Mathematics Grade 10 P1 Eastern Cape November 2019 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT NOVEMBER 2019 GRADE 10/GRAAD 10 MATHEMATICS P1/WISKUNDE V1 MARKING GUIDELINE/NASIENRIGLYN EXEMPLAR/EKSEMPLAAR MARKS/PUNTE: 100 This marking guideline consists of 8 pages./ Hierdie nasienriglyn bestaan uit 8 bladsye.

Downloaded from hlayiso.com 2 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2019) QUESTION 1/VRAAG 1 1.1 1.1.1 x=2 üx=2 (1) 1.1.2 x−20 üx−20 x2 üx2 Answer only: Full marks Slegs antwoord: Volpunte (2) 1.2 1.2.1 (a − 2)(a 2 + 2a + 4) üa 3 = a3 − 8 ü −8 (2) 1.2.2  a  a  2  + 1 − 1 üa  2  2  4 ü −1 a2 = −1 4 (2) 1.3 1.3.1 2x 2 − x − 6 ü (2 x + 3) = (2 x + 3)( x − 2) ü ( x − 2) (2) 1.3.2 (a − b)2 − 100c 2 ü (a − b − 10c) = (a − b − 10c)(a − c + 10c) ü (a − c + 10c) (2) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2019) MATHEMATICS P1/WISKUNDE V1 3 QUESTION 2/VRAAG 2 2.1 2.1.1 x( x + 5) = 0 üx=0 x = 0 or x = −5 ü x = −5 (2) 2.1.2 2 x + 1 3x + 1 = ü 4(2x + 1) = 3(3x + 1) 3 4 ü 8x + 4 4(2x + 1) = 3(3x + 1) ü 9x + 3 8x + 4 = 9 x + 3 ü x =1 x =1 (4) 2.2 2(4 − 3x)  20 ü 8 − 6x 8 − 6 x  20 ü − 6 x  12 − 6 x  12 ü x  −2 x  −2 (3) 2.3 a + b = 12 ………….(1) ü a = 12 − b 4a + 2b = 44 ……….(2) ü 4(12 − b) + 2b = 44 From (1)….. a = 12 − b ü 48 − 4b + 2b = 44 4(12 − b) + 2b = 44 üb = 2 48 − 4b + 2b = 44 ü a = 10 − 2b = −4 b=2 a = 10 OR From (1)….. b = 12 − a ü b = 12 − a 4a + 2(12 − a) = 44 ü 4a + 2(12 − a ) = 44 4a + 24 − 2a = 44 ü 4a + 24 − 2a = 44 2a = 20 ü a = 10 a = 10 üb = 2 b=2 OR 4a + 4b = 48 ……….(3) ü ü 4a + 4b = 48 (3) – (2) ü 2b = 4 2b = 4 ü b=2 b=2 ü a = 10 a = 10 OR 2a + 2b = 24 ……….(3) ü ü 2a + 2b = 24 (2) – (3) ü 2a = 20 2a = 20 ü a = 10 a = 10 ü b=2 b=2 (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 4 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2019) 2.4 Son/Seun Sipho ü 7 x + 25 Now/Tans x 7x ü 2(x + 25) In 25 years x + 25 7 x + 25 ü 2 x + 50 Equation: ü 5x = 25 Oor 25jaar 7 x + 25 = 2(x + 25) x=5 vergelyking üHis son is 5 years old/Sy 7 x + 25 = 2 x + 50 seun is 5 jaar oud 5x = 25 x=5 His son is 5 years old/Sy seun is 5 jaar oud. (5) [19] QUESTION 3/VRAAG 3 3.1 3.1.1 11 and/en14 ü for both 11 and 14 (1) 3.1.2 Tn = 3n − 4 ü 3n ü −4 (2) 3.1.3 T33 = 3(33) − 4 = 95 ü 3(33) − 4 ü 95 (2) 3.1.4 3n − 4 = 83 ü 3n − 4 = 83 3n = 87 ü n = 29 n = 29 (2) 3.1.5 3n − 4 = 116 ü 3n − 4 = 116 3n = 120 ü 3n = 120 n = 40 ü n = 40 (3) 3.2 3x + 2 − ( x + 3) = 6 x − 1 − (3x + 2) ü 3x + 2 − ( x + 3) 2 x − 1 = 3x − 3 ü 6 x − 1 − (3x + 2) x=2 ü 2x − 1 = 3x − 3 üx=2 (4) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2019) MATHEMATICS P1/WISKUNDE V1 5 QUESTION 4/VRAAG 4 4.1 Amount/Bedrag = 18, 18 x 3569 = R64 884, 42 üR64 884, 42 (1) 4.2 4.2.1 Loan/Lening = 0, 85 x 379 000 = ü0, 85 x 379 000 R322 150 üR322 150 OR/OF OR/OF (2) Loan/Lening = 379 000 – 0,15 x 379 000 ü379 000 – 0,15 x 379 000 = R322 150 üR322 150 (2) 4.2.2 A = P (1 + in) ü A = P (1 + in) A = 322150 (1 + 0,225  4) ü A = 322150 (1 + 0,225  4) A = R612 085 ü A = R612 085 (3) 4.2.3 Instalment/Paaiement ü 48 = 612085 = R12751,77 ü R12751,77 48 (2) 4.3 A = P(1 + i ) n ü A = P(1 + i )n 96714 ,02 = P(1 + 0,067 ) 6 ü 96714 ,02 = P(1 + 0,067 )6 96714 ,02 ü P = R 65539 ,47 P= (1,067 )6 P = R 65539 ,47 (3) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 6 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2019) QUESTION 5/VRAAG 5 5.1 5.1.1 y = −1 ü y = −1 (1) 5.1.2 f: ü x – intercept/afsnit ü y – intercept/afsnit g: üasymptote/asimptoot üorigin üshape/vorm and intersection (5) 5.1.3 (2;3) ü2 ü3 5.1.4 1 1 g (−1) = − ü g (−1) = − 2 2 g (1) = 1 ü g (1) = 1  1 1− −   1 mave =   = 2 3 1− −  1 − (− 1) 4 ü mave =  2 1 − (− 1) 3 ü mave = 4 (4) 5.2 5.2.1 m = tan 45  = 1 ü 1 (1) 5.2.2 y=x üy=x (1) 5.2.3 3 ü6 xy = 4  =6 2 6 y= x 6 üy= x (2) 5.2.4 ( A 6; 6 ) ü 6 ü 6 (2) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2019) MATHEMATICS P1/WISKUNDE V1 7 QUESTION 6/VRAAG 6 6.1 f: y = ax 2 + q ü 0 = a(−2) 2 + 8 0 = a(−2) 2 + 8 ü a = −2 4a = −8 ü y = −2 x 2 + 8 a = −2 y = −2 x 2 + 8 −4−0 g: m= = −2 ü m = −2 0 − (− 2) y = −2 x − 4 ü y = −2 x − 4 (5) 6.2 F (1;6) ü F (1;6) G (1;−6) ü G (1;−6) FG = 12 units/eenhede üFG = 12 units/eenhede (3) 6.3 y : y  8; y  R üü y : y  8; y  R (2) OR/OF (− ; 8 üü (− ; 8 (2) 6.4 −2 x  2 x  −2 ü and/en x  2 ü (2) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 8 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2019) QUESTION 7/VRAAG 7 7.1 7.1.1 2 2 P( A) = or/of 0,29 üü P( A) = or/of 0, 29 7 7 (2) 7.1.2 3 3 P(V ) = or/of 0,43 üü P(V ) = or/of 0, 43 7 7 (2) 7.1.3 4 4 P(C ) = or/of 0,57 ü P(C ) = or/of 0, 57 7 7 (1) 7.2 7.2.1 ü 2 Class/Klas = 30 ü3 ü9 R ü 16 S 2 3 9 16 (4) 7.2.2 16 8 16 8 = = 0,53 üü = = 0,53 30 15 30 15 (2) 7.2.3 (a) 14 7 14 7 = = 0,47 üü = = 0,47 30 15 30 15 (2) (b) 9 3 9 3 = = 0,3 üü = = 0,3 30 10 30 10 (2) [15] TOTAL/TOTAAL: 100 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief