Mathematical Literacy P2 Feb March 2014 Memo Eng hlayiso.com

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICAL LITERACY P2 FEBRUARY/MARCH 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 J Justification/reason This memorandum consists of 14 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 2 DBE/Feb.–Mar. 2014 NSC – Memorandum QUESTION 1 [31 MARKS] Ques Solution Explanation AS/L A 12.3.1 SF 1A circumference 1.1.1 Curved area of the cylinder = 2 × 3,14 × 70 cm × 140 cm 1SF substitution L3 = 61 544 cm2 CA 1CA curved area A Area of wrap = 1,06 cm × 61 544 cm2 M 2 1A increasing by 6% = 65 236,64 cm CA 1M concept OR 1CA area 6 M 2 2 OR Area of wrap: × 61544 cm = 3 692,64 cm 100 1M concept of % A ∴ Area of wrap = 61 544 cm2 + 3 692,64 cm2 1A increasing by 6% = 65 236,64 cm2 CA 1CA area (6) 12.3.1 2 1.1.2 Volume = 3,14 × (70 cm) × 140 cm SF 1SF substitution L3 = 2 154 040 cm3 CA 1CA simplification Total surface area = 2 × 3,14 × 70 cm(70 cm + 140 cm) = 439,6 cm × (210 cm) = 92 316 cm2 CA 1CA simplification Volume: Total surface area = 2 154 040 : 92 316 M 1M writing as a ratio = 23,333 : 1 ≈ 23 : 1CA 1CA ratio in required form ∴ Mathys' bales do conform. CA 1CA conclusion (6) 12.3.2 1.1.3 9 1SF substitution L4 Temperature in °F = × 55° + 32° SF 5 = 131° CA 1CA temperature in ° F CA 1CA verification No, his action was not correct. (3) 12.1.1 1.2 st 1 layer = 12 bales A 1A number of bales in L3 2nd layer = 5 bales 1st layer 3rd layer = 4 bales A 1A number of bales in 4th layer = 3 bales A 3rd layer 1A number of bales in last (4th) layer Total number of bales = 12 + 5 + 4 + 3M 1M adding = 24 CA 1CA simplification (5) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 3 DBE/Feb.–Mar. 2014 NSC – Memorandum Ques Solution Explanation AS/L 12.2.1 1 440 kg A 1A mass of each bale L2 1.3.1 Max number of days = 12kg/day × 10 A 1A consumption per 10 cows = 12 days CA 1CA time taken OR OR Consumption per 10 cows = 12 kg/day × 10 1A mass of each bale = 120 kg/day A 1 440 kg Max number of days = A 1A consumption per 10 120 kg/day cows = 12 days CA 1CA time taken (3) A 12.2.1 1.3.2 1 440 kg 1A correct values used L3 Max number of days = M 1M dividing 12 kg / day × number of cows 120 = number of cows CA 1CA simplified OR formula Using variables (3) 12.2.2 1.3.3 L3 1CA (1; 120) CA 3CA any other 3 points plotted correctly 1CA joining by means of a smooth curve CA CA CA CA (5) [31] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 4 DBE/Feb.–Mar. 2014 NSC – Memorandum QUESTION 2 [26 MARKS] Ques Solution Explanation AS/L 12.1.3 2.1 i = 0,072; n = 5 L3 A = R650 000(1 + 0,072)5 SF A 1A value of i = R920 210,7097 1SF substitution ≈ R920 210,71 CA 1CA price of bus (3) 12.2.1 2.2.1 Amount (in rand) A A 1A multiplying number by L4 = 400 × number of alumni members – 1 000 400 1A subtracting 1 000 OR Using symbols (2) 2.2.2 12.2. QUARTERLY CONTRIBUTION TOWARDS BUYING A L3 NEW SCHOOL BUS 14 000 12 000 A 10 000 A Amount (in rand) 8 000 A 6 000 A 4 000 A A 2 000 0 0 5 10 A 15 20 25 30 35 40 Number of alumni members 1A starting at (10 ; 4000) 1A for (20 ; 7 000) indicated by a circle 1A point (20 ; 8 000) 1A point (35; 13 000) 1A any other correct point between the 1A any other correct point between above two points the above two points 1A joining the points (7) 12.2.2 RG 8 600 + 1 000 M 2RG reading from graph L3 2.2.3 24 OR 400 OR = 24 CA 1M calculation 1CA solution (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 5 DBE/Feb.–Mar. 2014 NSC – Memorandum Ques Solution Explanation AS/L 12.1.3 2.3.1 Total amount deposited = R40 000 × 20 M 1M multiplying by 20 L3 = R800 000 CA 1CA amount deposited Total interest earned = R911 408,73 – R800 000 M 1M subtracting = R111 408,73 CA 1CA amount deposited quarterly (4) 12.1.2 2.3.2 Amount contributed by alumni 1A correct value for L2 (3) A A 18 members L3(3) = (400 × 18) × 4 + (400 × 25 – 1 000) × 12 1A value for 25 L4(2) + (400 × 35 – 1 000) × 4 members A 1A value for 35 A members = R28 800 + R108 000 + R52 000 1A R108 800 = R188 800 CA 1CA amount deposited R188 800 1M calculating % Percentage contribution = × 100%M R 800 000 1CA solution = 23,6 % CA His statement is not valid. O 1O conclusion (8) [26] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 6 DBE/Feb.–Mar. 2014 NSC – Memorandum QUESTION 3 [30 MARKS] Ques Solution Explanation AS/L 12.3.3 3.1.1 South East  A 2A correct direction L2 (2) 12.3.3 3.1.2 Exiting Hallmark, she must: A L3 * turn left and walk until she reaches the end of the fountain 1A first turn and * then turn right passing shop number 9 and then left direction towards entrance number 3 * then enter Cafe Teen on the right hand side A 1A destination OR OR Exiting Hallmark, she must: * walk straight passing entrance number 1  A 1A first turn and * then turn left at the corner and walk until she reaches the direction end of the fountain * then turn left passing shop number 11 and then right towards entrance number 3 * enter Cafe Teen on the right hand side  A 1A destination (2) 12.3.3 3.1.3 Cash 4 U  A 1A correct store L2 (1) 12.4.2 3.1.4 The names are not alphabetical  J 1J alphabetical order L4 The shops in the zones are not grouped together  J 1J numerical order (2) 12.4.5 3.1.5 4A 1A numerator L2 P(clothing shop) = 13 A 1A denominator (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 7 DBE/Feb.–Mar. 2014 NSC – Memorandum Ques Solution Explanation AS/L 12.3.1 3.2.1 13,9 m L2 (3) 2,3 m L3 (2) 10,4 m 5,8 m Total floor space = area of rectangle + area of trapezium 1 1M calculating = length × breadth + (sum of parallels) × height 2 height SF SF M 1 = 5,8 m × 10,4 m + (2,3 m + 10,4 m) × 8,1 m 2SF substitution into CA 2 CA correct formulae = 60,32 m + 51,44 m2 = 111,76 m2 CA 2 2CA simplifying 1CA total floor space OR 13,9 m OR 2,3 m 10,4 m 5,8 m Total floor space = area of rectangle + area of trapezium 1 = length × breadth + (sum of parallels) × height 2 SF SF M 1M calculating 1 = 13,9 m × 2,3 m + (13,9 m + 5,8 m) × 8,1 m height CA 2 2CA 2SF substitution = 31,97 m + 79,79 m2 = 111,76 m2CA 2CA simplification 1CA total floor space OR 13,9 m 2,3 m 10,4 m 5,8 m Total floor space = area of big rectangle + area of smaller rectangle + area of triangle 1 = length × breadth + length × breadth + × base × height 2 SF 1 SF M 1M calculating = 10,4 m × 5,8 m + 2,3 m × 8,1 + × 8,1 × 8,1 CA CA 2 2 height 2 2 = 60,32 m + 18,63 m + 32,81 m 2SF substitution 2CA simplification = 111,76 m2CA 1CA total floor space OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 8 DBE/Feb.–Mar. 2014 NSC – Memorandum Ques Solution Explanation AS/L 13,9 m 2,3 m 10,4 m 5,8 m Total floor space = area of rectangle – area of triangle 1 = length × breadth – × base × height 2 1M calculating SF 1 SF M height = 13,9 m × 10,4 m – × 8,1 m × 8,1m CA 2 2 2SF substitution = 144,56 m – 32,805 m 2CA 2CA simplification = 111,76 m 2 CA 1CA total floor space (6) 12.3.3 Note: The dist between the 2 entrances allow for 3.2.2 L4 ± 2 mm range The one horizontal measurement is 13,9 m 1A measuring the On the question paper Hallmark is 1,2 cmA side On the question paper the distance from the northern entrance 1A measuring the total door to the southern entrance door is 9,3 cmA length ∴ total distance = 9,3 × 13,9 M OR 1,2 cm : 13,9m 1M using scale and 1,2 M proportion ≈ 107,73 m 1cm = 11,583 m 1CA total distance ∴ total distance = 9,3 × 11,583 Note: A range of ≈ 107,72m values from 1 cm ∴ the distance is 110 metres CA to 1,4 cm will be accepted OR The one vertical measurement is 10,4 m 1A measuring the side On the question paper the side is 0,9 cm A 1A measuring the total On the question paper the distance from the northern entrance length door to the southern entrance door is 9,3 cm A 9,3 M ∴ total distance = × 10,4 M OR 0,9 cm : 10,4m 1M using scale and 0,9 proportion ≈ 107,47 m 1cm = 11,555.. m ∴ total distance = 9,3 × 11,556 = 107,47m 1CA total distance ∴ the distance is 110 metres CA Note: A range of values from 0,7 cm to 1,1 cm will be accepted OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 9 DBE/Feb.–Mar. 2014 NSC – Memorandum Ques Solution Explanation AS/L The other horizontal measurement is 5,8 m 1A measuring the side On the question paper Hallmark is 0,5 cm A 1A measuring the total On the question paper the distance from the northern entrance length door to the southern entrance door is 9,3 cm A ∴ total distance = 9,3 × 5,8 M OR 0,5 cm : 5,8 m M 1M using scale and 0,5 proportion ≈ 107,88 m 1cm = 11,6 m 1CA total distance ∴ total distance = 9,3 × 11,6 = 107,88m Note: A range of ∴ the distance is 110 metres CA values from 0,3 cm to 0,7 cm will be OR accepted The other vertical measurement is 2,3 m 1A measuring the side On the question paper Hallmark is 0,2 cm A 1A measuring the total On the question paper the distance from the northern entrance length door to the southern entrance door is 9,3 cm A 1M using scale and 9,3 ∴ total distance = × 2,3 M OR 0,2 cm : 2,3 m M proportion 0,2 ≈ 106,95 m 1cm = 11,5 m 1CA total distance ∴ total distance = 9,3 × 11,5 Note: A range of ≈ 106,95m values from 0,1 cm ∴ the distance is 110 metres CA to 0,4 cm will be accepted (4) 3.2.3 The area of the curtain = 3 × 4 = 12 m 2 A 1A curtain area 12.3.2 L4 The weigth of the curtain = 4,7 kg/ m 2 × 12 m 2 1CA curtain weight = 56,4 kg CA 1M multiplying Cost of a curtain material = R12,50/kg × 56,4 kgM 1CA cost of curtain = R705CA material 1O opinion The cost does NOT exceed R800.O (5) 12.4.4 3.3.1 Friday A 1A correct day L4 Data for week 1 only started on Friday J 1J explanation (2) 12.4.4 3.3.2 The number of people visiting the Mall on Friday, Saturday L4 and Sunday is the highest.  J 2J correct justification (2) A A 1A correct week 12.4.4 3.3.3 Week 4, Thursday 1A correct day L4 (2) [30] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 10 DBE/Feb.–Mar. 2014 NSC – Memorandum QUESTION 4 [38 MARKS] Ques Solution Explanation AS/L 12.1.1 4.1.1 Percentage of blacks = 79,6% A 1A correct percentage L3 Black population in 2011 = 79,6% of 51 770 560 M 1M using percentage 79,6 = × 51 770 560 100 = 41 209 365,76 CA 1CA black population ≈ 41 209 366 or 41 209 365 R 1R rounding (up or down) (4) 12.4.1 4.1.2 9,6 1M/A using percentage L2(3) Number of whites = × 44 819 778 M/A L3(2) 100 = 4 302 698,688CA 1CA white population 48,36 1M/A using percentage Number of white males = × 4 302 699 M/A 100 of white males = 2 080 785,086 ≈ 2 080 785 CA 1CA simplification Thandi’s calculation is NOT correct. J 1J verification (5) 12.4.4 4.1.3 Indian population in 2001 = 1 120 494 A 1A number of Indians L4 Indian population in 2011 = 1 294 264 A in 2001 J 1A number of Indians ∴ Thandi’s comment is not correct (the population in 2011 increased) 1J conclusion (3) 12.1.1 4.2.1 L3 Population in 2001 = 21 434 041 + 23 385 737 1A population in 2001 (a) = 44 819 778 A A = 44 819 778 – (14 365 288 + 2 215 211) = 28 239 279 CA 1CA simplification (2) 12.1.1 4.2.1 Male : female = 1 : 1,08 M OR 100 : 108M 1M ratio L4 (b) CA CA 100 1CA males 48 males and 52 females = × 100 1CA females 208 = 48 malesCA ∴ 52 femalesCA (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 11 DBE/Feb.–Mar. 2014 NSC – Memorandum Ques Solution Explanation AS/L 12.4.1 Dependency % (in 2011) 4.2.2 L2 n + m (a) = × 100% p 15 100 089 + 2 765 991 1SF substituting correct = × 100% SF 33 904 480 values = 52,695...% 1A simplification ≈ 52,70% A Dependency % (1996) n + m = × 100% p 13 766 443 + 1 934 664 1SF substituting correct = × 100% SF values 24 882 465 = 63,101...% 1CA simplification ≈ 63, 10% CA Difference = 63,10% – 52,70 % = 10,4% CA 1CA difference (5) 12.4.4 4.2.2 The dependency % decreased because there are more people L4 (b) in the category (P) 15 – 64 years. J 2J opinion OR Technology became more advanced. J OR Improved medication J OR Improvement in health J OR The receiving of social grants J OR Any other valid reason J (2) 12.4.3 4.3.1 Range = 1 290 – P M A 1M concept of range L3 569 = 1 290 – P M A OR P = 1 290 - 569 1A correct values used ∴ P = 721 CA = 721CA 1CA solution (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 12 DBE/Feb.–Mar. 2014 NSC – Memorandum Ques Solution Explanation AS/L 12.4.3 4.3.2 Mean M A 1A correct values used L3 814 + 921 + 1 201 + 1 290 + Q + 966 + 864 + 721 + 828 + 829 1M concept of Mean = 10 8 434 + Q = 10 8 434 + Q 936 = 10 Q = (936 × 10) – 8 434 = 9360 – 8 434 S 1S simplifying = 926 CA 1CA solution (4) 12.4.3 4.3.3 721; 814; 828; 829; 864; 921; 926; 966; 1 201; 1 290 M 1M arranging L3 864 + 921 1M concept of Median = M median 2 = 892,5 CA 1CA solution ≈ 893 (3) J 12.4.4 4.3.4 The sample is not representative of all the schools in South 2J reason L4 Africa. J The sample is too small compared to the number of schools 2J reason in the country. OR Any other suitable reasons. (4) [38] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 13 DBE/Feb.–Mar. 2014 NSC – Memorandum QUESTION 5 [25 MARKS] Ques Solution Explanation AS/L M 12.2.1 5.1.1 Loan amount = (Monthly payment ÷ loan factor) × 1 000 1M subject of formula L3 A 1A loan factor = (R17 550 ÷ 13,00) × 1 000 SF 1SF substitution = R1 350 000 CA 1CA solution (4) 12.1.3 5.1.2 She needs to have extra money available per month, for 2J reason L4 other expenses. J She will pay more on interest. J 2J reason OR Any other valid reason (4) 12.1.1 5.2.1 STL Bank: SF 1SF substitution 12.1.3 Monthly payment = (1 100 000 ÷ 1 000) × 13,91A 1A using correct factor 12.2.1 = R15 301 CA 1CA monthly payment L2 (3) 1M multiplying by 240 L3(2) ∴ Total repayment = R15 301 × 240 M 1CA final amount L4(3) = R3 672 240 CA Pragashni should rather take STL Bank’s deal. O 1O choice Although the interest rate is higher, the year term is shorter and the total repayment amount is R4 290 000 – R3 672 240 2J reason with = R617 760 less. J calculation OR OR SF A 1SF substitution Monthly payment (STL Bank) = (1 100 000 ÷ 1 000) × 13,91 1A using correct factor = R15 301 CA 1CA monthly payment SF 1SF substitution into Monthly payment (EP Bank) = (1 100 000 ÷ 1 000) × 13,00 formula = R14 300 CA 1CA monthly payment O 1O choice Pragashni should take EP bank his monthly instalment will 2J reason with be reduced by R15 301 – 14 300 = R1 001. J calculation (8) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 14 DBE/Feb.–Mar. 2014 NSC – Memorandum Ques Solution Explanation AS/L 12.1.3 5.2.2 Monthly payment M 1M manipulation 12.2.1 Loan factor = × 1 000 L4 Loan amount R13 255 = ×1 000 SF 1SF substitution R1 100 000 1CA factor = 12,05 CA CA CA ∴ the interest rate will be 14,25% over a period of 30 years 1CA interest 1CA period (5) 12.2.3 5.3 Line C represents a 16% interest rate. A 1A graph C L4 Line B represents a 14,25% interest rate. A 1A graph B The higher the interest rate, the higher your total repayment will be. J 2J reason OR OR The higher the interest rate, the steeper the graph. J 2J reason (4) [25] TOTAL: 150 Copyright reserved