# Mathematics Grade 9 November EC 2017 Memo (English) > Grade 9 Mathematics memorandum from the November EC 2017 (English), 11 pages. Read online or get the free PDF download, question paper linked when available. - Page: https://www.hlayiso.com/library/maths-gr9-memo-eng-nov2017/61c0a2e7-0643-4c28-9483-d6c68fe55bf8 - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 9 - Document type: Memorandum - Year: 2017 - Exam period: November EC - Curriculum: CAPS - Pages: 11 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 9](https://www.hlayiso.com/library/grade-9) › [Mathematics](https://www.hlayiso.com/library/grade-9/mathematics) › [Mathematics Grade 9 November EC 2017 Memo (English)](https://www.hlayiso.com/library/maths-gr9-memo-eng-nov2017/61c0a2e7-0643-4c28-9483-d6c68fe55bf8) ## About this document Mathematics · Grade 9 · November EC · 2017. Memorandum, 11 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com SENIOR PHASE GRADE 9 NOVEMBER 2017 MATHEMATICS MARKING GUIDELINE MARKS: 140 This marking guideline consists of 11 pages. ### Page 2 Downloaded from hlayiso.com 2 MATHEMATICS (EC/NOVEMBER 2017) INSTRUCTIONS AND INFORMATION 1. Give full marks for answers only, unless stated otherwise. 2. Accept any alternate correct solutions that are not included in the memorandum. 3. Underline errors committed by learners and apply Consistent Accuracy (CA). 4. THE FINAL MARK MUST BE CONVERTED TO 100. KEYS M Method CA Consistent Accuracy A Accuracy S Statement SF Substitution in Formula R Reason S/R Statement and Reason Ques. Mark allocation Total 1.1 A  (1) 1.2 D  (1) 1.3 C  (1) 1.4 A  (1) 1.5 B  (1) 1.6 B  (1) 1.7 C  (1) 1.8 D  (1) 1.9 B  (1) 1.10 A  (1) [10] Copyright reserved Please turn over ### Page 3 Downloaded from hlayiso.com (EC/NOVEMBER 2017) MATHEMATICS 3 QUESTION 2 [30 marks] Ques. Solution Mark allocation Total 5 2.1 0,000 014 6 = 1, 46 10 A Answer : 1 (1) 2.2.1 0, 06 y 4  0,1y 4 0, 06 y 4  0,1y 4  0,16y 4 A 16 4 16 4  y A 0,16 y 4 / y :1Mark  0, 4y 2 A OR 100 100 2 2 Answer: 1 Mark (2)  y A 5 2.2.2 A 3 x6 x2 x 2 :1Mark 2 0   x 2 A Answer : 1 Mark (4 x ) 1 (2) 2.2.3 4 1 2 8 2 (3x y ) 9 x y :1Mark x 2  x 1 y 2 x3 y 2 :1Mark 9x8 y 2 M Answer: 1 Mark  3 2 (3) x y M  9x11 CA 2.2.4 3( x  3)( x  3)  ( x  1)2 x2  9 :1Mark M M x2  2 x  1:1Mark  3( x  9)  ( x  2 x  1) 2 2 3x2  27  x2  2 x  1:1Mark  3x2  27  x2  2 x  1 A 2 x2  2 x  28 :1Mark (4)  2 x2  2 x  28 CA 2.2.5 1 2 1 1 13x 52 x 9 x 3 x 2 2 x  4 x   :1Mark 4 3 6 2 4 9 2 117 x  208x  162x :1Mark 13x 52 x 9 x M 36 :1Mark    71x 4 9 2 M :1Mark 117 x  208 x  162 x 36  36 M 71x  CA 36 (4) 2.3.1 2 x  6 x  36 2 2( x  3x 18) :1Mark 2  2( x  3x  18) A 2 ( x  6) :1Mark (3) A A ( x  3) :1Mark  2( x  6)( x  3) 2.3.2 9 x(5a  b)  2(b  5a) 9 x(5a  b)  2(5a  b) :1Mark M (5a  b) :1Mark  9 x(5a  b)  2(5a  b) (9 x  2) :1Mark A A  (5a  b)(9 x  2) (3) Copyright reserved Please turn over ### Page 4 Downloaded from hlayiso.com 4 MATHEMATICS (EC/NOVEMBER 2017) 2.4.1  2x  3 2 x  3  0 Answer: 1 mark Answer: 1 mark 3A 3 A x  or  x   2 2 (2) 2.4.2 3x  2 x  2  by LCM: 1Mark  7 3  3x  2  M  x2 9 x  6  7 x 14 :1Mark 21   21   Answer: 1 mark  7   3  3(3x  2)  7( x  2) 9 x  6  7 x 14 M (3)  2 x  8  x  4 CA 2.4.3 27.3x  1 1  3x  :1Mark 1 27  3x  M 27 3x  33 :1Mark 3  33 M x Answer: 1 mark  x  3 CA OR OR 27.3x  1 33 x  30 :1Mark 33.3x  1 3  x  0 :1Mark 33 x  30 M Answer: 1 mark 3  x  0M (3)  x  3 CA [30] Copyright reserved Please turn over ### Page 5 Downloaded from hlayiso.com (EC/NOVEMBER 2017) MATHEMATICS 5 QUESTION 3 [22 Marks] Ques. Solution Mark allocation Total 3.1 3.1.1 Figure 1 2 3 4 12 & 16: 1Mark Number of Triangles 4 8 12 16 (1) p  12 and q  16 A 3.1.2 4n :1Mark A Tn  4n (1) 3.1.3 120  4n M SF 120 : 1Mark n  30 CA n  30 :1Mark ∴30th figure A Answer : 1Mark (3) 3.2.1 y  2x  4 Let y  0 : 1 Mark 0  2 x  4M Answer : 1Mark x  2 CA (2) 3.2.2 y  4A 4 :1Mark (1) 3.2.3 3.2.3 & y  2x  4 3.3 x  int ercept :1Mark y  int ercept :1Mark (3) Straight Line : 1Mark A 3.3 x4 A A x  int ercept :1Mark (2) vertical / shape :1Mark A A 3.4 y  4A Answer: 1Mark (1) 3.5.1 1 1 A  1  1 M A  1  1:1Mark 3 3 A0 CA Answer: 1Mark (2) Copyright reserved Please turn over ### Page 6 Downloaded from hlayiso.com 6 MATHEMATICS (EC/NOVEMBER 2017) 3.5.2 1 1 (1)  1  B A (1)  1  B :1Mark 3 3 2 Answer : 1Mark B   CA 3 (2) Common difference = 3  (5)  2 Explanation : 1 Mark 3.6.1 y  intercept = -3 Since x = 0 A Answer : 1 Mark A y  2x  3 If ANSWER ONLY Full Marks (2) 3.6.2 21  2m  3 M Substitution : 1Mark m  12 CA Answer : 1Mark (2) [22] QUESTION 4 [12 marks] Ques. Solution Mark allocation Total 4.1  n r  M Formula : 1 Mark A  P 1   Substitution : 1Mark  100  R23413, 44 :1Mark 3  16  SF Answer : 1Mark A  15000 1    100  A  R23413, 44 CA Compound Interest  R 8 413, 44 CA (4) 4.2 Let the age of the son  x x and 36  x :1Mark  Age of the father  36  x M x  7 and 43  x :1Mark Son in 7 years time  x  7 4( x  7)  43  x :1Mark  Father in 7 years time  43  xM CA Answer:1Mark  4( x  7)  43  x Both ages:1Mark  4 x  28  43  x 5x  15 x  3 CA Son is 3 years old and the Father is 33 years old CA (4) 4.3 d  s  tM Formula/M :1Mark d  120km / h  3h 360km :1Mark d  360kmA 360km :1Mark d 90km / h t s Answer : 1Mark 360km t M 90km / h t  4h CA (4) [12] Copyright reserved Please turn over ### Page 7 Downloaded from hlayiso.com (EC/NOVEMBER 2017) MATHEMATICS 7 QUESTION 5 [14] Ques. Solution Mark allocation 5.1 A R Answer: 1 Mark 5.1.1 ∠ADC = 65˚ (corresponding ∠s, CD∥AB) Reason : 1 Mark (2) 5.1.2 ∠EBC = 65˚ (Given :∠ABE = ∠𝐸𝐵C) Answer: 1 Mark A R Reason: 1 Mark ∠BCD = 65˚(Alternate∠’s, EB∥CD) (2) 5.2 5.2.1 S R Statement: 1 Mark 2 x  35  3x  10  140  Exterior  of RST  Reason: 1 Mark Simplifying: 1 Mark 5x  25  140 A Answer: 1 Mark x  23 CA OR OR RTS  40 ( 's on a straight line=180 ) Statement: 1 Mark S R Reason: 1 Mark 2 x  35  3x  10  40  180 Sum of 3 ' s of RST=180  Simplifying: 1 Mark Answer: 1 Mark 5x  10  140 A x  23 CA (4) 5.2.2 QSR  2 x  35 Substitution/Method:1 Mark QSR  2(23 )  35 M Answer: 1 Mark QSR  81 CA (2) 5.3 5.3.1 A R Answer : 1 Mark QPR  35 Sum of 3 ' s of PQR=180  Reason : 1 Mark OR OR A R Answer : 1 Mark QPR  35  Complementary ' s  Reason : 1 Mark (2) 5.3.2 A R Answer : 1 Mark PSO  38  PO  OS , radii  Reason : 1 Mark OR OR A R Answer : 1 Mark PSO  38  PO  OS , radii  Reason : 1 Mark (2) [14] Copyright reserved Please turn over ### Page 8 Downloaded from hlayiso.com 8 MATHEMATICS (EC/NOVEMBER 2017) QUESTION 6 [11 marks] Ques. Solution Mark allocation Total 6.1 AD  DC  AB  BE Statement and reason: 1 mark  AC  AE..................................(1) Statement and reason: 1 mark Statement and reason: 1 mark In ∆𝐴𝐵𝐶and ∆ADE Statement and reason: 1 mark S/R 1 AC=AE [Proved at (1)] 2 ∠𝐴 = ∠𝐴 [Given] S/R 3 𝐴𝐵 = 𝐴𝐷 [Common] S/R ∴∆ABC≡ ∆ ADE SAS S/R (4) 6.2.1 MN 2 1 Statement : 1 mark   S Statement : 1 mark MK 4 2 MK 4 1 Statement : 1 mark   S Statement/Reason : 1 mark ML 8 2 NK 5 1   KL 10 2 S S/R MNK ||| MKL Corresponding sides are in proportion  (4) 6.2.2 KNM  65 ( 's on a straight line=180 )S/R Statement/Reason: 1 mark Answer : 1 mark A R Reason: 1 mark  MKL  65 [MNK ||| MKL] (3) [11] QUESTION 7 [18 marks] Ques. Solution Mark allocation Total 7.1 7.1.1 A  Area of 2's +Area of 3rectangles Substitution : 1Mark 1  SF Answer : 1Mark A  2   8cm  6cm   15cm 10cm  15cm  8cm  15cm  6cm 2  A  48cm  150cm  120cm2  90cm2 2 2 A  408cm2A (2) 7.1.2 V  Area of base  height M Formula : 1Mark 1 Substitution : 1Mark V   8cm  6cm 15cm SF Answer : 1Mark 2 V  360cm3 CA (3) Copyright reserved Please turn over ### Page 9 Downloaded from hlayiso.com (EC/NOVEMBER 2017) MATHEMATICS 9 7.2 7.2.1 AE 2  AB2  BE 2 [ Pythagoras] Substitution : 1Mark Answer : 1Mark AE 2  (5cm)2  (4cm)2 S AE 2  9cm2 AE  3cm CA (2) EC  3cm  AE  EC  3cm R 7.2.2 Answer : 1 Mark OR R Reason : 1 Mark EC  3cm  AE  EC  3cm;Diagonal BD of Kite bisects AC AC  6cm A (2) 7.2.3 BD  4cm  10cm  14cm Substitution : 1 Mark 1 42cm2 : 1 Mark Area of Kite ABCD  14cm  6cm  M Answer : 1Mark 2 1 Area of Kite ABCD   AC  BD  2 Area of Kite ABCD  42cm2 CA 3  Area of Quadrilateral PQRD    42  cm2 2  CA Area of Quadrilateral PQRD  63cm2 (3) 7.3 7.3.1 2 r  44 M 2 r  44 :1Mark 44 44 r r :1Mark 2 M 2 r  7cm CA Answer : 1Mark (3) 7.3.2 V   r 2  h M V   r 2  h :1Mark A   (7cm)2  44cm M A   (7cm)2  44cm :1Mark A  6773, 27cm3 CA Answer : 1Mark (3) [18] QUESTION 8 [10 marks] Ques Solution 8.1 8.1.1 x  5 :1Mark A A  x; y    x  5; y  2 y  2 :1Mark (2) 8.1.2 A A A ll A (6;2) : 1 Mark A (6;2) and B (0;  6) and C (8; 6) ll ll ll Bll (0;-6) : 1 Mark Cll (8; 6) :1Mark (3) Copyright reserved Please turn over ### Page 10 Downloaded from hlayiso.com 10 MATHEMATICS (EC/NOVEMBER 2017) 8.2 8.2.1 P(-2;2) ; Q(-3;-1) and R(0;2) and 8.2.2 2 Marks A ONE mark deducted for every incorrect plotting. (2) A A P//( 2;-2) : 1 Mark Q//(-1;-3) :1 Mark R//( 2;0) : 1 Mark A A (3) [10] QUESTION 9[13 marks] 9.1 9.1.1 Outcomes 3 outcomes: 1Mark 3 outcomes: 1Mark Outcomes Green Ruler [Blue Pencil;Green Ruler] Blue Pencil A Green Ruler [Blue Pencil;Green Ruler] White Ruler [Blue Pencil;White Ruler] Green Ruler [Red Pencil;Green Ruler] Red Pencil A Green Ruler [Red Pencil;Green Ruler] White Ruler [Red Pencil;White Ruler] (2) 9.1.2 2 1 Answer : 1 Mark P(Red Pencil and Green Ruler)  or or 0,33or 33% A (1) 6 3 9.1.3 P(White Pencil and Red Ruler)  0 or impossible A Answer : 1 Mark (1) Copyright reserved Please turn over ### Page 11 Downloaded from hlayiso.com (EC/NOVEMBER 2017) MATHEMATICS 11 9.2 9.2.1 3 points plotted:1Mark 3 points plotted:1Mark Test marks of learners 3 points lotted:1Mark 100 90 80 Naturalal Science mark 70 60 50 40 30 20 10 A A A 0 0 20 40 60 80 100 Maths mark (3) (15;90) : 1 Mark 9.2.2 (15;90) OR (90;40) OR (90;40) : 1 Mark (1) 9.2.3 There is a strong positive correlation. A Answer : 1 Mark OR Learners who perform well in Mathematics, generally perform well in OR Natural Science and learners who perform poorly in Mathematics generally perform poorly in Natural Science. A Answer : 1 Mark (1) 9.3 9.3.1 x  30 x  30 27  M 27  :1Mark 2 2 x  24 A Answer : 1 Mark (2) 9.3.2 300 CA from 9.3.1 Mean  10 300 :1Mark Mean  30 CA 10 Answer : 1 Mark (2) [13] TOTAL: 140 Copyright reserved Please turn over ## Related documents - [MATHS GR9 MEMO AFRK NOV2017](https://www.hlayiso.com/library/maths-gr9-memo-afrk-nov2017/4a0aa374-cbd1-4e2e-9cde-d218e99586e4) — Mathematics · Memo · 2017 - [Gr 9 Mathematics MEMO 2017](https://www.hlayiso.com/library/gr-9-mathematics-memo-2017/10d5916d-ccba-4bcb-9b95-58e0188825c6) — Mathematics · Memo · 2017 - [MATHS MEMO GR9 NOV2019 Afr D](https://www.hlayiso.com/library/maths-memo-gr9-nov2019-afr-d/2826bd2f-e14b-4756-98e7-adfc98792152) — Mathematics · Memo · 2019 · Afr - [MATHS GR9 MEMO NOV 2018 Afrikaans](https://www.hlayiso.com/library/maths-gr9-memo-nov-2018-afrikaans/cd89345a-48f5-44e9-9854-0fdbe414cbf0) — Mathematics · Memo · 2018 · Afr - [MATHS GR9 NOV2014 ENG MEMO](https://www.hlayiso.com/library/maths-gr9-nov2014-eng-memo/71e9d012-c73e-489a-8b32-0f9b3d23e4de) — Mathematics · Memo · 2014 · Eng - [MATHS GR9 NOV2014 AFR MEMO](https://www.hlayiso.com/library/maths-gr9-nov2014-afr-memo/93887a6e-ebe6-4b2e-b2d8-32f981c4d084) — Mathematics · Memo · 2014 · Afr ## More from Grade 9 Mathematics - [CAPS SP MATHEMATICS GR 7-9](https://www.hlayiso.com/library/caps-sp-mathematics-gr-7-9/ff259c2b-5f3b-46f9-ae34-66dd08358bbe) — CAPS · Eng - [FS Maths Grade 9 MOCK November 2025 P2 and Memo](https://www.hlayiso.com/library/fs-maths-grade-9-mock-november-2025-p2-and-memo/411ff601-2320-4073-a074-74427663bad7) — QP + Memo · 2025 - [KZN Zululand Maths Grade 9 March 2025 QP and Memo](https://www.hlayiso.com/library/kzn-zululand-maths-grade-9-march-2025-qp-and-memo/a5d0e368-52a3-4816-b3aa-3597ee4f5a9f) — QP + Memo · 2025 - [Maths Grade 9 June 2025 QP and Memo](https://www.hlayiso.com/library/maths-grade-9-june-2025-qp-and-memo/1c92dc3a-8522-46ea-9bc2-251685906fd8) — QP + Memo · 2025 - [2024 G9 Mock June Test Free State](https://www.hlayiso.com/library/grade-9-mathematics-2024-free-state-june-exam-question-paper/3628eaa4-69bb-4e63-a6b9-8c072d118cdb) — QP · 2024 - [2024 G9 Mock June Test Free State.docx](https://www.hlayiso.com/library/grade-9-mathematics-2024-free-state-june-exam-question-paper/7d033d41-fed7-4c04-a7cc-78d6c025aa9b) — QP · 2024 ## Collection - [Grade 9 past papers](https://www.hlayiso.com/grade-9/past-papers) --- Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. 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