Mathematical Literacy P1 Feb March 2014 Memo Eng hlayiso.com

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICAL LITERACY P1 FEBRUARY/MARCH 2014 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/Reading from a graph SF Correct substitution in a formula O Opinion/Example P Penalty, e.g. for no units, incorrect rounding off etc. R Rounding off This memorandum consists of 12 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 2 DBE/Feb.–Mar. 2014 NSC – Memorandum QUESTION 1 [27 MARKS] Ques Solution Explanation AS/L 12.1.1 24 × 345 1.1.1 45 % of 26,7 – L1 10 389 =12,015 – 0,8927467... A 1A calculation = 11,12225.... 11,12 CA 1CA rounding (2) A 12.1.1 1.1.2 1 068 267 A 1A fraction L1 17 1,068 = = =1 1A simplest form 1 000 250 250 (2) A 12.1.1 1.1.3 September 1970 A 1A Year L2 1A month (2) 12.1.1 1.1.4 R1 = €0,10717 L2 R1 × € 32 527 M/A 1M/A dividing €32 527 = € 0,10 717 = R303 508,4445 1CA simplification = R303 508,44 CA (2) 12.2.1 1.1.5 S(in metre) = 5(1,5)[ 1,5 – 1 ] SF 1SF substitution L1 = 3,75 CA 1CA distance (2) 12.4.5 1.1.6 18A 1A number of favourable L2 P(boy) = outcomes 42 A 3 1A number of possible = CA Outcomes 7 1CA simplification (3) 12.1.1 1.1.7 20 : 12 1A writing ratio L2 =5:3 1CA more trains ∴ 2 more trains per hour CA OR OR 60 Number of trains in peak periods = 12 = 5 A 1A number of trains in Number of trains more = 5 – 3 peak periods = 2 CA 1CA more trains (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 3 DBE/Feb.–Mar. 2014 NSC – Memorandum Ques Solution Explanation AS/L 12.3.2 1.2.1 1 000 g = 2,2 pound L2 2,2 ∴ 200 g = pounds A 1A divide by 5 5 = 0,44 pounds S 1S simplification (2) 12.3.2 1.2.2 10 tbsp = 125 m  OR 10 : 125 = 3 : x L1 125 × 3 1A times three x= A 1 tbsp = 12,5 m  A 10 = 37 ,5 mS ∴ 3 tbsp = 12,5 × 3 m  1S simplification = 37,5 m  S (2) 12.1.1 1.2.3 For 6 persons use 50 g 6 = 50 g L2 M/A 50 M/A OR 50 g × 9 1M/A using ratio ∴ 1 person use = 8,33 g 9= 6 6 ∴ 9 persons use 8,33 × 9 = 75 g CA 1CA amount of pine = 75 g CA nuts (2) 12.2.2 1.3.1 the breadth decreases A 2A correct answer L1 (1) 12.2.2 1.3.2 4 m A 2A breadth L1 (1) 12.3.2 1.3.3 6 m  A 2A side length L1 (2) A 12.3.1 1.3.4 400 cm 1A dividing and adding L2 Number of cabbages = + 1 at beginning 1CA number 16 cm = 25 + 1 = 26 CA (2) [27] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 4 DBE/Feb.–Mar. 2014 NSC – Memorandum QUESTION 2 [29 MARKS] Ques Solution Explanation AS/L 12.4.4 2.1.1 54%  RG 2RG percentage L1 (2) 12.4.4 2.1.2 Natural area lost = 70% – 34%  RG 1RG subtracting correct L1 = 36%  CA values 1CA area lost (2) M 12.1.1 127 909 1M calculating 2.1.3 Ave annual percentage rate = ×100% L1 9 474 740 percentage = 1,35 % per year CA 1CA percentage/annum (2) 158 + 160 12.4.3 2.2.1 Median = M 1M finding median L2 2 = 159 CA 1CA median (2) 12.4.3 2.2.2 6 athletes A 2A answer L2 (2) 5 A 12.4.5 2.2.3 P(less than 158) = 1A number less than 160 L2 12 A 1A total number of athletes ≈ 41,67 % CA 1CA % (3) 12.2.1 2.3.1 MHR female = 216 – (1,09 × 18) SF 1SF substitution L1 = 196,38 CA 1CA maximum heart rate (2) 12.2.1 202 −189,9 1SF substitution 2.3.2 Age = SF L1 0,55 = 22 CA 1CA age (2) 12.2.2 2.3.3 female A 1A answer L1 (a) (1) 12.2.2 2.3.3 186 beats per minute A 2A correct conclusion L1 (b) (2) 12.2.2 2.3.3 Female A 2A correct gender L1 (c) (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 5 DBE/Feb.–Mar. 2014 NSC – Memorandum Ques Solution Explanation AS/L 2.3.3 12.2.2 26 A 2A correct conclusion (d) L2 (2) 12.2.2 2.3.3 20 A 2A correct conclusion L2 (e) (2) RG 12.2.2 2.3.3 Difference in age = 22 – 18 RG 2RG correct values L2 (f) = 4 years CA 1CA correct conclusion (3) [29] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 6 DBE/Feb.–Mar. 2014 NSC – Memorandum QUESTION 3 [22 MARKS] Ques Solution Explanation AS/L M 12.1.1 3.1.1 A = R6,31 × 9 × 5 A 1M concept - multiplying L1(2) = R283,95 CA 1A correct values L2(1) 1CA simplification (3) 12.2.1 3.1.2 Weekly rate ×13 Monthly rate = L1 3 R303,30 × 13 SF 1SF substituting weekly rate = 3 = R1 314,30 CA 1CA simplification (2) SF SF 12.1.1 3.1.3 R316,80 - R303,30 1SF new rate L1 Percentage increase = × 100% 1SF old rate R303,30 ≈ 4,45 % CA 1CA percentage (3) 12.4.4. 3.2.1 Brazil  RT 1RT correct country L1 (1) M 12.1.1 3.2.2 Total = 10 017 + 5 526 + 0 + 91 916 + 84 + 9 631 tonnes 1M addition L1 = 117 174 tonnes  CA 1CA correct total (2)  RT 12.1.1 3.2.3 30,53 1RT correct percentage L1 Amount of peaches = × 1 200 000 tonnes  M 1 00 1M writing 1,2 million in full 1CA amount of peaches = 366 360 tonnes  CA produced (3) 2,5 12.1.1 3.3.1 Gauteng's production area = A 1A writing in fraction form 12.4.4 100 L1 1 =  CA 40 1CA simplification (2) 12.1.1 3.3.2 Percentage 1M subtracting from 60% L1 (Piketberg) = 60% – (12+20+11)% M 1A correct percentage = 17% A (2) A 12.4.4 3.3.3 Klein Karoo and Free State 11% 1A Klein Karoo L1 Wolsley/Tulbagh and Limpopo 12 % 1A Wolsley/Tulbagh A (2) 12.4.4 3.3.4 Ceres A 2A correct area L1 (2) [22] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 7 DBE/Feb.–Mar. 2014 NSC – Memorandum QUESTION 4 [23 MARKS] Ques Solution Explanation AS/L 12.1.1 4.1.1 A = 768 + 1 080 + 4 455 + 2 268 M 1M adding L1 = 8 571 CA 1CA correct value of A (2) 12.1.1 4.1.2 B × 3 × 5 × 9 = 4 455 M 1M multiplying and L1 B × 135 = 4 455 equating to 4 455 B = 33 CA 1CA correct value of B (2) A 12.1.1 4.1.3 36 × 3 × 2 × C = 1 080 1A correct number of L2 216 × C = 1 080 grades 1 080 C= M 1M dividing 216 C = 5 CA 1CA value of C (3) 12.1.1 4.2.1 The Book RT 1RT correct price L1 (1) 4.2.2 R1,80 RT 1RT median 12.4.3 (1) L1 12.4.4 4.2.3 1,52; 1,52; 1,50; 1,48; 1,47; 1,32; 1,32; 1,25; 1,10 A 2A correct order L1 (2) A A 12.4.3 1A R1,32 4.2.4 R1,32 and R1,52 L1 1A R1,52 (2) 12.4.3 4.2.5 Range = R8,99 – R7,68 M/A 1M/A subtracting extreme L1(1) = R1,31 CA values L2(1) 1CA correct range (2) M 12.4.3 4.2.6 1,70 + 1,73 + 1,75 + 1,75 + 1,80 + 1,92 + 1,99 + 2,05 + 2,15 1M finding mean L2 Mean = 9 16,84 = A 1A simplification 9 = 1,871111… CA 1CA mean ≈ R1,87 (3) 1R rounding to nearest 200 12.2.1 4.3.1 768 exercise books ≈ 800 exercise books 1CA number of packs L2 R 800 768 A OR Number of packs = OR Number of packs = 1A dividing 200 200 = 4 CA = 3,84 1R number of packs ≈4 R (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 8 DBE/Feb.–Mar. 2014 NSC – Memorandum Ques Solution Explanation AS/L A 12.2.1 4.3.2 R16 200 × 20 1A number per pack L2 Price per pack = SF 4 455 1SF substitution 1CA price per pack = R72,73 CA (3) [23] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 9 DBE/Feb.–Mar. 2014 NSC – Memorandum QUESTION 5 [26 MARKS] Ques Solution Explanation AS/L 12.3.4 5.1.1 C 3 A OR 3C 1A C L1 1A 3 (2) 12.3.1 5.1.2 Distance = 8 mm A 2A correct measurement L1 (2) 12.3.4 5.1.3 North East A 2A correct direction L2 (1) A A 12.3.4 5.1.4 R75 and R329 2A 1 mark for each road L1 (2) 12.4.3 5.2.1 Jackal A 1A correct predator L1 (1) 12.4.2 5.2.2 LOSS OF LIVESTOCK BY PREDATORS 4A any 4 bars L2 plotted correctly 1A all bars plotted correctly Unknown A 1M correct type of graph Stray Dogs A Leopards A M A Jackals Predator Caracals Bushpigs A Birds Baboons 0 10 20 30 40 % contribution to loss (6) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 10 DBE/Feb.–Mar. 2014 NSC – Memorandum Ques Solution Explanation AS/L A A A 12.3.1 5.3.1 Total length = 6 × 1,5 m + 8 × 1 m + 5 × 2 m 1A using 1,5 m L2 (a) = 9 m + 8 m + 10 m pieces = 27 m CA 1A using 1 m pieces 1A using 2 m pieces 1CA length (4) 12.3.1 5.3.1 Total area of mesh wire = 3 × B × H + 2 × L( H + B) SF 1SF substitute in L1 (b) = 3 × 1 m × 1,5 m + 2 × 2 m( 1,5 m + 1 m) formula = 4,5 m2 + 10 m2 S 1S simplify = 14,5 m2 CA 1CA surface area (3) 12.1.1 5.3.2 Total cost = R59,95 per m² × 699,3 m² M/A 1M/A multiplying L2 = R41 923,035 CA correct amounts ≈ R41 920 R 1CA solution 1R rounding (3) 12.1.1 5.3.3 Original 2 m becomes 3 m 1A using ratio L1 3 A A ∴ 1 m becomes m A OR 2 : 1 : 1,5 = 3 : 1,5 : 2,25 2 ∴ the height = 2,25 m CA 1CA height (2) [26] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 11 DBE/Feb.–Mar. 2014 NSC – Memorandum QUESTION 6 [23 MARKS] Ques Solution Explanation AS/L 12.3.1 6.1.1 Area = length × breadth L1 = 30 cm × 45 cm 1A solution = 1 350 cm2 A A 1A correct unit (2) 12.3.1 6.1.2 Perimeter = 2(length + breadth) SF L1 = 2(30 cm + 45 cm) S 1SF correct substitution = 2(75 cm) 1S simplification = 150 cm CA 1CA simplifying (3) 12.3.1 6.2.1 75 cm A 2A correct length L1 (2) 12.3.1 6.2.2 180 cm = 2 × 75 cm + 30 cm M 1M breaking down 180 cm L2 A A 1A number of tea towels ∴ She can make 8 tea towels and 12 dish cloths 1A number of dish cloths (3) 12.3.1 6.2.3 Area (in cm2 ) = 900 – (3)2 (4 - 3,14) SF 1SF substitution L1 = 900 – 7,74 S 1S simplification = 892,26 CA 1CA simplifying (3) A 12.1.1 6.3.1 Cost of the material = R45,00 × length of material (in metres) 2A formula L1 (2) 1M multiplying by 45 12.2.2 6.3.2 A = 5 × R45 M 1CA value of A L1(2) = R225 CA L2(2) 360 1M dividing by 45 B= M 1CA value of B 45 = 8 CA (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 12 DBE/Feb.–Mar. 2014 NSC – Memorandum Ques Solution Explanation AS/L 12.2.2 6.3.3 COST OF THE MATERIAL L2 500 450 400 A 350 300 A Cost (in rand) 250 A 200 150 100 50 A 0 0 1 2 3 4 5 6 7 8 9 10 Length (in metres) 1A (0;0) 1A (8;360) 1A any other point plotted correctly 1A joining the points with a straight line (CA for values of A and B only) (4) [23] TOTAL 150 Copyright reserved