Mathematical Literacy P2 Nov 2017 Memo Eng hlayiso.com

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICAL LITERACY P2 NOVEMBER 2017 MARKING GUIDELINES MARKS: 150 Symbol Explanation M Method MA Method with accuracy CA Consistent accuracy A Accuracy C Conversion S Simplification RT Reading from a table/ a graph / document/diagram SF Correct substitution in a formula O Opinion/Explanation P Penalty, e.g. for no units, incorrect rounding off, etc. R Rounding off NPR No penalty for rounding AO Answer only MCA Method with constant accuracy This marking guideline consist of 17 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 2 DBE/November 2017 NSC – Marking Guidelines QUESTION 1 [40 MARKS] Ques Solution Explanation T&L F 1.1.1 Decrease amount in thousands Decrease amount L2 = R32 187 × 4,402% M = R32 187 000 × 4,402% 1M % calculation ≈ R1 416,87 CA = R1 416 871,74 1CA decreased amount Communication Cost in thousands ≈ R1 417 000 = R32 187 – R1 416,87M Comm. Cost 1M subtracting = R30 770,13 = R32 187 000 – R1 417 000 = R30 770 R = R30 770 000 1R rounding OR OR Communication Cost in thousands 1M subtracting M = 32 187 – (4,402% × 32 187) M 1M % calculation = 32 187 – 1 416,87 = 30 770 CA R 1CA decreased amount 1R rounding OR OR M 100% – 4,402% = 95,598 % M 1M subtracting Communication Cost in thousand = R32 187 × 95,598% 1M % calculation = R30 770,12826 CA 1CA cost ≈ R30 770 R 1R rounding OR OR Communication Cost in thousands 1M subtracting M 1M adding all other values = R2 163 571 – R(67 257 + 640 601 + 69 866 + M 953 592 + 135 768 + 34 087 + 55 267 + 176 363) 1CA total for other values = R2 163 571 – R2 132 801 CA 1CA cost = R30 770 CA AO (4) F 1.1.2 Profits could decrease. O 2O explanation L4 OR Imported stock will cost more. O (2) F 1.1.3 342 534 RT For 2015: Percentage profit = × 100% 1RT correct values L4 2 250 041 SF 1SF substitution = 15,22345593% 1A percentage for 2015 A 360 651 For 2016: Percentage profit = × 100% 2 403 509 = 15,00518617 % A 1A percentage for 2016 The profit decreased O 1O comparison OR The profit nearly stayed the same. OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 3 DBE/November 2017 NSC – Marking Guidelines Ques Solution Explanation T&L NOTE: Calculated profit for 2015 is R343 002 thousand 343 002 1RT correct values Percentage profit = × 100% RT 1SF substitution 2 250 041 SF ≈ 15,24% A 1A percentage for 2015 For 2016: 360 651 Percentage profit = × 100% 2 403 509 1A percentage for 2016 = 15,00518617 % A The profit decreased O 1O comparison NPR (5) A MCA F 1.2 Income tax = R147 996 + 39% × R(663 000 – 550 100) 1A correct bracket L3 = R147 996 + 39% × R112 900 1MCA amount above = R147 996 + R44 031 S 1S simplification = R192 027 CA 1CA tax before rebate Total Income Tax (after rebates) M = R192 027 – R13 500 – R7 407 OR = R192 027 – R20907 1M subtracting both rebates = R171 120 CA 1CA tax after rebate (6) D 1.3 Increase number of donors for 2017 L3 = 110 000 × 9,6% = 10 560 M 1M calculating 9,6% Number of donors 2017 1CA calculating total = 110 000 + 10 560 donors for 2017 = 120 560 CA Increase number of donors for 2018 = 120 560 × 9,6% 1M calculating 9,6 % of = 11 573,76 M 2017 donors Number of donors 2018 = 120 560 + 11 573,76 = 132 133,76 1CA calculating donors ≈ 132 134 CA for 2018 OR OR Number of donors for 2017 1M multiplying correct =110 000 + (110 000 × 9,6%) M values = 120 560 CA 1CA calculating donors for 2017 Number of donors for 2018 1M multiplying correct % =120 560 + (120 560 × 9,6%) M to 2017 number = 132 133,76 1CA calculating number ≈ 132 134 CA for 2018 OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 4 DBE/November 2017 NSC – Marking Guidelines Ques Solution Explanation T&L OR OR Number of donors for 2017 1M multiplying and adding = 110 000 ×109,6% M percentages = 120 560 CA 1CA calculating total number for 2017 Number of donors for 2018 = 120 560 × 109,6% M 1M multiplying and adding = 132 133,76 correct % to 2017 number ≈ 132 134 CA 1CA calculating number for 2018 OR OR Number of donors for 2018 1M adding percentages M M M 1M multiplying correct = 110 000 × 109,6% × 109,6% numbers = 132 133,76 1M multiplying 109,6% twice ≈ 132 134 CA 1CA calculating number for 2018 NPR AO (4) D 1.4.1 Makes provision for other people who are not Asian, Black, L4 2O explanation Coloured or White. O OR Some donors don't indicate race. O OR The percentage of the races do not add up to 100%. O OR The other is ‘mixed’ race. O OR O They are from other countries. (2) O O D 1.4.2 As the years increase the percentage black donors increase. 2O increasing trend L4 (2) D 1.4.3 The number of donors are different every year. O 2O explanation L4 OR The graph represents percentages. O OR The percentages are rounded values. O OR O The graph shows that the bars’ heights are not the same. (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 5 DBE/November 2017 NSC – Marking Guidelines Ques Solution Explanation T&L D 1.4.4 The 2015 donors × 101,02% = 490914 L2 (a) 490 914 1MA dividing by 101,02% Number of donors = A OR 101,02% = 485 957,236… 1A number of donors ≈ 485 957 A NPR (2) CA from Q1.4.4 (a) D 1.4.4 % white = 100% – (8% + 38% + 5% + 2%) MA 1MA subtracting from 100% L3 (b) = 47% CA 1CA percentage Number of white donors = 485 957 × 47% MCA 1MCA % calculation = 228 399,79… ≈ 228 400 CA 1CA rounded number AO (4) P 1.5.1 P (Blood Type O ) L2 RT = (39 + 6)% 1RT correct two values 1A calculating probability = 45% OR 9 OR 0,45A (2) 20 P + 2A correct blood type L2 1.5.2 AB A (2) P 1.5.3 O L4 No, it is NOT most likely. 1O verification Can only receive blood from own blood group. O 2O explanation OR OR - P (O receiving blood from any donor) A = 1 1A numerator 8 A 1A denominator O ∴ It is NOT most likely. 1O verification (3) [40] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 6 DBE/November 2017 NSC – Marking Guidelines QUESTION 2 [37 MARKS] Ques Solution Explanation T&L O F 2.1.1 Inland prices have higher costs for transport / storage. 2O reason L4 OR O Coastal storages are close by and transport fees are lower. OR Fuel is imported via harbours. O OR Most refineries are along the coast. O (2) F 2.1.2 R2,67 A 1M multiplying L2 S = × R616,00 M OR R12,32 1A correct ratio = R133,50 CA 1CA storage cost OR R616,00M 1M dividing Number of litres = OR R12,32 = 50 A 1A litres OR S = 50ℓ × R2,67/ℓ = R133,50 CA 1CA storage cost OR OR R77 × R5,26 1A basic fuel price Basic fuel price = = R263 A R1,54 M 1M subtracting all from total CA 1CA storage cost S = R616 – R142,50 – R77,00 – R263,00 = R133,50 AO (3) M F 2.1.3 Number of litres consumed = 1 250 km × 7,3 ℓ ÷ 100 km 1M working with L4 = 91,25 ℓ A consumption rate Inland cost = 91,25 ℓ × R12,32/ ℓ 1A number of litres = R1 124,20 CA 1CA inland cost Coastal cost = 91,25 ℓ × R11,94/ ℓ = R1 089,525 1CA coastal cost ≈ R1 089,53 CA Statement is NOT valid. O 1O verification OR OR M 1M working with Litres consumed = 1 250 km ÷ 100 km × 7,3 = 91,25 A consumption rate 1A number of litres Difference in fuel price = R12,32 – R11,94 = R0,38 M 1M difference Difference in cost = R0,38/ ℓ × 91,25 ℓ ≈ R34,68 A 1A cost Statement is NOT valid. O 1O verification OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 7 DBE/November 2017 NSC – Marking Guidelines Ques Solution Explanation T&L OR OR Inland 1M working with Cost / 100 km = 7,3 ℓ × R12,32/ ℓ = R89,94 M consumption rate Number of 100km distances =1250 km ÷ 100 km = 12,5 A 1A cost Cost = 12,5 × R89,94 = R1 124,20 Coastal Cost / 100 km = 7,3 l × R11,94 = R87,16 Number of 100 km distances = 1250 km ÷ 100 km = 12,5 1A cost Cost =12,5 × R89,94 = R1 089,53 A 1M difference Difference = R1 124,50 – R1 089,53 = R34,67 M O 1O verification Statement is NOT valid. OR OR M 1M difference Difference = R12,32 – R11,94 = R0,38 Number of 100 km distances = 1 250 km ÷ 100 km = 12,5 1M multiplying with M M consumption rate Cost = R0,38 × 7,3 × 12,5 = R34,68 A 1M multiply with 12,5 Statement is NOT valid. O 1A cost 1O verification NPR (5) M F 2.2.1 R70,9 billion − R54 billion 1M % increase L2 % increase = × 100% A 1A correct values R 54 billion ≈ 31,296 % CA 1CA percentage OR OR R70,9 billion M 1M % increase × 100% = 131,2962% A 1A correct values R 54 billion % increase = 131,2962% – 100% 1CA percentage ≈ 31,296 % CA OR OR Using Trial & Error: 1M % calculation M A 1A increase amount R54 billion × 31,3% = R16,9 billion R16,9 billion + R54 billion = R70,9 billion 1CA percentage ∴ % increase = 31,3% CA NPR (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 8 DBE/November 2017 NSC – Marking Guidelines Ques Solution Explanation T&L F 2.2.2 7 + 118 = 125 A 1A adding ratio L3 values 7 × Total budgeted income = R70,9 billion A 1A using ratio values 125 7 M Total budgeted income = R70,9 billion ÷ 1M dividing by ratio 125 = R1 266,07 billion 1CA budget value ≈ R1 266 billion CA OR OR 7: 118 = R70,9 billion : x A 1A using proportion 7x = R70,9 billion ×118 R70,9 billion × 118 S x= 1S changing subject 7 ≈ R1 195,17 billion CA 1CA other revenues Total budgeted income = R1 195,17 billion + R70,9 billion = R1 266,07 billion ≈ R1 266 billion CA 1CA rounded value in billion (4) D 2.3.1 India  RT 2RT country L2 (2) D 2.3.2 0,02 0,52 0,63 0,91 1,12 1,23 2,03 2,17 2,97 3,62 4,11 1M use formula of L3 IQR IQR = Q 3 – Q 1 M 1A lower quartile A A 1A upper quartile = 2,97 – 0,63 1CA IQR = 2,34 CA AO [Accept 58 – 7 = 51] (4) D st O 2.3.3 Countries with high rankings are developed (rich, 1 world) as well 2O valid reason L4 as underdeveloped/developing (poor, 3rd world). OR O Countries with low rankings are developed (rich) as well as underdeveloped/ developing (poor). OR Counties listed are from all over the world (different continents). O OR O Rankings show the sample was chosen randomly. (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 9 DBE/November 2017 NSC – Marking Guidelines Ques Solution Explanation T&L 1RT reading both F 2.3.4 236,51 RT values L3 India: Mean Daily wage = MA 93,76% 1MA dividing by % ≈ 252,25 Rouble A 1A Indian day wage 237,35 SA: Mean Daily wage = 26,20% ≈ 905,92 Rouble A 1A SA day wage M 1M subtracting Difference = (905,92 – 252,25) Russian Rouble 1CA difference in = 653,67 Russian Rouble CA Rouble (6) D 2.3.5 Range = 425,52 – 21,44 A 1A range L4 = 404,08 Russian Rouble 1 Russian Rouble = 0,016 Euro M 1M multiplication ∴ 404,08 Russian Rouble = 404,08 × 0,016 Euro = 6,46528 Euro C 1C convert to Euro 1 South African Rand = 0,070 Euro C 1C convert to rand ∴ 6,46528 = R92,36 A 1A rand value 0,07 Learner solution is incorrect O 1O verification OR OR 0,016 C 1 Russian Rouble = Rand 1C dividing by 0,07 0,070 = R 0,2285714286 A 1A conversion factor Range = 425,52 – 21,44 A 1A range = 404,08 Russian Rouble = 404,08 × 0,2285714286 rand/rouble C 1C conversion = R92,36 A 1A rand value 1O verification Learner solution is incorrect O OR OR 1C conversion C CA Max. value to rand: 425,52 × 0,016 ÷ 0,07 = R97,26 1CA max value Min. value to rand: 21,44 × 0,016 ÷ 0,07 = R4,90 CA 1CA min value M 1M subtracting Range = R97,26 – R4,90 = R92,36 CA 1CA rand value Learner solution is incorrect. O 1O verification NPR (6) [37] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 10 DBE/November 2017 NSC – Marking Guidelines QUESTION 3 [40 MARKS] Ques Solution Explanation T&L A MP 3.1.1 33 Kwela Street A 2A correct number L2 1A correct street (3) M 3.1.2 Length 22 mm A (21 mm to 23 mm) 1A length L3 Width 9 mm A (8 mm to 10 mm) 1A width Scale 25 A mm = 30 m (24 mm to 26 mm) 1A measured scale 30 1M using the scale ∴ Length = × 22 m M 25 = 26,4 m CA 1CA length in m 30 1CA width in m Width = 9 × m = 10,8 m CA 25 OR OR A 1A measured scale Scale: 25 mm : 30 m (24 mm to 26 mm) 25mm : 30 000 mm 1M unit scale 1 : 1 200 M 1A length Length = 22 mm A (21 mm to 23 mm) 1A width Width = 9 mm A (8 mm to 10 mm) Actual length = 22 × 1 200 mm 1CA length in m = 26 400 mm = 26,4 m CA Actual width = 9 × 1 200 mm 1CA width in m = 10 800 mm = 10,8 mCA (6) CA from Q3.1.2 MP 3.1.3 On the enlarged map: L4 MCA 1MCA measured length Measured length = 62 mm (61mm to 64 mm) M CA Scaled length = 62 mm ÷ 5 = 12,4 mm ≠ 22 mm 1M dividing by 5 1CA simplification ∴ NOT valid O 1O verification OR OR On the enlarged map: 1A measured length A The measured width = 24 mm (23 mm to 26 mm) M 1M multiplying with 5 CA widths: 9 mm × 5 = 45 mm ≠ 24 mm 1CA simplification ∴ NOT valid O 1O verification OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 11 DBE/November 2017 NSC – Marking Guidelines Ques Solution Explanation T&L OR OR On the enlarged map: Measured length = 62 mmA (61mm to 64 mm) 1A measured Measured width = 24 mm (23 mm to 26 mm) 62 M 24 1M dividing Scale factor = OR width = 22 9 ≈ 2,82 CA ≈ 2,67 1CA scale factor ∴ Not valid 1O verification O (4) MA M 3.2.1 Length = 5 240 mm – 2 × 220 mm 1MA subtracting of thickness L3 = 4 800 mm CA 1CA internal length Width = 4 040 mm – 2 × 220 mm = 3 600 mm CA 1CA internal width MCA Floor area = 4 800 mm × 3 600 mm 1MCA substitution = 17 280 000 mm2 = 17 280 000 ÷ 1 000 000 C 1C conversion = 17,28 m2 CA 1CA internal area in m2 OR OR Length = 5 240mm = 5,24m C Width = 4 040mm = 4,04m 1C conversion of all values Wall thickness = 220mm = 0,22m MA 1MA subtracting thickness Interior Length = 5,24m – 2(0,22m) = 4,8mCA 1CA length Interior Width = 4,04m – 2(0,22m) = 3,6m CA 1CA width Floor Area = 4,8 m × 3,6 m MCA 1MCA substitution = 17,28m2 CA 1CA internal area in m2 (6) CA from Q3.2.1 M 3.2.2 Area of Ceiling board = 2 400 mm × 900 mm SF 1SF substitution L4 = 2 160 000 mm2 A 1A area of board 17 280 000 M 1M dividing Number of boards needed = 2 160 000 =8 1CA number of boards CA ∴ Need more than 7 O 1O deduction OR OR Number needed = 4 800 mm ÷ 2 400 mm M 1M dividing = 2 for length CA 1CA number length wise Number needed = 3 600 mm ÷ 900 mm = 4 for width CA 1CA number width wise Total needed = 2 × 4 = 8 CA 1CA number of boards ∴ Need more than 7 O 1O deduction OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 12 DBE/November 2017 NSC – Marking Guidelines Ques Solution Explanation T&L OR M L4 SF A 1SF substitution Area of one ceiling board = 2,4 m × 0,9 m = 2,16 m2 1A area of board M CA Total area coved by 7 boards = 2,16 m2 × 7 = 15,12 m2 1M multiplying 1CA total area ∴ Need more than 7 O 1O deduction (5) CA M SF 3.2.3 Length of cornice = 2 × (4 800 mm + 3 600 mm) 1CA values from Q 3.2.1 or RT L2 if reworked = 16 800 mm CA 1SF substitution 1CA length (3) CA from Q3.2.3 and Q3.2.2 F 3.2.4 16 800 ÷ 2 000 = 8,4 L4 CA Hence 9 lengths cornice needed. 1CA number of lengths A M Total cost = 8 × R91,44 + 9 × R53,64 1A using 2 correct prices = R731,52 + R482,76 1M multiplying = R1 214,28 CA 1CA cost The statement is correct. O 1O conclusion (5) MP 3.3.1 Above ground is a higher security risk O 2O reason L4 OR Safety reasons O OR Below the ground the cost will be less. O OR Above the ground it takes up space. O OR Underground, the water stays cooler/fresher than in direct sun/ lessen evaporation. O OR Aesthetic reasons. O OR O Below the ground for water to easily run into it. OR Less maintenance O (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 13 DBE/November 2017 NSC – Marking Guidelines Ques Solution Explanation Level M 3.3.2 8 000 ℓ = 8 000 000 cm3 L3 = 8 m3 C 1C Conversion Volume of a cylindrical tank = π × radius2 × length 8 m3 = 3,142 × radius2 × 2,9 m SF 1SF substitution 2 8 m3 1A change subject of formula (radius) = A 3,142 × 2,9 m = 0,87798239…S 1S simplification Radius = 0,87798239 ≈ 0,937 m CA 1CA radius Diameter = 1,874 m CA 1CA diameter OR OR Volume of a cylindrical tank = π × radius2 × length 8 000 000 cm3 = 3,142 × radius2 × 290 cm 1SF substitution SF 2 8 000 000 cm3 1A change subject of formula (radius) = A 3,142 × 290 cm = 8 779,8239… S 1S simplification Radius = 8779,8239 ≈ 93,7 cm CA 1CA radius Diameter = 187,4 cm CA 1CA doubling the radius = 1,874 m C 1C conversion to m NPR (6) [40] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 14 DBE/November 2017 NSC – Marking Guidelines QUESTION 4 [33 MARKS] Ques Solution Explanation T&L M 4.1.1 Dineo's maximum wind speed is 95 (MPH) L2 80,4672 1C conversion 95 MPH = × 95 km/h C 50 CA 1CA simplification = 152,887… km/h = 152,89 km/h R 1R rounding OR OR 50 mile = 80,4672 km 1 mile = 1,609344 km 1C conversion 95 MPH = 95 miles / hour × 1,609344 C 1CA simplification = 152,88768 km/h CA 1R rounding ≈ 152,89 km/h R OR OR 95 miles – 50 miles = 45 miles 50 miles = 80,4672 km 45 miles = x km x km = 80,4672 km × 45 miles ÷ 50 miles = 72,4205 km C 1C conversion Total distance = 80,4672 km + 72,4205 km = 152,887 km CA 1CA simplification ∴ 95 MPH = 152,89 km/h R 1R rounding AO (3) M&P 4.1.2 Measured distance between gridlines is 17 mm A 1A distance between L3 (5) Measured distance between P and Q is 39 A gridlines Meas M 1A distance P to Q L3 (3) 205,043 km MCA 1M using scale Actual distance = × 39 mm 1MCA using correct values 17 mm ≈ 470,39 km CA 1CA actual distance Distance = Ave. speed × time 470,39 km S 1S changing the subject of Ave. speed = SF 24 hours the formula ≈ 19,56 km/h CA 1SF substitution 1CA Ave speed (Accept 16 mm to 18 mm for gridlines and 38 mm to 42mm NPR for PQ distance) (8) OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 15 DBE/November 2017 NSC – Marking Guidelines Ques Solution Explanation T&L OR A 2A distance P to Q App. distance from P to Q is 2 13 gridlines 1M multiplying Distance = 2 13 × 205,043 km M A 1A using correct values = 478,4336667 km CA 1CA actual distance Distance = Ave. speed × time SF 1SF substitution 478,4336667 km = Ave. speed × 24 hours 1S changing the subject of Ave. speed ≈ 19,93 km/h CA S the formula 1CA ave. speed (Accept 2 16 up to 2 13 ) OR OR A 18 mm = 205,043 1A distance between 1 mm = 11,39 M gridlines A 1M unit scale Measured distance from the gridline to Q is 3 mm 1A distance to Q (2 to 4)mm Distance from P to Q M = 205,043 + 205,043 + 3 × 11,39 1M using scale = 444,256 km CA 1CA actual distance 444,256 km Ave. speed = SF S 24 hours 1SF substitution ≈ 18,51 km/h CA 1S changing the subject of the formula 1CA Ave speed NPR (8) D 4.2.1 10 RT 2RT correct value L2 (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 16 DBE/November 2017 NSC – Marking Guidelines Ques Solution Explanation T&L 4.2.2 Total number of storms per year according to affected world oceanic regions D 60 L2 50 40 A A A CA 30 Total number of storms 20 A 10 0 2015 2014 2013 2012 2011 2010 Year Indian Western Pacific Eastern Pacific North Atlantic 1A for 1st point 2A for the next 4 points correctly plotted 1A for the last point 1CA joining the points to form a broken line graph (5) D 4.2.3 North Atlantic RT 2RT correct region L2 (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 17 DBE/November 2017 NSC – Marking Guidelines Ques Solution Explanation T&L D (4) 4.2.4 Western Pacific: 1A number of storms WP F(4) Total storms = 39 + 30 + 52 + 34 + 40 = 195 A L4 1RT using amounts from Damages in million USD RT table = 10 200 + 8 410 + 22 800 + 6 080 + 10 600 = 58 090MCA 1MCA adding amounts North Atlantic: 1CA number of storms in Total storms = 12 + 9 + 13 + 19 + 19 = 72 CA NA Damages in million USD RT 1RT only using values to = 590 + 232 + 1510 + 75 000 +21 000 = 98 332 CA 2011 1CA amount of damage NOT valid statement, O 1O not valid O Western Pacific had the most storms but North Atlantic had 2O reason the greatest amount of damages. (9) D 4.3 Growth rate per 1 000 = 38,3 – 11,9 – 1,9 MA 1MA subtracting rates L2 = 24,5 CA 1CA growth rate 24,5 1MCA calculating ∴ percentage growth rate = × 100% MCA percentage (÷1 000 ×100) 1 000 = 2,45% CA 1CA simplification OR OR Percentage growth rate MA 1MA subtracting rates  38,3 11,9 1,9  =  − −  × 100% M 1M calculating percentage  1 000 1 000 1 000  1CA growth rate CA 24,5 = × 100% 1 000 1CA simplification = 2,45% CA AO (4) [33] TOTAL :150 Copyright reserved