Mathematical Literacy P1 June July 2015 Memo Eng hlayiso.com

Downloaded from hlayiso.com SENIOR CERTIFICATE EXAMINATION MATHEMATICAL LITERACY P1 2015 MEMORANDUM MARKS: 150 Symbol Explanation M Method MA Method with Accuracy CA Consistent Accuracy A Accuracy C Conversion D Define E Explain S Simplification RT/RG/RD Reading from a table or a graph or a diagram 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/Reason J Justification This memorandum consists of 15 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 2 DBE/2015 SCE – Memorandum KEY TO TOPIC SYMBOL: F = Finance; M = Measurement; MP = Maps, plans and other representations DH = Data handling; P = Probability QUESTION 1 [36] Ques Solution Explanation Topic F 1.1.1 R 360 ÷ R1,0746 M 1M identify R1,0746 L2 (a) = 335,008 kWh A 1A number of units MA 1MA adding New reading = 10,3 kWh + 335,0 kWh = 345,3 kWh OR MA OR Number of units purchased = 345,3 kWh – 10,3 kWh 1MA difference in = 335 kWh units Cost = 335 kWh × R1,0746 M 1M identify tariff = R359,99 ≈ R360 A 1A amount purchased (3) F 1.1.1 14 1MA multiply L2 (b) VAT amount = R 360 × MA 114 = R44,210526 A 1A VAT = R44,21  R 1R rounding OR OR R360 = 114% 𝑥 𝑥 is amount without VAT M 1M proportion method x = R360 × 100 ÷ 114 = R315,79 (excl. VAT) A 1A amount excl. VAT VAT amount = R360 – R315,79 = R44,21  A 1A amount of VAT Answer only Full marks (3)  MA F 1.1.2 Units used = 345,3 kWh – 250,7 kWh 1MA subtracting L1 = 94,6 kWh CA 1CA simplification Answer only Full marks (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 3 DBE/2015 SCE – Memorandum Ques Solution Explanation Topic MA F 1.1.3 Cost for first 500 units = 500 × R1,0746 1MA units × correct amount L3 = R537,30 CA per unit 1CA cost MA CA 1MA multiply with unit cost Next 60 units = 60 × R1,2208 = R73,25 of 60 units 1CA cost for 60 units M 1M adding Total cost = R537,30 + R73,25 1CA amount = R610,55 CA No penalty for rounding final answer (6) M F 1.1.4 Increase = R1,4809 × 13,5% 1M multiply by % L1 = R 0,1999215 M New tariff = R1,4809 + R 0,1999215 1M Adding increase = R1,6808215 CA 1CA new cost ≈ R1,6808 per unit OR OR M M New tariff = R1,4809 + 13,5% × R1,4809 1M 13,5% of R1,4809 = R1,6808215 CA 1M adding ≈ R1,6808 1CA new cost OR OR M New percentage = 100% + 13,5% = 113,5% 1M 113,5% M New tariff = R1,4809 × 113,5% 1M multiply by % = R1,6808215 CA 1CA answer with no ≈ R1,6808 rounding of values Answer only Full marks (3) F 1.2.1 8 A 2A number of instalments L1 (a) Accept 9 for full marks (2) F 1.2.1 R0,00 OR No amount OR None OR Nil A 2A arrear amount L1 (b) (2) RD F 1.2.1 R1321,21 M 1RD identify the correct L1 (c) Monthly interest = × 100% values R 4249,78 = 31,09 % A 1M calculate % 1A monthly interest (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 4 DBE/2015 SCE – Memorandum Ques Solution Explanation Topic MA A F 1.2.1 Total amount = R4 249,78 × 60 + R115 491,44 1MA multiply by 60 L1 (2) (d) CA 1A adding balloon L2 (2) = R254 986,80 + R115 491,44 payment = R370 478,24 CA 1CA simplification 1CA total amount (4) RT MA 1RT reading values F 1.2.2 R140 446,50 + R4 249,78 1MA adding correct L1 (a) = R144 696,28 A values 1A opening balance Answer only Full marks (3) F 1.2.2 Service Fee RT 2RT reading table L1 (b) Accept R57,00 full marks (2) E F 1.2.2 After the debit order was subtracted from the opening 1E subtracting debit L1 (c) balance, add the interest and E order add the service fee E 1E adding interest 1E service fee OR OR E E Add the service fee and interest to the opening balance 1E add service fee and then subtract the debit order. E 1E add interest 1E subtract debit order E OR E OR Add all the debits, subtract the credits from the 1E add all debits opening balance E 1E subtract credits 1E opening balance (3) [36] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 5 DBE/2015 SCE – Memorandum QUESTION 2 [29] Ques Solution Explanation Topic M 2.1.1 Capacity is the quantity that an empty container can 2E explanation L1 hold. E (2) MA M 2.1.2 W = 97 mm – 29 mm 1MA subtracting L1 = 68 mm CA 1CA value of W 1 Mark for answer 6,8 cm (2) A M 2.1.3 Volume = 75 mm × 68 mm × 210 mm SF 1SF substituting from L2 CA Q2.1.2 = 1 071 000 mm 3 1A for values 75 and 210 1CA volume 1 Penalty for wrong unit or mixing units (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 6 DBE/2015 SCE – Memorandum Ques Solution Explanation Topic M 2.1.4 Area (one face) = 75 mm × 210 mm L2 (3) = 15 750 mm 2 A 1A area of 1 side L3 (2) Area (one side) = 68 mm × 210 mm = 14 280 mm 2 CA 1CA area of side Area (top) = 75 mm × 68 mm = 5100 mm2 CA 1CA area of top Total surface area M 1M adding all areas = 2 × (15 750 mm2 + 14 280 mm2) + 5 100 mm2 = 65 160 mm2 = 651,6 cm2 CA 1CA area with unit OR OR Area = 7,5 cm × 21 cm = 157,5 cm 2 A 1A area of 1 side Area = 6,8 cm × 21 cm = 142,8 cm 2 CA 1CA area of side Area (top) = 7,5 × 6,8 = 51cm2 CA 1CA area of top Total surface area M 1M adding all areas = 2 × 157,5 cm + 2 × 142,8 cm 2 + 51cm2 2 1CA area with unit = 651,6 cm 2 CA OR OR M 1M adding all areas Lateral surface area = 2 × (7,5 cm + 6,8 cm) × 21 cm = 2 × 14,3 cm × 21 cm = 600,6 cm² CA 1CA area Top area = 7,5 cm × 6,8 cm = 51 cm² CA 1CA area of top M 1M total area Total surface area = 600,6 cm² + 51 cm² = 651,6 cm² CA 1CA area with unit (5) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 7 DBE/2015 SCE – Memorandum Ques Solution Explanation Topic M 2.2.1 1 000 m 1M divide L1 Number of cups = M 4 × 2 × 125 m = 1cup A 1A number of cups OR OR Volume = 125 mℓ × 2 = 250 mℓ A 1A total volume 1000 mℓ = 4 cups 250 m ∴ 250 mℓ = ×4 1000 m 1A number of cups = 1 cup A Answer only Full marks (2) MA M 2.2.2 Total volume = 2 × (125 mℓ + 720 mℓ) 1MA adding correct L1 = 1 690 mℓ CA values and multiply by 2 1CA total volume Answer only Full marks (2) M M 2.2.3 2 × 150 ÷ 1 000 kg 1M multiply by 2 L1 = 0,3 kg A 1A mass in kg Answer only Full marks (2) M 2.2.4 Elapsed time = 12:30 – 11:20 L2 = 1 hour 10 min A 1A elapsed time Time indicated on the recipe = 30 min + 15 min + 10 min 1MA adding time = 55 min MA indicated on recipe Extra time taken = 1 hour 10 min – 55 min = 15 min CA 1CA difference (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 8 DBE/2015 SCE – Memorandum Ques Solution Explanation Topic SF M o 2.2.5 Temperature (in F) = 180 × 1,8 + 32 1SF substitute L1 = 356 A 1A temp in ºF (2) M M 2.3.1 5 whole blocks + 5 half blocks + 6 quarter blocks 1M counting blocks L2 = 9 m² A 2A area Answer only Also accept any answer from 8 m² to 10 m² 2 marks (3) M 2.3.2 76 cm ÷ 100 = 0,76 m C 1C convert to m L2 Volume = 8 m² × 0,76 m SF 1SF substituting = 6,08 m 3 CA 1CA volume (3) [29] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 9 DBE/2015 SCE – Memorandum QUESTION 3 [22] Ques Solution Explanation Topic MP 3.1 Top view OR aerial view OR from above A 2A specific view L1 OR satellite view OR 2D top view OR (2) topographical view MP 3.2 G8, G9, G10 A 3A correct gates L1 (3) MP 3.3 Milner road A 2A road name L1 (2) MP 3.4 Left hand side OR south side A 2A correct side L1 (2) A MP 3.5 Zones 1, 3 and 4 or 5 A 1A for one of the zones L1 1A for any correct second zone (2) MP 3.6 AD OR DA A 2A name of assembly L2 point (2) MP 3.7.1 4 OR G4, G5, G6 & G7 OR 4-7 A 2A no. of entrances L1 (2) MP 3.7.2 South East A 2A direction L1 Accept East of South (2) P 3.8 4 A 1A numerator L1(1) = 0,57142… 1A denominator L2(2) 7 A 1R rounded percentage ≈ 57,1% R (3) E MP 3.9 To treat injured players or spectators 2E explanation L1 OR OR E Any other suitable explanation relating injury or 2E explanation medical related (2) [22] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 10 DBE/2015 SCE – Memorandum QUESTION 4 [32] Ques Solution Explanation Topic A A DH 4.1 Newlands, Cape Town 1A stadium L1 1A city (2) DH 4.2 New Zealand A 2A country L1 (2) DH 4.3 5 A 2A number L1 (2) P 4.4 3 A 1A numerator L2 12 A 1A denominator 1CA simplified fraction 1 CA = OR 0,25 4 Answer only Full marks (3) DH 4.5 Argentina RT 2RT reading table L1 (2) RT DH 4.6.1 33; 28; 27; 23; 13; 10 A 1RT reading table L1 1A descending order (2) DH 4.6.2 33 + 28 + 27 + 23 + 13 + 10 M 1M adding points and L2 (a) Mean = divide by 6 6 134 = S 1S simplify 6 = 22,333... CA 1CA mean from Q4.6.1 ≈ 22 points (3) DH 4.6.2 27 + 23 M L2 (b) Median = 1M median concept 2 = 25 CA 1CA median from Q4.6.1 Answer only Full marks (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 11 DBE/2015 SCE – Memorandum Ques Solution Explanation Topic DH 4.6.2 Range = 33 – 10 M 1M concept of range L2 (c) = 23 CA 1CA range from Q4.6.1 Answer only Full marks (2) DH 4.6.2 No mode CA 2CA mode from Q4.6.1 L2 (d) (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 12 DBE/2015 SCE – Memorandum Ques Solution Explanation Topic DH 4.7 L2 Points scored by and against the South African team during the 2014 Rugby Championship 35 30 A A 25 A Number of points scored 20 A 15 A 10 A 5 0 16-Aug 23-Aug 6-Sept 13-Sept 27-Sept 4-Oct Date of game 6A for each of the points plotted correctly and accurately 1 Penalty for not joining points. (6) (Accept 1 mark for every 2 bars in case of bar graph – max 3/6) DH 4.8.1 27 September A 2A date L1 (2) DH 4.8.2 13 days A 2A days L1 Accept 14 days (2) [32] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 13 DBE/2015 SCE – Memorandum QUESTION 5 [31] Ques Solution Explanation Topic DH 5.1.1 1 January 2014 RT 2RT L1 (2) DH 5.1.2 April RT 2RT L1 (2) MA RT F 5.1.3 Difference = (1 411 – 1 391) cent per litre 1MA subtract L1 CA 1RT reading from table = 20 cent per litre OR R0,20 c/l 1CA difference No penalty for unit omitted (3) DH 5.1.4 August A 2A August L1 Accept January (2) A A DH 5.1.5 August and September 1A August L1 1A September (2) RT F 5.1.6 1383 − 1361 SF 1RT reading from table L2 Percentage change = × 100% 1SF substitution 1383 = 1,5907 % CA ≈ 1,59 % R 1CA simplify 1R rounding OR RT OR SF 1RT reading from table 1377 − 1355 Percentage change = × 100% 1SF substitution 1377 = 1,5977 % CA 1CA simplify ≈ 1,60 % R 1R rounding OR RT OR 1401 − 1379 SF 1RT reading from table Percentage change = × 100% 1401 1SF substitution = 1,5703 % CA ≈ 1,57 % R 1CA simplify 1R rounding (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 14 DBE/2015 SCE – Memorandum Ques Solution Explanation Topic P 5.2.1 Monday OR 28/09/2015 RD 2RD reading diagram L1 (2) RD P 5.2.2 Barberton on 27/09/2015RD 1RD name of town L1 1RD date (2) DH 5.2.3 13ºC RD 2RD reading diagram L1 Accept 18ºC/13ºC (2) F 5.3.1 Cost price = R153,60 ÷ 24MA 1MA dividing correct L1 = R6,40 A values 1A cost price Answer only Full marks (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 15 DBE/2015 SCE – Memorandum Ques Solution Explanation Topic F 5.3.2 Profit per can = R9,00 – R6,40 L1 = R2,60 CA 1CA profit per can M 1M multiply by 96 Profit for 96 cans = R2,60 × 96 = R249,60 CA 1CA total profit OR OR M CA 1M multiply by 96 Profit for 96 cans = (96 × R9,00) – (96 × R6,40) 1CA cost price of 96 = R864 – R614,40 cans = R249,60 CA 1CA total profit OR OR M 1M multiply by 96 Profit for 96 cans = 96 (R9,00 – R6,40) 1CA profit per can = 96 (R2,60) CA = R249,60 CA 1CA total profit OR OR M 1M multiply by 24 Income for 1 case = R9,00 × 24 = R216 Profit on 1 case = R216 – R153,60 CA 1CA profit on 1 case = R62,40 Profit for 96 cans = R62,40 × 4 = R249,60 CA 1CA total profit (3) RG F 5.3.3 Selling price = R400 ÷ 40 M 1RG reading graph L1 (a) = R10 per can A 1M division 1A selling price OR OR RG 1RG reading graph Selling price = R200 ÷ 20 M 1M division = R10 per can A 1A selling price OR OR RG 1RG reading graph Selling price = R600 ÷ 60 M 1M division = R10 per can A 1A selling price Answer only Full marks (3) F 5.3.3 60 cans RG 2RG reading graph L1 (b) (2) [31] Copyright reserved