Mathematical Literacy Grade 11 P2 2018 Memo (Eng)

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 11 NOVEMBER 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/Reading from a 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 7 pages.

Downloaded from hlayiso.com 2 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2018) QUESTION 1 [31] Question Solution Explanation Topic and Level 1.1.1 Washing and drying = 18 × 16 1 MA Amount for L3 = R288  washing and drying F Staysoft = 5 = 3 3 18 1M Calculating 18 = ×  Staysoft 1 5 = R10,80  1CA Cost for Staysoft Ironing = 5 = 20 20 18 18 = × 1 5 = R72  1CA Cost for Ironing Total = 288 + 10,80 + 72 = R370,80  1CA Total (5) 1.1.2 (a) Helper amount 1A No. of days L4 No. of days = 9  F Amount = 9 × 150 1CA Amount for helper = R1 350  1CA Amount for wash Wash and dry = 15 × 5 × 16 and dry = R1 200  Stay soft = 5 = 3 3 75 75 = × 1CA Staysoft 1 5 = R45  Handwash = 2 × 5 × 25 1CA Hand wash = 250  Total amount for laundry = 250 + 45 + 1200 1CA Total amount = R1 495  Difference = R1 495 – R1 350 1CA Difference = R145  1O Invalid (8) Invalid laundry more expensive  1.1.2 (b) Probability = 3/31× 100  1A Numerator L2 1A Denominator P = 9,7%  1CA Answer as a % to 1 decimal place (3) Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2018) MATHEMATICAL LITERACY P2 3 1.2.1 3 1M Dividing L3 Across the length = 0,5  1C Conversion to metres Maps =6 2,5 Across the breadth = 0,5 =5 1CA Breadth/Width Number of containers fitting = 6 × 5 1CA Number of containers = 30  1O Opinion (5) 10 can fit  1.2.2 (a) Volume of container L3 V = π × (radius)2 × height M = 3,142 × 25 cm × 25 cm × 70 cm  1 Calculating radius = 137 462,5 cm3 = 137 462,5 cm3  1CA Volume No. of litres = 137 462,5 ÷ 1 000 = 137,4625 litres  1C Litres NPR (3) 1.2.2 (b) Capacity for 5 containers CA from 1.2.2 (a) L4 = 137,4625 litres × 5 1MA Litres for 5 containers M = 687,3125 litres  Water for 5 days of hand wash 1MA Amount of washing Washing = 30 × 5 = 150 kg  1M Number of litres 25 kg = 100 litres 100 150 1CA Total litres 150 = 1 × 25  = 600 litres  1 O Valid (5) Statement is valid  1.2.3 Some material cannot be washed using machine.  2A Reason or explanation L4 OR D Clothes can be damaged  Accept any other relevant reason. (2) [31] Copyright reserved Please turn over

Downloaded from hlayiso.com 4 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2018) QUESTION 2 [22] Question Solution Explanation Topic and Level 2.1 Amount for editing = 97 × 100  L4 1M Multiply correct = R 9 700  F values 𝑛𝑜𝑟𝑚 𝑡𝑖𝑚𝑒 1A Amount for editing Typist = × rate × no. of documents 60 28 = × 195 × 100  1SF Substituting 60 = R9 100  1A Correct values 105 9 700 1CA Amount for typist %= × 9 100  OR 9 100 x 100 100 = R9 555  = 106,59% – 100% 1M Increasing by 5% = 6,59% 1CA Increased amount Not valid, more than 5%  1O Invalid (8) 2.2 2.2.1 31,7; 33,7; 35; 36,3; 36,4; 37; 39,2; 40,6; 41; 42,3; 1RT Reading the correct L2 44,1 high and low values. D Range = 44,1 g – 31,7 g  1M Subtraction = 12,4 g  1CA Range (3) 2.2.2 Guest 1: Play energy drink = 29,2 g 1RT Correct values for L4 Powerade = 38,8 g  guests M Total = 68 g  1CA Total number for grams for guest 1 Guest 1: 5g = 1 teaspoon 1 68  68 = × 1CA Teaspoons sugar 1 5 = 13,6 teaspoons  guest 1 Guest 2: Appletiser = 31,7g Coca-Cola = 35g Total = 66,7 g  1CA Total no of grams for guest 66,7 Guest 2: 5 1CA Teaspoons sugar = 13,34 teaspoons  guest 2 Statement not valid. Energy drink contains more sugar.  1O Invalid (6) Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2018) MATHEMATICAL LITERACY P2 5 2.2.3 L2 5M One mark for each D Amount of sugar in grams for correct bar different drinks 45 Amount of sugar in grams 40 35 30 25 20 15 10 5 0 Different Drinks (5) [22] QUESTION 3 [20] Question Solution Explanation Topic and Level 3.1 3.1.1 Distance from Durban to Nelspruit = 689 km L3 1RT Distance Maps 𝐷 and Speed = M 𝑇 689 1SF Substituting correct 110 =  𝑡𝑖𝑚𝑒 values 110 × time = 689 689 1S Change subject of Time =  110 the formula = 6,26 hrs  1CA Answer in hours = 6hrs 16 min  1C Converting 6,26 hrs to minutes 1CA Hours and min (6) 3.1.2 Ascending order L2 58; 273; 292; 342; 475; 532; 656; 1050; 1120; 1RT Correct values Maps 1463  1M Ascending order and D 475+532 1M Concept of median Median =  1CA Median 2 = 503,5  (4) Copyright reserved Please turn over

Downloaded from hlayiso.com 6 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2018) 3.1.3 Pretoria: 1M Adding all correct L2 and 1120+273+342+292+532+58+1050+656+1463+475 values L3 10 6261  Maps = 10  1M Divide by 10 and D = 626,1 km  1CA Mean Port Elizabeth:  𝟏𝟏𝟐𝟎 + 𝟏𝟑𝟗𝟑 + 𝟏𝟑𝟕𝟑 + 𝟏𝟏𝟐𝟐 + 𝟕𝟓𝟐 + 𝟏𝟎𝟔𝟐 + 𝟑𝟎𝟎 + 𝟗𝟐𝟕 + 𝟕𝟓𝟔 + 𝟔𝟑𝟓 𝟏𝟎 1RT Adding all correct 9 440 values = 944 km  1CA Mean 10 Difference = 944 – 626,1  = 317,9 km  1M Subtraction 1CA Difference (7) 3.1.4 Distance = 998 km  1RT Correct distance L2 Cost = 70 × 998  1M Multiplying by 70 F and = 69860 cents Maps = 698,60 1R Nearest rand =R699  OR 998 × 0,7  1RT Correct distance = 698,60  1M Multiplying by 70 = R699  1R Nearest rand (3) [20] QUESTION 4 [27] Question Solution Explanation Topic and Level 4.1 Area of circle = 𝝅 × 𝐫𝐚𝐝𝐢𝐮𝐬 × 𝐫𝐚𝐝𝐢𝐮𝐬 1A Radius L3 = 3,142 × 0,45 × 0,45 1SF Substitution M = 0,636255 m2  1CA Area of centre 𝝅 × 𝒓𝒂𝒅𝒊𝒖𝒔 × 𝒓𝒂𝒅𝒊𝒖𝒔 Area of semicircle = 𝟐 3,142 ×4,9×4,9 = 2 = 37,71971 m2  1CA Area of semi-circle Difference = 37,71971 - 0,636255 1CA Difference = 37,08 m2  NPR (5) 4.2.1 169 seats  3A No of seats (3) L2 Maps 4.2.2 4 1A Numerator L2 15  1A Denominator (2) P Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2018) MATHEMATICAL LITERACY P2 7 4.2.3 CA from 4.2.1 Right side = 171  1A No of seats on right L4 Total amount = 171 × 150 side F = R25 650  1 CA Amount for right seats Left side = 169 × 150 1CA Amount for left = 25 350  seats Middle = 255 108 Cost of 1 seat in the middle = × 150 1A Amount for a middle 100 = R162  seat Total amount = 255 × 162 1CA Total amount for = R41 310  middle seats Total amount for all seats = 25 650 + 25 350 + 41 310  1M Adding all values = R92 310  1CA Total amount Statement invalid  1O Invalid (8) 4.3.1 Grade 11 boys = 191 – (54 + 73) OR 137 – 73 L3 = 191 – 127  = 64 1M Subtracting D = 64  1CA No. of boys in Gr. 11 Total number of learners = 158 + 137 + 99 1CA Total number of = 394  learners 64 1M Multiply by 100 %= × 100  1CA Percentage 394 = 16,2%  NPR (5) 4.3.2 From Grade 10 to Grade 12 the number of learners 2O Trend L4 is decreasing.  D Learners failed  OR 2O Reason Learners dropped out  OR Accept any other relevant reason (4) [27] TOTAL: 100 Copyright reserved Please turn over