Grade 12 NSC Mathemetical Literacy P1 Preperatory 2016 Possible Answers hlayiso.com

Downloaded from hlayiso.com PREPARATORY EXAMINATION 2016 MEMORANDUM MATHEMATICAL LITERACY P1 (10601) Codes Explanation M Method MA Method with Accuracy CA Consistent Accuracy A Accuracy C Conversion D Define J Justification / Reason / Explain S Simplification Reading from a table OR a graph OR a diagram OR a map RT / RD / RG 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 NP No penalty for rounding-off OR omitting units 14 pages

Downloaded from hlayiso.com 10601/16 GAUTENG DEPARTMENT OF EDUCATION PREPARATORY EXAMINATION – 2016 MATHEMATICAL LITERACY (First Paper) MEMORANDUM KEY TO TOPIC SYMBOL: F = Finance; M = Measurement; MP = Maps, Plans and other representations DH = Data Handling; P = Probability QUESTION 1 [47] Ques Solution Explanation Level 1.1.1 Gross monthly income 1M Method 1A Answer L1 = R14 872 + R780 M F Answer only full marks = R15 652 A (2) 1.1.2 Annual income = R15 652 x 12 1MA Calculating annual income = R187 824 MA 1M Subtraction 1MA Multiplication by Taxable income = Annual income ‒ Annual 12 Pension 1A Answer L3 M MA F = R187 824 – (R1 112,03 x 12) = R187 824 – R13 344,30 = R174 479,70 A (4) OR R15 652 – R1 112,03 MA 1MA Deduct / Minus 1A Answer = R14 539,97 A 1MA Multiplication by 12 1A Answer Annual = R14 537,97 x 12 MA = R174 479,70 A (4) 2

Downloaded from hlayiso.com 10601/16 Ques Solution Explanation Level 1.1.3a Medical Tax Credits = (Tax Credit of Main 1MA Addition and Multiplication member + Tax Credits of dependants) x 12 by 12 1A Answer = (R257 + R257) x 12 MA = R6 168 A L2 F OR (R257 x 2) x 12 MA = R6 168 A (2) 1.1.3b Tax payable = Annual tax ‒ Rebate ‒ Medical 1RT Reading correct value credits 1MA Multiply by 2 and 12 RT MA 1S Simplification both = (18% × R174 479,70) ‒ R12 726 ‒ (R257 × 2 × 12) brackets L2 1CA Answer F = R31 406,35 ‒ R12 726 ‒ R6 168 S CA from Q1.1.2 = R12 512,35 CA (3) 1.1.3c Monthly PAYE = R12 512,35 ÷ 12 1CA Answer 1R Rounding = R1 042,695833 L1 CA from Q1.1.3a F ≈ R1 042,70 CA (2) 1.1.4a Total deductions 1MA Addition = R1 042,70 + R1 112,03 + R1 500 + 1A Answer R148,72 MA L1 Answer only ful arks F = 3 803,45 A (2) 1.1.4b Nett monthly income = R15 652 – R3 803,45 1MA Subtraction MA 1A Answer TL1 F = R11 848,55 A Answer only full marks (2) 1.1.5a 1% A 2A Answer TL1 (2) F 1.1.5b Total amount for the month paid over for UIF 1MA Knowing to double the amount = R148,72 x 2 OR R148,72 + R148,72 1A Answer TL1 MA F Answer only full marks = R297,44 A (2) 3

Downloaded from hlayiso.com 10601/16 1.1.6 Annual medical tax credit 1MA Adding R172 and Multiply by 12 = [(R257 x 2) + R172] × 12 MA 1A Answer TL2 = R686 x 12 Answer only full marks F (2) = R8 232 A 1.1.7a Annual amount saved 1A Answer = Monthly amount saved x 12 Answer only full marks L1 = R3 000 x 12 F = R36 000 A (1) 1.1.7b Simple interest 1A Answer = Amount saved x interest rate x term 1CA Answer = R36 000 × = R1 224 A Amount saved = R36 000 + R1 224 = R37 224 CA OR L1 F R36 000 x M 1M Multiply by 103,4 % 1CA Answer = R37 224 CA OR R36 000 + (R36 000 x ) M 1M Addition 3,4% 1CA Answer = R37 224 CA (2) 1.2.1 Amount to be financed 1MA Subtract deposit = Purchase Price ‒ (Deposit + Admin Fee + 1MA Adding correct Registration/Licence Fee) values L2 MA MA 1A Answer F = R104 995 ‒ (R10 500 + R1 010 + R788) (3) = R96 293,00 A 4

Downloaded from hlayiso.com 10601/16 Ques Solution Explanation Level 1.2.2 Amount to be financed 1MA Adding correct = Purchase Price + Admin Fee + amounts Registration/Licence Fee 1A Answer 1MA Addition and = R104 995 + R1 010 + R788 MA multiplication 1CA Answer year 1 = R106 793 A 1CA Answer year 2 Year 1: R106 793 + ( ) MA L2 F = R106 793 + R12 815,16 = R119 608,16 CA Year 2: = R119 608,16 + ( ) = R119 608,16 + R14 352,9792 = R133 961,14 CA (5) 1.2.3 Percentage Deposit 1MA Correct values in MA fraction = M 1M Multiply by 100% 1A Answer L1 1R Rounding F = 10,0004… % A (4) = 10% R 1.3.1 Interest rate: 2O Definition A percentage of a sum of money charged by the bank for lending money to a person. O L1 OR F Any similar definition (2) 1.3.2 Percentage increase 2SF Substitution = 1S Simplification 1CA Answer = SF L2 F S (3) = 12,5% CA 5

Downloaded from hlayiso.com 10601/16 Ques Solution Explanation Level 1.3.3 Value B 1MA Addition = R140,00 + R179,39 + R170,86 + R584,79 + 1A Answer L1 R380,98 MA F Answer only full marks = R1 465,02 A (2) 1.3.4 Vat 2017 = R1 655,55 × MA 1MA Correct value 1MA Multiply by 14% L1 (2) F = R231,78 1.3.5 Additional amount to budget 1M Calculating 12% = R1 887,33 ‒ R1668,86 MA 1A Answer L1 F = R218,47 A Answer only full marks (2) [47] 6

Downloaded from hlayiso.com 10601/16 QUESTION 2 [29] Ques Solution Explanation Level 2.1.1(a) Length = 24 inches x 2,54 C 1C Conversion = 60,96 cm A 1A Answer length 1A Answer breadth Width = 12 inches x 2,54 1A Answer height = 30,48 cm A Depth = 10 inches x 2,54 L2 = 25,4 cm A (4) M 2.1.1(b) Volume = L × B × H 1SF Substitution SF 1CA Answer = 60,96 cm × 30,48 cm × 25,4 cm = 47 194,74432 cm3 L2 = 47 194,74 cm3 CA M (2) 2.1.2a Triangular containers to fit in length 1MA Division with correct MA values = (60,96 cm ÷ 10 cm) × 2 MA 1MA Multiply by 2 L2 1AR Answer M = 12,192 (3) ≈ 12 A 2.1.2b Triangular containers to fit in width 1MA Division and multiply with correct values = (30,48 cm ÷ 10 cm) × 2 MA 1A Answer 1R Rounding L2 = 6,096 M ≈6 A R (2) 2.1.3 Number of cardboard boxes 1MA Division 1AR Answer correctly = rounded Answer only full marks = MA L1 M = 10,4166 … boxes ≈ 10,5 boxes OR 10½ boxes AR (2) 2.2.1 80 : 5 MA 1MA Correct ratio L1 1S Simplification M 16 : 1 S (2) 7

Downloaded from hlayiso.com 10601/16 Ques Solution Explanation Level 2.2.2 P(Sandwich Spread) = × 100% M 1M Multiply by 100% 1MA Correct values in MA fraction 1A Answer L1 = 3,472 …% L ≈ 3,47% A NP for rounding (2) 2.3 The time that it will take Mrs. Madiba to reach 1MA Addition the school. 1A Answer L1 = 10:50 + 45 min MA M = 11:35 A (2) 2.4.1 97th Percentile RG 2RG Read from graph L1 (2) M 2.4.2 150 cm ÷ 100 = 1,5 m C 1C Conversion 1SF Substitution 1M Manipulating formula 1CA Answer 2J Justification 27 kg/m2 = SF L2 M 27 kg/m2 × 2,25 m2 = Mass (in kg) M 60,75 kg = Mass CA Yes J Mrs. Madiba’s opinion is correct. J (6) 2.4.3 Suzy’s new BMI = 21,8 kg/m2 RG 2RG Reading from the L1 graph M (2) [29] 8

Downloaded from hlayiso.com 10601/16 QUESTION 3 [20] Ques Solution Explanation Level 3.1.1 Total distance = 89 km RM 2RM Reading from map L1 (2) MP 3.1.2a Drummond RM 2RM Reading from map L1 MP 11:30 OR Half past 11 RM (2) 3.1.2b Approximately 45 km RM 1RM Reading form map L1 (1) MP 3.1.3 Between Botha’s Hill and Pinetown RM 2RM Reading from map L1 OR MP Botha’s Hill to Winston Park RM (2) 3.1.4 Average Speed = 1SF Substitution 1CA Answer 89 km NP for rounding = SF L1 6,5 h MP = 13,6923 … km/h ≈ 13,69 km/h CA (2) 3.1.5 Just after Drummond is an uphill 5A One mark per Slight uphill on Botha’s hill A description Small downhill down Botha’s hill A L2 Steepest sloping uphill A MP Steepest sloping downhill A Long downhill run A (5) 3.2.1 1 : 500  RM 2RP Reading from plan L1 (2) MP 3.2.2 Every 1 unit on the plan represents 500 units 2RP Reading from plan in actual dimensionsRP No units of L1 measurements necessary MP e.g. cm, mm, m. (2) 3.2.3 South Western Wall = 9,8 cm RP 2RP Reading from plan L1 (2) MP [20] 9

Downloaded from hlayiso.com 10601/16 QUESTION 4 [35] Ques Solution Explanation Level 4.1.1 12 MA 2MA Adding the number of ages L1 (2) D 4.1.2 6 hours 11 minutes 59 sec  RT 2RT Reading from table OR L1 06:11:59 (2) D 4.1.3 Time range = 5h 38m 01s ‒ 3h 38m 10s 1MA Subtracting correct MA values 1A Answer = 1h 59m 51s A L1 Answer only full marks D (2) 4.1.4 Mean age = 1MA Concept of mean 1M Denominator MA 1A Answer = M L2 D = 19,1 ≈ 19 years A (3) 4.2.1 Mode = 292 minutes A 1A Correct Mode L1 (1) D 4.2.2 Mean = 193 + 220 +…+ 372 1A Mean 9 1MA Mean time 1C Conversion 2 478 = 9 = 275,33 minutes A L3 D 275,33 Mean time = MA 60 = 4,5888 … hours (3) = 4 hours 35 minutes C 10

Downloaded from hlayiso.com 10601/16 Ques Solution Explanation Level 4.2.3 228; 237; 250; 251; 281; 292; 292; 306; 329; 338 1MA Ascending order MA 1MA Concept of median 1A Answer Median = 281 + 292 MA L2 2 D = 286,5 minutes A (3) 4.2.4 Range = Highest ‒ Lowest 1MA Concept of range 1A Answer = 372 ‒ 193 MA L1 Answer only full marks D = 179 minutes A (2) 4.3.1 Baised AA 2RG Reading from graph L1 (2) D 4.3.2 (a) Only a definite number of values is possible. The data takes only certain 2J Justification L1 values. E.g. the number of learners in a D class. (One cannot have ‘half a learner’.) (2) (b) Information can be measured continuously. An infinite number of 2J Justification L1 steps. (2) D 11

Downloaded from hlayiso.com 10601/16 4.4.1 L2 D 1A 50 1A 100 6A for the 6 points correctly plotted 1A 100 1A 150 1A 75 1A 175 (6) 4.4.2 Profit = R8,00 x 26  M 1M x 26 = R208,00  AA 2A Answer Answer only full L1 marks D (3) 4.4.3 Tuesday  and Thursday  RT 2RG Reading from table L1 (2) D [35] 12

Downloaded from hlayiso.com 10601/16 Question 5 [19] Ques Solution Explanation Level 5.1.1 1 299 men RT 1RT Reading from table (1) L1 D 5.1.2 2 508 men  RT 1RT Reading from table L1 (1) D 5.1.3 Drinkers : High Blood Pressure 1MA Ratio with correct values 2552 : 688 MA 1AR Answer and rounding L1 3.709 :1 D 4 :1 AR (2) 5.1.4 P(No High Blood Pressure) = 2 508 1MA Numerator L2 3 119 MA 1A Denominator D (2) 5.2.1 P = Current reading ‒ Consumption 1MA Subtraction 1A Answer = 125 334 kℓ ‒ 34,5 kℓ MA L1 Answer only full marks M =125 299,5 A (2) 5.2.2 ≤6kl free 1M First 2 charges 9 × R8,60 = R77,40 1M Multiply by 9,5 10 × R9,60 = R96.00 M 1CA Answer L1 9,5 × R10,60 =R100,70 MA F Total (Q) = R274.10 CA (3) 5.2.3a 1M Calculation of first three ≤4kl free tariffs 3 × R4,67 = R14.01 1MA Dividing amount 8 × R9,94 = R79,52 remaining by R14,60 R93,53 1CA Answer L3 FM Remaining amount = R181,60 ‒ R93,53 = R88,07 M (3) 5.2.3b Remaining kℓ = 1MA Calculate remaining kl used 1CA Answer = 6,03 kℓ MA L3 R = 4kℓ + 3kℓ + 8kℓ + 6,03kℓ FM = 21.03 kℓ CA (2) 13

Downloaded from hlayiso.com 10601/16 Ques Solution Explanation Level 5.2.3c S = 14% of (R181,60+ R274,10) MA 1MA 14% of correct amounts 1S Simplification of S= × R455,70 S bracket 1A Answer L1 M S = R63,798 (3) S = R63,80 A 5.2.4 T = R63,80 + R 455,70 M 1M Addition 1CA Answer L1 T = R519,50 CA M Answer only full marks (2) [19] TOTAL: 150 14

10601 / 16 Vr Oplossing Verduideliking Vlak 5.2.3c S = 14% of (R181,60+ R274,10) MA 1MA 14% van korrekte bedrae 1S Vereenvoudiging van S= × R455,70 S hakies 1A Antwoord L1 S = R63,798 M S = R63,80 A (3) 5.2.4 T = R63,80 + R 455,70 M 1M Optel 1CA Antwoord L1 T = R519,50 CA M Slegs antwoord volpunte (2) [19] TOTAAL: 150 14 Downloaded from hlayiso.com

10601 / 16 VRAAG 5 [19] Vr Oplossing Verduideliking Vlak 5.1.1 1 299 mans RT 1RT Lees vanaf tabel L1 (1) D 5.1.2 2 508 mans  RT 1RT Lees vanaf tabel L1 (1) D 5.1.3 Drinkers : Hoë bloeddruk 1MA Verhouding met korrekte waardes 2 552 : 688 MA 1AR Antwoord en afronding L1 3.709 :1 D 4 :1 AR (2) 5.1.4 P(Nie hoë bloeddruk) = 2 508  1MA Noemer 3 119 1MA Teller L1 MA (2) D 5.2.1 A = Huidige lesing ‒ Verbruik 1MA Aftrek 1A Antwoord = 125 334 kℓ ‒ 34,5 kℓ MA L1 Slegs antwoord volpunte M =125 299,5 A (2) 5.2.2 1M Eerste 2 koste ≤6kl Gratis 1M Vermenigvuldig met 9,5 9 × R8,60 = R77,40 M 1CA Antwoord L2 10 × R9,60 = R96.00 M F 9,5 × R10,60 = R100,70 MA Totaal (Q) = R274.10 CA (3) 5.2.3a ≤4kl Gratis 1M Berekening van eerste drie 3 × R4,67 = R14.01 M kostes 8 × R9,94 = R79,52 1CA Antwoord R93,53 CA 1CA Antwoord L3 FM Oorblywende bedrag = R181,60 ‒ R93,53 = R88,07 CA (3) 5.2.3 b 1MA Bereken oorblywende kl Oorblywende kℓ = gebruik 1CA Antwoord = 6,03 kℓ MA L3 FM R = 4kℓ + 3kℓ + 8kℓ + 6,03kℓ = 21.03 kℓ CA (2) 13 Downloaded from hlayiso.com

10601 / 16 4.4.1 L2 D 1A 50 1A 100 6A vir die 6 punte korrek geplot 1A 100 1A 150 1A 75 1A 175 (6) 4.4.2 Wins = R8,00 x 26 M 1M Vermenigvuldig met 26 = R208,00 AA 2A Antwoord L1 Slegs antwoord volpunte D (3) 4.4.3 Dinsdag RT 2 RT Lees vanaf tabel L1 Donderdag RT (2) D [35] 12 Downloaded from hlayiso.com

10601 / 16 Vr Oplossing Verduideliking Vlak 4.2.3 228; 237; 250; 251; 281; 292; 292; 306; 329; 338 1MA Stygende volgorde MA 1MA Konsep van mediaan 1A Antwoord Mediaan = 281 + 292 MA L2 2 D = 286,5 minute A (3) 4.2.4 Omvang = Hoogste ‒ Laagste 1MA Konsep van omvang 1A Antwoord = 372 ‒ 193 MA L1 Slegs antwoord volp nte D = 179 minute A (2) 4.3.1 Vooroordeeld AA 2A Antwoord L1 (2) D 4.3.2 (a) Slegs ʼn definitiewe aantal waardes is moontlik. Die data neem sekere waardes bv. die aantal student in ʼn klas (jy kan nie ʼn halwe student hê nie) J 2J Regverdiging (2) L1 (b) Inligting kan gemeet word op ʼn D continuum skaal. Oneindigende aantal stappe. J 2J Regverdiging (2) 11 Downloaded from hlayiso.com

10601 / 16 VRAAG 4 [35] Vr Oplossing Verduideliking Vlak 4.1.1 12 MA 2MA Optel aantal ouderdomme L1 (2) D 4.1.2 6 ure 11 minute 59 sek  RT 2RT Lees vanaf tabel OF L1 06:11:59 D (2) 4.1.3 Omvangstyd = 5h 38m 01s ‒ 3h 38m 10s 1MA Aftrek van korrekte MA Waardes 1A Antwoord = 1h 59m 51s A L1 Slegs antwoord volpunte D (2) 4.1.4 Gem. = 1MA Konsep van gemiddeld 1M Teller MA 1A Antwoord = M L2 D = 19,1 ≈ 19 jaar A (3) 4.2.1 Modus = 292 minute A 1A Korrekte Modus L1 (1) D 4.2.2 Gemiddeldd = 193 + 220 +…+ 372 1A Gemiddeld 9 1MA Gemiddelde tyd 1C Herleiding = = 275,33 minute A L3 D Gemiddelde tyd = MA = 4,5888… ure = 4 ure 35 minute C (3) 10 Downloaded from hlayiso.com

10601 / 16 VRAAG 3 [20] Vr Oplossing Verduideliking Vlak 3.1.1 Totale afstand = 89 km RM 2RM Lees vanaf kaart L1 (2) MP 3.1.2a Drummond RM 2RM Lees vanaf kaart L1 11:30 OF Half 12 RM (2) MP 3.1.2b Ongeveer 45 km RM 1RM Lees vanaf kaart L1 (1) MP 3.1.3 Tussen Botha’s Hill en Pinetown RM 2RM Lees vanaf kaart L1 OF MP Botha’s Hill na Winston Park RM (2) 3.1.4 Gemiddelde spoed = 1SF Vervanging 1CA Antwoord 89 km NP vir afronding = SF 6,5 h L1 MP = 13,6923 … km / h ≈ 13,69 km / h CA (2) 3.1.5 Net na Drummond is ʼn opdraande 5A Een punt per beskrywing Klein opdraande op Botha’s hill A Klein afdraande vanaf Botha’s hill A L2 Steilste opdraande helling A MP Steilste afdraande helling A Lang afdraande hardloop A (5) 3.2.1 1 : 500  RM 2RP Lees vanaf plan L1 (2) MP 3.2.2 Elke 1 eenheid op die plan verteenwoordig 1A 1 eenheid 500 eenhede in die werklikheid. RP 1A 500 eenhede Geen meeteenhede mag L1 gegee word, bv. cm, mm, MP m. (2) 3.2.3 Suid-Westelike muur = 9,8 cm RP 2RP Lees vanaf plan Bevestig en meet finale L1 kopie MP (2) [20] 9 Downloaded from hlayiso.com

10601 / 16 Vr Oplossing Verduideliking Vlak 2.2.2 P(Sandwich Spread) = × 100% MA 1M Vermenigvuldig met 100% 1MA Korrekte waardes in = 3,472 …% Breuk L1 1A Antwoord P ≈ 3,47% A NP vir afronding (2) 2.3 Die tyd wat dit Mev. Madiba sal neem om die 1MA Optel skool te bereik: 1A Antwoord L2 = 10:50 + 45 min MA M = 11:35 A (2) 2.4.1 97ste Persentiel RG 2RG Lees vanaf grafiek L1 (2) M 2.4.2 150 cm ÷ 100 = 1,5 m C 1C Herleiding 1SF Vervanging 1M Manipuleer formule 1CA Antwoord 1J Regverdiging 27 kg / m2 = SF L2 M 27 kg / m2 × 2,25 m2 = Massa (in kg) M 60,75 kg = Massa CA Mev. Madiba se opinie is korrek. J (6) 2.4.3 Suzy se nuwe LMI = 21,8 kg / m2 RG 2RG Lees vanaf grafiek Aanvaar 21,7 en 21,9 L1 M (2) [29] 8 Downloaded from hlayiso.com

10601 / 16 VRAAG 2 [29] Vr Oplossing Verduideliking Vlak 2.1.1a Lengte = 24 duim x 2,54 C 1C Herleiding = 60,96 cm A 1A Antwoord lengte 1A Antwoord wydte Wydte = 12 duim x 2,54 1A Antwoord diepte L2 = 30,48 cm A M Diepte = 10 duim x 2,54 = 25,4 cm A (4) 2.1.1 Volume = Lengte x Wydte x Diepte 1SF Vervanging SF 1CA Antwoord = 60,96 cm × 30,48 cm × 25,4 cm L2 = 47 194,74432 cm3 M = 47 194,74 cm3 CA (2) 2.1.2a Driehoekige houers wat in lengte inpas 1MA Deel met korrekte MA waardes = (60,96 cm ÷ 10 cm) × 2 MA 1MA Vermenigvuldig met 2 1AR Antwoord L2 = 12,192 M ≈ 12 A (3) 2.1.2b Driehoekige houers wat in wydte inpas 1MA Deel en vermenigvuldig met korrekte waardes = (30,48 cm ÷ 10 cm) × 2 MA 1A Antwoord 1R Afronding L2 = 6,096 M ≈6 A R (2) 2.1.3 Aantal kartonbokse 1MA Deling 1AR Antwoord korrek = afgerond Slegs antw ord volpunte = MA L1 M = 10,4166 … bokse ≈ 10,5 bokse OF 10½ bokse AR (2) 2.2.1 80 : 5 MA 1MA Korrekte verhouding 1S Vereenvoudiging L1 16 : 1 S M (2) 7 Downloaded from hlayiso.com

10601 / 16 Vr Oplossing Verduideliking Vlak 1.3.3 Waarde B 1MA Optel = R140,00 + R179,39 + R170,86 + R584,79 + 1A Antwoord R380,98 MA L1 Slegs antwoord volpunte F = R1 465,02 A (2) 1.3.4 BTW 2017 = R1 655,55 × MA 1MA Korrekte waarde 1MA Vermenigvuldig met 14% L1 F = R231,78 (2) 1.3.5 Addisionele bedrag by begroting 1M Bereken 12% = R1 887,33 ‒ R1668,86 MA 1A Antwoord L1 F = R218,47 A Slegs antwoord volpunte (2) [47] 6 Downloaded from hlayiso.com

10601 / 16 Vr Oplossing Verduideliking Vlak 1.2.2 Bedrag wat gefinansier moet word 1MA Optel korrekte = Kosprys + Admin Fooi + bedrae Registrasie / Lisensie Fooi 1A Antwoord 1MA Optel en = R104 995 + R1 010 + R788 MA vermenigvuldig 1CA Antwoord jaar 1 = R106 793 A 1CA Antwoord jaar 2 Jaar 1: R106 793 + ( ) MA L3 F = R106 793 + R12 815,16 = R119 608,16 CA Jaar 2: = R119 608,16 + ( ) = R119 608,16 + R14 352,9792 = R133 961,14 CA (5) 1.2.3 Persentasie Deposito 1MA Korrekte waardes in MA breuk = M 1M Vermenigvuldig met 100% L1 1A Antwoord F = 10,0004… % A 1R Afgeronde Antwoord = 10% R (4) 1.3.1 Rentekoers: 2O Definisie ʼn Persentasie van ʼn bedrag geld wat deur die bank gehef word wanneer geld aan ʼn persoon geleen word. L1 O F OF Enige soortgelyke definisie s (2) 1.3.2 Persentasie verhoging 2SF Vervanging 1S Vereenvoudiging = 1CA Antwoord = SF L2 F S = 12,5% CA (3) 5 Downloaded from hlayiso.com

10601 / 16 Vr. Oplossing Verduideliking Vlak 1.1.6 Jaarlikse mediese belastingkrediet 1MA Optel R172 en Vermenigvuldig met 12 = [(R257 x 2) + R172] × 12 MA 1A Antwoord L2 = R686 x 12 Slegs antwoord volpunte F = R8 232 A (2) 1.1.7a Jaarlikse bedrag gespaar 1A Antwoord = Maandelikse bedrag gespaar x 12 = R3 000 x 12 Slegs antwoord vol L1 punte F = R36 000 A (1) 1.1.7b Enkelvoudige rente 1A Antwoord = Bedrag gespaar × Rentekoers x Termyn 1CA Antwoord = R36 000 × = R1 224 A Bedrag gespaar = R36 000 + R1 224 = R37 224 CA OF L1 F R36 000 x M 1M Vermenigvuldig met 103,4 % 1CA Antwoord = R37 224 CA OF R36 000 + (R36 000 x ) M = R37 224 CA 1M Optel 3,4% 1CA Antwoord (2) 1.2.1 Bedrag wat gefinansier moet word 1MA Trek deposito af = Kosprys ‒ (Deposito + Admin Fooi + 1MA Optel korrekte waardes Registrasie / Lisensie Fooi) 1A Antwoord MA MA L2 = R104 995 ‒ (R10 500 + R1 010 + R788) F = R96 293,00 A (3) 4 Downloaded from hlayiso.com

10601 / 16 Vr Oplossing Verduideliking Vlak 1.1.3a Mediese Belastingkrediete = (Belasting 1MA Optel en vermenigvuldig Krediet van Hooflid + Belastingkrediet van met 12 afhanklike) x 12 1A Antwoord = (R257 + R257) x 12 MA L2 = R6 168 A F OF (R257 x 2) x 12 MA = R6 168 A (2) 1.1.3b Belasting betaalbaar = Jaarlikse belasting ‒ 1RT Lees korrekte waarde Korting ‒ Mediese krediet 1 MA Vermenigvuldig met 2 en RT 12 = (18% × R174 479,70) ‒ R12 726 ‒ R6 168 1S Vermeenigvuldig met L2 albei hakies F = R31 406,35 ‒ R12 726 ‒ R6 168 SF 1CA Antwoord = R12 512,35 CA CA van V1.1.2 (3) 1.1.3c Maandelikse PAYE = R12 512,35 ÷ 12 1A Antwoord 1R Afronding = R1 042,695833 A L1 CA van V1.1.3a F ≈ R1 042,70 R (2) 1.1.4a Totale aftrekkings 1MA Optel = R1 042,70 + R1 112,03 + R1 500 + 1A Antwoord R148,72 MA L2 Slegs antwoord volpunte F = 3 803,45 A (2) 1.1.4b Netto maandelikse inkomste = 1MA Aftrek R15 652 – R3 803,45 1A Antwoord MA L1 Slegs antwoord volpunte F = R11 848,55 A (2) 1.1.5a 1% A 2A Antwoord L1 (2) F 1.1.5b Totale bedrag aan WVF vir die maand 1MA Weet dat bedrag verdubbel oorbetaal word 1A Antwoord L1 = R148,72 x 2 OF R148,72 + R148,72 F MA Slegs antwoord volpunte = R297,44 A (2) 3 Downloaded from hlayiso.com

10601 / 16 GAUTENGSE DEPARTEMENT VAN ONDERWYS VOORBEREIDENDE EKSAMEN – 2016 WISKUNDIGE GELETTERDHEID (Eerste Vraestel) MEMORANDUM SLEUTEL TOT ONDERWERP SIMBOOL: F = Finansies; M = Meting; MP = Kaarte, planne en ander voorstellings DH = Datahantering; P = Waarskynlikheid VRAAG 1 [47] Vr Oplossing Verduideliking Vlak 1.1.1 Bruto maandelikse inkomste 1M Metode 1A Antwoord = R14 872 + R780 M L1 Slegs antwoord volpunte F = R15 652 A (2) 1.1.2 Jaarlikse inkomste = R15 652 x 12 1MA Bereken jaarlikse inkomste 1M Aftrek = R187 824 MA 1MA Vermenigvuldig met 12 1A Antwoord Belasbare inkomste = Jaarlikse inkomste ‒ Jaarlikse pensioen L3 M MA F = R187 824 – (R1 112,03 x 12) = R187 824 – R13 344,30 = R174 479,70 A OF R15 652 – R1 112,03 MA 1MA Aftrek 1A Antwoord = R14 539,97 A 1MA Vermenigvuldig met 12 1A Antwoord Jaarliks = R14 537,97 x 12 MA = R174 479,70 A (4) 2 Downloaded from hlayiso.com

VOORBEREIDENDE EKSAMEN 2016 MEMORANDUM WISKUNDIGE GELETTERDHEID (EERSTE VRAESTEL) (10601) Kodes Verduideliking: M Metode MA Metode met Akkuraatheid CA Konstante Akkuraatheid A Akkuraatheid C Herleiding D Definieer J Regverdiging / Rede / Verduideliking S Vereenvoudiging RT / RD / RG Lees vanaf ʼn tabel OF ʼn grafiek OF ʼn diagram OF ʼn kaart OF ʼn plan F Kies die korrekte formule SF Vervang in ʼn formule O Opinie P Straf, bv. geen eenhede, verkeerde afronding ens. R Afronding NP Geen straf vir afronding OF eenhede uitgelaat 14 bladsye Downloaded from hlayiso.com