Mathematical Literacy Grade 12 P1 2018 Memo (Eng)

NATIONAL SENIOR CERTIFICATE GRADE 12 SEPTEMBER 2018 MATHEMATICAL LITERACY P1 MARKING GUIDELINE MARKS: 150 Symbol Explanation M Method MA Method with accuracy CA Consistent accuracy A Accuracy C Conversion S Simplification RT/RG/RM Reading from a table/Reading from a graph/Reading from a map F Choosing the correct formula SF Substitution in a formula J Justification P Penalty, e.g. for no units, incorrect rounding off etc. R Rounding Off/Reason AO Answer only NPR No penalty for rounding This marking guideline consists of 10 pages.

2 MATHEMATICAL LITERACY P1 (EC/SEPTEMBER 2018) QUESTION 1 [30 MARKS] INTEGRATED QUESTION Question Solution Explanation/Marks AO: FULL MARKS Lev. 1.1 1.1.1 Inflation is the increase in the price of goods 2A Explanation F and services over time.A L1 OR The decrease in the purchasing value of the money over time. ✓✓A (2) 1.1.2 Difference = R125 – R120 = R5 M 1M Difference F = 5 × 100 = 500 cents A 1CA In cents (2) L1 1.1.3 112 . 5 1M Multiply112,5 by 125 and F  125 Price = 100 M dividing by 100 L1 1CA Price = R140,60 CA OR OR 1M For profit 12,5 1CA Price Profit = 100 × 125 = R15,625M OR NPR Price = 125 × 1,12 (Concept of money) Price = 125 + 15,625 = R140,625 = 140, 625 = R140,60 CA (2) = R140,60 1.2 1.2.1 90 1MA Divide by 60 and F Cost of a dozen =  12  MA 60 multiply by 12 L1 = R18,00  CA 1CA Cost OR 60 Dozens = 12 =5 1MA Number of dozens 90 Cost of a dozen = 5 1CA Cost (2) = R18  1.2.2 (( 60  75 )  90 ) 1M Subtraction for profit FL1  100 % Profit = 90 M 1M Fraction 45/90 multiply 45 by 100 = 90 x 100% M 1CA Percentage = 50% CA (3) 1.3 Time = 24:00 – 16:25 = 7h 35 min M 1M For 7h 35min From midnight to 3:30 am = 3h30 min M 1M For 3h 30min ∴ Total time = 7 hours 35 min + 3h30min 1CA Time M = 11h 05 min CA (3) L1 1.4 2  1000 1M Correct numerator and Plastic cups = M 275 denominator = 7,2727 1A Rounding down = 7 A (2) ML1 Copyright reserved Please turn over

(EC/SEPTEMBER/2018) MATHEMATICAL LITERACY P1 3 1.5 1.5.1 2𝑐𝑚 × 75 𝑚 1M for Numerator Length = 2M 100 𝑐𝑚 1M for Denominator ML1 1CA With correct unit = 1,5 m  (3) 1.5.2 Scale = 100 : 7 500 2M 1M Ratio form = 1 : 75  1CA 1C Conversion 1CA Answer MP (3) L1 1.6 1.6.1 72,67% = Nelson Mandela District  A 2A District (2) DL1 1.6.2 Districts are: Chris Hani East, OR Tambo 1A First 2 districts Coastal, A 1A Second 2 districts Amathole East and Amathole West A (2) DL 1 1.6.3 Nelson Mandela, Sarah Baartman OR Tambo 1M Identifying 7 best Inland, Chris Hani West, Alfred Nzo West, Joe performing districts in Gqabi, Buffalo City M order 1CA for middle district ∴ 𝑀𝑖𝑑𝑑𝑙𝑒 𝑑𝑖𝑠𝑡𝑟𝑖𝑐𝑡 = CHRIS HANI WEST CA (2) DL1 1.6.4 26 ,1 1M Correct values Candidates failed = × 313 030 M 100 1CA Candidates = 81 701 CA OR 73,9 Passed = 100 × 313 030 1MA Candidates passed = 231 329 M 1CA Candidates failed Number failed = 313 030 – 231 329 Accept 81 700 (2) = 81 701CA DL1 [30] Copyright reserved Please turn over

4 MATHEMATICAL LITERACY P1 (EC/SEPTEMBER 2018) QUESTION 2 [46 MARKS] FINANCE Question Solution Explanation/Marks AO: FULL MARKS Lev. 2.1 2.1.1 Value of A = R3 250 + R4 500 + R1 200  M 1M Addition = R8 950  A 1A Value of A OR Value of A = R7 950 + R1 000  M 1M Addition = R8 950 A 1A Value of A (2) L2 2.1.2 Value of B = R1 440 – R1 500 M 1M Subtracting correct = -R60  A values Value of B = R1 440 – R1 500 M 1A Negative answer = (R60)  A 1M Subtracting correct values 1A Answer in brackets (2) 2.1.3 Value of C = R2 600 – R200 M 1M Subtraction = R2 400  CA 1CA Value of C OR Value of C = R5 780 – (R880 + R1 000 + R1 500) M 1M Subtraction = R2 400 CA 1CA Value of C (2) L1 2.1.4 Delivery A 2A Correct answer (2) L1 2.2 2.2.1 Annual gross salary = R65 000 ×12 M 1M Multiply by 12 = R780 000 CA 1CA Annual salary (2) L1 2.2.2 Mrs John’s annual pension fund contribution: 7 ,5 1M Multiply by 12 = ( 100 × 65 000) ×12 M CA Pension contr. = R58 500 CA (2) L1 2.2.3 Medical Aid contribution = R1 050 ×12 M 1M Multiply by 12 = R12 600 A 1A Contribution (2) L1 2.2.4 75 1M Correct values Mrs John’s performance bonus = 100 ×65 000 M 1M Multiply by 75% L1 = R48 750 A (2) Copyright reserved Please turn over

(EC/SEPTEMBER/2018) MATHEMATICAL LITERACY P1 5 2.2.5 Annual taxable income CA from 2.2.1, 2.2.2 = R780 000 + R48 750 – (58 500 + R12 600) M and 2.2.3 = R828 750 – R71 100 M 1M Total income = R757 650 CA 1M Total contributions 1M Subtraction 1CA Taxable income (4) L2 2.3 2.3.1 R701 301 and above RT 2RT Correct group (2) L1 2.3.2 Rebate Mrs John will receive = R13 500 + R7 407 M 1M Adding correct = R20 907 A rebates 1A Total rebate (2) L1 2.3.3 Actual tax = Income tax calculated on taxable CA from 2.2.5, 2.3.1 income Rebate and 2.3.2 = 206 964 + 41 % of the amount above 701 300 – 1SF Substitution R20 907 1S Simplification = 206 964 + 0,41 × (757 650 – 701 300) –20 907 SF 1S Simplification = 206 964 + 0,41 × 56 350 – 20 907 S 1CA Nearest rand = 206 964 + 23103,50 – 20 907 S (4) = R209 160,50 CA L3 2.3.4 Net annual salary = Annual taxable income – CA from 2.2.5 and 2.3.3 Actual annual tax 1SF Substitution L1 = R757 650 – R209 160,50 SFM 1M Subtraction = R548 489,50 CA 1CA Net salary (3) 2.4 2.4.1 15 , 5 L2 st Interest 1 year = 100 × 400 000 = R62 000 A 1A Interest 2nd year amount = R400 000 + R62 000 = R462 000 S 1S New amount 2nd year 15 , 5 1S Interest in 2nd year nd Interest 2 year = 100 × R462 000 1A Total interest = R71 610S (4) Total interest = R62 000 + R71 610 = R133 610  A Copyright reserved Please turn over

6 MATHEMATICAL LITERACY P1 (EC/SEPTEMBER 2018) 2.4.2 Phone D: 100 1M Cost without VAT Cost excluding VAT = 115 × 1 750 1M Subtraction = R1 521,74 M 1A VAT value VAT = R1 750 – R1 521, 74 M = R228,26 A OR 1M Fraction 15 1M Multiplication VAT = 115M × 1 750 M 1A VAT value (3) = R228,26 A L2 2.4.3 Phone D : Phone E 3:2 60 : E M 1M Ratio form 60  2 E= 3 1CA Number of phones = 40 CA (2) 2.4.4 Total cost = 60 × 1 750 + 40 × 2 000 CA from 2.4.3 = 105 000 M + 80 000 M 1M Cost for phones D L1 = R185 000  CA 1M Cost for phones E 1CA Total cost (3) 2.4.5 1: 0,52709 1M Conversion 185 000 : Total cost in CYN 1A Total cost L2 Total cost = 185 000 × 0,52709 M 1A Answer in Yuan = CYN 97511,65 AA (3) [46] Copyright reserved Please turn over

(EC/SEPTEMBER/2018) MATHEMATICAL LITERACY P1 7 QUESTION 3 [25 MARKS] MEASUREMENT Ques. Solution Explanation/Marks AO: FULL MARKS Level 3.1.1 Length = 4 880 mm ÷ 1 000 C 1C Divide by 1000 = 4,88 m A 1A Metres (2) L1 3.1.2 Distance A = 3 (150 mm) M 1M Multiplication by 3 = 450 mm A 1A Distance (2) L1 3.1.3 Height of the wall = 2,1 m + 450 mm ÷ 1 000 C CA from 3.1.2 = 2,1 m + 0,45 m S 1C Conversion = 2,55 m CA 1S Simplification 1CA Height (3) L1 3.1.4 Area = Length × Width CA from 3.1.1 4,88 1M Dividing by 2 = 2  × 2,1 m  M 1M Multiplication = 5,124 m2 CA Unit 1CA Area 1A Unit (4) L2 3.1.5 Area covered by bricks CA from 3.1.1 and = Area of garage – Area of double door 3.1.3 = (2,55 m × 5,18 m)  – (4,88 m × 2,1 m) MM 1M Area of garage = 13,209 m2 – 10,248 m2 S 1M Area of double door = 2,961 m2 CA 1M Subtraction 1S Simplification 1CA Area covered by bricks (5) L2 3.2.1 Height of the bricks and cement = (12 x 2) + 76 mm M 1M Multiplication by 2 = 100 mm CA and addition 1CA Height (2) L1 3.2.2 2 500 CA from 3.2.1 Number of rows of bricks = 100 M M 1M Conversion 1M Division = 25 CA 1CA Number of rows of bricks (3) L1 3.2.3 Volume = 23 cm ×11 cm × 7,6 cm SF C 1SF Substitution = 1 922,8  A cm3 A 1C Conversion 1A Volume 1A Units (4) L2 [25] Copyright reserved Please turn over

8 MATHEMATICAL LITERACY P1 (EC/SEPTEMBER 2018) QUESTION 4 [31 MARKS] DATA HANDLING Ques. Solution Explanation/Marks AO: FULL MARKS Level 4.1. Range = 3,316 kg – 0,182 kg RT M 1RT Correct values 1M Subtraction = 3,134 kg CA 1CA Range (3) L2 4.2 1,668 kg A 2A Median (2) L2 4.3 Average = 1,26 ×2 +1,371 ×9 +1,668 ×8 + 1,746 ×4 + 1,849 ×8 + 2,163 1M Adding +2,333 + 3,128 ×2  M 1M Concept of mean 60,731 = 35 M 1R Average to = 1,735 nearest kg = 2 kg R (3) L2 4.4 22  A 2A Explanation (2) L1 4.5 8 1M Fraction value Probability = 35 M × 100% M 1M Multiply by 100 = 22,9 % CA 1CA Percentage (3) L1 Copyright reserved Please turn over

(EC/SEPTEMBER/2018) MATHEMATICAL LITERACY P1 9 4.6 L2 4M For the first 4 bars plotted correctly 1M For the next 2 bars plotted correctly 1M For the last 2 bars plotted correctly (6) 4.7 0,182; 0,182; 0,182; 0,309; 0,729; 0,729; 0,729; 0,856; 0,856; 0,856; 0,856 1M Correct values L1 0,936; 2,448; 2,448; 2,449; 3,038; 3,316; 3,316; 3,316; 3,316; 3,316; 3,316 1M Value or Q2 position of median Q3 1CA Q3 0,856+0,936 Q2=Median= 2 = 0,896  MM Q3=3,316  CA (3) 4.8 3,316 kg  2A RT Modal L1 value (2) 4.9 8 1M Fraction L1 % of total fishes = 57 × 100 M 1M Multiply by 100 = 14,04 %  CA 1CA Percentage (3) 4.10 3rd hour  2RT Correct hour L1 (2) 4.11 1,1,1,1,2,3,3,4,6  2M Arrangement L1 (2) [31] Copyright reserved Please turn over

10 MATHEMATICAL LITERACY P1 (EC/SEPTEMBER 2018) QUESTION 5 [18 MARKS] MAPS, PLANS and OTHER REPRESENTATIONS Ques Solution Explanation/Marks AO: FULL MARKS Level 5.1 There is no seat for Lundi here A 2A No seat MP (2) L1 5.2 South A 2A Direction L1 (2) 5.3 A8  RP 1RP First seat L1 A11  RP 1RP Second seat A15  RP 1RP Third seat (3) 5.4 35 RP 2RP Available seats L1 (2) 5.5 B14  RP 2RP Asi’s seat no. L1 (2) 5.6 Row J  RP 2RP Furthest row L1 (2) 5.7 Side BB  RP 2RP Correct side L1 (2) 5.8 20 1A Numerator P P(seat from side AA) = A A 194 1A Denominator L2 1CA Answer = 0,103 CA (3) [18] TOTAL: 150 Copyright reserved Please turn over