Mathematics Grade 9 ANA 2013 Memo (English)

Downloaded from hlayiso.com Page 1 of 10 ANNUAL NATIONAL ASSESSMENT 2013 GRADE 9 MATHEMATICS MEMORANDUM MARKS: 140 This memorandum consists of 10 pages. Important Information • This is a marking guideline. In instances where learners have used different but mathematically sound strategies to solve the problems, they (learners) should be credited. • Unless otherwise stated, learners who give a correct answer only, should be awarded full marks. • Underline errors committed by learners and apply Consistent Accuracy (CA) marking. KEY M Method mark CA Consistent Accuracy mark A Accuracy mark QUESTION 1 1.1 B 1.2 A 1.3 B 1.4 B 1.5 C Give 1 mark for each 1. correct answer. 1.6 C 1.7 D 1.8 D 1.9 A 1.10 D [10] QUESTION 2 2.1 6𝑥 5 − 15𝑥 3 = 6𝑥 − 5𝑥M 6𝑥: 1 mark 𝑥 4 3𝑥 2 −5𝑥: 1 mark = 𝑥 CA Answer: 1 mark or 18𝑥 5 −15𝑥 5 3𝑥 5 18𝑥 5 −15𝑥 5 4 𝐌 = 4M : 1 mark 3𝑥 3𝑥 3𝑥 4 5 = 𝑥 CA 3𝑥 : 1 mark 3𝑥 4 Answer: 1 mark (3 ) 2.2 𝑥(𝑥 + 2) − (𝑥 − 1)(𝑥 − 3) 𝑥 2 + 2𝑥: 1 mark = 𝑥 2 + 2𝑥 − (𝑥 2 − 4𝑥 + 3) M 𝑥 − 4𝑥 + 3: 1 mark 2 = 𝑥 2 + 2𝑥 − 𝑥 2 + 4𝑥 − 3M Simplification: 1 mark = 6𝑥 − 3CA Answer: 1 mark (4) Grade 9 English Mathematics Memo 1

Downloaded from hlayiso.com Page 2 of 10 2.3 3 �225𝑥 4 − �125𝑥 6 = 15𝑥 2 𝐌 − 5𝑥 2 𝐌 15: 1 mark = 10𝑥 2 𝐂𝐀 𝑥 2 : 1 mark −5: 1 mark 𝑥 2 : 1 mark 10𝑥 2 : 1 mark (5) 2.4 2𝑥 + 1 𝑥 + 2 1 − − 4 2 4 Same denominator: 1 mark 2𝑥+1−2(𝑥+2)−1 Note: = M 4 If learners treat the expression as an equation 0 out of 4. If learners leave out denominator the maximum mark is 3 out of 4. 2𝑥+1−2𝑥−4−1 = M −2𝑥 − 4: 1 mark 4 −4 Simplifying numerator: 1 mark (4) = M 4 = −1CA Answer: 1 mark [16] QUESTION 3 3.1 6𝑎3 − 12𝑎2 + 18𝑎 Common factor 6𝑎 : 1 mark = 6𝑎(𝑎2 − 2𝑎 + 3) A 𝑎2 − 2𝑎 + 3:1 mark (2) 3.2 7𝑥 2 − 28 = 7(𝑥 2 − 4) A 7(𝑥 2 − 4): 1 mark = 7(𝑥 − 2)(𝑥 + 2) A 7(𝑥 − 2)(𝑥 + 2): 1 mark (2) Note: If learners give answer as [4] �√7𝑥 − √28��√7𝑥 + √28� , then give 1 mark out of 2. QUESTION 4 4.1 3𝑥 − 1 = 5 3𝑥 = 6 M Add 1 on both sides: 1 mark 𝑥 = 2 CA Answer: 1 mark (2) Grade 9 English Mathematics Memo 2

Downloaded from hlayiso.com Page 3 of 10 4.2 2(𝑥 − 2)2 = (2𝑥 − 1)(𝑥 − 3) Squaring a binomial: 1 mark 2(𝑥 2 − 4𝑥 + 4) = 2𝑥 2 − 7𝑥 + 3M Product of 2 binomials: 1 mark 2𝑥 2 − 8𝑥 + 8 = 2𝑥 2 − 7𝑥 + 3M 2𝑥 2 − 8𝑥 + 8: 1 mark Answer: 1 mark (4) 𝑥 = 5 𝑪𝑨 4.3 2𝑥 − 3 𝑥 + 1 3𝑥 − 1 + = 2 3 2 ×6 3(2𝑥 − 3) + 2(𝑥 + 1) = 3(3𝑥 − 1) M Multiply LHS and RHS by 6: 1 mark 6𝑥 − 9 + 2𝑥 + 2 = 9𝑥 − 3M Multiplying out: 1 mark 8𝑥 − 7 = 9𝑥 − 3M Simplification: 1 mark 𝑥 = −4 CA Answer: 1 mark (4) 4.4 𝑥 3 = 64 or 𝑥 3 = 64 𝑥 3 = 43 M 3 𝑥 = √64 M Calculation: 1 mark 𝑥 = 4 A 𝑥 = 4 A Answer: 1 mark (2) [12] QUESTION 5 5.1 19 and 23 A 19: 1 mark 23: 1 mark (2) 5.2 𝑇𝑛 = 4𝑛 + 3 𝐀 4𝑛: 1 mark or 𝑇𝑛 = 7 + 4(𝑛 − 1) 𝐀 +3: 1 mark or 7: 1 mark 4(𝑛 − 1): 1 mark (2) 5.3 𝑇𝑛 = 4(50) + 3𝐌 Substituting 50 for 𝑛: 1 mark = 203𝐂𝐀 Answer: 1 mark (2) Note: Give full marks if learner has correctly substituted in his/her “incorrect” general term from 5.2. [6] Grade 9 English Mathematics Memo 3

Downloaded from hlayiso.com Page 4 of 10 QUESTION 6 6.1 432 Time= 96 ℎ𝐌 Formula/ Substitution: 1 mark 36 = ℎ 8 1 = 4 ℎ or 4 ℎ 30 𝑚𝑖𝑛𝐂𝐀 Answer: 1 mark (2) 2 or Speed x time = distance 96 km/h x time = 432 km𝐌 432 𝑘𝑚 Time = = 4,5 ℎ𝐀 96 𝑘𝑚/ℎ 6.2 𝐴 = 𝑃(1 + 𝑛𝑖) M Formula: 1 mark 𝐴 = 𝑅3 500(1 + 3(0,06)) 𝐌 = 𝑅3 500(1,18) Substitution: 1 mark = 𝑅4 130,00 CA Calculation: 1 mark 𝑆. 𝐼 = 𝑅4 130 − 𝑅3 500𝐌 Subtraction: 1 mark = 𝑅630 𝐂𝐀 or S.I answer: 1 mark 𝑃.𝑛.𝑟 𝑆. 𝐼 = 100 𝐌 or Formula: 1 mark 𝑅3 500(3)(6) 𝑆. 𝐼 = M 100 Substitution (𝑃, 𝑛, 𝑟): 3 marks = 𝑅630 CA Answer: 1 mark (5) 6.3 𝐴 = 𝑃(1 + 𝑖)𝑛 M Formula: 1 mark = 𝑅7 500(1 + 0,13)3 𝐌 = 𝑅7 500(1,13)3 Substitution 𝑖 = 0,13: 1 mark = 𝑅10 821,73𝐂𝑨 Substitution 𝑃 & 𝑛: 1 mark or Year 1: 𝑅7 500 × 13% = 𝑅975,00 Answer: 1 mark (4) Year 2: 𝑅8 475,00 × 13% = 𝑅1 101,75 Year 3: 𝑅9 576,75 × 13% = 𝑅1 244,98 M The amount will be 𝑅10 821,73CA [11] Grade 9 English Mathematics Memo 4

Downloaded from hlayiso.com Page 5 of 10 QUESTION 7 7.1.1 A B C 𝑥 co-ordinate 0 2 4 1 mark per 𝑦 co-ordinate −2 0 2 pair of coordinates of each point (3)    𝑥: 1 mark 7.1.2 𝑦 = 𝑥 − 2 A −2: 1 mark (2) 7.2.1 Y x=1 4 0 1 2 X y = -2x + 4 𝑦 − intercept: 1 mark  𝑥 − intercept: 1 mark per graph +  Labelling graph: 1 mark per graph +  (5) 7.2.2 (1; 2) CA 1 mark for 𝑥-value Note: Give full marks if learner has correctly identified the 1 mark for 𝑦-value (2) point of intersection of his/her graphs. [12] Grade 9 English Mathematics Memo 5

Downloaded from hlayiso.com Page 6 of 10 QUESTION 8 Note: Penalty for leaving out reasons: only deduct 1 mark for this entire question. 8.1.1 𝑇�1 = 𝑃 �1 = 25° (∠𝑠 opp. equal sides of ∆)A Correct statement with reason: 1 mark (1) 8.1.2 �2 = 50° 𝑀 (ext ∠ of ∆𝑀𝑃𝑇) A or Correct statement with reason: 1 mark (1) (suppl. ∠𝑠 on a str line) A 8.1.3 𝑅� + 𝑇 �2 = 130° (sum of ∠𝑠 of ∆= 180°A Correct statement with reason: 1 mark But 𝑅� = 𝑇 �2 (∠𝑠 opp. equal sides of ∆)A Correct statement with reason: 1 mark = 65°A Answer: 1 mark (3) 8.2.1 𝐵𝐷 + 𝐷𝐸 = 𝐶𝐸 + 𝐷𝐸 A Answer: 1 mark (1) 8.2.2 ∆𝐴𝐶𝐷 ≡ ∆𝐴𝐵𝐸 (𝑠∠𝑠) A Correct statement and reason: 1 mark NOTE: Order of the vertices should be correct (1) 8.3 In ∆𝐾𝑁𝑄 and ∆𝑀𝑃𝑄 𝑄� = 𝑄� ( common) A Correct statement with reason: 1 mark 𝑁𝑄 = 𝑃𝑄 (given) A Correct statement with reason: 1 mark 𝐾𝑄 = 𝑀𝑄 (given)A Correct statement with reason: 1 mark ∴ ∆𝐾𝑁𝑄 ≡ ∆𝑀𝑃𝑄 (𝑠∠𝑠) 𝐀 Correct deduction with reason: 1 mark (4) 8.4.1 In ∆𝑄𝑃𝑁 and ∆𝐿𝑀𝑁 𝑁� =𝑁 � ( common angle) A Correct statement with reason: 1 mark � 𝑃1 = 𝑀 � (corr∠𝑠, QP || LM) A Correct statement with reason: 1 mark 𝑄� 1 = 𝐿� (corr∠𝑠, QP || LM) A Correct statement with reason: 1 mark Correct deduction and reason: 1 mark R ∴ ∆𝑄𝑃𝑁 ||| ∆𝐿𝑀𝑁 (∠∠∠ ) A Note: Do not penalize learners if they � =𝑁 leave out the third condition (𝑁 �). (4) 8.4.2 𝑄𝑃 𝑃𝑁 𝑄𝑁 = 𝑀𝑁 = 𝐿𝑁 (prop sides of similar ∆s)A Correct statement with reason:1 mark 𝐿𝑀 3 𝑃𝑁 Substitution: 1 mark = A 8 16 𝑃𝑁 = 6 𝑐𝑚 A Answer: 1 mark Note: Answer only give 3 marks (3) [18] Grade 9 English Mathematics Memo 6

Downloaded from hlayiso.com Page 7 of 10 QUESTION 9 9.1 and Y 9.3 A B X O B '' B' A '' A' Correct drawing of ∆𝐴′ 𝑂𝐵′ : 2 marks A Correct drawing of ∆𝐴′′ 𝑂𝐵′′ : 2 marks𝐀 (4) 9.2 𝐵′ (4; −2) CA NOTE: Give full marks if learner has written the coordinates of B’ from his/her triangle. Answer: 1 mark (1) 9.4 𝐴′ 𝐴′′ = 4 unitsCA NOTE: Give full marks if learner has given the correct length of his/her A’A” . Answer: 1 mark (1) [6] Grade 9 English Mathematics Memo 7

Downloaded from hlayiso.com Page 8 of 10 QUESTION 10 10.1.1 Area of ring = 𝜋𝑅 2 − 𝜋𝑟 2 𝐌/𝐀 Formula: 1 mark = 𝜋(𝑅 2 − 𝑟 2 ) Subtraction: (2) 1 mark 10.1.2 Area of ring or 𝜋(14 + 8)(14 − 8) 𝑐𝑚2 M Substitution: = 𝜋(142 ) − 𝜋(82 ) 𝑐𝑚2 M 1 mark = 132𝜋 𝑐𝑚2 𝐂𝐀 = 𝜋(22)(6) 𝑐𝑚 2 = 132𝜋 𝑐𝑚2 CA Answer: 1 mark (2) 10.2.1 𝑄𝑇 = 𝑇𝑅 = 24 𝑐𝑚  (∆𝑃𝑄𝑇 ≡ ∆𝑃𝑅𝑇) A Correct statement with (2) reason: 2 marks 10.2.2 In ∆𝑃𝑄𝑇: 𝑃𝑇 2 = (252 − 242 ) 𝑐𝑚2 (Pythagoras) M Correct statement with reason: = (625 − 576) 𝑐𝑚2 M or (25+24)(25-24) 𝑐𝑚2 M 2 marks = 49 𝑐𝑚2 𝑃𝑇 = 7 𝑐𝑚M Calculations: 1 mark (4) Answer: 1 mark 10.2.3 Area ∆𝑃𝑄𝑅 = 𝑏𝑎𝑠𝑒×ℎ𝑒𝑖𝑔ℎ𝑡 M 1 or = (𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡) Formula: 1 mark 2 2 = (48)(7) 𝑐𝑚2 M Substitution: 1 mark 2 = (24)(7) 𝑐𝑚2 Answer: 1 mark (3) = 168 𝑐𝑚2 CA 10.2.4 Volume = Area of base × heightM Formula/substitution: 1 mark = 168 𝑐𝑚2 × 80 𝑐𝑚 = 13 440 𝑐𝑚3 CA Answer: 1 mark (2) 10.2.5 Surface area = 2(𝐴𝑟𝑒𝑎 ∆𝑃𝑄𝑅) + 2(𝐴𝑟𝑒𝑎 𝑃𝑅𝑆𝑊) + 𝐴𝑟𝑒𝑎 𝑄𝑅𝑆𝑈M Formula: 1 mark = 2(168) 𝑐𝑚2 + 2(80 × 25) 𝑐𝑚2 + 80(48) 𝑐𝑚2 M Substitution: 3 marks = 336 𝑐𝑚2 + 4 000 𝑐𝑚2 + 3 840 𝑐𝑚2 = 8 176 𝑐𝑚2 CA Answer: 1 mark (5) [20] Grade 9 English Mathematics Memo 8

Downloaded from hlayiso.com Page 9 of 10 QUESTION 11 11.1 Mark x f f.x 1 2 2 2 3 6 3 4 12  4 6 24 5 7 35  𝑓. 𝑥 values: 4 marks 6 9 54  7 4 28 8 3 24  9 2 18 (4) 11.2 Number of learners =Σf =40 A Answer: 1 mark (1) 11.3 ∑ 𝑓𝑥 The mean mark = ∑ 𝑓 M Formula: 1 mark 203 Substitution: 1 mark = 40 M Answer: 1 mark = 5,075CA Note: If answer is given as 5 then (3) give full marks. 11.4 9 % of learners = × 100M Correct fraction: 1 mark 40 = 22,5CA Answer: 1 mark (2) NOTE: If answer is given as 22,5 then give full marks. [10] QUESTION 12 12.1 Stem Leaves 13 7  14 5679  15 033366788 16 355  17 037 Ordered table: 5 marks (5)  12.2.1 Range = (177 − 137) 𝑐𝑚 = 40 𝑐𝑚A Answer: 1 mark (1) 12.2.2 Mode = 153 𝑐𝑚A Answer: 1 mark (1) 12.2.3 Median = 156 𝑐𝑚A Answer: 1 mark (1) 12.2.4 14 A Answer: 1 mark (1) [9] Grade 9 English Mathematics Memo 9

Downloaded from hlayiso.com Page 10 of 10 QUESTION 13 13.1 𝑃(𝐺) = 5 A Answer: 1 mark (1) 12 13.2 4 𝑃(𝑊) = 12 = 3A 1 Answer: 1 mark (1) 13.3 𝑃(𝑊) = 3 A Answer: 1 mark (1) 11 [3] QUESTION 14 Let 𝑥 boys play soccer and hockey 150 + (130 − 𝑥) = 200 M or 130 + (150 − 𝑥) = 200 M Correct statement: 1 mark 280 − 𝑥 = 200M 280 − 𝑥 = 200 M Calculation: 1 mark 𝑥 = 80 A 𝑥 = 80 A Answer: 1 mark or Total number of boys who play hockey and soccer Correct statement: 1 mark = 150 + 130 = 280M Calculation: 1 mark But this is 80 more than the number of boys in grade 9 which means 80 boys must Answer: 1 mark (3) play both soccer and hockey M [3] TOTAL: 140 Grade 9 English Mathematics Memo 10