Mathematical Literacy P1 Feb March 2015 Memo Eng hlayiso.com

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICAL LITERACY P1 FEBRUARY/MARCH 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 table/Reading from graph/Reading from 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 Reason RO Rounding off J Justification KEY TO TOPIC SYMBOLS: F = Finance; M = Measurement; MP = Maps, Plans and other representations DH = Data Handling; P = Probability This memorandum consists of 11 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 2 DBE/Feb.–Mar. 2015 NSC – Memorandum QUESTION 1 [35] Ques Solution Explanation Topic F 1.1.1 This is the 8th month of the new financial year for 2E explanation L1 which she receives a salary advice E (2) D F 1.1.2 Gross Income is the Income Earned before any L1 deductions are made. 2D definition (2) F 1.1.3 R 500 1M multiply with 100% L1 Pecentage = × 100% M R 7 952 1A %UIF contribution ≈ 6,2877…% A (2) Also accept 6,29% OR 6,3% F 1.1.4 7,5 M 1M calculating 75% L1 × R7 952 = R596,40 M/A 1M/A calculating 100 accurate value OR OR M 596,40 M × 100% = 7,5% 1M correct fraction 7952 1M multiply with 100% (2) F 1.1.5 R5 981,67 RD 1RD total pension L1 (2) M F 1.1.6 Hourly rate = R7 452 ÷ 172,5 1M Division by 172,5 L1 = R43,20 A 1AHourly rate (2) M F 1.1.7 Difference in rate : R120,45 – R75,80 1M subtraction L1 = R44,65 A 1A difference in rate (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 3 DBE/Feb.–Mar. 2015 NSC – Memorandum Ques Solution Explanation Topic F 1.2.1 Total income (in rand) 1A R2,50 L2 A A 1A No of blocks of = 2,50 × number of blocks of fudge fudge OR A A 1A × R2,50 Total income (in rand) = 2,50 × x 1A variable with ( x = number of blocks of fudge) explanation (2) M F 1.2.2 B = R30 ÷ R2,50 1M multiplying by L2 R2,50 = 12 A 1A simplify AO (2) M F 1.2.3 R24,99 ÷ 2.5 = R9,996 1M dividing by 2,5 L1 (a) ≈ R10,00 M 1A cost OR OR M Shanté took the cost price of the 2,5 kg sugar and M 1M Cost Price divided it by the quantity to determine the price of 1 kg 1M dividing by 2,5 of sugar. (2) M M 1.2.3 Number of batches = 1 000 ÷ 250 1M division by 250 L1 (b) = 4 A 1A no of batches AO (2) F 1.2.3 100 mℓ ÷ 5 = 20 mℓ M 1M dividing by 5 L1 (c) C= R0,95 × 20 = R11,80 CA 1CA cost of item OR OR M 100 1M correct fraction C= × R0,59 = R11,80 CA 1CA cost of item 5 OR OR 100 : 5 1M ratio C : 0,59 M C = R100 × 0,59 ÷ 5 = R11,80 CA 1CA simplify AO (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 4 DBE/Feb.–Mar. 2015 NSC – Memorandum Ques Solution Explanation Topic F 1.2.3 Cost of one block of fudge = R40,50 ÷ 54 M 1M division L1 (d) = R0,75 A 1A cost price AO (2) F 1.2.4 R30 RG 2RG Reading from L1 (a) graph (2) 1.2.4 F Income and expenses for making one batch of fudge (b) L2 140 120 A 100 Amount in rand 80 A 60 40 A A 20 0 A 0 10 20 30 40 50 60 Number of blocks of fudge 1A point (0;0) 3A plotting of any other 3 correct points 1A joining the points (5) F 1.2.5 Break-even point – it is the point where the income 2E explanation of L1 and and expenses are exactly the same. E point on intersection OR No profit or loss is made E Explanation only (without using the word break-even point) Full marks (2) [35] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 5 DBE/Feb.–Mar. 2015 NSC – Memorandum QUESTION 2 [26] Ques Solution Explanation Topic M 2.1.1 Radius = 8,5 cm ÷ 2 = 4,25 cm M 1M radius L2 Volume of a cylinder = 3,142 × 4,252 × 10,5 cm3 SF 1SF substitution = 595,899 cm3 CA 1CA volume ≈ 595,9 cm3 A 1A unit in cm3 (4) M 2.1.2 M 1M subtracting 500 L3 Volume of empty space = 595,9 – 500 cm3 1CA volume = 95,9 cm3  CA (2) M 2.1.3 500 cm 3 1SF substitution L2 Height of motor oil in can = SF 3,142 × 4,25 (cm)2 500 cm 3 A 1A simplification = 56,752375 ≈ 8,8 cm CA 1CA height (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 6 DBE/Feb.–Mar. 2015 NSC – Memorandum Ques Solution Explanation Topic M 2.2.1 1  SF L2 Area of a triangle = × 980 × 1 200 mm² 1SF substitution 2 = 588 000 mm²  CA 1CA area of triangle (2) M 2.2.2 Area of trapezium side L2 = (2 × 588 000) + 2 088 000 mm²  SF 1SF substitution S = 1 176 000 + 2 088 000 mm² 1S simplification = 3 264 000 mm2 A 1A area C Total area in m² = 3 264 000 ÷ 1 000 000 1C conversion = 3,264  CA 1CA total area (5) M M 2.2.3 11,676 − 2 × 3,264 2 1M subtraction L3 Area of slanted side = m 1M division by 2 2 M = 2,574 m²  CA 1CA area (3) M M 2.3.1 Total area = 11,676 × 25 m² 1M multiply by 25 L1 = 291,9 m²  CA 1CA total area (2) M M 2.3.2 Total number of coats = 25 × 2 1M multiply L1 = 50 A 1A coats of paint (2) M M 2.3.3 Minimum number of tins = 585 ÷ 25 tins 1M division by 25 L1 = 23,352 tins  CA 1CA simplification ≈ 24 tins  R 1R rounding up (3) [26] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 7 DBE/Feb.–Mar. 2015 NSC – Memorandum QUESTION 3 [21] Ques Solution Explanation Topic A A MP 3.1 Perdeberg and Petrusburg 1A Perdeberg L1 1A Petrusburg (2) MP 3.2 South East A 2A Directions L1 (2) MP 3.3 165 km 1SF substitution L2 Time = SF 1A simplification 97,3 km / h = 1,695 hours A But 0, 695 hours × 60 minutes C 1C multiply × 60 = 41,7 minutes A 1A minutes 1CA time Time ≈ 1 hour 42 minutes CA (5) RD RD MP 3.4 Provincial road number 31 and 64 1RD Road 31 L1 1RD Road 64 (2) MP 3.5 Phillippolis A 3A finding the correct L2 town (3) RD M M MP 3.6 Distance = 145 – (39 + 19 + 33 + 12) km 1M Identify 145 km L2 = 42 km A 1M subtracting 1M adding distances 1A distance AO (4) MP 3.7 5,4 cm on map = 2,7 km in reality L3 2,7 km × 100 000 = 270 000 cm C 1C convert km to cm 5,4 : 270 000 M 1M write as a ratio 1 : 50 000 S 1S simplify (3) [21] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 8 DBE/Feb.–Mar. 2015 NSC – Memorandum QUESTION 4 [36] Ques Solution Explanation Topic M  A DH 4.1.1 300 ; 256; 249; 182; 173; 169; 163; 155; 145; 144; 141 1 M descending L1 order 1A arrange all (2) DH 4.1.2 Jacques Kallis  A 2A name of player L1 (2) DH 4.1.3 Mean = M L2 300 + 256 + 249 + 182 + 173 + 169 + 163 + 155 + 145 + 144 + 141 1M adding of values 11  M 1M division by 11 2077 = 11  CA ≈ 188,8181 1CA mean Also accept 189 runs (3) DH 4.1.4 145 1SF substitution L2 Strike rate = × 100 SF 121 = 119,83  A 1A strike rate rounded in context (2) P 4.1.5 5 A 1A numerator L2 11  A 1A denominator (2) DH 4.2.1 C  A 2A L1 (2) DH 4.2.2 E  A 2A L1 (2) DH 4.2.3 A  A 2A L1 (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 9 DBE/Feb.–Mar. 2015 NSC – Memorandum Ques Solution Explanation Topic M 1M number format DH 4.3.1 1 100 000 – 1 098 959 = 1 041 A 1A difference L1 CA 1CA identify year Therefore 2007 is the closest AO (3) DH 4.3.2 2005  RT 2RT reading from L1 table (2) 4.3.3 DH (a) 33,5 L1 P= × 572 600 M 1M % of 572 600 100 ≈ 191 821 A 1A value P AO (2) DH 4.3.3 178 373A 1A numerator L1 (b) Q = × 100 1A denominator 559 631A ≈ 31,9 A 1A percentage AO RT M (3) DH 4.3.4 559 631 – 178 373 = 381 258 CA 1RT correct values L1 1M subtracting 1CA no of deaths (3) DH 4.3.5 2004 RT and 2006 RT 1RT 2004 L1 1RT 2006 (2) DH 4.3.6 2003 RT and 2010 RT 1RT 2003 L1 1RT 2010 (2) RT DH 4.3.7 579 371 : 1 109 926 M 1RT reading correct L1 values 1A correct ratio (2) [36] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 10 DBE/Feb.–Mar. 2015 NSC – Memorandum QUESTION 5 [32] Ques Solution Explanation Topic M F 5.1.1 Amount = R9 247,95 – R4 000 1M subtracting L1 = R5 247,95 A 1A amount (2) F 5.1.2 R 350 A M 1A correct fraction L1 (a) x 100 % = 3,5% A 1M multiply by 100% R10 000 1A percentage (3) M F 5.1.2 Total monthly amount = R764,84 + R75,00 + R20,50 1M adding L1 (b) = R860,34 A 1A simplify (2) RT M F 5.1.2 Total amount of loan = R764,84 × 36 months 1RT reading values L2 (c) = R27 534,24 CA 1M multiply M 1CA simplify Interest = R27 534,24 – R10 000 1M subtract = R17 534,24 CA 1CA interest (4) F 5.2.1 RD M 1RT reading values L1 Amount = R149 995,00 – R25 000 1M subtract = R124 995,00 CA 1CA amount (3) F 5.2.2 Total monthly repayments 1M multiplying L1 = R4 068,06 × 36 M CA 1CA correct amounts = R146 450,16 (2) RD M F 5.2.3 Difference = R5 819,44 – R4068,06 1RD reading values L1 = R1 751,38 A 1M subtracting 1A difference (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 11 DBE/Feb.–Mar. 2015 NSC – Memorandum Ques Solution Explanation Topic M M 5.3.1 Width = 5 inch ÷ 0,394 cm 1M dividing by 0,394 L2 = 12,69 cm A 1A simplification Length = 7 inch ÷ 0,394 cm = 17,77 cm A 1 A simplification (3) M M 5.3.2 Length = 17,77 – 15 cm 1M subtracting L1 = 2,77 cm CA 1CA length Width = 12,69 – 10 cm = 2,69 cm CA 1 CA width (3) D 5.4.1 30 – 39 years A 2Adetermining the L2 modal age group (2) P RT A L2 5.4.2 Age group 80+ 1RT reading table (a) 1A age group (2) P 5.4.2 2 953 490 RT 1RT reading L2 Probability = (b) 25 362194 RT numerator 1RT reading ≈ 0,12 CA denominator 1CA decimal fraction (Also accept 0,1 or 0,116) (3) [32] Copyright reserved