Mathematics Grade 9 Paper 1 November EC 2016 Memo (English)

Downloaded from hlayiso.com SENIOR PHASE GRADE 9 NOVEMBER 2016 MATHEMATICS MEMORANDUM MARKS: 100 This memorandum consists of 9 pages.

Downloaded from hlayiso.com 2 MATHEMATICS (EC/NOVEMBER 2016) NOTE:  This is marking guideline. In instances where learners have used different mathematically sound strategies to solve the problems, they should be credited.  Underline errors committed by learners and apply Consistent Accuracy (CA) marking. KEY M Method mark CA Consistent Accuracy mark A Accuracy mark S Statement R Reason S/R Statement and Reason QUESTION 1 [10 marks] Ques. Mark Allocation Total 1.1 B  1 mark for each correct answer (1) 1.2 C  (1) 1.3 A  (1) 1.4 B  (1) 1.5 A  (1) 1.6 D  (1) 1.7 C  (1) 1.8 A  (1) 1.9 C  (1) 1.10 C  (1) [10] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2016) MATHEMATICS 3 QUESTION 2 [26 marks] Ques. Solution Mark Allocation Total −7 2.1 6,74 × 10  A Answer: 1 mark (1) 3 2.2.1 √𝑥 3 + 𝑥 0 𝑥 : 1 mark AA +1: 1 mark 𝑥+1 (2) 2.2.2 √0,03 𝑥 8 + 0,01𝑥 8 8 √0,04 𝑥 : 1 mark A Answer: 1 mark √0,04 𝑥 8 0,2 𝑥 4 A (2) 2.2.3 (2𝑑 2 𝑒)2 2 4 2 2 𝑑 𝑒 : 1 mark (4𝑑 −3 𝑒 −2 )−1 22 𝑑 −3 𝑒 −2 : 1 mark M M Answer: 1 mark 2 4 2 2 −3 −2 2 𝑑 𝑒 ×2 𝑑 𝑒 16𝑑  A (3) 2 2 2.2.4 2(𝑥 + 2) − 2(𝑥 + 1)(𝑥 + 2) 𝑥 + 4𝑥 + 4 : 1 mark M M 𝑥 2 + 3𝑥 + 2 : 1 mark 2(𝑥 2 + 4𝑥 + 4) − 2( 𝑥 2 + 3𝑥 + 2) 2𝑥 2 + 8𝑥 + 8 − 2𝑥 2 − 6𝑥 − 4 2𝑥 2 + 8𝑥 + 8 − 2𝑥 2 − 6𝑥 − 4 A : 1 mark 2𝑥 + 4  CA Answer: 1 mark (4) 2.3.1 𝑥 2 + 5𝑥 − 24 𝑥 + 8: 1 mark A A 𝑥 − 3: 1 mark (𝑥 + 8)(𝑥 − 3) (2) 2.3.2 2(𝑎 − 𝑏) − 𝑏 + 𝑎 +1(𝑎 − 𝑏: 1 mark 2(𝑎 − 𝑏) + 1(𝑎 − 𝑏)  M (2 + 1)(𝑎 − 𝑏): 1 mark (2 + 1)(𝑎 − 𝑏) M Answer : 1 mark 3(𝑎 − 𝑏)  A (3) 2.4.1 4𝑥 − 10 = 6 4𝑥 = 16: 1 mark 4𝑥 = 16 M Answer : 1 mark 𝑥=4 A (2) 2.4.2 3𝑥 − 10 2𝑥 − 5 × LCD: 6: 1 mark = 2 3 9𝑥 − 30 = 4𝑥 − 10: 1 mark 3𝑥−10 2𝑥−5 Answer: 1 mark 6×( ) =6×( ) M 2 3 9𝑥 − 30 = 4𝑥 − 10 A 5𝑥 = 20 𝑥=4  CA (3) Copyright reserved Please turn over

Downloaded from hlayiso.com 4 MATHEMATICS (EC/NOVEMBER 2016) 2.4.3 𝑥2 = 4 (𝑥 + 2)(𝑥 − 2) = 0: 1 mark (𝑥 + 2)(𝑥 − 2) = 0  M Answer: 1 mark 𝑥 = −2 or 𝑥 = 2 A (2) 5 5 2.4.4 3𝑥 = 96 2 : 1 mark 𝑥 5 = 32 Answer: 1 mark 𝑥 5 = 25 M 𝑥=2 A (2) [26] QUESTION 3 [8 marks] Ques. Solution Mark Allocation Total 3.1.1 16  A Answer: 1 mark (1) 3.1.2 5𝑛: 1 mark AA 5𝑛 + 1 +1: 1 mark (2) 3.2.1 𝑦=𝑥 A Answer: 1 mark (1) 3.2.2 𝑥 = −2 A Answer: 1 mark (1)  𝑦 = −2𝑥 + 1   𝑦 intercept : 𝑦 = +1 A 1 𝑥 intercept : 𝑥 = 2 A label A (3) [8] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2016) MATHEMATICS 5 Question 4 [12 marks] Ques. Solution Mark Allocation Total 4.1 𝑝. 𝑛. 𝑖 = 𝑆𝐼 M Formula: 1 mark 5 000 × 𝑛 × 0,12 = 1 800 M Substitution: 1 mark Answer: 1 mark 𝑛=3 A OR 𝐴 = 𝑃(1 + 𝑛𝑖) M 6 800 = 5 000(1 + 0,12𝑛) M 1.36 = 1 + 0,12𝑛 0,36 = 0,12𝑛 𝑛 = 3 A (3) 4.2 Let the numbers be 𝑎 and 𝑏 M Any method: 1 mark 𝑎 + 𝑏 = 143 𝑎−𝑏 =7 75: 1 mark 𝑎 =𝑏+7 𝑏 + 7 + 𝑏 = 143 68: 1 mark 2𝑏 = 136 𝑏 = 68 𝑎 = 68 + 7 𝑎 = 75 A A The numbers are 75 and 68 OR 𝑎 + 𝑏 = 143 𝑎−𝑏 =7 ∴ 2𝑎 = 150 (adding the 2 equations) 𝑎 = 75 75 + 𝑏 = 143 𝑏 = 68 (3) Copyright reserved Please turn over

Downloaded from hlayiso.com 6 MATHEMATICS (EC/NOVEMBER 2016) 4.3 S = 10 Answer: 2 marks PENS PENCILS 10 − 7 = 3 boxes A (2) 4.4 𝑑 = 𝑠 × 𝑡 M Formula/method: 1 mark 𝑑 = 100 𝑘𝑚/ℎ × 3ℎ A 𝑑 = 300 𝑘𝑚 300 𝑘𝑚: 1 mark 𝑑 300 𝑘𝑚 𝑆=𝑡 : 1 mark 2 ℎ𝑟𝑠 300𝑘𝑚 Av. Speed = M Answer: 1 mark 2ℎ𝑟𝑠 = 150 𝑘𝑚/ℎ CA OR Speed 100 km x km Time 3 hrs 2 hrs 2x = 300 x = 150 km/h (4) [12] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2016) MATHEMATICS 7 QUESTION 5 [16 marks] Ques. Solution Mark Allocation Total 5.1.1 ∠𝐴𝐶𝐵 = ∠𝐷𝐶𝐹 = 32° (Vert. opp. ∠′𝑠) Statement and ∠𝐸𝐵𝐶 = ∠𝐴𝐶𝐵 = 32° (Alt. ∠′𝑠, EB ∥ DA) reason: 1 mark each (2) ′ 5.1.2 ∠𝐶𝐴𝐵 + ∠𝐴𝐵𝐸 = 180° ( Co int. ∠ 𝑠 ∶ 𝐸𝐵||𝐷𝐴 ) S/R Statement and ∠𝐶𝐴𝐵 = 180° − 65° M reason: 1 mark ∠𝐶𝐴𝐵 = 115°A Substitution: 1 mark Answer: 1 mark OR ∠𝐶𝐴𝐵 + ∠𝐴𝐶𝐵 + ∠𝐴𝐵𝐶 = 180° (∠′𝑠 of a ∆) S/R ∠𝐶𝐴𝐵 = 180° − (32° + 33°) [∠𝐴𝐵𝐶 = 65° − 32°]M ∠𝐶𝐴𝐵 = 180° − 65° ∠𝐶𝐴𝐵 = 115° A (3) 5.2.1 ∠𝐴 + ∠𝐴𝐵𝐶 = ∠𝐵𝐶𝐸 ( Ext ∠ of a ∆) S/R Statement and reason: 1 mark (2𝑥 − 48°) + (𝑥 + 14°) = 116° M 3𝑥 − 34° = 116° Substitution: 1 mark 3𝑥 = 150° A 𝑥 = 50° Answer: 1 mark OR S/R ∠𝐴 + ∠𝐴𝐵𝐶 + + ∠𝐴𝐶𝐵 = 180° (∠′𝑠 of a ∆) (2𝑥 − 48°) + (𝑥 + 14°) + 64° = 180° M 3𝑥 + 30° = 180° 3𝑥 = 150° 𝑥 = 50° A (3) 5.2.2 ∠𝐴= 2𝑥 − 48° Substitution: 1 mark = 2(50°) − 48° M = 100° − 48° Answer: 1 mark = 52° A (2) 5.2.3 ∠𝐴𝐵𝐶 = 50° + 14° = 64° Correct statement: ∠𝐴𝐶𝐵 = 180° − 116° = 64° 1 mark S R ∆𝐴𝐵𝐶 is an isosceles triangle (∠𝐴𝐵𝐶 = ∠𝐴𝐶𝐵) Correct Reason: 1 mark (2) 5.3.1 S R Correct statement: ∠𝐴𝐵𝐶 = 40° (Complementary ∠′𝑠) 1 mark Correct Reason: 1 mark (2) 5.3.2 S R Correct statement: ∠𝐴𝐷𝑂 = 32° (AO =OD / radii) 1 mark Correct Reason: 1 mark (2) [16] Copyright reserved Please turn over

Downloaded from hlayiso.com 8 MATHEMATICS (EC/NOVEMBER 2016) QUESTION 6 [11 marks] Ques. Solution Mark Allocation Total 6.1 STATEMENT REASON Correct 𝐾𝐿 = 𝐾𝑁 Given A statement 𝐿𝑀 = 𝑁𝑀 Given A with reason: 1 𝐾𝑀 = 𝐾𝑀 Common A mark each ∴ ∆KLM ≡ ∆KNM. 𝑆𝑆𝑆A (4) 6.2.1 STATEMENT REASON Correct 𝐴̂ = 𝑃̂ Alt .∠’s, AB ∥ PQ A statement 𝐵̂ = 𝑄̂ Alt .∠’s, AB ∥ PQ A with reason: 1 𝐴𝑂̂𝐵 = 𝑃𝑂̂𝑄 Vert. opp. ∠’s  A mark each ∴ ∆ ABO ||| ∆PQO. 𝐴𝐴𝐴  A (4) 6.2.2 𝑂𝑄 𝑂𝑃 (Corr. sides are proportional) Statement and reason: 1 mark = S/R 𝑂𝐵 𝐴𝑂 𝑥 12 𝑐𝑚 : 𝑥 12 𝑐𝑚 = 1 mark 5 𝑐𝑚 6 𝑐𝑚 = A 5 𝑐𝑚 6 𝑐𝑚 Answer: 1 mark 𝑥 = 𝑂𝑄 = 10 𝑐𝑚 CA (3) [11] QUESTION 7 [8 marks] Ques. Solution Mark Allocation Total 7.1 14 𝑐𝑚 : 1 mark 𝑑 = 7 × 2 = 14𝑐𝑚 M 𝜋𝑟 2 𝑠 2 − : 1 mark 2 𝜋𝑟 2 196 𝑐𝑚2 − 77𝑐𝑚2 : 1 mark Area of the shaded part = 𝑠 2 − Answer: 1 mark 2 M 22 × 49 = 14 × 14 − 7 2 = 196 𝑐𝑚2 − 77 𝑐𝑚2 A = 119 𝑐𝑚2 CA (4) 7.2 2𝑙 + 2 𝑏 = 16 M 2𝑙 + 2 𝑏 = 16 : 1 mark 𝑙+ 𝑏=8 𝑏 =8−𝑙 𝑙 × 𝑏 = 15 : 1 mark 𝑙 × 𝑏 = 15 M 5 :1 mark 𝑙(8 − 𝑙) = 15 8𝑙 − 𝑙2 = 15 3 : 1 mark 𝑙2 − 8𝑙 − 15 = 0 (𝑙 − 5)(𝑙 − 3) = 0 𝑙 = 5 or 𝑙 = 3 A 𝑏 = 3 or 𝑏 = 5 A (4) [8] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2016) MATHEMATICS 9 Question 8 [9 marks] Ques. Solution Mark Allocation Total 8.1.1 H GH G T GT RH R H T RT H YH Y T YT H WH W WT T H BH  A T B BT (2) Key: G− Green, R − Red, Y − Yellow, B − Black, W − White, H − head, T − tail 2 marks for correct tree diagram. 8.1.2 5 1 Answer: 1 mark 10 = 2 A (1) 8.1.3 2 1 Answer: 1 mark = A 10 5 (1) 8.2.1 5 Answer: 1 mark A 8 (1) 8.2.2 A A 1 : 1 mark 1 4 = 25% 25% : 1 mark (2) 4 8.2.3 A A 5 : 1 mark 5:1 1 : 1 mark (2) [9] TOTAL: 100 Copyright reserved Please turn over