Mathematics Grade 9 November EC 2018 Memo (English)

Downloaded from hlayiso.com SENIOR PHASE GRADE 9 NOVEMBER 2018 MATHEMATICS MARKING GUIDELINE MARKS: 140 This marking guideline consists of 13 pages.

Downloaded from hlayiso.com 2 MATHEMATICS (EC/NOVEMBER 2018) 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 marking guideline. 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 QUESTION 1 [10 marks] Ques. Mark allocation Total 1.1 C  (1) 1.2 A  (1) 1.3 B  (1) 1.4 A  (1) 1.5 D  (1) 1.6 B  (1) 1.7 C  (1) 1.8 A  (1) 1.9 C  (1) 1.10 D  (1) [10] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2018) MATHEMATICS 3 QUESTION 2 [25] Ques. Solution Mark allocation Total 2.1 1 042 000 000 = 1,042 10 A Answer: 1Mark (1) 9 2.2.1  2 7z  3z 2   4 z 3    3 2   14 z 3 2   2 7z   M    : 1 Mark  3z 2   4 z 3    3 7z   3 2  4z  M 2  3z 2  4z2 3 : 1 Mark 2  5 z 5 2 CA / z 3 3 3 Answer: 1 Mark (3) 2.2.2 2( x  3)  3( x  1)(2 x  5) 2 M M ( x2  6 x  9) : 1 Mark  2( x2  6 x  9)  3(2 x2  3x  5)  2 x 2  12 x  18  6 x 2  9 x  15 (2 x2  3x  5) : 1 Mark  4 x 2  3 x  33 CA Answer: 1 Mark (3) 2.2.3 2  2x y  1 2  2   2y   3y   2  : 1 Mark 2  3 xy   2y   2  M  3 xy  2  3 xy 2  2   : 1 Mark  3 xy 2   2y    M  2y  Answer: 1 Mark 9 x2 y 2  CA 4 OR 2 OR  2 x 1 y   2   3y  2  2 x 2 y 2  2 4 M 3 y 2  2 x 2 y 2 1 2 2 : 1 Mark x y 32 y 4  4 M 1 1 2 2 9 x y 9 4 : 1 Mark  x 2 y 2 CA 1 4 9 Answer: 1 Mark OR OR Copyright reserved Please turn over

Downloaded from hlayiso.com 4 MATHEMATICS (EC/NOVEMBER 2018) Ques. Solution Mark allocation Total 2  2 x 1 y   2  1  3y  2 : 1 Mark 1  2x y  1  2 M  2   2x y  1  3y   2   3y  1 1 : 1 Mark  2 2 2 M 2 x 2 y 2 2 2 x y 32 y 4 32 y 4 Answer: 1 Mark 9 x 2 y 2 CA  4 (3) 2.2.4  y  0 3 13 x : 1 Mark 169 x 6   99   p q x 4 : 1 Mark 3 12 Answer: 1 Mark x M 13x3 1  x 4M 13  A x (3) 2.3.1 ax 2  5ax  6a a( x  5x  6) : 1 Mark 2 M ( x  3) : 1 Mark  a( x2  5x  6) ( x  2) : 1 Mark CA  a ( x  3)( x  2) (3) 2.3.2 (2 x  3 y)  (3 y  2 x) x 2 M (2 x  3 y)  (2 x  3 y) x2 : 1 Mark  (2 x  3 y)  (2 x  3 y) x 2  (2 x  3 y)(1  x2 )CA  (2 x  3 y )(1  x)(1  x)CA (2 x  3 y)(1  x2 ) : 1 Mark (2 x  3 y )(1  x)(1  x) : 1 Mark (3) 2.4.1 x 2x  3  by LCM: 1 Mark  1 Answer: 1 Mark 2 3  x   2x  3  M 6    6    6 1 2  3   3x  4 x  6  6 7 x  0  x  0 CA (2) Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2018) MATHEMATICS 5 Ques. Solution Mark allocation Total 2.4.2 x  x  12 2 ( x  4)( x  3) : 1 Mark  x  x  12  0 2 Both solutions: 1 Mark  ( x  4)( x  3)  0M CA  x  4 or x  3 (2) 2.4.3 2 1 5 : 1 Mark 5x2  25 Answer: 1 Mark 2 M 5  5 x 2  x  2  2  x  4CA (2) [25] Copyright reserved Please turn over

Downloaded from hlayiso.com 6 MATHEMATICS (EC/NOVEMBER 2018) QUESTION 3 [26] Ques. Solution Mark allocation Total 3.1 3.1.1 q = 17 and r = 25 : 1 Mark Shape 1 2 3 4 … 25 Number of rectangles 5 9 13 17 … 101 A q  17 and r  25 (1) 3.1.2 A A 4n : 1 Mark Tn  4n  1 1 : 1 Mark (2) 3.1.3 Tn  4n  1 205  4n  1SF Tn =205 : 1 Mark n  51 Answer: 1 Mark  Shape number 51 has 205 rectangles.CA (2) 3.2.1 n 2 : 1 Mark A Tn  n2  1A 1 : 1 Mark (2) 3.2.2 Tn  n  1 2 SF (n  10) : 1 Mark T10  (10)2  1 SF Answer: 1 Mark Tn  101CA (2) 3.3.1 1 SF x = -2 : 1 Mark A x2 2 Answer: 1 Mark 1 SF A   2  2 2 A  1 CA (2) 3.3.2 SF x = B : 1 Mark 1 SF Answer: 1 Mark  B  2  4 2 B  4 CA (2) Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2018) MATHEMATICS 7 Ques. Solution Mark allocation Total 3.4.1 Common difference = 7  (10)  3 Explanation: 1 Mark y  intercept = -1 Since x = 0 A Answer: 1 Mark A If ANSWER ONLY y  3x  1 Full Marks OR y  y1 OR m 2 x2  x1 m = 3: 1 Mark 7  (10) m Answer: 1 Mark 2  (3) If ANSWER ONLY m3 A Full Marks y  3x  1 A (2) 3.4.2 y  3x  1 SF both values : 1 Mark 8  3q  1 SF Answer: 1 Mark q  3CA (2) 3.5.1 y  x  3 Answer: 1 Mark 0  x  3 x  3A (1) 3.5.2 y  x  3 x  int ercept : 1 Mark A y  int ercept : 1 Mark Straight Line : 1Mark A A (3) Copyright reserved Please turn over

Downloaded from hlayiso.com 8 MATHEMATICS (EC/NOVEMBER 2018) Ques. Solution Mark allocation Total 3.6.1 Monday to Tuesday A Answer: 1 Mark OR Saturday to Sunday A (1) 3.6.2 15 packets of sweets sold A Answer: 1 Mark OR 10 packets of sweets sold A (1) 3.6.3 Thursday to Saturday A Answer: 1 Mark (1) 3.6.4 The decrease varied. A Answer: 1 Mark (1) 3.6.5 The sales were constant.A Answer: 1 Mark OR No increase or decrease in the sales. A (1) [26] QUESTION 4 [12] Ques. Solution Mark allocation Total 4.1 SI  P.i.n M Formula: 1 Mark 720  1800.i.5SF Substitution: 1 720 Mark i 1800  5 Answer: 1 Mark 720 r 100 1800  5 r  8% CA (3) 4.2 Let the breadth of the original playground  x  The length of the original playground  x  1 2  x  1  x  : 1 Mark  The perimeter of the original playground  2  x  1  x  M 4 x  2 : 1 Mark  4 x  2 M The length of the new playground  x  4 2  x  4  x  1 : 1 Mark The breadth of the new playground  x  1  The perimeter of the new playground  2  x  4  x  1 M 4 x  6 : 1 Mark  4 x  6M Answer: 1 Mark  The difference in perimeter  4 x  6   4 x  2   4meters CA (5) 4.3 Total distance travelled  210km Distance & Time : 1 Mark Total time travelled  2,5hours M Dis tan ceM Formula : 1 Mark Average Speed  Time Substitution : 1 Mark 210km  SF Answer: 1 Mark 2,5hours  84km / h CA (4) [12] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2018) MATHEMATICS 9 QUESTION 5 [19] Ques. Solution Mark allocation Tot 5.1 A R Answer: 1 Mark 5.1.1 x  85  Alternate  ' s; AC//HF  Reason: 1 Mark (2) 5.1.2 A Answer: 1 Mark R y  95  Co-interior  ' s  180 ; BF//CD Reason: 1 Mark (2) 5.2 5.2.1 Statement and reason: S/ R Q1  55  Corresponding  ' s; MN//QP  1 Mark Statement: 1 Mark S R Reason: 1 Mark 55  2 x  5  3x  40  180 3  ' s of PQO  180  Answer: 1 Mark 5 x  90  180 x  18 CA (4) 5.2.2 MON  2 x  5 Substitution/Method:1 Mark  2(18 )  5 M Answer: 1 Mark  31CA (2) 5.3.1 Statement and reason:1 Mark ADB  ABD  110 [3 's of ABD=180 ]S/R Statement and reason:1 Mark But ADB  ABD[ADB is isosceles with AD  AB]S/R Answer: 1 Mark  ADB  55 A (3) 5.3.2 S R Statement :1 Mark BDC  55 [Diagonal of rhombus ABCD bisect 's] Reason:1 Mark Answer: 1 Mark ADC  110 A OR OR S R DAE  ADC  180 [Co-interior 's =180 ; DC//AE] Statement :1 Mark 70  ADC  180 [Co-interior 's =180 ; DC//AE] Reason:1 Mark ADC  110 A Answer: 1 Mark (3) 5.3.3 S Statement :1 Mark R DBE  125 [Exterior  of ABD] Reason:1 Mark (2) 5.3.4 The opposite side of a rhombus are parallel.R Reason:1 Mark OR DC//AER (1) [19] Copyright reserved Please turn over

Downloaded from hlayiso.com 10 MATHEMATICS (EC/NOVEMBER 2018) QUESTION 6 [12] Ques. Solution Mark allocation Total 6.1 In PMN and NOP : 1. MPN=PNO [ Alternate 's; MP //NO] S/R Statement and reason: 1 Mark Statement and reason: 1 Mark 2. MNP=NPO [ Alternate 's; MN //PO]S/R Statement and reason: 1 Mark 3. PN=PN [Common]S/R Statement and reason: 1 Mark  PMN  NOP[S]S/R Statement and reason: 1 Mark  PM = NOS/R (5) 6.2.1 In DBE and FCE : Statement and reason: 1 Mark 1. E1 =E3 [Vertically Opposite 's] S/R Statement and reason: 1 Mark Statement and reason: 1 Mark 2. D1 =F [Alternate 's; AB//CF] S/R Statement and reason: 1 Mark 3. B=C2 [Alternate 's; AB//CF] S/R  DBE lll FCE[] S/R Note: The learner can use any 2 pairs of equal angles in his/her proof and then simply state that the last pair of corresponding angles are equal because the sum of 3 angles of a triangle is equal to 180˚. (4) 6.2.2 DB BE Statement and reason:  [ DBE lll FCE ] S/R 1 Mark FC CE DB 8 Substitution: 1 Mark  SF Answer: 1 Mark 5 10 DB  4cm CA (3) [12] QUESTION 7 [13] Ques. Solution Mark allocation Total 7.1.1 Total Surface Area of pipe closed Substitution: 1 Mark  2 r  2 r  h 2 25000,79434cm2:1 Mark SF 2513, 274123cm2:1 Mark  2 18  2 18 120 2 CA Answer: 1 Mark  15607, 4323cm2 Total Surface Area of circular ends  2 r 2  2 15 2  1413,716694cm2 A Total Surface Area of pipe open both sides  15607, 4323cm2 1413,716694cm2  14193,72cm2 CA (4) Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2018) MATHEMATICS 11 Ques. Solution Mark allocation Total 7.1.2 Total Volume of pipe closed Substitution: 1 Mark   r2  h 122145,1224cm3 : 1 Mark   18 120 SF 2 84823,00165cm3 : 1 Mark Answer: 1 Mark  122145,1224cm3CA Total Volume of hole   r 2  h   15 120 2  84823,00165cm3A Total Volume of pipe ( open on both sides)  122145,1224cm3  84823,00165cm3  37322,12cm3 CA (4) 7.2 FC  52  42 Theorem of Pythagoras FC  3cm : 1 Mark FC  3cm A 12cm 2 : 1 Mark 1 36 cm2 : 1 Mark Area of ABC =  6cm  4cm 2 28,27433388 cm2 : 1 Mark  12cm 2 CA Answer: 1 Mark Area of square ACDE = 6cm  6cm = 36 cm 2 CA Area of Circle =   (3cm)2 = 28,27433388 cm2 CA Area of ALL the shaded sections = 12 cm2  (36cm2  28, 27433388cm2 ) = 19,73 cm2 CA (5) [13] Copyright reserved Please turn over

Downloaded from hlayiso.com 12 MATHEMATICS (EC/NOVEMBER 2018) QUESTION 8 [10] Ques. Solution Mark allocation Tot. 8.1 8.1 and Al  1; 2  : 1 Mark 8.2 Bl  0,5; 2  : 1 Mark and A A C l  1;0,5  : 1 Mark 8.3 (3) 8.2 A A A Al /  4; 2  : 1 Mark A A B / /  4; 1 : 1 Mark A A C / / 1; 2  : 1 Mark A (3) 8.3 𝑃(−3; 0) : 1 Mark 𝑄(−2; 0) : 1 Mark 𝑅(−3; −1): 1 Mark 𝑆(−4; −1): 1 Mark (4) [10] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2018) MATHEMATICS 13 QUESTION 9 [13] Ques. Solution Mark allocation Total 9.1.1 a = 19 A Value of a: 1 Mark b = 90 Value of b and c: 1 Mark A c = 61 (2) 9.1.2 a) 150 3 A Numerator: 1 Mark P  Girl   / Denominator: 1 Mark 250 5 A (2) 9.1.2 b) 76 38 A Numerator: 1 Mark P  A boy owning a cell phone   / Denominator: 1 Mark (2) 250 125 A 9.1.2 c) 61 A Answer: 1 Mark P  A learner with no cell phone   (2) 250 A 9.2.1 The mean tends to be shifted upwards if there are Statement: 1 Mark extreme values. S Reason: 1 Mark In this case, there are a few higher salaries, so the Reason: 1 Mark mean is shifted upwards,R while the median shows that half of the salaries will be below R5 225. R Accept any other logical explanation. (3) 9.2.2 A The median: 1 Mark The median is generally a better indicator of the real Reason: 1 Mark situation when the data is not evenly spread out.R Accept any other logical explanation. (2) [13] TOTAL: 140 Copyright reserved Please turn over