Mathematical Literacy P2 Nov 2014 Memo Eng hlayiso.com

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICAL LITERACY P2 NOVEMBER 2014 MEMORANDUM MARKS: 150 Symbol Explanation M Method M/A Method with accuracy CA Consistent accuracy A Accuracy C Conversion S Simplification RT/RG Reading from a table/Reading from a graph SF Correct substitution in a formula O Opinion/Example P Penalty, e.g. for no units, incorrect rounding off, etc. R Rounding off NPR No penalty for rounding This memorandum consists of 20 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 2 DBE/November 2014 NSC – Memorandum QUESTION 1 [38 MARKS] Ques Solution Explanation A L4 1.1.1 The data is discrete, because the violent incidents is 1A correct type counted/whole numbers/integral values /categorised O 1O reason (2) * L3 1.1.2 Total number of incidents involving boys = 13 + 12 + 18 + 11 + 10 + 16 = 80  S 1S total number of boys Total number of incidents involving girls = 7 + 3 + 4 + 7 + 5 + 19 RG 1RG reading from graph = 45  CA 1CA total number of girls Difference = 80 – 45 = 35 CA 1CA difference OR OR Total for boys and girls = 20+15+22+18+15+35 = 125  S 1S Total number of boys and girls Total for boys = 13 + 12 + 18 + 11 + 10 + 16 = 80  S 1S Total number of boys Number of girls = 125 – 80 = 45  CA 1CA number of girls Difference = 80 – 45 = 35  CA 1CA Difference OR OR The total of the differences between boys and girls  A  A  A 2A Positive differences = 6 + 9 + 14 + 4 + 5 – 3 1A for negative 3 = 35  CA 1CA the differences Max 2 marks if part data used Answer only full marks (4) * This question must not be marked in Limpopo. The paper will be marked out of 143 and scaled and then the candidates’ total mark will be up-scaled to 150 marks Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 3 DBE/November 2014 NSC – Memorandum Ques Solution Explanation * L3(1) 1.1.3 Cyber bullying A 1A/RG reading from L4(2) graph Girls avoiding physical violence. O OR Girls are afraid of confrontation and fighting O 2O explanation OR O Easier to express their emotions/feelings on social media (3) 1.2.1 L2 Range = Highest value – Lowest value 5 = 18 – A  M 1M concept of range 1CA value of A A = 13  CA M OR OR A = 18 – 5 = 13 CA 1M concept of range using 5 1CA value of A Answer only full marks (2) NB: Answer from Q L2 1.2.2 13 + 14 × 4 + 15 × 5 + 16 × 10 + 17 × 13 + 18 × 7  M 1.2.1 Mean = 40  A 1M adding all 40 values 1A dividing by 40 651 = 1CA Simplification 40  CA = 16,275 NPR Answer only full marks (3) * This question must not be marked in Limpopo. The paper will be marked out of 143 and scaled and then the candidates’ total mark will be up-scaled to 150 marks Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 4 DBE/November 2014 NSC – Memorandum Ques Solution Explanation A L2 1.2.3 15 + 16 1A identifying the correct B= = 15,5  CA 2 values 1 CA value of B [If only B = 15 then one mark and If answer only B=23 then M one mark] 16 + 17 C= = 16,5  CA 1 M concept of median 2 1 CA value of C D = 17  CA 1 CA value of D Answer Only full marks (5) L2 A 1.2.4 30 1A 30 grade 9 boys P= 40  A 1A no. of boys 40 1CA decimal = 0,75  CA Answer Only full marks (3) L4 1.2.5 The grade 9 boys are too old for their grade. J 2J reason OR Social: J Need recognition / low self- esteem / identity crisis. OR Economic: To gain favours from others. J OR Educational: They are frustrated by their lack of progress. J OR Environmental factors/ emotional factors J OR J Contextual factors/ No parental control/Peer pressure OR J Violent community / child headed family/gang related (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 5 DBE/November 2014 NSC – Memorandum Ques Solution Explanation 1.3.1 Total cost in Rand 1A constant cost A A A 1A 15 persons = 300 for the first 15 passengers + 50 × the number 1A number of persons more of persons more than 15 A than 15 1A multiply by the rate R50 OR OR Total cost (in Rand) 1A constant cost A A A 1A using 15 persons A = 300 + (the number of persons – 15 ) × 50 1A using a variable with explanation 1A multiply by the rate R50 OR OR Total cost (in Rand) 1A constant cost A A A 1A using 15 persons = 300 + (n – 15 persons) × 50 1A using a variable with A explanation Where n is the number of persons more than 15 1A multiply by the rate R50 OR OR Total cost (in Rand) 2A – 450 A A 1A number of persons = (number of persons)× 50 – 450 A 1A multiply by the rate R50 (4) L3 1.3.2 SF (a) 900 = 300 + (n – 15 persons) × 50 1SF Substituting in formula (n – 15 persons) × 50 = 600 n – 15 persons = 12 n = 27 A 1A Maximum number OR OR 2 RT Max number of RT passengers 27 [Both 25 and 27 one mark and 25 only, no marks] (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 6 DBE/November 2014 NSC – Memorandum Ques Solution Explanation 1.3.2 NB: Use CA from Q1.3.2(a) L3 (b) 10 learners + 1 teacher 10 learners + 1 teacher MA 2MA working with ratio 4 learners + 1 teacher ∴ 24 learners and 3 teachersA 1A Number of teachers 24 : 3 CA 1CA ratio in correct order = 8: 1 CA 1CA simplified ratio OR OR 1 educator for 10 learners  MA 1MA working with ratio 1 ∴ × 27 = 2,454545... teachers  CA 1CA number of teachers 11 ∴ 3 teachers  R 1R Rounding up And 24 learners 24 : 3  CA 1CA ratio in correct order 8: 1  CA 1CA simplified ratio (5) L4 1.3.3 There is only one double six.  A 1A probability of double six There is 6 combinations of seven. A 1A probability of seven ∴ Mr Boitumelo has a larger probability than Miss Ansie to accompany the learners.  O 1O explanation OR OR A 1 1A probability of double six P (double six) = ≈ 2,8% 36 6 1 A P (seven) = = ≈ 16,7% 1A probability of seven 36 6 ∴ Mr Boitumelo has a larger probability than Miss Ansie to accompany the learners.  O 1O explanation (3) [38] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 7 DBE/November 2014 NSC – Memorandum QUESTION 2 [33MARKS] Ques Solution Explanation L3 R500 2.1.1 Volume of petrol = litre 1M dividing by R14,04 M R14,04/ ℓ = 35,61253561 litre  A 1A volume Distance each model can travel with 35,613 ℓ of petrol: Sonic 1.6 : 35,613 ×100 km ≈ 531,54 km  CA 1CA distance 6,7 35,613  CA 1CA distance Aveo 1.6 : ×100 km ≈ 487,85 km 7,3 ∴ Sonic 1.6 will travel a greater distance.  O 2O conclusion OR OR M R500 1M dividing by Volume of petrol = = 35,613 ℓ  A R14,04/ ℓ R14,04/ 1A volume Finding distance using consumption rate for each model: 100 km Sonic rate = = 14,925 km/ℓ 6,7 Distance = 14,925 km/ℓ × 35,613 ≈ 531,5 km  CA 1CA distance 100 km Aveo rate = = 13,70 km/ℓ 7,3 Distance = 13,70 km/ℓ × 35,613 ≈ 487,9 km  CA 1CA distance ∴ Sonic 1.6 will travel a greater distance.  O 2O conclusion [Correct conclusion only 2 marks] (6) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 8 DBE/November 2014 NSC – Memorandum Ques Solution Explanation L4 2.1.2 Number of stops and the length of stopping while the engine is running.  O 1O any FIRST correct OR factor The driving pattern of the driver for example fast acceleration and hard breaking. O OR 1O for any SECOND O correct factor Driving at high speeds with open windows OR Use of the air conditioner.  O OR The condition of the car with relation to tyre pressure, load, etc.  O OR O Condition of the road surface, and the slope of the road. O OR Mechanical fault / condition / Electronic damage OR Load and number of passengers in vehicle  O OR Traffic congestion  O (2) 2.1.3 Sonic 1M dividing by 12 Monthly petrol cost (in Rand) 1A multiply petrol price M A MA 1MA multiply by 35 000 6,7 consumption rate = ×14,04 × = 2 743,65 CA 12 100 1 CA petrol cost Sonic Total running cost(in Rand) = 2 743,65 + 2 657,00 = 5 400,65 CA 1CAtotal running cost for Aveo the Sonic Monthly petrol cost (in Rand) 35 000 7,3 = ×14,04 × = 2 989,35 CA 1 CA petrol cost Aveo 12 100 Total running cost(in Rand) = 2 989,35 + 1 942,00 = 4 931,35 CA 1CA total running cost for the Aveo ∴ Aveo 1.6 is more economical.  O 1O conclusion OR [3 out of 8 marks if petrol cost ignored] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 9 DBE/November 2014 NSC – Memorandum Ques Solution Explanation L4 2.1.3 Sonic 1.6 M 1M multiplying by 12 Cont. Instalment cost per year = 12 × R 2 657 = R 31 884  MA 1MA multiply by 6,7 Petrol cost per year = 35 000 km × × R14,04/ ℓ  A consumption rate 100 km 1A multiply petrol price = 2 345 × R14,04 = R 32 923,80  CA 1CA petrol cost Sonic Total running cost for the year = monthly instalments for 12 months + petrol cost per year = R 31 884 + R 32 923,80 =R 64 807,80  CA 1CA total running cost for the Sonic Aveo 1.6 Instalment cost per year = 12 × R 1 942 = R 23 304 7,3 Petrol cost per year = 35 000 km × × R14,04/ ℓ 100 km = 2 555 × R14,04 = R 35 872,20  CA 1 CA petrol cost Aveo Total running cost per year = monthly instalments for 12 months + petrol cost per year = R 23 304 + R 35 871,20 1CA total running cost for =R 59 176,20  CA the Aveo The Aveo 1.6 is more economical.  O 1O conclusion MA OR OR R14,04 / ℓ × 6,7 = R94,068  A 1MA multiply by consumption rate Sonic: R94,068 : 100 1A multiply petrol price x : 35 000 1 CA petrol cost Sonic ∴x = R32 923,80 CA M 1M multiplying by 12 Total running cost = R32 923,80 + 12 × R2 657 1CAtotal running cost for = R64 807,80  CA the Sonic Aveo : R14,04 / ℓ × 7,3 = R102,492 R102,492 : 100 y : 35 000 1 CA petrol cost Aveo ∴y = R35 872,20  CA Total running cost = R35 872,2 + 12 × R1 942 1CA total running cost for = R59 176,20 CA the Aveo 1O conclusion ∴ Aveo 1.6 is more economical.  O (8) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 10 DBE/November 2014 NSC – Memorandum Ques Solution Explanation  RG L2 2.2.1 Age 6 to 7 years. 2RG the age [6 or 7 one mark] [Including other intersection points ONLY one mark] (2) L4 2.2.2 Growth is a continuous phenomenon. O 1O any FIRST correct reason OR  O 1O for any SECOND Growth is affected by many factors like nutrition and health. correct reason OR  O It is influenced by genetic makeup inherited from parents. OR This graph is for average heights. O OR Physical disabilities will influence height  O (2) L2 RG 2.2.3 Between 4 and 6 years 1RG reading from graph Between 11 and 14 years RG 1RG reading from graph [5 and 13 only one mark] (2) L4 2.2.4 Boys stay longer than girls in childhood. RG 2RG comparing childhood stage Both girls and boys remain the same in pre-adolescence.RG 1RG comparing pre- adolescence Girls stay longer in adolescence. RG 2RG comparing adolescence OR OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 11 DBE/November 2014 NSC – Memorandum Ques Solution Explanation 2.2.4 Cont. Childhood Girls stay in childhood stage: 7 years RG 2RG number of years in Boys stay in childhood stage: 9 years childhood Pre-adolescence Girls stay in pre-adolescent stage: 2 years Boys stay in pre-adolescent stage: 2 years RG 1RG number of years in pre-adolescence Adolescence Girls stay in adolescent stage: 6 years Boys stay in adolescent stage: 4 years RG 2RG number of years in adolescence (5) L4 2.2.5 The girls’ height slows down/stabilizes/levels/evens out. O 2O trend OR O The girls’ growth rate relating to height decreases. [0 marks or 2 marks] [Trend relating to girls only] (2) L3 2.2.6 Height in inches 1C conversion C = 165 × 0,3937 1A accuracy = 64,9605 A  CA 2CA conclusion The boy’s height is above the average height for boys [Range 62 to 65] OR OR Height in cm = 63 C 1C conversion 0,3937 = 160,02  A 1A accuracy  CA The boy’s height is above the average height for boys 2CA conclusion [Range 157 to 165] (4) [33] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 12 DBE/November 2014 NSC – Memorandum QUESTION 3 [34 MARKS] Ques Solution Explanation Note: Afrikaans scripts to be marked differently L3 3.1.1 1MA annual salary  MA Annual salary = R 20 416,67 × 12 = R 245 000,04 Pension = R 245 000,04 × 6 % = R 14 700 ,00  CA 1CA pension Taxable amount without bonus 1CA subtracting the pension  CA = R 245 000,04 – R 14 700,00 = R 230 300, 04 Taxable annual income  CA 1 CA taxable annual income = R230 300,04 + R20 416,67 = R250 716,71 OR OR Monthly pension = R20 416,67 × 6% = R1 225  MA 1MA pension Monthly taxable salary = R20 416,67 – R1 225 = R19 191,67  CA 1CA subtracting the pension  MA 1MA annual salary Annual taxable income = R19 191,67 × 12 + R20 416,67 = R250 716,71  CA 1 CA taxable annual income OR OR Annual taxable income  MA  MA 1MA multiplying by 13 = (13 × R 20 416,67) – (12 × R 20 416,67 × 6%) 1MA calculating the pension = R 265 416,71 – R14 700  CA 1CA subtracting the pension = R250 716,71  CA 1 CA taxable annual income [Pension omitted lose 2 marks] [Bonus omitted lose 1 mark] (4) A  SF NB: Amount from Q3.1.1 L3 3.1.2 Rate of tax = R 29 808 + 25% × (R250 716,71 – R 165 600) 1A for correct tax bracket = R 29 808 + R 85 116,71 × 25% 1SF for substituting into the = R 29 808 + R 21 279,18 formula = R 51 087,18  CA S 1S simplification Annual tax after rebate = R 51 087,18 – R 12 080,00 1CA for tax amount = R 39 007,18  CA 1CA for tax amount after rebate NPR (5) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 13 DBE/November 2014 NSC – Memorandum Ques Solution Explanation L3 3.1.3  CA 1CA for tax value per Monthly Tax = R 39 007,18 ÷ 12 = R 3 250,60 month Net monthly salary = Monthly salary – pension – monthly tax M 1M for subtracting both = R 20 416,67 – R 1 225 – R 3 250,60 values = R 15 941,07  CA 1CA net salary [CA only if a monthly salary is used] OR OR Annual salary after tax = Annual salary – pension – annual tax M 1M for subtracting both = R245 000,04 – R 14 700,00 – 39 007,18 values = R 191 292,86  CA 1CA annual salary R191 292,86 ∴ Net monthly salary = 12 = R15 941 ,07  CA 1CA monthly salary [dividing by 12] (3) 3.2.1 Amount if inflation rate was used for increase 1A correct amount from L3(4) A M table L4(1) = R44,8 billion × 105,77% 1M percentage increase = R47,38496 billion  CA 1CA increased amount M 1M comparing This amount is less than the amount which was allocated, therefore 1O stating that she is her claim was valid.  O correct OR OR Amount if inflation rate was used for increase 1A correct amount from A M = R44 800 000 000 × 105,77% table 1M percentage increase = R47 384 960 000  CA 1CA increased amount M This amount is less than the amount which was allocated, therefore 1M comparing her claim was valid.  O 1O stating that she is correct OR OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 14 DBE/November 2014 NSC – Memorandum Ques Solution Explanation 3.2.1 Cont. Difference = R47,9 billion – R44,8 billion  A 1A correct amount from = R3,1 billion  M table Percentage increase 1M subtracting correct values R3,1 billion = × 100%  MA R44,8 billion 1MA calculating the = 6,919642857 % percentage increase ≈ 6,9 %  CA 1CA for rounding off Her claim is valid. O 1O stating that she is Note correct [Word billion must be there when subtracting and not for %] (5) * CA from Q3.2.1 L3(3) 3.2.2 Department of National Defence percentage growth from 2013/14 L4(2) to 2014/15 is 6,9%  CA 1CA correct percentage South African national budget percentage growth from 2013/14 to 2014/15  M/A 1M/A using correct values R1,25 trillion − R1,15 trillion = × 100% M 1M calculating growth R1,15 trillion 1CA calculating average = 8,69565174 %  CA % 1O Stating that the Dr Khoza’s statement is correct. O increase is greater (5) L3 3.2.3 Amount 2013/14 = 8,1% × R 41,6 billion + R41,6 billion  M 1M for increasing by 8,1% = R3,3639 billion + 41,6 billion 1CA the amount = R44,9696 billion  CA Amount 2014/15 = 5,9% × R 44,9696 billion + R44,9696 billion = R2,6532064 billion + 44,9696 billion  M 1M for increasing by 5,9% = R 47,6228064 billion  CA 1CA the amount OR OR M  CA 1M for increasing by 8,1% Actual amount = R 41,6 billion ×108,1% = R 44,9696 billion 1CA the amount 1M for increasing by 5,9% M  CA 1CA the amount R 44,969 6 billion × 105,9% = R 47,622 806 4 billion NPR or R47 622 806 400 [Penalty 1 mark if billions omitted] (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 15 DBE/November 2014 NSC – Memorandum Ques Solution Explanation L4 3.2.4 Difference =R48 billion - R47,9 billion = R 0,1 billion. 1O for identifying In reality the difference is not 0,1 O the difference of 0,1 but an amount of R100 000 000 (one hundred million)  O 1O For knowing Example: that 0,1 billion is R 47,9 billion rounded R48 billion implies that there will be an over 100 000 000 allocation of R100 million  O 1O suitable example must be chosen (3) L4 3.3.1 A visual representation is more understandable (make sense of) for the 2O reason general public than a table with values only.  O OR A visual representation is easier to read than text or table consisting of values.  O OR The actual values are in billions and trillions which many people don’t understand, where in these graphs percentages are used which are more understandable.  O (2) O L4 3.3.2 A bar graph (multiple/compound) is more appropriate to display this 1O identifying the data type of graph  O The bar graph will allow for a much more-in-depth analysis of the 2O for explaining trends in the collection of tax between the different categories over a the advantage of a period of time. bar graph OR OR Line or broken line graph O 1O identifying the type of graph The two lines will allow for a much more-in-depth analysis of the trends in the collection of tax between the different categories over a 2O for explaining period of time.  O the advantage of a broken line graph (3) [34] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 16 DBE/November 2014 NSC – Memorandum QUESTION 4 [45 marks] Ques Solution Explanation L2 4.1.1(a) A A CA 1A correct row number M15 and M16 1A seat number 1CA second seat number [15 and 16 two marks] (3) A A L2 4.1.1(b) 24 × 2 = 48 seats 1A 24 seats 1A total number of seats (2) RT MA RT * seats from Q 4.1.1 (b) L3 4.1.1(c) Total income in OR = (72×78) + ( 388 × 48) +( 83 × 42) 1MA adding the values + (81 × 28) + (112 × 15) + (82 × 10) 1RT cost zone A and B S RT 1RT cost for zone C and D = 5 616 +18 624 + 3 486 +2 268 +1 680 + 820 1RT cost for zone E and F 1S simplification = 32 494 CA 1CA answer [One mark for every 2 zones] (6) L4 4.1.2(a) Cost for 1 zone B ticket = 48 OR A 1A cost of ticket = R27, 2183 × 48 = R 1 306,48 C 1C convert OR to Rand Cost in Euro for one flight ticket = 492, 29 492,29 Cost in OR for one flight ticket = M 1M convert Euro to OR 1,87126 = 263,08 Cost in Rand for one flight ticket = 263,08 × R 27, 2183 M 1M convert OR to Rand = 7 160, 59 CA 1CA cost of one ticket Total cost per person = R 1 306,48 + R 7 160, 59 = R 8 467,07 CA 1CA calculating total cost per person Total cost for two = R 8 467,07 × 2 = R 16 934,14 CA 1CA calculating total cost for two people OR OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 17 DBE/November 2014 NSC – Memorandum Ques Solution Explanation A 4.1.2(a) Cost for 2 zone B tickets = 2 × 48 OR = 96 OR 1A cost for one ticket (cont.) = R27, 2183 × 96 1C conversion = R2 612, 96 C Cost for 2 flight tickets = 2 × €492, 29 A = €984, 58 1A 2 flight tickets R27,2183× 984,58 M 2M convert Euro to €984, 58 = rand 1,87126 = R14 321, 15 CA 1CA cost of 2 tickets in rand Total cost = R2 612, 96 + R14 321, 15 = R16 934, 11 CA 1CA total cost OR OR A Cost for Zone B tickets: 2 × 48 OR = 96 OR A 1A cost for one ticket 1A cost of 2 tickets 2× 492,29 C 1C conversion to OR Flight tickets in OR = 1,87126 = 526,1588448 CA 1CA ticket price Total cost: 526,1588448 + 96 = 622,1588448 CA 1CA total cost Cost in Rand = 622,1588448 × 27,2183 C 1C convert OR to Rand = 16 934,11 CA 1CA cost in rand (7) L2 4.1.2(b) Time leaving Johannesburg + flight time 1A adding = 20h30 +11h25 = 31h55 A CA 1CA correct time Time in South Africa when they arrived: 07:55 or 7.55 am or [If written as 07h55 five minutes to eight in the morning one mark only] Answer only full marks (2) 4.2.1 South westerly ( SW) A 2A correct direction L2 OR South, south westerly (SSW) (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 18 DBE/November 2014 NSC – Memorandum Ques Solution Explanation 4.2.2 L4 O This chart only shows distances from Muscat. OR O They don’t lie in the same direction. O OR This is not a map / strip chart. 2O opinion (2) 4.2.3 RT M 1RT correct value L2 Muscat to Sydney ≈ 3 349km × 3,5 1M multiplication by 3 349 ≈ 10 716,8 to 11 721, 5km CA 1CA correct distance [Range of values 3,2 to 3,5] [3 or 4 then max 2 marks] (3) L4 4.3.1 TSA = P × H + K A SF 1A total area of = 8 × 110 mm × 250 mm + 58 423 mm2 panels = 220 000 mm2 + 58 423 mm2 1SF substitution in = 278 423mm2 S formula = 0,278 423 m2 C 1S simplification For 0,07 m2 one needs 100mℓ of paint 1C conversion to 100 m2 ∴ 1 m2 one need mℓ M 1M Method 0,07 = 1 428,57 mℓ ∴ 0,278423 m2 need = 1428,571429 × 0,278423 = 397,7471429 mℓ ≈ 397,75 mℓ 1CA paint needed CA Two coats = 2 × 397, 75mℓ for 1 coat = 795, 49 mℓ CA 1CA paint needed 795,49 m for 2 coats Number of spray cans = 250 m = 3,18184 ≈4 CA 1CA rounding up Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 19 DBE/November 2014 NSC – Memorandum 4.3.1 OR OR L4 Cont. TSA = P × H + K 1A total area of A C SF panels = 8 × 0,110 m× 0,250m + 0,058 423 m2 1C conversion to m2 = 0,22 m2+ 0,058 423 m2 1SF substitution in = 0,278 423 m2 S formula 1S simplification For 0,07 m2 one needs 100mℓ of paint 100 1M method ∴ 1 m2 one need mℓ M 0,07 = 1 428,57 mℓ ∴ 0,278423 m2 need = 1428,571429 × 0,278423 = 397,7471429 mℓ ≈ 397,75 mℓ CA Two coats = 2 × 397, 75mℓ 1CA paint needed = 795, 49 mℓ for 1 coat CA 1CA paint needed 795,49 m for 2 coats Number of spray cans = = 3,1819 250 m ≈ 4 CA 1CA rounding up OR OR TSA = P × H + K A C SF 1A total area of = 8 × 0,110 m× 0,250m + 0,058 423 m2 panels 1C conversion to = 0,22 m2+ 0,058 423 m2 m2 = 0,278 423 m2 S 1SF substitution in A formula 1 spray can covers = 0,07 × 2,5m2 1S simplifying = 0,175 CA 1A spray rate per can 0,2784823 1CA simplification Number of cans = ×2 M 0,175 1M for two coats = 3,1819 ≈ 4 CA 1CA rounding up Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 20 DBE/November 2014 NSC – Memorandum Ques Solution Explanation 4.3.1 OR OR cont. TSA = P × H + K 1A total area of panels A SF 1SF substitution in formula = 8 × 110mm × 250mm + 0,058423m2 1C conversion to m2 = 8 × 0,11m × 0, 25m + 0,05423m2 C = 0,22 m2 + 0,058423m2 = 0,278423m2 S 1S simplification 100 ml covers 0,07 m2 100 × 0,278423 ∴ 0,28m2 will need = mℓ M 1M method 0,07 = 397,7471429mℓ = 397,75mℓ CA 1CA paint needed for 1 coat Two coats = 2 × 397, 75mℓ = 795, 49 mℓ CA 1CA paint needed for 2 coats 795,49 m Number of spray cans = =3,181 ≈ 4 CA 1CA rounding up 250 m (8) 4.3.2 MA L2 Height = 240 mm × 164 1MA correct height = 39 360 mm CA 1CA correct answer in mm = 39, 36 meters C 1C conversion ∴ The height of the actual tower is approximately 39, 4m OR OR MA C 1MA correct height Height = 25cm – 1cm = 24 cm = 0,24 m 1C conversion 1CA correct answer in m Actual height = 0,24 × 164 = 39,36 m CA NPR (3) 4.4 A L2 1. Mount the vertical poles to the kick base and 1A for the vertical poles fasten with the screws. A 1A for the screws A 2. Slide the three glass panels into the vertical poles. 1A glass panels A 3. Place the top aluminium frame on top and fasten 1A for the top frame with screws. A 1A Screws A 1A interior standards 4. Screw the interior standards onto the aluminium framing and insert the brackets. A 1A brackets [Single word answers not acceptable.] (7) [45] TOTAL: 150 Copyright reserved