Mathematics Grade 9 DEAF 2012 Memo (English)

Downloaded from hlayiso.com Page 1 of 10 ANNUAL NATIONAL ASSESSMENT 2012 GRADE 9 MATHEMATICS MEMORANDUM (DEAF LEARNERS) MARKS: 140 This memorandum consists of 9 pages. Important Information  This is a marking guideline. In instances where learners have used different but mathematically sound strategies to solve the problems they should be credited.  Unless stated otherwise, learners who give a correct answer only, should be awarded full marks.  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 × 10 A Answer: 1 mark (1 ) 2.2.1 (3 ) + 2 27 +2 : 1 mark = 27 + 2  A = 29  CA Answer = 29 : 1 mark (2) 2.2.2 a²b² 4a²bc Answer: 2 marks × ac² 20b³ : 1 mark = M : 1 mark (2 ) =  Grade 9 Mathematics Memorandum (Deaf Learners) 1

Downloaded from hlayiso.com Page 2 of 10 2.3.1 4ab (5 ² ² + 2 −3) Distributive law: 1 mark each term = 20 +8  − 12  A 20 : 1 mark 8 : 1 mark −12 : 1 mark (3 ) 2.3.2 (2 − 1) − ( + 1)( − 1) Expanding: 1 mark Simplifying: 1 mark =4 −4 +1 − +1  M Answer: 1 mark (3) =3 −4 +2  CA 2.4.1 8 ³+4 ² Common factor: 1 mark (2 + 1): 1 mark (2) = 4 ²  (2 + 1) A 2.4.2 9 ²−36 ² = 9( ²−4 ²)  M = 9( − 2 )( + 2 ) A OR = (3 − 6 )(3 + 6 )  A = 3( − 2 )3( + 2 ) Common factor: 1 mark = 9( − 2 )( + 2 ) A Difference of two squares: 1 mark (2 ) 2.4.3 − −2 +2 = ( − ) − 2( − ) M/A Common factor t: 1 mark = ( − ) ( − 2)  M/A Common factor 2: 1 mark Common factor( − ): 2 marks OR = ( − 2) − ( − 2) = ( − 2)( − ) (4) 2.5.1 3( + 6) = 12 Dividing by 3: 1 mark + 6 = 4 M Answer:1 mark = −2 CA OR OR −18 both sides:1 mark Answer:1 mark (2 ) 3 + 18 = 12 3 = −6 M = −2 CA 2.5.2 −2 = 0 Common factor: 1 mark ( − 2) = 0 M Answer: 2 marks (1 mark per factor) (3 ) = 0 or = 2 CA Grade 9 Mathematics Memorandum (Deaf Learners) 2

Downloaded from hlayiso.com Page 3 of 10 2.5.3 Multiplying by 6: 1 mark – =1 Expanding: 1 mark Answer: 1 mark (3 ) 2( + 1) − ( − 1) = 6 M 2 + 2 − + 1 = 6 CA = 3 CA 2.5.4 2 = 32 2 : 1 mark Applying exponential law:1 mark 2 =2 M Answer: 1 mark (3 ) + 1 = 5 M/ CA = 4 CA [30] QUESTION 3 3.1 3 Number of boys = 4  × 240 = 180 M /CA 3 : 1 mark 4 Answer: 1 mark (2) 3.2 Speed =  M/ A Formula / Substitution: 1 mark 40 /ℎ = × 60 M 0,225 1 mark (3) OR 0,225 × 60 minutes Answer: 1 mark = 13,5  CA 3.3 (1 + )  M/ A = OR Formula: 1 mark = 12 000(1 + 0,065)  M Year 1: 12 000 x 6,5% = 780 Substitution: 1mark = 16 441,04 CA 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 = 1+  There will be R16 441,04 in his account , = 12 000 1 +  = 16 441,04 3.4.1 15% cash = × 15 000 Answer: 1 mark (1) = 2 250 A 3.4.2 = (1 + ) CA from 3.4.1 Formula: 1 mark = 12 750[1 + 2(0,1)] = 15 300 Substitution: 2 marks OR Answer: 1 mark OR =  Formula: 1 mark Grade 9 Mathematics Memorandum (Deaf Learners) 3

Downloaded from hlayiso.com Page 4 of 10 Substitution: 1mark ( )( ) =  Answer: 1 mark (4) Answer: 1 mark = 2 250 Total amount= 12 750 + 2 250 = 15 300 3.4.3 Monthly payments = 15 300 ÷ 24 CA from 3.4.2 15 300 ÷ 24: 1 mark = 637,50 Answer: 1 mark (2) = 15 300 [15] QUESTION 4 4.1 18  23 A Answer: 2 marks 1 mark per term (2) 4.2 Add 5 to the last term to get the next term. A Correct answer: 1 mark (1) 4.3 = 5 − 2 A Answer: 2 marks (2) 4.4 = 5 − 2 CA from 4.3 Substitution into the formula: 1 mark 38 = 5 − 2 CA Solving equation: 1 mark 40 = 5  M Answer: 1 mark (3) ∴8 = A [ ] Grade 9 Mathematics Memorandum (Deaf Learners) 4

Downloaded from hlayiso.com Page 5 of 10 QUESTION 5 5.1.1 Parallel A Answer: 1 mark (1) 5.1.2 = −2 A Answer: 1 mark (1) 5.1.3 4 A Answer: 1 mark (1) 5.1.4 Non-linear  A Answer: 1 mark (1) 5.2.1     X-intercept and Y-intercept: 2 marks per graph + Gradient: 1 mark per graph+ Correct labelling of graph: 1 mark per graph + (8) Grade 9 Mathematics Memorandum (Deaf Learners) 5

Downloaded from hlayiso.com Page 6 of 10 5.2.2 At point of intersection:−2 + 1 = − 2 Equation:1 mark −3 = −3 Answer: 1 mark (2) =1 = (1) − 2 = −1 ∴ (1; −1) OR = −2 + 1 = 1; = (−2(1) + 1) = −1 = −2 = 1; = 1− = −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 + 3 + 4 = 180°  (TQ S is a straight angle) M/ A Statement: 1 mark 9 = 180° Reason: 1 mark = 20° A (3) Answer: 1 mark 6.1.2 = 3  (alternate Ls ;PQ|| RS) M Statement: 1 mark = 60°  CA for 6.1.1 Reason: 1 mark Answer: 1 mark (3) 6.1.3 =  (corresponding Ls; PQ|| RS M =2 ° = 40° A OR + + = 180°  (sum of Ls of ∆ = 180 ) M 4 + + = 180° 80° + 60° + = 180° = 180° − 140 = 40° A OR + = +  (ext L of = sum of int opp Ls) M 2 +3 = + 5 = 60° + Statement: 1 mark 100° = 60° + Reason: 1 mark = 40° A Answer: 1 mark (3) 6.2 ∆ ≡∆  (S, <, S) A Statement with vertices in the correct order: 1 mark Reason : 1 mark (2) 6.3.1 In ∆ ABD and ∆ ACD Correct statement with reason: 1 mark AB = AC (given)  A Correct statement with reason: 1 mark Correct statement with reason: 1 mark BD = CD (given) A Correct deduction and reason : 1 mark (4) AD = AD ( common side) A ∆ ABD ≡ ∆ ACD (s, s, s) A OR Grade 9 Mathematics Memorandum (Deaf Learners) 6

Downloaded from hlayiso.com Page 7 of 10 In ∆ ABD and ∆ ACD  ABD =  ACD (opposite angles of a kite) AB = AC (given)  BD = CD (given) ∆ ABD ≡ ∆ ACD (s,  , s) 6.3.2 In ∆ ABE and ∆ ACE AB = AC (given)  A AE = AE (common side) A = corr. <s of congr. ∆s/ bisecting diagonals)  A ∆ ABE ≡ ∆ ACE (s, L, s) A OR In ∆ ABE and ∆ ACE AB = AC (given)  Correct statement with reason: 1 mark = (L opp = sides of )  Correct statement with reason: 1 mark = corr. <s of congr. ∆s) or BE=EC (AD bisects BC) Correct statement with reason: 1 mark  Correct deduction and reason: 1 mark (4) ∆ ABE ≡ ∆ ACE (L, L,s) 6.3.3 + = 180°( B C is str L) A Correct statement with reason: 1 mark Correct statement with reason: 1 mark But = (corr. <s of congr. ∆s) A Conclusion: 1 mark (3) ∴ = = 90° A OR BC  AD  (diagonals of a kite)  6.3.4 AE is perpendicular to BC A Answer: 1 mark (1) 6.4 =  (proportional sides of similar ∆ ) A Proportional sides and reason: 2 marks substitution: 1 mark =  Answer: 1 mark (4) ( )( ) = = 9  [27] QUESTION 7 7.1 ) (−1; 1) ) ′(1; 1) A Answer:2 marks 1 mark per pair of co-ordinates (2) 7.2 ) (1; 7) ) ′(1; 1) A Answer:2 marks 1 mark per pair of co-ordinates (2) Grade 9 Mathematics Memorandum (Deaf Learners) 7

Downloaded from hlayiso.com Page 8 of 10 7.3 The perimeter = (3 + 1,5 + 1 + 1 + 2 + 0,5) =9  Length of reduced sides: 1 mark M/A Answer: 1 mark (2) OR (6+3+2+2+4+1)÷2= 9 units 7.4 Area of fig. P: Area of reduced fig = 4: 1 M/A Answer: 2 marks OR (Dimensions halved: Area of original) (2) Area of fig P to Area of reduced Fig = 1:0,25 or 1: 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 Cylinder 3 faces No vertices 2 edges Shape of the faces: Circles and a rectangle (6) 8.2 The surface area of a rectangular prism is: Formula/ substitution into the correct = 2( × ) + 2( × ℎ) + 2( × ℎ) formula: 3 marks = 2(7,2 × 5) + 2(7,2 × 3,32) + 2(5 × 3,32)  M/ A Calculation: 1 mark (5) = (72 + 47,808 + 33,2)  CA Answer: 1 mark Wrong formula: 0 marks = 153, 01  CA OR 2 + 2( + )ℎ = 2(7,2 × 5) + 2(7,2 + 5)3,32 = 72 + 81,01  = 153,01  8.3 Volume of a cylinder = 1 000 Formula & substitution: 2 marks ℎ = 1 000  M Calculation: 1 mark (2,82) ℎ = 1 000 M Answer: 1 mark (4) ℎ = 40,0268 ≈ 40  CA [15] Grade 9 Mathematics Memorandum (Deaf Learners) 8

Downloaded from hlayiso.com Page 9 of 10 QUESTION 9 9.1.1 Class Interval Tally marks Frequency 1 mark per interval 1 – 10 ___ 0 11 – 20 2  II 21 – 30 12  IIII IIII II 31 – 40 9  IIII IIII 41 – 50 7  IIII II A (4) 9.1.2 14 Labelling: 2 marks Correct bars: 2 marks 12 10  8 A Frequency  6 4 2 0 1 - 10 11- 20 21 - 30 31 - 40 41 - 50 (4) Number of marks  9.2.1 a) Range = 10 − 5 = 5  A Range: 1 mark (2) b) 5 5 6 7 8 9 9 9 9 10 Median: 1 mark Median = 8 A 9.2.2 Mode = 5 Answer: 1 mark (1) 9.2.3 Mean = M Formula:1 mark Answer: 1 mark (2) = 6 A [13] TOTAL: 140 Grade 9 Mathematics Memorandum (Deaf Learners) 9

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