MLIT P2 S12 MEMO Eng hlayiso.com

Downloaded from hlayiso.com Province of the EASTERN CAPE EDUCATION NATIONAL SENIOR CERTIFICATE GRADE 12 SEPTEMBER 2012 MATHEMATICAL LITERACY P2 MEMORANDUM MARKS: 150 Symbol Explanation M Method MA Method with accuracy CA Consistent accuracy A Accuracy C Conversion S Simplification RT/RG Reading from a table/Reading from a graph 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 This memorandum consists of 9 pages.

Downloaded from hlayiso.com 2 MATHEMATICAL LITERACY P2 (Memo) (SEPTEMBER 2012) QUESTION 1 LO 1 AS 1.1 1.1.1 Amount for deposit = 0,15 x 5 989  1:M 12.1.1 = R 898,35  1:A LO 1 AS 1.1.2 Monthly payments = 7 508,70 12.1.1 30  1:M = R 250,29  1:A LO 1 AS 1.1.3 Total Amount = Deposit + Total of 30 months 12.1.1 = R 898,35 + (R250,29 x 30)  = R 898,35 + R 7 508,70 1: M = R 8 407,05  1: CA LO 1 AS 1.1.4 A = R 7 508,70 12.1.1 P = R 5 989 – R 898,35 1:CA: = R 5 090,65  Calculating P- value n = 30 months = 2,5 years  1:A 2,5 years i=? A 1+ni = /P 1 + 2,5i = 7 508,70/ 5 090,65 1:SF 1 + 2,5i = 1,474998281 1:S (A/P) 2,5i = 0,4749982812 1:S (-1) i = 0,1899993125 1:S (/2.5) i = 19%  1:CA (x100) LO 1 AS 1.1.5 A=? 12.1.1 P = R5 989 n = 2,5 x 2 =5  1:A Calculating n i = 11,25 / 100 = 0,1125 / 2 = 0,05625  1:A Calculating i A = P(1+i)n = 5 989(1+0,05625)5  1:SF = 5 989(1,05625)5 = 5 989(1,31472103)  1:CA for = R 7 873,86  Simplifying 1:CA LO 1 AS 1.1.6 Option 2:  It will cost R 533,19 less than 1:J (Option) 12.1.2 Option 1 1:R

Downloaded from hlayiso.com (SEPTEMBER 2012) MATHEMATICAL LITERACY P2 (Memo) 3 LO 3 AS 1.2 1.2.1 Actual length in cm = 40 x 2,5 cm  1:M (x2,5) 12.3.3 = 100 cm  1:A LO 3 AS 1.2.2 Area of television in m2 = 97 cm x 58,7 cm  1:SF (Correct 12.3.1 5 693,9 cm2  values) = 10 000  1:A = 0,57 m2  1:C OR 1:A = 97 cm x 58,7 cm  100 100 = 0,97 m x 0,587 m  = 0,57 m2  LO 3 AS 1.2.3 Length = 97 cm + 10 cm + 10 cm 12.3.1 = 117 cm  Width = 58,7 cm +10 cm + 10 cm 1:A = 78,7 cm  (Calculating length) Area for mounting = 117 cm x 78,7 cm 9 207,9 cm2  1:A = 10 000 (Calculating = 0,92 m2  width) OR = 117 cm x 78,7 cm  1:C 100 100 1:CA = 1,17 m x 0,787 m = 0,92 m2  1:C Area of wall = 0,8 m2 1:A Area on the wall will be too small for the television to be mounted.  1:J LO 3 AS 1.3 Isidingo = 30 minutes – (45 seconds x 5) 1:C (convert 12.3.2 = 30 – (0,75 minutes  x 5) 45s to min) = 30 – 3,75  1:M = 26,25 minutes viewing time OR 26 min 15 sec 1:A “Sewende Laan” = 30 minutes – (0,5 x 4) = 30 – 2  1:M = 28 minutes viewing time  1:A “Sewende Laan” will give maximum viewing time.  1:CA [37]

Downloaded from hlayiso.com 4 MATHEMATICAL LITERACY P2 (Memo) (SEPTEMBER 2012) QUESTION 2 LO 4 AS 2.1 2.1.1 Number of learners interviewed = 6+10+8+14+2  1:RG 12.4.1 = 40 learners  1:A LO 4 AS 2.1.2 % learners interviewed = 40/1030 x 100  1:M 12.4.3 = 3,88 %  1:A = 3,9 %  1:CA (1dec. place) LO 4 AS 2.1.3 No  12.4.4 He only conducted the survey in his class  1:A OR It is not representative of the whole school 1:R LO 4 AS 2.1.4 No  1:A 12.4.4 Favourite chocolate in my class  1:A (subs. school with class) LO 4 AS 2.1.5 It must be representative of all the learners in the 12.4.4 school across the different grades, race, age, gender.  2:R LO 4 AS 2.1.6 40% learners interviewed = 40/100 x 1030  12.4.3 OR = 0,4 x 1030  1:M = 412 learners  1:A LO 2 AS 2.2 2.2.1 2008  1:RG 12.2.1 The graph from 2007 to 2008 is decreasing  1:R LO 2 AS 2.2.2 2010 – 2011  1:RG 12.2.1 The gradient is the steepest  1:R LO 2 AS 2.2.3 Price in 2008 = R3,70 12.2.3 1,057  = R3,50  OR 1: RG Price in 2 008 = R3,70 (correct %) 105,7 x 100 1:M = R3,50  1:A LO 2 AS 2.2.4 % change in 2011 = R 4,50 – R3,99 12.2.1 R3,99 x 100  = 0,51 3,99 x 100  2:MA = 12,8%  1:A

Downloaded from hlayiso.com (SEPTEMBER 2012) MATHEMATICAL LITERACY P2 (Memo) 5 LO 3 AS 2.3 2.3.1 128 mm = 12,8 cm  1:C (convert 12.3.2 Volume = l x b x h mm to cm) = 12,8 cm x 2,5 cm x 1,5 cm  1:SF(correct = 48 cm3  values) 1:A (in cm3) OR 2,5 cm = 25 mm ; 1,5 cm = 15 mm 1:SF (correct Volume = l x b x h values) = 128 mm x 25 mm x 15 mm  1:C(convert = 48 000 mm3/ 1 000  mm3 to cm3 = 48 cm3  1:A(in cm3) LO 3 AS 2.3.2 Surface area = 2 x Area of base + perimeter of 1:SF(correct 12.3.1 base x height values) = 2(12,8 cm x 2,5 cm)+2(12,8cm+2,5 cm)x1,5 cm 2:S (area of = 64 cm2 + 30,6 cm x 1,5 cm base and = 64 cm2 + 45,9 cm2  perimeter) = 109,9 cm2  1:A (x 30,6 and 45,9) 1:CA LO 3 AS 2.3.3 Surface Area of wrapping=SA of bar+(12,5%of bar) 1:M (12,5% 12.3.1 = 109,9 cm2 + (0,125 x 109,9 cm2) of 109,9) CA = 109,9 cm2 + 13,7375 cm2  1:A (13,7375) = 123,6375 cm2  1:MA (add) = 123,64 cm2  1:R (round to OR 123,64) = 1,125  x 109,9 cm2 = 123,6375 cm2  = 123,64 cm2  [35]

Downloaded from hlayiso.com 6 MATHEMATICAL LITERACY P2 (Memo) (SEPTEMBER 2012) QUESTION 3 LO 1 AS 3.1 3.1.1 (a) R 600 000  12.1.1 1:RT LO 1 AS (b) Transfer Duty 1:F(correct 12.1.1 = R 12 000 + 5% of the amount above R 1 000 000 formula)  1:A(amount > = 12 000 + 0,05 x 200 000  1 mil) = 12 000 + 10 000  1:A(5%) = R 22 000  1:M(add) 1:A LO 1 AS 3.1.2 (a) Monthly repayments 12.1.1 = Bond amount in 1 000 x factor = 1 200  x 9  = R 10 800  OR Monthly repayments = Bond Amount x factor 1 000 1:A(amount = 1 200 000 x 9 in 1 000) 1 000 2:RT(correct = 1 200  x 9  factor) = R 10 800  1:CA LO 1 AS (b) Monthly salary = R 450 000 1:M (/12) 12.1.3 12  1:A = R 37 500  2:MA (0,3) 30% of monthly salary = 37 500 x 0,3  R 11 250  OR 30% of salary = 450 000 x 0,3  = 135 000  12  = R 11 250  Yes she will qualify  1:J LO 1 AS (c) Monthly repayments 12.1.1 = Bond amount in 1 000 x factor = 1 200  x 8,39  = R 10 068  OR Monthly repayments = Bond Amount x factor 1 000 = 1 200 000 x 8,39 1 000 1:A(amount = 1 200  x 8,39  in 1 000) = R 10 068  2:RT(correct The monthly repayment is R 732 (R 10 800 – factor) R10 068) less than the 20 year period  1:A OR The longer the period, the less the monthly repayment  1:R

Downloaded from hlayiso.com (SEPTEMBER 2012) MATHEMATICAL LITERACY P2 (Memo) 7 LO 1 AS (d) Final Amount = monthly repayment x number 1:SF 12.1.1 of payments 1:C (convert = 10 800 x 240  20yrs to = R 2 592 000  months) 1:CA LO 1 AS (e) Interest = Final Amount – Loan Amount 12.1.1 = R 2 592 000 – R 1 200 000  1:M = R 1 392 000  1:A LO 2 AS 3.2 3.2.1 Interest  12.2.3 Interest is calculated on a new principal amount every month  1:A OR Interest is compounded monthly  2:R (Accept any other logical explanation) LO 2 AS 3.2.2 Loan  12.2.3 Loan amount is decreasing  1:A OR 2:R Interest is calculated on a smaller amount  (Accept any other logical explanation) LO 2 AS 3.2.3 Accept 2022 – 2023  12.2.1 2:A LO 2 AS 3.2.4 With a home loan compounded interest (monthly) 12.2.3 apply, while with vehicle finance simple interest apply  2:A [35]

Downloaded from hlayiso.com 8 MATHEMATICAL LITERACY P2 (Memo) (SEPTEMBER 2012) QUESTION 4 LO 2 AS 4.1 4.1.1 R 3 000  12.2.1 OR Total Cost = 300 x 10 = R 3 000  OR = 150 x 20 = R 3 000  OR = 100 x 30 = R 3 000  OR = 75 x 40 = R 3 000  OR = 60 x 50 = R 3 000  2:A LO 2 AS 4.1.2 A  = p n  OR 12.2.1 A  = p x n  OR 3000  = p n  OR n = A / p  OR p = A / n  3:F LO 2 AS 4.1.3 A: A = pn 12.2.1 3000 = p x 20 1:M p = 3000/20  1:A = R 150  B: A = pn 3000 = 75n n = 3000/75  1:M = 40 children  1:A LO 2 AS 4.1.4 3 000 = pn 12.2.1 3 000 = p x 45 p = 3 000/45  1:M = R 66,67  1:A LO 2 AS 4.1.5 As the number of children increase,  the price per 12.2.3 child decrease 2:A LO 2 AS 4.1.6 350 12.2.2 300 250 200 Price per child 150 100 50 0 5: 1 mark for 0 10 20 30 40 50 each point Number of children plotted correctly

Downloaded from hlayiso.com (SEPTEMBER 2012) MATHEMATICAL LITERACY P2 (Memo) 9 LO 3 AS 4.2 4.2.1 Diameter = 2,7 cm x 2 1:A(diameter) 12.3.1 = 5,4 cm  1:A(2mm to Length of toy = 5,4 cm – 0,2 cm  cm = 5,2 cm  1:M 1:A LO 3 AS 4.2.2 Volume of inner ball = (4/3)πr3 1:SF(correct 12.3.1 = (4/3) x 3,14 x (2,7 cm)3  values) = (4/3) x 3,14 x 19,683 cm3  1:S(2,73) = 82,41 cm3  1:A(in cm3) LO 4 AS 4.3 4.3.1 2 years 3 years 4 years 5 years Total 5:A 12.4.5 Boys 5 3 6 8 22 1 mark for Girls 4 8 5 6 23 each missing Total 9 11 11 14 45 value LO 4 AS 4.3.2 Average age of children 12.4.1 = (9x2) + (11x3) + (11x4) + (14x5)  45 = 18 + 33 + 44 + 70 45 = 165  2:M 45 1:S = 3,67 years  1:A LO 4 AS 4.3.3 (a) P(girl aged 5 years) = 6 /45  1:A 12.4.6 = 0,133  OR 13,3%  (numerator) 1:A (denominator) 1:A LO 4 AS (b) P(boy be invited) = 22 /45  1:A 12.4.6 = 0,489  OR 48,9%  (numerator) 1:A (denominator) 1:A LO 4 AS (c) P(boy or girl aged 2 years be invited) 1:A 12.4.6 = 9/45  (numerator) = 0,2  OR 20%  1:A (denominator) 1:A [43] TOTAL: 150