L4 M |
| 2.1.3 | Super-sized tuna tin = 3 167,136 791,784 ✓ = 4 times bigger✓ Suggested price for the super-sized tuna tin = R11,99 x 4✓ = R47,96✓ | CA from 2.1.2 1M Dividing 1CA Times bigger 1M Multiplication 1CA Price (4) | L4 M & F |
| 2.2 | Box A Across the length = 1 000 𝑚𝑚 240 𝑚𝑚 ✓✓ = 4,16… ≈ 4 tins✓ Across the width = 500 𝑚𝑚 240 𝑚𝑚 = 2,08 ≈ 2 tins✓ Height = 200 𝑚𝑚 70 𝑚𝑚 = 2,85 ≈ 2 tins✓ Number of tins in Box A = 4 x 2 x 2 = 16 tins✓ | CA from 2.1.2 1C Diameter cm to mm 1M Dividing 1CA Number of tins across length 1CA Number of tins across width 1CA Number of tins on top of each other 1CA Number of tins in box (6) | L3 M |
QUESTION 3 [26]
| Ques. | Solution | Explanation | Level |
| 3.1 3.1.1 | The strip chart is not drawn to scale ✓✓ | 2A Reason (2) | L4 M&P |
| 3.1.2 | Distance = 203 + 180 = 383 km ✓ OR Distance = (662 – 459) + 180 ✓ = 203 + 180 = 273 km ✓ | 1RM Correct distances 1CA Distance 1RM Correct values 1CA Distance (2) | L2 M&P |
| 3.1.3 | ‘R’ stands for Regional Routes,✓ ‘N’ stands for National Routes or freeways.✓ | 1A Regional route 1A National route (2) | L2 M&P |
| 3.1.4 | Distance from Aliwal North to Harrismith, including Colesberg = 74 + 56 + 69 + 36 + 36 + 69 + 56 + 247 + 131 + 102✓✓✓✓ = 876 km✓ | 3RM Correct distances 1M Adding 1CA Distance (5) | L3 M&P |
| 3.1.5 | Time spent on the Regional route 𝑇𝑖𝑚𝑒 = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑆𝑝𝑒𝑒𝑑 𝑇𝑖𝑚𝑒 = 322✓✓ 80 ✓ = 4,025 hrs Time spent on the national routes 𝑇𝑖𝑚𝑒 = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑆𝑝𝑒𝑒𝑑 𝑇𝑖𝑚𝑒 = 554✓ 100 = 5,54 hrs ✓ Time spent for travelling and pitstops = 4,025 hrs + 5,54 hrs + 1,5 hrs✓✓ = 11,065 hrs✓ Statement not valid | CA from 3.1.4 1M Changing subject of formula 1SF Correct values 1CA Hours 1SF Correct values 1CA Hours 1A Time for pit stops 1M Adding 1CA Total time 1O Not valid (9) | L2 & L3 & L4 M& M& P |
| 3.2 | Total operating costs = [Fixed cost + (Petrol factor x petrol price + Service and Repair cost + Tyre cost)] x distance travelled = [526 + (8,03 x 12,87 + 22,73 + 16,70) x 876✓ = (526 + 103,3461 + 22,73 + 16,70) x 876✓✓ = 668,7761 c x 876✓ = 585 847,8636 c✓ = R5 858,48✓ | CA from 3.1.4 1SF Correct values 1S Fuel 1M Adding 1M Multiply 1S Answer in cents 1CA Answer in Rand (6) |
QUESTION 4 [19]
| Ques. | Solution | Explanation | Level |
| 4.1.1 | Amount for Student Service = 18,9 - (5,6 + 2,5 + 2,4 + 1,1 + 1,9 + 2,4)✓ = 18,9 – 15,9✓ = $3 000 000 OR $3 million✓ | 1RG Correct values 1M Subtract from 18,9 1CA Amount NB Penalise with 1 mark if not written in millions and 1 mark for incorrect unit (3) | L2 DH |
| 4.1.2 | Salaries and Benefits 2014/2015 = 18,8 x 100%✓ 70,7 = 26,59123055%✓ Salaries and Benefits 2015/2016 = 16,6 x 100% 43,4 = 38,24884793%✓ Difference in % = 38,24884793% – 26,59123055%✓ = 11,65761738✓ = 11,7%✓ | 1MA Correct values 1CA Percentage 1CA Percentage 1M Subtracting 1CA Difference 1CA % to 1 decimal place (6) NB Penalise with 1 mark if not written in millions and 1 mark for incorrect unit | L2 F |
| 4.1.3 | Financial Aid does not appear in the 2014/2015 pie chart✓✓ | 2A Explanation | L4 P |
| 4.1.4 | Amounts do add up to $70,7 million OR $43,4 million✓✓ | 2A Explanation | L4 DH |
| 4.2 | First year = 35 000 x 1,075 = R37 625✓ Second year = 37 625 x 1,075 = R40 446,88✓ Third year = 40 446,88 x 1,0775 = R43 581,51✓ Statement is not valid✓ OR Final Amount = 35 000 x 1,075 x 1,075 x 1,0775✓✓✓✓ = R43 581,51✓ Statement not valid✓ | 1M Correct % 1CA Amount 1CA Amount 1M Correct % 1CA Amount 1O Not valid (6) | L3 & L4 F |
TOTAL: 100