Maths Lit P2 Gr12 Memo June 2018 English hlayiso.com

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 JUNE 2018 MATHEMATICAL LITERACY P2 MARKING GUIDELINE MARKS: 100 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 This marking guideline consists of 8 pages.

Downloaded from hlayiso.com 2 MATHEMATICAL LITERACY P2 (EC/JUNE 2018) QUESTION 1 [36] Ques. Solution Explanation Level 1.1.1 Total actual expenditure value for 2017 RT M 1RT Correct values L2 = R62 459,75 + R125 000,05 + R63 241,20 +R200 541,65 1M Addition F = R451 243,10 OR Total actual expenditure value for 2017 1RT Correct values = R461 864,70 – R10 621,60 RT M 1M Subtraction (2) = R451 243,10 1.1.2 Actual value is the amount of money that was either 2A Explain actual L4 received or spent.A value F Budgeted value is the amount of money that is predicted to 2A Explain be either received or spent A budgeted value OR Amount of money planned to cover all expenses. Accept any logical explanation. (4) 1.1.3 Teaching resources RT 1RT L2 & School bought most of the teaching resources the previous 2R Reason L4 year R F OR They received teaching resources from donors R Accept any other valid reason. (3) 1.1.4 Disagree, because the schools budget for 2018 shows a 2A Explanation L4 negative balance .A F OR From 2016, the balance decreased. A Accept any other explanation. (2) 1.1.5 Percentage increase for 2017 1F Correct formula L4 = R164 535,70 − R149 567,00  100% F SF 1SF Correct values F R149 567,00 1CA Percentage = 10% CA 1SF Correct values 1CA Percentage Percentage increase for 2018 1O Valid R180 976,00 − R164 535,70 =  100% SF R164 535,70 = 9,99% ≈ 10% CA Statement is valid O (6) Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/JUNE 2018) MATHEMATICAL LITERACY P2 3 1.1.6 School fee amount in 2015 1MA 2016 value L2 = R149 567,00 MA divided by 1,1 F 1,1 1A 2015 School fee = R135 970,00 A OR School fee amount in 2015 1MA 2016 value = R149 567,00 MA divided by 1,1 110% 1A 2015 School fee (2) = R135 970,00 A 1.2.1 Mean of Excelsior = 15+ 50 + 43 + 34 +19 + 67+ 29 + 87 + 94 + 79 + 96 + 99 + 43 M 1M Concept of L3 & L4 755 13 M mean DH = 1M Divide by 13 13 = 58,08% OR 58,1% OR 58% CA 1CA Mean Mean of Whittlesea 25 + 27 + 32 + 38 + 40 + 45 + 53 + 59 + 60 + 67 + 75 + 78 + 84 + 89 + 91 + 97 1MA Add and = 16 MA divide by 16 960 = 16 1CA Mean = 60% CA Statement is valid O 1O Valid NPR (6) 1.2.2 IQR for Excelsior L2, L3 & 1M Arrange L4 15; 19; 29; 34; 43; 43; 50; 67; 79; 87; 94; 96; 99 M 1A Concept of DH median Quartile 2 (Median) = 50% A 1MA Correct values divided by 2 29+34 MA 1CA Q1 Quartile 1 (Lower) = 2 = 31,5% CA 87 + 94 Quartile 3 (Upper)= 2 1CA Q3 = 90,5% CA 1M Concept of IQR = 90,5% – 31,5% M IQR = 59% CA 1CA IQR 1O Incorrect Learner’s solution is incorrect O (8) Copyright reserved Please turn over

Downloaded from hlayiso.com 4 MATHEMATICAL LITERACY P2 (EC/JUNE 2018) 1.2.3 14 P( at least 75%) = 29 A 1A Numerator L2 A 1A Denominator P = 0,35897… 1CA Rounding (3) = 0,359 CA 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. L4 Accept any other logical explanation 2A Reason (2) D 2.1.2 Volume of the original tin = 𝜋  radius2  height A 1A Radius L3 & = 3,142  6 cm  6 cm x 7 cm SF 1SF Substitution L4 = 791,784 cm3 CA 1CA Volume M Volume of the super-sized tin = 𝜋  radius2  height A = 3,142  12 cm  12 cm x 7 cm 1A Radius = 3 167,136 cm3 CA 1CA Volume Not valid O 1O Not valid The volume of the super-sized tin is not double the volume of the original tin. O 1O Explanation (7) 2.1.3 Super-sized tuna tin CA from 2.1.2 L4 3 167,136 1M Dividing M& = 791,784 M 1CA Times bigger F = 4 times bigger CA Suggested price for the super-sized tuna tin = R11,99  4 M 1M Multiplication = R47,96 CA 1CA Price (4) Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/JUNE 2018) MATHEMATICAL LITERACY P2 5 2.2 Box A CA from 2.1.2 L3 1 000 𝑚𝑚 1C Diameter cm to M Across the length = 240 𝑚𝑚 C M mm = 4,16… 1M Dividing ≈ 4 tins CA 1CA Number of tins 500 𝑚𝑚 across length Across the width = 240 𝑚𝑚 = 2,08 1CA Number of tins ≈ 2 tins CA across width 200 𝑚𝑚 Height = 70 𝑚𝑚 1CA Number of tins = 2,85 on top of each other ≈ 2 tins CA Number of tins in Box A = 4  2  2 1CA Number of tins in box (6) = 16 tins CA QUESTION 3 [26] Ques. Solution Explanation Level 3.1 3.1.1 The strip chart is not drawn to scale A 2A Reason L4 (2) M&P 3.1.2 Distance = 203 + 180 RM 1RM Correct L2 = 383 km CA distances M&P OR 1CA Distance Distance = (662 – 459) + 180 RM 1RM Correct values = 203 + 180 1CA Distance = 273 km CA (2) 3.1.3 ‘R’ stands for Regional Routes, A 1A Regional route L2 ‘N’ stands for National Routes or freeways. A 1A National route M&P (2) 3.1.4 Distance from Aliwal North to Harrismith, including Colesberg 3RM Correct L3 RM RM RM distances M&P = 74 + 56 + 69 + 36 + 36 + 69 + 56 + 247 + 131 + 102 M 1M Adding = 876 km RM 1CA Distance (5) Copyright reserved Please turn over

Downloaded from hlayiso.com 6 MATHEMATICAL LITERACY P2 (EC/JUNE 2018) 3.1.5 Time spent on the Regional route CA from 3.1.4 L2 & 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 L3 & 𝑇𝑖𝑚𝑒 = 1M Changing L4 𝑆𝑝𝑒𝑒𝑑 subject of formula M& 322 1SF Correct values M& 𝑇𝑖𝑚𝑒 = M SF 80 1CA Hours P = 4,025 hrs CA Time spent on the national routes 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑇𝑖𝑚𝑒 = 𝑆𝑝𝑒𝑒𝑑 1SF Correct values 554 1CA Hours 𝑇𝑖𝑚𝑒 = SF 100 = 5,54 hrs CA Time spent for travelling and pitstops 1A Time for pit = 4,025 hrs + 5,54 hrs + 1,5 hrs A M stops = 11,065 hrs CA 1M Adding 1CA Total time Statement not valid O 1O Not valid (9) 3.2 Total operating costs CA from 3.1.4 = [Fixed cost + (Petrol factor  petrol price + Service and 1SF Correct values Repair cost + Tyre cost)]  distance travelled 1S Fuel 1M Adding = [526 + (8,03  12,87 + 22,73 + 16,70)  876 SF 1M Multiply = (526 + 103,3461 + 22,73 + 16,70)  876 S M 1S Answer in cents = 668,7761 c  876 M 1CA Answer in = 585 847,8636 c S Rand (6) = R5 858,48 CA Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/JUNE 2018) MATHEMATICAL LITERACY P2 7 QUESTION 4 [19] Ques. Solution Explanation Level 4.1.1 Amount for Student Service 1RG Correct values L2 = 18,9 - (5,6 + 2,5 + 2,4 + 1,1 + 1,9 + 2,4) RG 1M Subtract from DH = 18,9 – 15,9 M 18,9 = $3 000 000 OR $3 million CA 1CA Amount NB Penalise with 1 mark if not written in millions and 1 mark for incorrect unit (3) 4.1.2 Salaries and Benefits 2014/2015 1MA Correct values L2 18,8 = 70,7  100% MA 1CA Percentage F = 26,59123055% CA 1CA Percentage Salaries and Benefits 2015/2016 16,6 = 43,4  100% = 38,24884793% CA 1M Subtracting 1CA Difference Difference in % = 38,24884793% – 26,59123055% M 1CA % to 1 decimal = 11,65761738 CA place = 11,7% R (6) NB Penalise with 1 mark if not written in millions and 1 mark for incorrect unit 4.1.3 Financial Aid does not appear in the 2014/2015 pie chart A 2A Explanation L4 P 4.1.4 Amounts do add up to $70,7 million OR $43,4 millionA 2A Explanation L4 DH Copyright reserved Please turn over

Downloaded from hlayiso.com 8 MATHEMATICAL LITERACY P2 (EC/JUNE 2018) 4.2 First year = 35 000  1,075 1M Correct % L3 & = R37 625 CA 1CA Amount L4 F Second year = 37 625  1,075 = R40 446,88 CA 1CA Amount M Third year = 40 446,88  1,0775 1M Correct % = R43 581,51 CA 1CA Amount Statement is not valid O 1O Not valid OR M M M M Final Amount = 35 000  1,075  1,075  1,0775 = R43 581,51 CA Statement not valid M (6) TOTAL: 100 Copyright reserved Please turn over