Mathematics Grade 9 MAINSTREAM 2012 Memo (English)

Downloaded from hlayiso.com Page 1 of 10 ANNUAL NATIONAL ASSESSMENT 2012 GRADE 9 MATHEMATICS TEST MEMORANDUM MARKS: 140 This memorandum consists of 9 pages. Important Information x This is a marking guideline. In instances where learners have used different but mathematically sound strategies to solve the problems they should be credited. x Unless stated otherwise, learners who give a correct answer only, should be awarded full marks. x Apply consistency accuracy and underline errors committed by learners. KEY M Method mark CA Consistent Accuracy mark A Accuracy mark QUESTION 1 1.1 C 1.2 B 1.3 C 1.4 C 1.5 B Give 1 mark for each 1. correct answer. 1.6 B 1.7 C 1.8 C 1.9 C 1.10 B [10] QUESTION 2 2.1 3,56 × 10ି଺ A Answer: 1 mark (1 ) 2.2.1 (3‫)ݔ‬ଷ + 2‫ ݔ‬ଷ 27‫ ݔ‬ଷ +2‫ ݔ‬ଷ : 1 mark = 27‫ ݔ‬ଷ + 2‫ ݔ‬ଷ 9 A = 29‫ ݔ‬ଷ 9 CA Answer = 29‫ ݔ‬ଷ : 1 mark (2) 2.2.2 a²b² 4a²bc Answer: 2 marks × ସ௔ర ௕య ௖ ac² 20b³ : 1 mark ૝ࢇ૝ ࢈૜ ࢉ ଶ଴௔௕మ ௖ మ = 9M ࢇ૜ ૛૙ࢇ࢈૛ ࢉ૛ ૞ࢉ : 1 mark ࢇ૜ (2 ) = ૞ࢉ9 Grade 9 Mathematics Memo 1

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Downloaded from hlayiso.com Page 2 of 10 2.3.1 4ab (5ܽ²ܾ² + 2ܾܽ െ3) Distributive law: 1 mark each term = 20ܽଷ ܾ ଷ 9 + 8ܽଶ ܾ ଶ 9 െ 12ܾܽ9 A 20ܽଷ ܾ ଷ : 1 mark 8ܽଶ ܾ ଶ : 1 mark െ12ܾܽ: 1 mark (3 ) 2.3.2 (2‫ ݔ‬െ 1)ଶ െ (‫ ݔ‬+ 1)(‫ ݔ‬െ 1) Expanding: 1 mark Simplifying: 1 mark = 4‫ ݔ‬ଶ െ 4‫ ݔ‬+ 19(െ‫ ݔ‬ଶ + 19 M Answer: 1 mark (3) = 3‫ ݔ‬ଶ െ 4‫ ݔ‬+ 29 CA 2.4.1 8‫݌‬³ + 4‫݌‬² Common factor: 1 mark (2‫ ݌‬+ 1): 1 mark (2) = 4‫݌‬²9(2‫ ݌‬+ 1)9 A 2.4.2 9‫݌‬²െ36‫ݍ‬² = 9(‫݌‬²െ4‫ݍ‬²) 9M = 9(‫ ݌‬െ 2‫ ݌()ݍ‬+ 2‫)ݍ‬9 A OR = (3‫ ݌‬െ 6‫()ݍ‬3‫ ݌‬+ 6‫ )ݍ‬9 A = 3(‫ ݌‬െ 2‫)ݍ‬3(‫ ݌‬+ 2‫)ݍ‬ Common factor: 1 mark = 9(‫ ݌‬െ 2‫ ݌()ݍ‬+ 2‫)ݍ‬9 A Difference of two squares: 1 mark (2 ) 2.4.3 ‫ ݔݐ‬െ ‫ ݕݐ‬െ 2‫ ݔ‬+ 2‫ݕ‬ = ‫ ݔ(ݐ‬െ ‫)ݕ‬9 െ 2(‫ ݔ‬െ ‫)ݕ‬9 M/A Common factor t: 1 mark = (‫ ݔ‬െ ‫)ݕ‬9(‫ ݐ‬െ 2) 9/ M/ A Common factor 2: 1 mark Common factor(‫ ݔ‬െ ‫)ݕ‬: 2 marks OR = ‫ ݐ(ݔ‬െ 2)9 െ ‫ ݐ(ݕ‬െ 2)9 = (‫ ݐ‬െ 2)9(‫ ݔ‬െ ‫)ݕ‬9 (4) 2.5.1 3(‫ ݔ‬+ 6) = 12 Dividing by 3: 1 mark ‫ ݔ‬+ 6 = 49 M Answer:1 mark ‫ = ݔ‬െ29 CA OR OR െ18 both sides:1 mark Answer:1 mark (2 ) 3‫ ݔ‬+ 18 = 12 3‫ = ݔ‬െ69 M ‫ = ݔ‬െ29 CA 2.5.2 ‫ ݔ‬ଶ െ 2‫ = ݔ‬0 Common factor: 1 mark ‫ ݔ(ݔ‬െ 2) = 09 M Answer: 2 marks (1 mark per factor) (3 ) ‫ = ݔ‬09 or ‫ = ݔ‬29 CA Grade 9 Mathematics Memo 2

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Downloaded from hlayiso.com Page 3 of 10 2.5.3 ௫ାଵ ௫ିଵ Multiplying by 6: 1 mark – = 1 Expanding: 1 mark ଷ ଺ Answer: 1 mark (3 ) 2(‫ ݔ‬+ 1) െ (‫ ݔ‬െ 1) = 69 M 2‫ ݔ‬+ 2 െ ‫ ݔ‬+ 1 = 69 CA ‫ = ݔ‬39 CA 2.5.4 2௫ାଵ = 32 2ହ : 1 mark Applying exponential law:1 mark 2௫ାଵ = 2ହ9 M Answer: 1 mark (3 ) ‫ ݔ‬+ 1 = 59 M/ CA ‫ = ݔ‬49 CA [30] QUESTION 3 3.1 3 Number of boys = 4 9 × 240 = 1809 M /CA 3 : 1 mark 4 Answer: 1 mark (2) 3.2 ௗ௜௦௧௔௡௖௘ Speed = 9M/ A Formula / Substitution: 1 mark ௧௜௠௘ ଽ௞௠ 40݇݉/݄ = ଽ ଽ ௧௜௠௘ ‫ݎ݋‬0,225: 1 mark ‫݁݉݅ݐ‬ × 60݉݅݊‫ݏ݁ݐݑ‬9 M ସ଴ ସ଴ (3) OR 0,225 × 60 minutes Answer: 1 mark = 13,5݉݅݊‫ݏ݁ݐݑ‬9 CA 3.3 ‫(ܲ = ܣ‬1 + ݅)௡ 9 M/ A OR Formula: 1 mark = ܴ12000(1 + 0,065)ହ 9 M = ܴ16441,049 CA Year 1: 12 000 x 6,5% = 780 Substitution: 1mark Year 2: 12 780 x 6,5% = 830,70 (3) Year 3: 13 610,7 x 6,5% = 884,70 Answer: 1 mark OR Year 4: 14 495,4 x 6,5% = 942,20 ௡ Year 5: 15 437,6 x 6,5% = 1 003,44 ௥ ‫ ܲ = ܣ‬ቀ1 + ቁ 9 There will be R16 441,04 in his account ଵ଴଴ ଺,ହ ହ = ܴ12000 ቀ1 + ቁ 9 ଵ଴଴ = ܴ16441,049 3.4.1 ଵହ Answer: 1 mark (1) 15% cash= × ܴ15000 ଵ଴଴ = ܴ22509 A 3.4.2 ‫(ܲ = ܣ‬1 + ݊݅)9 CA from 3.4.1 Formula: 1 mark = ܴ12 7509[1 + 2(0,1)]9 = ܴ153009 Substitution: 2 marks OR Answer: 1 mark ௉௡௥ OR ܵ‫= ܫ‬ 9 ଵ଴଴ Formula: 1 mark ோଵଶ ଻ହ଴9(ଶ)(ଵ଴) Substitution: 1mark = 9 ଵ଴଴ Answer: 1 mark Answer: 1 mark (4) = ܴ2250 Total amount= ܴ12750 + ܴ2250 Grade 9 Mathematics Memo 3

Downloaded from hlayiso.com Page 4 of 10 3.4.3 Monthly payments= ܴ15300 ÷ 249 CA from 3.4.2 ܴ15300 ÷ 24: 1 mark = ܴ637,509 Answer: 1 mark (2) = ܴ153009 [15] QUESTION 4 4.1 189ܽ݊݀239 A Answer: 2 marks 1 mark per term (2) 4.2 Add 5 to the last term to get the next.9 A Correct answer: 1 mark (1) 4.3 ܶ௡ = 5݊9 െ 29 A Answer: 2 marks (2) 4.4 ܶ௡ = 5݊ െ 2 CA from 4.3 Substitution into the formula: 1 mark 38 = 5݊ െ 29 CA Solving equation: 1 mark 40 = 5݊9 M Answer: 1 mark (3) ‫ ׵‬8 = ݊9 A [ૡ] Grade 9 Mathematics Memo 4

Downloaded from hlayiso.com Page 5 of 10 QUESTION 5 5.1.1 Parallel9 A Answer: 1 mark (1) 5.1.2 ‫ = ݕ‬െ29 A Answer: 1 mark (1) 5.1.3 49 A Answer: 1 mark (1) 5.1.4 Non-linear 9 A Answer: 1 mark (1) 5.2.1 9 9 9 9 9 1 9 2 9 9 X-intercept and Y-intercept: 2 marks per graph99 +99 Gradient: 1 mark per graph9+9 Correct labelling of graph: 1 mark per graph9 +9 (8) 5.2.2 At point of intersection:െ2‫ ݔ‬+ 1 = ‫ ݔ‬െ 29 Equating:1 mark െ3‫ = ݔ‬െ3 Substitution: 1 mark (2) ‫ = ݔ‬1 ܽ݊݀‫( = ݕ‬1) െ 2 = െ19 Grade 9 Mathematics Memo 5

Downloaded from hlayiso.com Page 6 of 10 OR ‫ = ݕ‬െ2‫ ݔ‬+ 1 ‫ = ݔ݂ܫ‬1; ‫( = ݕ‬െ2(1) + 1) =  െ1 ‫ ݔ = ݕ‬െ 2‫ = ݔ݂ܫ‬1; ‫ = ݕ‬1െ=  െ1 Therefore, this point lies on both graphs and it must be the point of intersection [14] . QUESTION 6 6.1 6.1.1 2‫ ݔ‬+ 3‫ ݔ‬+ 4‫ = ݔ‬180°9(TQ ෡ Sisastraightangle)9 M/ A Statement: 1 mark 9‫ = ݔ‬180° Reason: 1 mark ‫ = ݔ‬20°9 A (3) Answer: 1 mark 6.1.2 ‫ = ݕ‬3‫ ݔ‬9 (alternate ‘ s ;PQ|| RS)9 M Statement: 1 mark ‫ =ݕ‬60° 9 CA for 6.1.1 Reason: 1 mark Answer: 1 mark (3) 6.1.3 ܵመ = ܳ෠ଵ 9 (corresponding ‘ s; PQ|| RS9 M ‫ = ݖ‬2‫ݔ‬° ‫ = ݖ‬40°9 A OR ܳ෠ + ܴ෠ + ܵመ = 180° 9 (sum of ‘ s of ο= 180଴ )9M 4‫ ݔ‬+ ‫ ݕ‬+ ‫ = ݖ‬180° 80° + 60° + ‫ = ݖ‬180° ‫ = ݖ‬180° െ 140 ‫ = ݖ‬40°9 A OR ܳ෠ଵ + ܳ෠ଶ = ܴ෠ + ܵ ෡ 9 (ext ‘ of = sum of int opp ‘ s)9M 2‫ ݔ‬+ 3‫ ݕ = ݔ‬+ ‫ݖ‬ 5‫ = ݔ‬60° + ‫ݖ‬ Statement: 1 mark 100° = 60° + ‫ݖ‬ Reason: 1 mark ‫ = ݖ‬40°9 A Answer: 1 mark (3) 6.2 οܸܵܶ ‫ ؠ‬ο‫ܥܤܣ‬9(S, ‘ , S)9 A Statement with vertices in the correct order: 1 mark Reason : 1 mark (2) 6.3.1 In ¨$%'DQG¨$&' Correct statement with reason: 1 mark AB = AC (given) 9 A Correct statement with reason: 1 mark Correct statement with reason: 1 mark BD = CD (given)9 A Correct deduction and reason : 1 mark (4) AD = AD ( common side) 9A ¨ ABD Ł¨$&D (s, s, s)9 A OR In ¨$%'DQG¨$&' ‘ ABD = ‘ ACD (opposite angles of a kite) AB = AC (given) 9 BD = CD (given)9 ¨ ABD Ł¨$&' V ‘ , s)9 Grade 9 Mathematics Memo 6

Downloaded from hlayiso.com Page 7 of 10 6.3.2 In ¨$%(DQG¨$&( AB = AC (given) 9 A AE = AE (common side)9 A ‫ܣ‬መଵ = ‫ܣ‬መଶ ( corr. ‘ s of congr. οs/ bisecting diagonals) 9 A ¨ ABE Ł¨$&( V, ‘ , s)9 A OR In ¨$%(DQG¨$&( AB = AC (given) 9 Correct statement with reason: 1 mark ‫ܤ‬෠ଵ = ‫ܥ‬መଵ (L‫ݏ‬opp = sides of ) 9 Correct statement with reason: 1 mark ‫ܣ‬መଵ = ‫ܣ‬መଶ (corr. ‘ s of congr. οs) or BE=EC (AD bisects Correct statement with reason: 1 mark BC)9 Correct deduction and reason: 1 mark (4) ABE Ł¨$&(( ‘ , ‘ ,s) 9 6.3.3 ‫ܧ‬෠ଵ + ‫ܧ‬෠ଶ = 180°( B‫ܧ‬෠ C is str L)9 A Correct statement with reason: 1 mark Correct statement with reason: 1 mark But ‫ܧ‬෠ଵ = ‫ܧ‬෠ଶ (ܿorr. ‘ of congr. οs)9 A Conclusion: 1 mark (3) ‫ܧ ׵‬෠ଵ = ‫ܧ‬෠ଶ = 90°9 A OR Correct statement: 2 marks BC A AD 99 (diagonals of a kite) 9 Reason: 1 mark 6.3.4 AE is perpendicular to BC9 A Answer: 1 mark (1) 6.4 ஺஻ ஻஼ Proportional sides and reason: 2 marks = 9 (proportionalsidesofsimilarο‫)ݏ‬9 A ா஽ ஽ி substitution: 1 mark ஺஻ ଵହ௖௠ = 9 A Answer: 1 mark ଺௖௠ ଵ଴௖௠ (4) (ଵହ)(଺) ‫= ܤܣ‬ ܿ݉ ଵ଴ ‫ = ܤܣ‬9ܿ݉9 A [27] QUESTION 7 7.1 ‫ܦ‬ᇱ (െ1; 1)9‫ܧ‬Ԣ(1; 1)9 A Answer:2 marks 1 mark per pair of co-ordinates (2) 7.2 ‫ܣ‬ᇱ (1; 7)9‫ܤ‬Ԣ(1; 1)9 A Answer:2 marks 1 mark per pair of co-ordinates (2) 7.3 The perimeter= (3 + 1,5 + 1 + 1 + 2 + 0,5)‫ = ݏݐ݅݊ݑ‬9‫ݏݐ݅݊ݑ‬99 Length of reduced sides: 1 mark M/A Answer: 1 mark (2) OR Grade 9 Mathematics Memo 7

Downloaded from hlayiso.com Page 8 of 10 (6+3+2+2+4+1)÷2= 9 units 7.4 Area of fig. P: Area of reduced fig= 4: 199 M/ A Answer: 2 marks ଵ OR (Dimensions halved: Area ସ of original) (2) ଵ Area of fig P to Area of reduced Fig = 1:0,25 or 1: ସ OR Area of fig P to Area of reduced Fig = 10:2,5 [8] QUESTION 8 8.1 3-D figure Name of figure No. of No. of vertices No. of 1 mark faces edges for each answer Cylinder9 3 faces9 No vertices9 2 edges9 Shape of the faces: Circles9 and a rectangle9 (6) 8.2 The surface area of a rectangular prism is: Formula/substitution into the correct = 2(݈ × ܾ) + 2(݈ × ݄) + 2(ܾ × ݄) formula: 3 marks = 2(7,2 × 5)9 + 2(7,2 × 3,32)9 + 2(5 × 3,32)݉ଶ 9 M/ A Calculation: 1 mark (5) = (72 + 47,808 + 33,2)݉ଶ 9 CA Answer: 1 mark = 153, 01݉ଶ9 CA Wrong formula: 0 marks OR 2݈ܾ + 2(݈ + ܾ)݄9 = 2(7,2 × 5) + 2(7,2 + 5)3,3299 = 72݉ଶ + 81,01݉ଶ 9 = 153,01݉ଶ9 8.3 Volume of a cylinder= 1000ܿ݉ଷ Formula: 1 marks ߨ‫ ݎ‬ଶ ݄ = 1000ܿ݉ଷ99 M Substitution: 1 mark ߨ(2,82)ଶ ݄ = 1000ܿ݉ଷ9 M Calculation: 1 mark (4) Answer: 1 mark ݄ = 40,0268ܿ݉ ൎ 40ܿ݉9 CA [15] Grade 9 Mathematics Memo 8

Downloaded from hlayiso.com Page 9 of 10 QUESTION 9 9.1.1 Class Interval Tally marks Frequency 1 mark per correct interval 1 – 10 ___ 0 11 – 20 2 9 II 21 – 30 12 9 IIII IIII II 31 – 40 IIII IIII 9 9 41 – 50 IIII II 7 9 A (4) 9.1.2 14 Labelling: 2 marks Correct bars: 2 marks 12 A 10 8 Frequency 9 6 4 2 0 1 - 10 11- 20 21 - 30 31 - 40 41 - 50 (4) Number of marks 9 9.2.1 Range= 10 െ 5 = 59 A Range: 1 mark Median: 1 mark (2) 55678ૡ999910 Median = 89 A 9.2.2 Mode= 59 Answer: 1 mark (1) ହା଻ା଻ାହାହା଻ାହାହା଼ା଺ 9.2.3 Mean = 9M Formula:1 mark ଵ଴ Answer: 1 mark (2) = 69 A [13] TOTAL: 140 Grade 9 Mathematics Memo 9

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