Mathematics Grade 9 ANA 2014 Memo (English)

Downloaded from hlayiso.com Page 1 of 9 ANNUAL NATIONAL ASSESSMENT 2014 GRADE 9 MATHEMATICS MEMORANDUM 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 (learners) should be credited. • Unless stated otherwise, learners who give a correct answer only, should be awarded full marks. • Underline errors committed by learners and apply Consistent Accuracy (CA) marking. • Do not penalise learners if units of measurement are omitted. KEY M Method mark CA Consistent Accuracy mark A Accuracy mark QUESTION 1 1.1 D 1.2 C 1.3 B 1.4 A 1.5 D 1. Give 1 mark for each correct answer. [10] 1.6 C 1.7 C 1.8 D 1.9 B 1.10 D QUESTION 2 2.1 2,07 × 10−7 A Answer: 1 mark (1) 2.2.1 3 �73 − (−3)2 = √73 − 9 3 Calculation: 1 mark 3 = √64 M Answer: 1 mark (Answer only: 1 mark) = 4CA (If the answer is a decimal number, no mark ) (2) 2.2.2 √100 < √110 < √121 M √100 and √121: 1 mark 10 < √110 < 11 A Answer: 1 mark (2) (Answer only: 1 mark) 2.2.3 3×59 = 3 × 52 M 57 Applying exponential law: 1 mark = 75CA Answer: 1 mark (Answer only: 1 mark) (2) 2.3 −2 1, 03 × 10 + 3, 8 × 10 −3 = 0,0103 + 0,0038 M Simplification: 1 mark = 0,0141 CA Answer: 1 mark (Answer only: 1 mark) or or = 10−2 (1,03 + 0,38) M or = 10−3 (10,3 + 3,8) M Taking out common factor: 1 mark = 10−2 (1,41) = 10−3 (14,1) = 0,0141 CA = 0,0141 CA Answer: 1 mark (Answer only: 1 mark) (2) [9] Grade 9 Mathematics Memo Page 1 of 9

Downloaded from hlayiso.com Page 2 of 9 QUESTION 3 3.1 2(𝑥 + 2)2 − (2𝑥 − 1)(𝑥 + 2) Squaring a binomial: 1 mark = 2(𝑥 2 + 4𝑥 + 4) − (2𝑥 2 + 3𝑥 − 2) M Product of 2 binomials:1 mark = 2𝑥 2 + 8𝑥 + 8 − 2𝑥 2 − 3𝑥 + 2 M Simplification: 1 mark Answer: 1 mark = 5𝑥 + 10 CA or or = (𝑥 + 2)(2(𝑥 + 2) − (2𝑥 − 1)) M Taking out common factor (𝑥 + 2): = (𝑥 + 2)(2𝑥 + 4 − 2𝑥 + 1) M 1 mark 2(𝑥 + 2) − (2𝑥 − 1):1 mark = (𝑥 + 2)(5) or 5𝑥 + 10 CA Simplification: 1 mark Answer: 1 mark (4) 3.2 15𝑥 2 𝑦 3 +9𝑥 2 𝑦3 or 15𝑥 2 𝑦 3 9𝑥 2 𝑦 3 Simplification: 1 mark + Answer: 1 mark 8𝑥 2 𝑦 3 8𝑥 2 𝑦 3 8𝑥 2 𝑦 3 24𝑥 2 𝑦 3 15 9 = M = + M 8𝑥 2 𝑦 3 8 8 24 = 3 CA = 8 (2) = 3 CA 3.3 𝑥 2 −4𝑥 𝑥 2 −2𝑥−8 𝑥(𝑥 − 4): 1 mark 𝑥(𝑥−4) = (𝑥−4)(𝑥+2) M (𝑥 − 4)(𝑥 + 2): 1 mark 𝑥 𝑥 : 1 mark = CA 𝑥+2 𝑥+2 (3) 3.4 𝑥2 + 2𝑥 2 − 7𝑥 2 2 3 6 3𝑥 2 + 4𝑥 2 − 7𝑥 2 :1 mark 3𝑥 2 +4𝑥 2 −7𝑥 2 = M Common denominator 6: 1 mark 6 Answer: 1 mark 0 = (If the expression is treated as an 6 equation, no mark i.e × by 𝟔) = 0 CA or or 𝑥2 2𝑥 2 7𝑥 2 + − 2 3 6 3𝑥 2 4𝑥 2 7𝑥 2 = + −  M Common denominator: 2 marks 6 6 6 Answer: 1 mark 0 = (If the expression is treated as an 6 equation, no mark i.e × by 𝟔) = 0  CA (3) 3.5 6𝑥 2 3𝑦 3 × 7𝑥𝑦 2𝑥 9𝑦2 9𝑦 2 : 1 mark = A 7 7: 1 mark or or 18𝑥 2 𝑦 3 = 14𝑥 2 𝑦 9𝑦2 9𝑦 2 : 1 mark = A 7: 1 mark (2) 7 [14] Grade 9 Mathematics Memo Page 2 of 9

Downloaded from hlayiso.com Page 3 of 9 QUESTION 4 4.1 3𝑥 2 𝑦 − 9𝑥𝑦 2 + 12𝑥 3 𝑦 3 3𝑥𝑦: 1 mark = 3𝑥𝑦(𝑥 − 3𝑦 + 4𝑥 2 𝑦 2 ) A 𝑥 − 3𝑦 + 4𝑥 2 𝑦 2 : 1 mark (2) 4.2 2(𝑥 + 𝑦) − 𝑡(𝑥 + 𝑦) (𝑥 + 𝑦): 1 mark (2 − 𝑡): 1 mark (2) = (𝑥 + 𝑦)(2 − 𝑡) A 4.3 4𝑥 2 − 𝑦 2 (2𝑥 − 𝑦): 1 mark = (2𝑥 − 𝑦)(2𝑥 + 𝑦) A or (2𝑥 + 𝑦)(2𝑥 − 𝑦) A (2𝑥 + 𝑦): 1 mark (2) 4.4 𝑥 2 − 11𝑥 + 18 (𝑥 − 9): 1 mark = (𝑥 − 9)(𝑥 − 2)) A or (𝑥 − 2)(𝑥 − 9)A (𝑥 − 2): 1 mark (2) [8] QUESTION 5 5.1 (𝑥 − 2)2 + 3𝑥 − 2 = (𝑥 + 3)2 𝑥 2 − 4𝑥 + 4: 1 mark 𝑥 2 − 4𝑥 + 4 + 3𝑥 − 2 = 𝑥 2 + 6𝑥 + 9 M 𝑥 2 + 6𝑥 + 9: 1 mark 𝑥 2 − 𝑥 + 2 = 𝑥 2 + 6𝑥 + 9 Simplification: 1 mark Answer: 1 mark (4) −7𝑥 = 7  𝑥 = −1 CA 5.2 𝑥 2 − 5𝑥 − 6 = 0 Factors: 1 mark (𝑥 − 6)(𝑥 + 1) = 0 𝐌 𝐨𝐫 (𝑥 + 1)(𝑥 − 6) = 0𝐌 𝑥 − 6 = 0 or 𝑥 + 1 = 0 (𝑥 + 1) = 0 or (𝑥 − 6) = 0 𝑥 = 6 or 𝑥 = −1CA 𝑥 = −1 or 𝑥 = 6CA Both answers: 1 mark (2) 5.3 𝑥+2 𝑥−3 − =0 3 4 × by 12: 4(𝑥 + 2) − 3(𝑥 − 3) = 0 M Multiplying by LCD: 1 mark 4𝑥 + 8 − 3𝑥 + 9 = 0 M Simplification: 1 mark Answer: 1 mark (3) 𝑥 = −17CA [9] Grade 9 Mathematics Memo Page 3 of 9

Downloaded from hlayiso.com Page 4 of 9 QUESTION 6 6.1 Position in the 1 2 3 4 5 sequence (𝑛) Term (𝑇𝑛 ) 1 8 27 64A 125A (2) 6.2 𝑇𝑛 = 𝑛 A 3 Answer: 1 mark (1) 6.3 𝑇𝑛 = 512 Substitution : 1 mark 𝑛3 = 512 M Answer: 1 mark 3 𝑛3 = 83 or 𝑛 = √512 (If 𝑻𝒏 ≠ 𝒏𝟑 1 mark for substitution from 6.2) (2) ∴ 𝑛 = 8 A [5] QUESTION 7 7.1. Gradient = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑦 𝑣𝑎𝑙𝑢𝑒 M 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥 𝑣𝑎𝑙𝑢𝑒 Formula: 1 mark 5−0 = 0−1 M Substitution : 1 mark = 5 Answer: 1 mark −1 Answer only: 3 marks = −5 A (3) 7. 2 𝑦 = −5𝑥 + 5 CA −5: 1 mark +5: 1 mark (2) 7.3 Gradient = −5 (parallel lines have equal gradients) CA Answer: 1 mark (1) [6] QUESTION 8 8.1 Decreased mass = 240 − � 15 × 240� 𝑘𝑔 100 = (240 − 36) 𝑘𝑔 A 36: 1 mark = 204 𝑘𝑔 A Answer: 1 mark or or Decreased mass = 85% of 240 𝑘𝑔 17 17 = � × 240� 𝑘𝑔  A : 1 mark 20 20 = 204 𝑘𝑔  A Answer: 1 mark (2) 8.2 Number of litres= 420 𝑘𝑚 CA 420 𝑘𝑚 : 1 mark 12 𝑘𝑚/𝑙 12 𝑘𝑚/𝑙 = 35 A Answer: 1 mark (2) 8.3 Number of boys: number of girls : total number = 5 ∶ 6 ∶ 11 5 : 1 mark 11 Number of boys = 𝐀 × 44M 5 × 44: 1 mark 11 (3) = 20 A Answer: 1 mark Grade 9 Mathematics Memo Page 4 of 9

Downloaded from hlayiso.com Page 5 of 9 8.4 Neither direct nor indirect proportion. A Neither direct nor indirect proportion: 1 mark Reason: Area ≠ a constant x length and Area x length ≠ a constant.  Reason: 1 mark Note: Direct proportion with explanation (as the side of the square increases the area increases): 1 mark (If answer is indirect proportion, no mark) (2) 8.5 𝑆𝐼 = 𝑃.𝑛.𝑟 M Formula: 1 mark 100 𝑅4 000(𝑛)(3) Substitution 𝑆𝐼: 1 mark 𝑅840 = 100 M 84 000 = 12 000𝑛 M Substitution 𝑃: 1 mark 84 000 Substitution 𝑟: 1 mark 𝑛= 12 000 𝑛 = 7 A Calculation: 1 mark Number of years= 7 Answer: 1 mark or or 𝐴 = 𝑃 + 𝑆𝐼 𝑅 4840 : 1 mark 𝐴 = 𝑅 4840 A 𝐴 = 𝑃(1 + 𝑛𝑖) M 𝑅4 840 = 𝑅4 000(1 + 𝑛 � �) M 3 Formula: 1 mark 100 𝑅4 840 = 𝑅4 000(1 + 𝑛(0,003)) Substitution 𝑃: 1 mark 𝑅4 840 = 𝑅4 000 + 𝑅120𝑛 M Substitution 𝑟: 1 mark 120𝑛 = 840 𝑛 = 7 CA Calculation: 1 mark Number of years= 7 Answer: 1 mark (6) 8.6 𝐴 = 𝑃(1 + 𝑖)𝑛 M Formula: 1 mark = 𝑅600(1 + 0,06)2 M = 𝑅600(1,06)2 Substitution 𝑃 : 1 mark = 𝑅674,16A Substitution 𝑖 = 0,06: 1 mark Answer: 1 mark (4) [19] QUESTION 9 9.1.1 � + 𝐹� = 90° or their sum is 90°A 𝐷 Answer: 1 mark (1) 9.1.2 180°A Answer: 1 mark (1) 9.1.3 360°A Answer: 1 mark (1) 9.1.4 parallel A Answer: 1 mark (1) 9.1.5 equalA Answer: 1 mark (1) Grade 9 Mathematics Memo Page 5 of 9

Downloaded from hlayiso.com Page 6 of 9 9.2 Statement Reason 𝐵�1 = 𝐶̂1 = 65° corr. ∠𝑠 and 𝐴𝐵 ∥ 𝑇𝐶 A Correct statement : ½ mark Reason: ½ mark 𝐶̂2 = 𝐴̂ = 43° alt. ∠𝑠 and 𝐴𝐵 ∥ 𝑇𝐶 A Correct statement : ½ mark Reason: ½ mark Correct statement : ½ mark 𝐵�2 = 180° − 65° ∠s on a str. line or adj. suppl. ∠s Reason: ½ mark (3) A = 115° 9.3 Statement Reason 𝐵� = 𝐶̂ = 𝑥 ∠𝑠 opp. equal sides of ∆  A Correct statement : ½ mark Reason: ½ mark 𝐴̂ + 𝐵� + 𝐶̂ = 180° sum of ∠𝑠 of ∆= 180° A Correct statement : ½ mark 𝐴̂ + 2𝑥 = 180° A Reason: ½ mark 𝐴̂ = 180° − 2𝑥 Substitution: ½ mark Answer: ½ mark (3) [11] QUESTION 10 10.1 Statement Reason ∆𝐷𝐸𝐹 ≡ ∆𝑃𝑄𝑅 A s ∠ 𝑠 A Correct statement : 1 mark Reason: 1 mark (2) 10.2 Statement Reason In ∆𝑀𝑁𝑇 and ∆𝑀𝑁𝑃 Correct statement : 1 mark 𝑀𝑇 = 𝑀𝑃 A equal radii A Reason: 1 mark 𝑀𝑁 = 𝑀𝑁 A common Correct statement : 1 mark Correct statement : 1 mark �1 = 𝑁 𝑁 �2 = 90° A given or 𝑀𝑁 ⊥ 𝑃𝑇 A Reason: 1 mark ∴ ∆𝑀𝑁𝑇 ≡ ∆𝑀𝑁𝑃 A 90° hyp. s A Correct statement : 1 mark Reason: 1 mark ∴ 𝑃𝑁 = 𝑁𝑇 corr. sides of congruent ∆𝑠 or Reason: 1 mark (8) ∆𝑠 are congruent  A Grade 9 Mathematics Memo Page 6 of 9

Downloaded from hlayiso.com Page 7 of 9 or Statement Reason In ∆𝑀𝑃𝑁 and ∆𝑀𝑇𝑁 𝑃� = 𝑇� A ∠𝑠 opp. equal radii A Correct statement : 1 mark Reason: 1 mark �1 = 𝑁 𝑁 �2 = 90° A given or 𝑀𝑁 ⊥ 𝑃𝑇 A Correct statement : 1 mark Reason: 1 mark 𝑀𝑁 = 𝑀𝑁 A common Correct statement : 1 mark ∴ ∆𝑀𝑃𝑁 ≡ ∆𝑀𝑇𝑁 A ∠ ∠ s A Correct statement : 1 mark Reason: 1 mark ∴ 𝑃𝑁 = 𝑁𝑇 corr. sides of congruent ∆𝑠 or (8) ∆𝑠 are congruent  A Reason: 1 mark 10.3.1 Statement Reason 𝐵𝐹 = 𝐶𝐸 given A Correct statement : ½ mark Reason: ½ mark 𝐵𝐹 + 𝐹𝐶 = 𝐶𝐸 + 𝐹𝐶 A Adding FC on both sides : 1 mark (2) ∴ 𝐵𝐶 = 𝐸𝐹 10.3.2 Statement Reason In ∆𝐴𝐵𝐶 and ∆𝐷𝐸𝐹 𝐴𝐶 = 𝐷𝐹 given A Correct statement : ½ mark Reason: ½ mark 𝐴𝐵 = 𝐷𝐸 given A Correct statement : ½ mark Reason: ½ mark Correct statement : ½ mark 𝐵𝐶 = 𝐸𝐹 proved A Reason: ½ mark Correct statement : 1 mark ∴ ∆𝐴𝐵𝐶 ≡ ∆𝐷𝐸𝐹 A 𝑠𝑠𝑠  A Reason: 1 mark (5) 10.3.3 Statement Reason corr. ∠𝑠 of congruent ∆𝑠 or ∆𝐴𝐵𝐶 ≡ ∆𝐷𝐸𝐹 A Correct reason:1 mark (1) Grade 9 Mathematics Memo Page 7 of 9

Downloaded from hlayiso.com Page 8 of 9 10.3.4 Statement Reason 𝐴𝐵 ∥ 𝐸𝐷 A corr. ∠𝑠 are equal  A Correct statement : 1 mark Reason: 1 mark Accept 𝐴𝐵 = 𝐸𝐷 A given  A (2) 10.4.1 Statement Reason In ∆𝐴𝐵𝐷 and ∆𝐴𝐶𝐸 𝐴̂ = 𝐴̂ A common A Correct statement : 1 mark Reason: 1 mark 𝐵� = 𝐶̂  A given A Correct statement : 1 mark �1 = 𝐸�1 Reason: 1 mark 𝐷 sum of .∠𝑠 of ∆= 180 ∆𝐴𝐵𝐷 ||| ∆𝐴𝐶𝐸  A ∠ ∠ ∠ A Correct statement : 1 mark Reason: 1 mark (6) 10.4.2 Statement Reason 𝐴𝐵 𝐵𝐷 𝐴𝐷 = = A proportional sides of similar ∆𝑠 A 𝐴𝐶 𝐶𝐸 𝐴𝐸 Correct statement : 1 mark 𝐵𝐷 9 Reason: 1 mark = M Substituting 𝐶𝐸 = 21, 21 7 𝐴𝐷 = 9, 𝐴𝐸 = 7: 1 mark 9(21) 𝐵𝐷 = 𝑐𝑚  M 7 Making 𝐵𝐷 subject of formula: 1 mark 𝐵𝐷 = 27 𝑐𝑚  A Answer: 1 mark (5) [31] QUESTION 11 11.1 In ∆𝐴𝐵𝑇: 𝐴𝐵2 = 𝐴𝑇 2 + 𝐵𝑇 2 M Pythagoras Formula/substitution: 1 mark (5)2 = 𝐴𝑇 2 + (3)2 25 = 𝐴𝑇 2 + 9 𝐴𝑇 2 = 16 𝑐𝑚2 A Calculations: 1 mark Answer: 1 mark 𝐴𝑇 = 4 𝑐𝑚 A or or In ∆𝐴𝐵𝑇: Formula/substitution: 1 mark 𝐴𝑇 2 = 𝐴𝐵2 − 𝐵𝑇 2 M Pythagoras = (25 − 9) 𝑐𝑚2 = 16 𝑐𝑚2  A Calculations: 1 mark (3) Answer: 1 mark 𝐴𝑇 = 4 𝑐𝑚 A Grade 9 Mathematics Memo Page 8 of 9

Downloaded from hlayiso.com Page 9 of 9 11.2.1 Perimeter of ADCT= (4 + 12 + 5 + 9) 𝑐𝑚 CA from 11.1 = 30 𝑐𝑚CA Answer: 1 mark (1) 11.2.2 Area of ADCT or Area of ADCT CA from 11.2.1 (𝐴𝐷+𝑇𝐶)×𝐴𝑇 1 = M = (sum of parallel sides)× height 2 (12+9)×4 2 Formula: 1 mark = 𝑐𝑚2 M M Substitution: 1 mark 2 (12+9)×4 = 42 𝑐𝑚2 CA = 𝑐𝑚2 M Answer: 1 mark 2 = 42 𝑐𝑚2 𝐂A (3) 11.3 2𝜋𝑟 = 52 𝑐𝑚M Formula/substitution: 1 mark 𝜋𝑟 = 26 𝑐𝑚 Answer: 1 mark 𝟐𝟔 Formula/substitution: 1 mark 𝑟 = 8,276 … 𝑐𝑚 or 𝑟 = 𝑐𝑚 A 𝝅 Rounded off answer: 1 mark If 𝑟 is rounded off maximum: Area = 𝜋𝑟 2 M 3 marks = 215,1774 … 𝑐𝑚2 Accept 215,29 or 215,09 ≈ 215,18 𝑐𝑚2 CA (4) 11.4 𝑘 = 2A (1) [12] QUESTION 12 12.1 2 𝑥 = �√8 + √2� 2 2 = �√8� + 2√8√2 + �√2� M Squaring a binomial: 1 mark = 8 + 8 + 2 M = 18 CA Simplifying: 1 mark Answer: 1 mark (Answer only: 1 mark) or or 2 𝑥 = �√8 + √2� 2 Simplifying: 1 mark 𝑥 = �2√2 + √2� M 2 𝑥 = �3√2�  M Simplifying: 1 mark Answer: 1 mark 𝑥 = 18 CA (Answer only: 1 mark) (3) 12.2 1 � 𝑥=3 Squaring both sides: 1 mark √ 1 1 = 9 M = 81: 1 mark √𝑥 𝑥 1 = 81 M Answer: 1 mark 𝑥 1 (Answer only: 1 mark) 𝑥= 81 CA or or 1 Squaring both sides: 1 mark � 𝑥=3 −1 √ 1 𝑥 4 = 3: 1 mark = 9 M 1 𝑥 = : 1 mark √𝑥 81 −1 𝑥 4 = 3M 𝑥 = 3−4 (Answer only: 1 mark) 1 (3) 𝑥 = 𝐂A 81 [6] Total: 140 Grade 9 Mathematics Memo Page 9 of 9