Mathematical Literacy P1 Nov 2015 Memo Eng hlayiso.com

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICAL LITERACY P1 NOVEMBER 2015 MEMORANDUM MARKS: 150 Codes Explanation M Method MA Method with Accuracy CA Consistent Accuracy A Accuracy C Conversion D Define J Justification/Reason/Explain S Simplification RD Reading from a table OR a graph OR a diagram OR a map 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 OR omitting units This memorandum consists of 17 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 2 DBE/November 2015 NSC – Memorandum KEY TO TOPIC SYMBOL: F = Finance; M = Measurement; MP = Maps, Plans and other representations DH = Data Handling; P = Probability QUESTION 1 [38] Ques Solution Explanation Level L1 1.1.1 67 × 2 + 16 MA 1MA multiply by 2 = 150 CA and adding 16 1CA simplifying Answer only full marks (2) M A L1 1.1.2 Cost = R225,00 × 152 = R34 200 1M multiply by R225 1A for 152 OR OR M Number of persons = R34 200÷ R225 = 152 A 1M divide by R225 (150 guests + bridal couple) 1A number of persons OR OR M A Cost per person = R34 200 ÷ 152 = R225 1M divide by 152 1A cost per person (2) L1 M 1.1.3 R66 450 1M correct fraction % Reception costs = × 100% R125 000 = 53,16% CA 1CA percentage Answer only full marks NP – rounding (2) L1 1.1.4 Flowers and decor = 1,8% × R125 000 M 1M percentage = R2 250 A 1A amount Answer only full marks (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 3 DBE/November 2015 NSC – Memorandum Ques Solution Explanation Level L2 1.1.5 Rand value = GHS 30 000 ÷ 0,32253 M 1M divide ≈ R93 014,60 A 1A correct rounding Shortfall = R125 000 – R93 014,60 M 1M subtraction = R31 985,40 CA 1CA amount OR OR Cedi value = R125 000 × 0,32253 MA 1MA multiply = GHS 40316,25 M Shortfall = GHS 40 316,25 – GHS 30 000 1M subtraction = GHS 10 316,25 A 1A shortfall amount Rand value = GHS 10 316,25 ÷ 0,32253 = R31 985,40 CA 1CA amount Answer only full marks NP – rounding (4) L1 A 1.1.6 14 1A multiply by 14% × R1 349 = R188,86 100 M 1M adding amount Cost including VAT = R1 349 + R188,86 = R1 537,86 A 1A amount with VAT M 1M multiply by 0,32253 Selling price in cedi = R1 537,86 × 0,32253 1CA value to nearest ≈ 496 CA cedi OR OR A 1A working with 14% VAT inclusive cost = R1 349 × 1,14 M 1M multiply by 1,14 = R1 537,86 A 1A amount with VAT 1M multiply by 0,32253 Selling price in cedi = 1 537,86 × 0,32253 M 1CA value to nearest ≈ 496 CA cedi OR OR 1M multiply by 0,32253 Price in cedi = 1 349 × 0,32253 M 1A cedi price = 435,09 A Selling price including VAT in cedi 1A working with 14% = 435,09329 × 1,14 A M 1M multiply by 1,14 ≈ 496 CA 1CA value to nearest cedi Answer only full marks (5) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 4 DBE/November 2015 NSC – Memorandum Ques Solution Explanation Level 1.1.7 A J L1 • Photographer (video) to create memories of the 1A wedding expense L2 wedding day 1J explanation • Wedding attire – usually special wedding attire are required • Wedding contract to pay for the lawyer’s fees for drawing up the contract • Gifts as a token for members who serve • DJ to provide for the music at the reception (accept any valid wedding expense with an explanation ) (2) L1 1.2.1 Employee works and receives money for the work 1D employee done D Employer is a person or institution that hires workers and pays wages/salary for work done D 1D employer (2) L1 1.2.2 Unemployment Insurance Fund D 2D expanding (2) L1 1.2.3 R15 521 A 2A amount (2) A L1 1.2.4 No 1A correct statement No amount allocated E 1E reason (2) L1 1.2.5 Monthly tax credit = R2 760 ÷ 12 MA 1MA divide correct = R230 CA value by 12 1CA monthly tax credit Answer only full marks (2) L1 1.2.6 A = R13 909 + R20 013 + R8 640 M 1M correct values = R42 562 CA 1CA total deductions Answer only full marks (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 5 DBE/November 2015 NSC – Memorandum Ques Solution Explanation Level L1 1.2.7 Gross non-retirement funding income = R15 521 + R26 188 + R8 640 MA 1M using the correct = R50 349 values/codes/words 1A addition OR Adding the amounts with source codes 3605, 3713 and 3810 OR Adding the annual payment other allowances and medical aid contributions (2) L2 1.2.8 Remaining monthly contributions A 1A R13 909 = R13 909 – R4 975,25 1CA subtracting = R8 933,75 CA R4 975,25 M 1M dividing the Average monthly contribution = R8 933,75 ÷ 7 A remaining amount = R1 276,25 CA 1A by 7 1CA pension per month (only if division by 4,5,6,7) Answer only full marks (5) [38] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 6 DBE/November 2015 NSC – Memorandum QUESTION 2 [31] Ques Solution Explanation Level L3 SF 2.1.1 Total area of a rectangular piece = 30 cm × 12 cm 1SF substitution = 360 cm²  A 1A simplifying M Off-cut piece = 360 cm² – 355,25 cm² 1M subtraction = 4,75 cm²CA 1CA area of off-cut Total off-cut piece for both sides = 4,75 cm² × 2 M 1M multiply by 2 = 9,5 cm² CA 1CA area of off-cut OR OR M SF Total area of 2 rectangular pieces = 2 × 30 cm × 12 cm 1SF substitution = 720 cm²  A 1M multiply by 2 1A simplifying Area of both sides of stocking = 355,25 cm² × 2 M 1M multiply by 2 = 710,5 cm² M Total off-cut piece = 720 cm² – 710,5 cm² 1M subtraction = 9,5 cm² CA 1CA area of off-cut OR OR Total off-cut area M SF M = (2 × 30 cm × 12 cm) – (355,25 cm² × 2) 1SF substitution  A M 1M multiply by 2 = 720 cm² – 710,5 cm² = 9,5 cm²CA 1M multiply by 2 1A simplifying 1M subtraction 1CA area of off-cut Answer only full marks (6) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 7 DBE/November 2015 NSC – Memorandum Ques Solution Explanation Level L2 2.1.2 SF 1  1 SF substitution Area of a triangle =  × 3 cm × 5 cm  2  = 7,5 cm² A 1A simplifying Area of 6 triangles = 7,5 cm² × 6 M 1M multiply by 6 = 45 cm² CA 1CA total area OR OR SF 1 SF substitution 1  Area of triangles =  × 3 cm × 5 cm  × 6 M 1M multiply by 6 2  A 1A simplifying = 7,5 cm²× 6 1CA total area = 45 cm² CA Answer only full marks NP -units (4) L2 2.1.3 Time taken = 9 × 18 minutes = 162 minutes MA 1MA time in minutes = 2 h 42 min OR 2,7 h C 1C converting time Finishing time = 08:25 + 2h42M 1M adding = 11:07 CA 1CA finishing time correct notation Answer only full marks Two marks for 11: 𝒙𝒙 (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 8 DBE/November 2015 NSC – Memorandum Ques Solution Explanation Level M A L2 2.2 Number of reels along length = 195 mm ÷ 23mm 1M dividing length by = 8,4782… diameter ≈ 8 R 1A diameter 1R number rounded Number of reels along breadth = 120 mm ÷ 23mm down = 5,2173… ≈ 5 R 1R number rounded down Total = 5 × 8 = 40 CA 1CA total number Full marks for Total = 5 × 8 = 40 Max of 2 marks if divided by circle’s area Max of 3 marks if divided by square area 1 mark for area of rectangle only (5) L2 2.3.1 Painted surface area of the lid A SF 1A radius = 3,142 × 3,6 cm (3,6 + 2 × 0,9) cm C 1SF substitution ≈ 61 cm² CA 1C conversion 1CA surface area to nearest cm2 OR OR Painted surface area of the lid A SF 1A radius = 3,142 × 36 mm (36 + 2 × 9) mm 1SF substitution = 6108,05 mm² CA 1CA surface area to ≈ 61 cm² C nearest cm2 1C conversion Max of 3 marks if inner radius used Max of 2 marks if units are mixed (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 9 DBE/November 2015 NSC – Memorandum Ques Solution Explanation Level L2 2.3.2 Capacity = 75% 250 mℓ M 1M multiply by 75% = 187,5 mℓ CA 1CA capacity in mℓ Volume = 187,5 cm³ Height of the water in the jar Volume of the water (in cm 3 ) = π × radius 2 187,5 cm 3 = SF 3,142 × (3,25 cm) 2 2SF substitution 187,5 cm 3 2 = 33,187375 cm = 5,6497… cm CA 1CA simplification ≈ 6 cm R 1R nearest cm OR OR Volume of the water (in cm 3 ) = π × radius 2 250 cm 3 2SF substitution = SF 3,142 × (3,25 cm) 2 250 cm 3 2 = 33,187375 cm = 7,532… cm CA 1CA simplification Height of the water in the jar = 75% 7,532...cm M 1M multiply by 75% = 5,6497… cm CA 1CA height of water ≈ 6 cm R 1R nearest cm Answer only full marks (6) M 1M multiply by 2 L1 2.3.3 1 2 1 2× = = A 1A fraction 16 16 8 2 Accept 16 Answer only full marks (2) [31] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 10 DBE/November 2015 NSC – Memorandum QUESTION 3 [24] Ques Solution Explanation Level L1 3.1.1 Exit 3 RD 2RD reading from plan (2) A J L1 3.1.2 No, there is no power outlet available in that seat 1A answer 1J reason (2) RD L2 3.1.3 C 109 RD 1RD correct row 1RD correct seat number (2) L1 3.1.4 Total seats = seats one side + seats in middle + seats other side = (3+2×6+ 3×7 + 6×8 +5)+(8 +13 + 11×14 + 6) + (3 + 5 + 6 + 3×7 + 5×8) 3MA adding correct MA MA MA number of seats in each = 89 + 181 + 75 section = 345 CA 1CA total seats Answer only full marks Max 2 marks if answer only 344 or 346 (4) L1 3.1.5 104 and 110 RD 2RD seat numbers (2) A L2 3.1.6 Number of seats with access to a power supply = 52 1A counting seat 1CA numerator 52 CA 1CA writing as a Probability = denominator from 3.1.4 345 CA 27 9 OR 345 115 54 18 OR OR 345 115 Max 2 Answer only full marks (3) L1 3.2.1 14 times RD 2RD reading from map [Free State 15 times] If 13 one mark (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 11 DBE/November 2015 NSC – Memorandum Ques Solution Explanation Level L1 3.2.2 Distance = 94,7 km – 76 km MA 1MA subtracting from = 18,7 km A 94,7 1A distance Answer only full marks (2) L1 3.2.3 Blue Hills RD 2RD reading from map (2) RD RD L1 3.2.4 WP 4, WP 5, WP 6 RD 3RD reading from map OR OR WP3 to WP4 , WP 4 to WP5 , WP5 to WP6 RD 3RD reading from map 2 marks for W4 to W6 (3) [24] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 12 DBE/November 2015 NSC – Memorandum QUESTION 4 [30] Ques Solution Explanation Level J L1 4.1.1 The data for the global regions is qualitative. 2J explanation OR OR The global regions cannot be expressed as numerical data J 2J explanation (2) L1 4.1.2 5% RT and 8% RT 3RT Correct modal % Two marks for first correct answer, one mark for second correct answer (3) L2 4.1.3 7+8 2M for adding correct Median = % M values and dividing by 2 2 = 7,5% CA 1CA answer Answer only full marks (3) RT L1 4.1.4 Total usage = 3% + 8% + 11% = 22% CA 1RT correct values 1CA total Answer only full marks (2) M L1 4.1.5 2% + 9% + 23% + 22% = 56% CA 2M Adding all correct values. Note: 1CA total Candidates that add the 4% of the Middle East is also correct. Answer only full marks Answer only 60% full marks (3) L1 4.1.6 16% RG 2RG correct value (a) (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 13 DBE/November 2015 NSC – Memorandum Ques Solution Explanation Level 4.1.6 WORLD POPULATION AND MEANS OF COMMUNICATION (b) PERCENTAGES PER GLOBAL REGION 30 Percentage world population Percentage Internet communication Percentage cell phone communication A 20 PERCENTAGES A 10 A A A A 0 A B C D E F G H I J K L GLOBAL REGIONS 1A mark for every TWO points plotted correctly L2 (1 × 6) (6 ) (Penalty of one mark if points are not joined) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 14 DBE/November 2015 NSC – Memorandum Ques Solution Explanation Level L1 4.1.7 South Asia OR I RD 2RD reading from graph or table (2) MA L1 4.2.1 Rural Number = 7 095 476 818 × 48%A 1MA multiplying with % 1A 48 % = 3 405 828 873 A 1A persons OR OR MA Urban number = 7 095 476 818 × 52% 1MA multiplying with % = 3 689 647 945 A 1A urban number Rural = 7 095 476 818 – 3 689 647 945 = 3 405 828 873 A 1A persons Answer only full marks (3) L1 4.2.2 Social networking users 1 856 680 860 SF 1SF dividing the correct = × 100% value by 7 095 476 818 7 095 476 818 1CA answer in % = 26,167…% CA Answer only full marks NP - rounding (2) L1 4.2.3 6 572 950 124 A 2A for correct digits (2) [30] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 15 DBE/November 2015 NSC – Memorandum QUESTION 5[27] Ques Solution Explanation Level MA F 5.1.1 M = 2 925 + 1 970 + 1 963 + 1 568 + 1 700 1MA adding all values L1 + 1 817 + 1 342 + 2 118 = 15 403 CA 1CA value of M Answer only full marks Full marks for 15 404 Penalty of one if given as 1 000’s (2) F 5.1.2 Value for both N M 1M subtracting from total L2 = 12 898 – ( 2 394 + 1 302 + 1 405 + 1 490 + 1 311 + R1 756) 1CA cost for both = 3 240 CA 1M dividing by 2 M 1CA amount R 3 240 Each received = = R1 620 CA 2 OR OR Sibiya:A M 1A for R1 970 M N = R1 970 – R349 – R1 = R1 620 CA 1M for subtracting R349 1M for subtracting R1 1CA total Sibiya OR Magome:A M M OR N = R1 963 – R342 – R1 = R1 620 CA 1A for R1 963 1M for subtracting R342 1M for subtracting R1 1CA total Magome Answer only full marks Penalty of one if given as 1 000’s (4) M CA D 5.1.3 Range = R2 925 000 – R 1 342 000 = R1 583 000 1M concept of range L2 1CA range Answer only full marks Penalty of one if not given as 1 000’s (2) F 5.1.4 Songelwa : Magome = 30 : 342 L1 = 5 :57 A 1A correct values = 1 : 11,4 CA 1CA form NP - rounding (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 16 DBE/November 2015 NSC – Memorandum Ques Solution Explanation Level F 5.1.5 Sibiya: L2 Increase = R1 970 000 – R1 872 000 M = R98 000 Phillips: Increase = R1 700 000 – R1 625 000 2M subtracting any two = R75 000 M of Sibiya, Phillips, Mabilane Mabilane: Increase = R2 118 000 – R2 032 000 = R86 000 M Magome: Increase = R1 963 000 – R1 861 000 = R102 000 A 1A amount for Magome Magome received the greatest increase CA 2CA correct person Full marks if only Magome was calculated correctly with conclusion (5) D 5.1.6 Mabunda MD A 2A the correct person L1 Penalty one mark if an extra name is added (2) P 5.2.1 100% A 2A correct % L1 Accept 100 (2) P 5.2.2 14 A 1A numerator L2 P= 18 A 7 1A denominator = CA 1CA simplification 9 OR OR M 1M subtracting from 1 4 7 P=1– = 1A denominator 18 A 9 CA 1CA simplification Answer only full marks (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 17 DBE/November 2015 NSC – Memorandum Ques Solution Explanation Level A D 5.3 Growth 1 st year = 4 705 306 × 5% L3 ≈ 235 265 M 1A calculating 5% Total after the 1st year = 4 705 306 + 235 265 1M adding = 4 940 571 CA 1CA first year total Growth 2nd year = 4 940 571 × 5,9% = 291 493 OR 291 494 CA 1CA calculating 5,9% of total Total after 2nd year = 4 940 571 + 291 493 = 5 232 064 OR 5 232 065 CA 1CA 2nd year total OR OR 100% + 5% = 105% A 1A increasing with 5% Total after 1st year = 4 705 306 × 105%M 1M percentage = 4 940 571,3 CA calculation 1CA first year total 100% + 5,9% = 105,9% Total after 2nd year = 4 940 571,3 × 105,9% CA 1CA increasing with = 5 232 065,007 5,9% ≈ 5 232 065 CA 1CA 2nd year total, rounded OR OR Total after 2nd year M A M A 1M percentage = 4 705 306 × 105% × 105,9% calculation = 5 232 065,007 1A increasing by 105% ≈ 5 232 065 CA 1M percentage calculation 1A increasing by 105,9% 1CA 2nd year total, rounded Answer only full marks (5) [27] TOTAL: 150 Copyright reserved