Mathematical Literacy Grade 11 P2 North West June 2017 Memo

NATIONAL SENIOR CERTIFICATE GRADE 11 MATHEMATICAL LITERACY P2 JUNE 2017 MARKING GUIDELINES MARKS: 75 SYMBOL EXPLANATION M Method M/A Method with accuracy CA Consistent accuracy A Accuracy C Conversion D Definition S Simplification RT/RG/RD Reading from a table/Reading from a graph F Choosing the correct formula SF Correct substitution in a formula O Opinion/Example P Penalty, e.g. for no units, incorrect rounding off etc. R Rounding off J Justification/Reason NP No penalty for rounding OR omitting units This marking guideline consists of 6 pages

Mathematical Literacy/P2 2 NW/June 2017 NSC – Grade 11 MARKING GUIDELINES QUE. SOLUTIONS EXPLANATIONS TL QUESTION 1 [32 MARKS] 1.1.1 150 g × 50 1M Multiplying by 2 = 7 500 g 50 = 7,5 kg 1S simplification 1C answer in kg (3) 1.1.2 0,25 kg = 250g 1C conversion 4 1M method number of loaves to be baked =  = 33,333 1No.of loaves = 33 Yes, comment is valid 1O opinion OR OR 1M method 7,5g × 50 = 375 g 1S simplification = 0,375 kg 1C conversion Yes, comment is valid 1O opinion (4) 1.2.1 A = R400 + R3,50 × 120 1M method 2 = R400 + R420 1A answer for A = R820 1M method 1A answer for B OR (4) A = R680 + R140 = R820 B= = 160 loaves 1.2.2 Cost = R400 + R3,50 × number of loaves 1A R400 + 2 1A product (2) 1.2.3 Cost = R400 + R3,50 × 50 1SF substitution 2 = R575 1A answer (2)

Mathematical Literacy/P2 3 NW/June 2017 NSC – Grade 11 1.2.4 Cost and Income of Loaves of bread 2000 1800 1600 Income 1400 1A (0,400) Amount in rand 1A (280,1 380) 1200 Cost 1A joining the 1000 points 800 1A labelling the graph 600 1A any other 400 correct point 200 (5) 3 0 0 40 80 120 160 200 240 280 Loaves of bread 1.2.5 (160, R960) 1A number of 2 loaves 1A amount (2) 1.2.6 Cost of 200 loaves = R1 100 1RT cost 3 Income from 200 loaves = R1 200 1RT income Profit = R1 200 – R1 100 = R100 1A profit (3) 1.2.7 1M percentage 4 1A answer Percentage = × 100% 1Conclusion = 9,09% (3) Thus, claim is not valid OR  = R110 Thus, claim is not valid 1.3 Total time = 30 hours  1A total time 4 15 1M method Then × 30 hours = 4,5 hours 1CA expected time 100 Time given to languages = 4 hours which is less than expected for languages Thus, the time allocated is not according the staff agreement 1O opinion OR Total time = 30 hours  (4)

Mathematical Literacy/P2 4 NW/June 2017 NSC – Grade 11 Then 100% = 13,3% Thus, claim is not valid QUESTION 2[21 MARKS] 2.1.1 Total monthly cost of electricity 1M method 3 = 324kWh × 117,86cents 1A in cents =38 186,64 cents 1A in rand = R381,87 (3) 2.1.2 Total monthly cost of electricity 1M block 1 3 = 350kWh × 117,86 cents+ 244,8× 163,19 cents 1M block 2 =41 251 cents + 39 948,912 cents 1S simplification = 81 199,912 cents 1A in cents =R812,00 1A in rand (5) 2.1.3 Too many appliances used that consumes a lot of electricity 2A answer 4 OR (2) Heaters used for warming up the house OR High consumption of water from the geyser Accept any relevant reason 2.2.1 1SF substitution 2 1A answer BMI = 1R rounding (3) BMI =  = 28,25096981 kg/m2 = 28,3 kg/m2  2.2.2 Overweight 2CA answer 2 (2) 2.2.3 Diet 2A reason 4 OR (2) Ill-health OR Use of weight-loss products Accept any other relevant reason 2.2..4 ˚C= (˚F – 32˚)÷1,8 1SF temp. in ˚C 3 43˚=(˚F – 32˚) ÷1,8 1S simplification

Mathematical Literacy/P2 5 NW/June 2017 NSC – Grade 11 ˚F = 43˚×1,8 +32˚ 1A answer ˚F = 109,4˚ 1R rounding ˚F = 110˚ (4) Penalise for rounding QUESTION 3[22 MARKS] 3.1.1 1RT using correct 4 values Percentage = × 100 1M method = 69,48051948 1A answer No, the claim is not valid 1O opinion OR (4) × 443 520 = R310 464 No, the claim is not valid 3.1.2 R 10 047 1M method 4 Average cost =  27 1A answer = R372,1111 (2) = R370 3.1.3 57 120 1M method 4 Overseas donations =  0,08 1A answer = ¥ 714 000 1O opinion Thus, the claim is correct (3) 3.1.4 Deliver food parcels 2O opinion 4 OR Fuel OR Maintenance (2) OR Going to buy garden material OR Asking for donations OR Attending board meetings Accept any relevant reason 3.2.1 Volume = 100×60×30 1SF substitution 3 = 180 000 cm3  1S simplification = 180 litres 1C conversion (3) 3.2.2 180 litres 1M method 4 Number of fish=  7 litres 1A no. of fish = 25,7142857 1O opinion = 25 OR No, 26 fish cannot fit in the tank 1M method 1A volume needed OR 1O opinion (3) Volume for 26 fish= 26 × 7 litres [2 out of 3 if no. of fish is 26] = 182 litres No, the volume needed is less than the volume of the tank. 3.3 180 litres = 180 × 1,7598 pint 1M method 3 = 316,764 pint 1A in pint

Mathematical Literacy/P2 6 NW/June 2017 NSC – Grade 11 316,764 pints = 316,764 ×0,125 gallons 1M method = 39,5955 gallons 1A in gallons = 40 gallons 1CA answer (5) QUESTION PAPER ANALYSIS QUESTION TL2 TL3 TL3 QUESTION 1 1.1.1 3 1.1.2 4 1.2.1 4 1.2.2 2 1.2.3 2 1.2.4 5 1.2.5 2 1.2.6 3 1.2.7 3 1.3 4 QUESTION 2 2.1.1 3 2.1.2 5 2.1.3 2 2.2.1 3 2.2.2 2 2.2.3 2 2.2.4 4 QUESTION 3 3.1.1 4 3.1.2 2 3.1.3 3 3.1.4 2 3.2.1 3 3.2.2 3 3.3 5 TOTAL MARKS 18 25 32 PERCENTAGES 24 33 43