Mathematics Grade 9 November EC 2017 Memo (English)

Downloaded from hlayiso.com SENIOR PHASE GRADE 9 NOVEMBER 2017 MATHEMATICS MARKING GUIDELINE MARKS: 140 This marking guideline consists of 11 pages.

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

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

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

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

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

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

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

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

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

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