Mathematical Literacy P1 May June 2016 Memo Eng hlayiso.com

Downloaded from hlayiso.com SENIOR CERTIFICATE EXAMINATIONS MATHEMATICAL LITERACY P1 2016 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 OR a graph SF Correct substitution in a formula J Reason/Explain/Decision P Penalty, e.g. for no units, incorrect rounding off etc. R Rounding off NPR No penalty for rounding This memorandum consists of 13 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 2 DBE/2016 SCE – Memorandum QUESTION 1 [44] Tolerance range 2 marks Ques Solution Explanation T&L J J L1 1.1.1 It is the outstanding (still owing) balance of the previous 2 J explanation month's account. OR J J Opening balance for new month. (2) L1 1.1.2 Aug : 19 days A 2A correct number of Sep : 9 days A days per month CA Therefore total number of days lapsed = 19 + 9 = 28 1CA Total Answer only For 28 : 3 marks For 27 or 29 : 1 mark (3) L1 1.1.3 Total Basic Levy = R2 105,89 + R2 158,50 MA 1MA adding levies = R4 264,39 A 1A total amount Answer only full marks (2) L1 1.1.4 New reading – previous reading 1A identify the reading (a) = 1 190 786 kWh – 1 158 957 kWh A M 1M subtracting correct = 31 829 kWh order (if dividing max 1 ) (2) A M L1 1.1.4 31 829 × 0,6303 1A identifying the (b) = R20 061,8187 values ≈ R20 061,82 1M multiply by 0,6303 (2) MA L2 1.1.5 R2 105,89 + R2 158,50 + R20 061,82 + R24 781,93 1MA adding all = R49 108,14 A amounts 1A total before VAT 14 M 1M calculating 14% VAT = × R49 108,14 VAT 100 ≈ R6 875,14 OR OR 1M calculating 14% R6 875,14 VAT Amount = M 1A amount before VAT 14% = R49 108,14 A 1MA adding all = R2 105,89 + R2 158,50 + R20 061,82 amounts + R24 781,93 (3) A Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 3 DBE/2016 SCE – Memorandum Ques Solution Explanation T&L L1 1.1.6 24 781,93 MA 1MA dividing B = 137 = R180,89 A 1A tariff Answer only full marks (2) L2 1.1.7 C M 1M adding and = 2 105,89 + R2 158,50 + R20 061,82 + R24 781,93 subtracting 0,03 + 6 875,14 – 0,03 = 55 983,25 CA 1CA account total OR OR M C = 49 108,14 + 6 875,14 – 0,03 1M adding and = 55 983,25 CA subtracting 0,03 CA value from1.1.5 Answer only full marks (2) L1 1.1.8 To round down the amount due to the non-availability of 1c and 2c coins. J 2J explanation OR Rounding down to 5c (2) CA from Q1.1.7 L3 Monthly interest rate = 10% ÷ 12 M 1.1.9 1M divide by 12 1 Interest after 1 month = × R55 983,25 120 ≈ R466,527 A 1A 1st month's interest Amount payable after 1 month (November 15) = R55 983,25 + R466,527 M 1M adding interest ≈ R56 449,777 CA 1CA value after 1 month 1 Interest after 2 months = × R56 449,77 120 ≈ R470,415 Amount payable after 2 months (Dec 15) = R56 449,777 + R470,415 1CA value after 2 ≈ R56 920,19 CA months OR OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 4 DBE/2016 SCE – Memorandum Ques Solution Explanation T&L M CA from Q1.1.7 1.1.9 Monthly interest rate = 10% ÷ 12 1M divide by 12 Amount payable after 1 month (November 15) 1A monthly interest  1 A  1M calculating = × R55 983,25  + R55 983,25 M interest and adding  120  1CA value after 1 ≈ R56 449,777 CA month Amount payable after 2 months (by 15 Dec) 1CA value after 2  1  months = × R56 449,777  + R56 449,78  120  (Max 3 marks if interest rate is not ≈ R56 920, 19 CA monthly) (5) M L1 1.1.10 New three-phase commercial levy = R2 105,89 + R50,00 1M adding R50 to a (a) = R2 155,89 A levy 1A simplification Answer only full marks (2) L2 MA A 1.1.10  12,2  1MA calculating (b) New tariff per kWh =  × R0,6303  + R0,6303 percentage of tariff  100  1A adding 0,6303 = 0,0768966 + R0,6303 1CA tariff per kWh ≈ R 0,7072 CA OR OR A MA 1A percentage  112,2  increase New tariff per kWh =  × R0,6303   100  1MA calculating percentage of tariff ≈ R 0,7072 CA 1CA tariff NPR Answer only full marks (3) L1 1.2.1 Income is less/smaller than expenditure J 2J terminology used ( income & OR expenditure) Expenditure is more/bigger than income J more than /exceeds 2J less/smaller than OR 2J shortfall Amount of shortfall from income. J (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 5 DBE/2016 SCE – Memorandum Ques Solution Explanation T&L L1 1.2.2 The municipality showed a surplus. J 1J decision (from the subtraction) A = R65 771 447 – R28 490 095 1MA finding = R37 281 352 MA differences (2) L1 1.2.3 Six million, nine hundred and seventy nine thousand, nine hundred and nine rand A 2 A correct number and wording. (If six million, five hundred and thirty thousand seven hundred and eighty five rand : Max 1 mark) (2) L1 1.2.4 Department B A 2A answer (2) L2 1.2.5 % difference Expenditure 2014 − Expenditure 2013 = × 100% Expenditure 2013 SF 1SF substitute correct R33 031 602 − R30 645 928 = × 100% values from table R30 645 928 1CA simplify ≈ 7,784636183% CA 1R rounding ≈ 8% R Answer only full marks (3) A P 1.2.6 3 1A numerator L2 P= × 100% 1A denominator 7 A ≈ 42,86% CA 1CA % Answer only full marks NPR (3) [44] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 6 DBE/2016 SCE – Memorandum Question 2 [28] Tolerance range 1 mark Ques Solution Explanation T&L 2.1.1 L1 (a) Length of rectangular area to be cleared 1MA adding MA 250 mm × 2 = 1 430 mm + 250 mm × 2 CA 1CA length = 1 930 mm Width of rectangular area to be cleared 1A width (CA 1170) = 1 420 mm A (Counting bricks: Accept Length 1370 + 500 = OR 1870 mm and Width 1650 mm) 2 marks for width and 1 mark for length Answer only full marks (3) 2.1.1 CA from Q2.1.1 (a) L2 (b) SF 1SF substitute correct Total area = 1 930 mm × 1 420 mm values = 1,93m × 1,42 mC 1C conversion CA 1CA area in m² ≈ 2,7406 m² OR OR SF Total area = 1 930 mm × 1 420 mm 1SF substitute correct values = 2 740 600 mm² CA 1CA area in mm² ≈ 2,7406 m² C 1C conversion OR OR 1C conversion Total area (in m2) C SF 1SF correct values = ( 1,73 + 0,250 × 2 ) × (0,92 + 0,25 × 2) substituted = 1,93 1,42 1CA area in m² = 2,7406 CA NPR (3) C M L1 MA 2.1.2 Length of A = 2 × 220 mm + 3 × 10 mm 1C converting 1M adding mortar = 470 mm 1MA mortar measure (Accept 450 mm) (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 7 DBE/2016 SCE – Memorandum Ques Solution Explanation T&L MA L1 2.1.3 Width of a cement slab = 2 12 × 22 cm + 2 cm 1MA multiply length of (a) one brick by 2 12 and = 57 cm CA adding 2 cm (or 20mm) 1CA width Answer only full marks (2) C SF L2 2.1.3 Volume of one cement slab = 92 cm × 57 cm × 3,5 cm 1SF correct values (b) substituted from (a) = 18 354 cm³ CA 1C conversion 1CA volume in cm³ Answer only full marks (3) M L2 2.2.1 Height = [1 800 mm – ( 2 × 40) mm] ÷ 10MA 1M subtracting 80 1MA divide by 10 = 172 mm CA 1CA height in mm Answer only full marks (3) M L1 2.2.2 Side length = 2 025 cm 2 = 45cm A 1M square root (a) 1A side length Answer only full marks (2) M L2 2.2.2 Total floor area = 2 025cm² × 15 = 30 375 cm² 1M area multiplied by 15 (b) 1CA area in m² = 3,0375 m² CA NPR Answer only full marks (2) L2 A A 2A 2.2.3 3  1A 3,142 (a) Area of circle = 3,142 ×  cm  1A correct radius 2  1A squaring 2 = 7,0695 cm (3) CA 45 cm from Q2.2.2(a) L3 2.2.3 SF M 1SF correct values (b) Surface area = 180 cm × 45 cm – 10 × 7,0695 cm2 1M subtracting = 8 100 cm2 – 70,695 cm2 CA 1CA simplification = 8 029,305 cm2 CA 1CA total surface area (4) [28] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 8 DBE/2016 SCE – Memorandum QUESTION 3 [24] Tolerance range 0 marks Ques Solution Explanation T&L L1 3.1.1 ORPEN Gate RD 2RD reading from map (2) L1 3.1.2 R537, R536 ,R36 , R532 RD 2D reading from map (2) L1 3.1.3 R40 RD 2RD reading from map (2) [KZN do not mark this question.] L2 3.1.4 Lydenburg RD 3RD reading from map (3) L1 3.1.5 North West RD 2D reading from map (2) L1 3.2.1 Lifts A A 2A for 1st feature Escalators A 1A for 2nd feature Stairs/ Steps P for INCORRECT features added (3) L1 3.2.2 Clockwise RD 2RD reading from plan [Eastern Cape do not mark this question] (2) A L1 3.2.3 S124 A 1A for S 1A correct number (accept 1024) (2) L3 3.2.4 20 mm : 5 m A 1A ratio in different = 20 mm : 5 000 mm C units 20 5 000 1C converting to the = mm : mm same units 20 20 = 1 mm : 250 mm Scale = 1 : 250 CA 1CA scale (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 9 DBE/2016 SCE – Memorandum Ques Solution 3.2.5 Store B Store C Entrance/ Café Store A Exit Hallway Revolving Store D Lifts Service door Rest area counter Escalators Entrance/ Exit Restaurant Store S Café Store E A Store X Hallway Restaurant Store C W Store F Restaurant B Store Y Store G Hallway Restaurant Store Store Store M D V N Store Store Store H L O Store Store Store Store Store R T U Store Z K P Store I Store Store J Q Hallway Rest area Café Hallway Hallway Mall office Lifts Escalators Emergency exit Emergency exit (Source:www.edrawsoft.com) 3.2.5 2A route to ANY exit L2 1A shortest route (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 10 DBE/2016 SCE – Memorandum Question 4 [24] Tolerance range 1 mark Ques Solution Explanation T&L A L1 4.1 CONTINUOUS. 1A continuous The data represents mass ( in kilogram) which can be 1J explanation J expressed in smaller fractional units. (2) A L1 4.2 Other meat 2A item 46% CA 1CA percentage (Accept Beef –7 % then Max 2 marks) (3) A 1A correct value from table L1 4.3 6,7 kg × 49 320 500 M 1M multiply by 49 320 500 CA = 330 447 350 kg. 1CA total in kg Answer only full marks (3) M L1 4.4 M = 43,8 – (13,8 + 3,7 + 3,6 + 22,4) 1M subtracting = 43,8 – 43,5 A 1A 43,5 = 0,3 CA 1CA value of M Answer only full marks (3) A L1 4.5 Fish and seafood 2A identifying fish and seafood (2) L1 4.6 A CA A 1A Correct position – 46,0% ; –7,0% ; –5,0% ; 109% ; 119,0% . – 46% 1CA position of the –7% and –5% 1A arrangement of the positive percentages (If Other meat ; beef ; mutton ; poultry ; pork max 2 marks) Penalty 1 mark if in descending order (3) A 4.7 No mode 2A correct answer L1 (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 11 DBE/2016 SCE – Memorandum Ques Solution Explanation T & L L2 4.8 Consumption of different food items in South Africa from 1994 to 2009 35 30 25 20 A 1999 kg/year 2004 2009 15 10 A A A A A 5 0 Total eggs Poultry Pork Meat, other Total fish and other Mutton Beef Total offal seafood Food items 1A for each bar plotted correctly (for the last bar - mark any bar below 10 as correct) (6) [24] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 12 DBE/2016 SCE – Memorandum Question 5 [30] Tolerance range 0 marks Ques Solution Explanation T&L A F 5.1.1 5 365 : 112 043 MA 1MA writing as a ratio L1 1A correct values ≈ 1 : 20,884 CA 1CA form 1:… NPR (3) F 5.1.2 R150, R200 and R300 A 2A correct values L1 (2) MA F 5.1.3 9 288 1MA correct values L1 % savings = × 100% M 1M percentage 202 714 1CA % savings ≈ 4,58 % CA Answer only full marks (3) F 5.1.4 Fixed expense A 2A answer L1 (2) F 5.1.5 R126 696 – R112 043 M 1M subtract correct values L1 1CA difference = R14 653 CA Answer only full marks (2) A DH A 5.2.1 Charles and David Koch 2A Charles Koch L1 1A David Koch (3) M A DH 5.2.2 $79,2 billion – $15,7 billion 1A correct values / names L2 1M subtraction = $63,5 billion CA 1CA solution including billions (3) D A 5.2.3 1A arranging values L2 40,1 ; 40,6 ; 41,7 ; 42,9; 42,9 ; 54,3 ; 64,5; 72,7; 77,1; 79,2 M 1M concept of median 42,9 billion + 54,3billion Median = $ 2 = $48,6 billion CA 1CA median (No penalty omitting billion) (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 13 DBE/2016 SCE – Memorandum Ques Solution Explanation T&L D 5.2.4 Mean (in billions$) M L2 3,9 + 6,7 + 3,3 + 7,4 + 15,7 + 4,0 + 6,3 + 6,3 + 3,1 + 4,0 1M concept of mean = 1A dividing by 10 10 A 60,7 = 10 = 6,07 CA 1CA simplification (No penalty omitting billion) (3) P A 5.2.5 2 1A numerator L2 P (south african <7 ) = 1A denominator 10 A 1 = CA 1CA simplified fraction 5 (3) D 5.2.6  6 300 000 000  L2 = R  M 1M dividing by rate  0,0606  1CA simplification = R103 960 396 000 CA = R 103960,3960 million R 1R rounding ≈ R103 960 million OR R103 960 000 000 OR OR $6,3 billion = $6 300 million 1M dividing by rate $6 300 million M 0,0606 1CA simplification CA = R 103960,3960 million 1R rounding ≈ R103 960 million OR R103 960 000 000 R (3) [30] Copyright reserved