GR12 MLIT P2 Sept2017 MEMO Eng hlayiso.com

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 SEPTEMBER 2017 MATHEMATICAL LITERACY P2 MARKING GUIDELINES MARKS: 150 Codes Explanation M Method MA Method with Accuracy CA Consistent Accuracy A Accuracy C Conversion D Define J Justification/Reason/Explain S Simplification RD Reading from a table OR a graph OR a diagram OR a map OR a plan F Choosing the correct formula SF Substitution in a formula O Opinion P Penalty, e.g. for no units, incorrect rounding off, etc. R Rounding Off NPR No penalty for rounding OR omitting units AO Answer only This marking guideline consists of 8 pages.

Downloaded from hlayiso.com 2 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2017) QUESTION 1 [31] Ques. Solution Explanation Level 1.1.1 Emerald – R 19 089 + 1A Emerald F L2 Onyx R 23 551 +  1A Onyx (2) 1.1.2 Emerald = member + adult + child 1RT Correct values F = R2 477 + R1 761 + R914  L2 = R5 152  1CA Add values Onyx = member + adult + child 1RT Correct values = R3 587 + R2 362 + R1 149  = R7 098  1CA Add values Difference = 7 098 – 5 152  1M Subtracting = R1 946  1CA Difference (6) 1.1.3 Government subsidy = R5 152 – R2 530  1MA Difference F = R2 622 L2 2 622 1M ×100 Government % = 5 152 × 100  1CA % Rounded to = 50,89% 1decimal place (3) = 50,9% 1.1.4 It is important for people to be healthy.  2O Importance D OR L4 Accept any other relevant reason. (2) 1.2.1 Volume = π × r2 × h  1M Calculating radius M = 3,142 × 0,225 × 0,225 × 0,84  1 Conversion L3 = 0,1336 m3 1SF Substitution in 70 formula Volume of traditional beer = 100 × 0,1336 1MA Finding 70% = 0,09353 m3  OR 1MA Finding 70% Height of container = 0,7 × 0,84 = 0,588 m  1M Calculating radius Volume = π × r2 × h 1A Conversion = 3,142 × 0,225 m × 0,225 m × 0,588 m  1SF Substitution in = 0,09353 m3 formula (4) 1.2.2 Length of store room = 2 m = 200 cm  1M Conversion M L4 Number of the containers along the length 1M Dividing by 45 = 200 / 45  = 4,444… 1CA Number of = 4 containers  containers across the length Width of store room = 1,5 m = 150 cm Number of the containers along the width 1CA Number of = 150 / 45 containers across the = 3,333… width = 3 containers  Number of containers in total 1CA Total number of =4×3 containers = 12 containers  Statement is invalid  1O Invalid (6) Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/SEPTEMBER 2017) MATHEMATICAL LITERACY P2 3 Ques. Solution Explanation Level 1.3.1 Amount before increase = 336000/ 106,5% 1M Dividing F = R315 492,96 1M Using 106,5% L2 1CA Amount (3) 1.3.2 Bonus = 336 000 / 12 1M Monthly salary F = 28 000  L3 105,8 1M Using 5,8% Year 1 = 100 × 28 000  1CA Amount = R 29 624  106,5 Year 2 = 100 × 29 624 1MA Finding 6,5% = R 31 549,56  1CA Amount (5) QUESTION 2 [42] Ques. Solution Explanation Level 2.1.1 10 hours = 10 × 60 1MA minutes in 10 M = 600 minutes  hours L2 1 page = 26 minutes 600 minutes = 600 / 26  1M Dividing by 26 = 23 pages  1CA Number of pages Supposed to develop 23 papers therefore 20 papers are below norm time.  1O Conclusion (4) 2.1.2 2015 𝑅𝑎𝑡𝑒 − 2013 𝑅𝑎𝑡𝑒 D % increase = × 100 2013 𝑅𝑎𝑡𝑒 L3  169,30−147,36 1M Difference = × 100  1M × 100 147,36 = 14,89 %  Accept 14,9% 1CA answer % (3) 2.1.3 Amount of developing material: M&F 𝑁𝑜𝑟𝑚 𝑡𝑖𝑚𝑒 L4 × rate for developing × number of pages 60 1SF Substituting 26 = 60 × 169,30 × 161  correct values = 11 811,50  1S Simplification For 10 employees = 11 811,50 × 10 = 118 114,9667  1CA For 10 people Km travelled = 35×2×2×7+2×25×3×7+12×2×5×7 1M Calculating = 980 + 1 050 + 840 distance = 2 870 km  1CA Total distance Transport = rate for transport × number of km 1M Multiplying rate Amount = 2 870 × 2,82  per km = R8 093,40  1CA Amount Total amount = 118 114,9667 + 8093,40 = R 126 208,37  1CA Total Amount Balance = R130 000 - R 126 208,37 = R 3 791,63  1CA Difference Statement invalid; Balance less than R4 000  1O Invalid (10) Copyright reserved Please turn over

Downloaded from hlayiso.com 4 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2017) Ques. Solution Explanation Level 2.2.1 USA: F 46 075,25 + 33% of amount above 189 300  1 correct tax bracket L3 = 46 075,25 + 0,33 (350 500 – 189 300)  1 SF = 46 075,25 + 0,33 × 161 200 = 46 075,25 + 53 196 1 simplification 99 271,25  1M dividing by 12 = 12  1CA (5) = 8 272,60 dollars  2.2.2 Income in South Africa L4 = $350 500 × 14,11 = R 4 945 555  1 conversion Income Tax 1F Choosing correct = 208 587 + 41% of amount above 701 300  tax bracket = 208 587 + 0,41 × (4 945 555 - 701 300) = 208 587 + 0,41 × 4 244 255 = 208 587 + 1 740 144,55  1S Simplification = 1 948 731,55 – 13 257  1M Subtract rebate = 1 935 474,55 / 12 = R 161 289,55  1CA Monthly Tax =R161 289,55/14,11 1C Answer in Dollars = 11430,87 dollars Statement is valid  1O Valid (7) 2.2.3 People in the higher tax brackets are feeding the 2R Reason D government bills.  L4 OR It is discouraging for people occupying higher positions and earning higher salaries.  OR People need to be treated equally Accept any other valid reason (2) 2.3 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 M Speed = 𝑡𝑖𝑚𝑒 L3 Time taken = 08:55 – 06:00 1M Travelling time = 2 hours 55 minutes  Less time spent in Nanaga 1M Difference in time = 2 hrs 55 min – 0 h 30 min  1C Conversion = 2 hrs 25 minutes = 2,416666…hrs  311 1SF Substitution Speed = 2,416666  1CA Speed = 128,69 km/h 1O Opinion (6) They travelled above the speed limit  2.4 School B  has performed better. The mean of School B is 1M Choosing school D higher, meaning 50% of the class were able to get 56.  2 O First reason L4 Minimum mark in School B is higher.  2 O Second reason (5) Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/SEPTEMBER 2017) MATHEMATICAL LITERACY P2 5 QUESTION 3 [30] Ques. Solution Explanation Level 3.1.1    3RT (1mark for every M&P 58 + 279 + 45 + 455 + 232 + 303 + 280 + 49 + 498  three correct values) L2 = 2 199 km  1M Adding 1CA Answer (5) 3.1.2 N1 – South Africa  4 (1 mark for route M&P N4 – South Africa  and country). L2 A3 – Botswana  If only roads A2 – Botswana mentioned max 2 B6 – Namibia If only countries mentioned max 2 (4) 3.1.3 OPTION 1: F Accommodation = R 1 550 L4 Breakfast = R 95 × 4 1MA Cost of = R 380  breakfast Total amount = R 1 550 + R 380 = R 1 930  1CA Total cost OPTION 2: Accommodation with breakfast = R 550 × 4 = R 2 200  1MA Total cost Difference = 2 200 – 1 930 = R 270  1CA Difference Not true, they would save R 270  1O Invalid (5) 3.1.4 Probability of getting a self-catering unit at no extra cost 1A Numerator P 5 1A Denominator L2 = 8 × 100 1CA % = 62,5%  1 R Round to = 63%  nearest % (4) 3.1.5 Distance travelled 1M Adding distances M&P = 58+98+41+41+550+105+105+738 1 CA Total distance L4 = 1 736 km 1MA Finding the Difference = 2199 – 1736  difference = 463 km 1CA distance Statement is valid  1O Valid (5) 3.2.1 Percentage achievement in Mathematical Literacy is 2 O Describing the D decreasing from 2013 to 2016 trend (2) L4 3.2.2 Maths decreased from 2013 to 2015 1O Description for D Maths increased from 2015 to 2016 Mathematics for 2013 L4 to 2015 1O Description for Mathematics for 2015 to 2016 (2) 3.2.3 Mathematics = 265 810 – 263 903 1A Increase in D = 1 907 Mathematics L3 Mathematical Literacy = 388 845 – 361 865 1A Decrease in = 26 980 Mathematical Literacy Ratio = 1 907 : 26 980 1CA Ratio (3) Copyright reserved Please turn over

Downloaded from hlayiso.com 6 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2017) QUESTION 4 [47] Ques. Solution Explanation Level 4.1.1 C=2×π×r 1SF Substituting M 157,1 = 2 × 3,142 × r  correct formula L3 157,1 = 6,284r 1S Simplification 157,1 1CA Calculate the r = 6,284  radius = 25 cm  1C Convert to metres = 0,25 m (4) 4.1.2 D = 0,25 × 2 M = 0,5 m  1CA finding diameter L3 Total height = 1 + 0,75 + 0,5 = 2,25 m  1CA total height Space without decoration = 4 – 2,25 = 1,75 m  1CA space without 1,75 decoration From top and from bottom = 2 = 0,875 m  1CA (4) 4.1.3 Area for red paint = area of rectangle + area of triangle 1M Using correct M&F =ℓ × 𝒷+½b×h formulae L4 = 1,5 m × 1 m + ½ × 0,75 m × 0,75 m  1SF Substituting = 1,5 m + 0,28125 m = 1,78125 m2  1CA Area of shaded For 15 decorations = 1,78125 × 15 parts = 26,71875  1CA area for 15 Two coats = 26,71875 × 2 decorations = 53,4375 m2  1CA for 2 coats 53,4375 Litres of paint needed = 8 = 6,6796875  6,6796873 1MA litres of paint 5 litres = needed 5 = 1,33 = 2 (5 litres of paint)  1MA number of 5 litre Area for white paint tins = πr2 + ½ b × height = 3,142 × 0,25 × 0,25 + ½ × 0,75 × 0,75 = 0,196375 + 0,28125 = 0,477625  1CA area for white For 15 decorations = 0,477625 × 15 paint = 7,164375 For 2 coats = 7,164375 × 2 = 14,32875 14,32875 Litres needed = 8 = 1,79 1CA number 5 litre = 1 (5 litre)  tins 1MA cost for white White paint = R499  paint Red = 505 × 2 1MA cost for red = R 1 010  paint Argument not valid, cost of red paint is not twice that of 1O Not valid (12) white paint  Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/SEPTEMBER 2017) MATHEMATICAL LITERACY P2 7 Ques. Solution Explanation Level 4.2.1 North East 2A Direction (2) M&P L2 4.2.2 𝑠𝑢𝑚 𝑜𝑓 𝑣𝑎𝑙𝑢𝑒𝑠 D Mean = 21 1SF Substitution L3 22+33+34+30+25+29+23+23+22 26,762 = +30+29+30+28+24+25+25+58  21 26,762 × 21 = 432 + 5B  1M Mean value ×21 562 – 432 = 5B 130 = 5B 1S Simplification 130 =B 5 26 °C = B  1CA Value of B (4) 4.2.3 Minimum temperatures: CA from 4.2.2 D L3 Lower quartile (Q1) = 8 Upper quartile (Q3) = 11 Interquartile range = 11 – 8 1MA Finding IQR for =3 min. temp. Maximum temperatures: 22 22 23 23 24 25 25 25 26 26 26 26 26 28 29 29 30 30 30 1M Ascending order 33 34  Median = 26  1CA Median 24+25 Lower quartile = 2 = 24,5  1MA Finding Q1 29+30 Upper quartile = 2 = 29,5  1MA Finding Q3 Interquartile range = 29,5 – 24,5 1CA Finding IQR =5 Difference = 5 – 3 =2 1CA Difference (7) Copyright reserved Please turn over

Downloaded from hlayiso.com 8 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2017) Ques. Solution Explanation Level 4.2.4 CA from 4.2.2 and D Compound Bar Graph for the Box-and- 4.2.3 L2 Whisker values for the minimum and maximum temperatures 5M (1 mark for every set of bars plotted 40 correctly) Minimum- and Maximum temperature values 35 1M Correct graph 30 25 20 15 Minimum 10 Maximum 5 0 Box-and- whisker values (6) 4.2.5 Probability that a temperature is ≥ 28 °C 1A Numerator P 8 1A Denominator L2 = 21 1R To 3 decimal = 0,380952381 places (3) = 0,381  4.2.6 Measured distance = 6,6 cm 1MA Measure on map M&P (accept 6,4 - 6,8) L3 = 6,6 cm : 1 045 km 1M Ratio = 6,6 : 104 500 000 1C Converting to cm =1 : 15 833 333,33 1S Simplification 1R Round to nearest = 1 : 16 000 000  million (5) TOTAL: 150 Copyright reserved Please turn over