GAUTENG DEPARTMENT OF EDUCATION GAUTENGSE DEPARTEMENT VAN ONDERWYS PROVINCIAL EXAMINATION PROVINSIALE EKSAMEN JUNE / JUNIE 2017 GRADE / GRAAD 9

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1 GAUTENG DEPARTMENT OF EDUCATION GAUTENGSE DEPARTEMENT VAN ONDERWYS PROVINCIAL EXAMINATION PROVINSIALE EKSAMEN JUNE / JUNIE 2017 MATHEMATICS WISKUNDE MEMORANDUM 8 pages / bladsye 1

2 GAUTENG DEPARTMENT OF EDUCATION GAUTENGSE DEPARTEMENT VAN ONDERWYS PROVINCIAL EXAMINATION PROVINSIALE EKSAMEN MATHEMATICS / WISKUNDE QUESTION 1 / VRAAG 1 Q/V ANSWER / ANTWOORD MARK ALLOCATION / PUNTE- TOEKENNING 1.1. B 1 mark / punt 1.2. B 1 mark / punt 1.3. B 1 mark / punt 1.4. C 1 mark / punt 1.5. D 1 mark / punt 1.6. C 1 mark / punt 1.7. C 1 mark / punt 1.8. D 1 mark / punt 1.9. D 1 mark / punt 1.10 A 1 mark / punt 2

3 QUESTION 2 / VRAAG a 2 b + 3ab 2 + 2a 2 b 4ab 2 a 2 b ab (x + y) + 4(3x 2y) 4(2x 3y) = 2x + 2y + 12x 8y 8x + 12y = 6x + 6y (2a 2 b 3 ) 2 (2a 2 b) a 6 b 1 = 4a 4 b 6 x 8a -6 b 3 4a 6 b -1 = 32a -2 b 9 4a 6 b -1 = 8b 10 a x3 64 = -3x (x 2) = 3x 4 5x 10 = 3x 4 5x 3x = x = 6 x = x 1 = 81 3 x 1 = 3 4 x 1 = 4 x = x 3 + x 4 = 1 4x + 3x = 12 7x = 12 x = x ( 2xy) = ( 1) 2 (2( 1)(2)) 3 = 1 ( 4) 3 = 1 ( 64) = 65 = 4a 4 b 6 x 8a -6 b 3 4a 6 b -1 = 8a -2 b 9 a 6 b -1 = 8b 10 a 8 3,5 x 2x + 2y + 12x 8y 8x + 12y 1 mark for / 1 punt vir 6x + 6y 4a 2 b 6 x 8a -6 b 3 32a - 4 b 9 8b 10 a 10 3x for 4 5x 10 for 2x = 6 for x = 3 for 3 4 x 1 = 4 1 mark for / 1 punt vir x = 5 4x + 3x = 12 7x = 12 x = mark for substitution / punt vir vervanging 1 ( 64) 65 3

4 QUESTION 3 / VRAAG 3 Q/V ANSWER / ANTWOORD MARK ALLOCATION/ x , x ,00 = R ,00 Deposit / Deposito A = P(1 + i x n) = R (1 + 0,12 x 3) = R Total amount = R R Totale bedrag = R , A = P (1 + i) n = (1 + 0,09) 3,5 = R , Option 1, it is a cheaper option than Option 2 / Opsie 1, dit is ŉ goedkoper opsie as opsie 2 (or any reasonable explanation) ,1923 = R 5, S$ 1 = R 5,20 (a) S$ 550 = R 5,20 x 550 = R (b) Number of DVDs = R Aantal DVD s = 28,6 28 R mark formula / punt vir formule R R R R mark formula correct / punt vir korrekte formule ( 1 + 0,09) 3,5 R ,50 1 mark for Option 1 / punt vir Opsie 1 1 mark for reason / punt vir rede R5,20 R 5,20 x 550 R R

5 QUESTION 4 / VRAAG T6 = 11,50 2 marks for / punte vir 11, Add 0,25 to the previous term to get the next term / plus 0,25 by vorige term om volgende term te kry 2 marks for answer / 2 punte vir antwoord 4.3 Tn = = 0,25 n = 0,25 n n = 60 25=0,25 n = 0,25 n n = 60 QUESTION 5 / VRAAG C1 = D2 = 20 o Alt. < ; DE//BC Verwisselende hoeke DE//BC D3 = 180 o 68 o 20 o Angles of triangle / Hoeke van driehoek = 180 o D3 = 92 o B1 + B2 = D1 = 68 o ADC = D1+D2 = 68 o + 20 o = 88 o Corresponding / Ooreenstemmend <; DE//BC OR/OF = 88 1 for / vir 20 o 1 for reason / vir rede B1 + B2 = D1 = 68 o Corresponding / Ooreenstemmend <; DE//BC ADC = 88 o Angles on straight line / Hoeke op reguit lyn A = C2 = (180 o 88) 2 A = 46 o C3 = B1 + B2 + A Exterior angle of triangle / Buitehoek van ŉ driehoek = 68 o + 46 o = 114 o C1 + C2 + C3 = 180 o Angles on a straight line / Hoeke op reguitlyn = 180 o C3 = 180 o 20 o 46 o = 114 o 5 A = 46 o 1 mark for / punt vir C3 = B1 + B2 + A and reason / en rede 1 mark for / punt vir C3 = 114 o C1 + C2 + C3 = 180 o Angles on a straight line / Hoeke op ŉ reguityn = 180 o for C3 = 180 o 20 o 46 o C3 = 114 o

6 5.2. In MNO and PNO: MO = OP MN = PN ON = ON MNO PNO Radii of circle / Radiusse van sirkel Given / Gegee Common side / Algemene newe S;S;S MO = OP Radius of circle / Radius van sirkel MN = PN Given / Gegee ON = ON Common side / Algemene newe MNO PNO S;S;S 5.3 In OPQ and / en OSR: O1 = O2 Vert. opp < P = S Alt < e ; PQ // RS Q = R Alt < e ; PQ // RS OPQ /// OSR <;<;< x x x 4 + 5x 4 = 360 o Angles of quad / Hoeke van vierhoek = 360 o 18x = 360 o x = 20 o 1 mark for / 1 punt vir O1 = O2 Vert. opp < 1 mark for/ 1 punt vir P = S Alt < e ; PQ // RS 1 mark for/ 1 punt vir Q = R Alt < e ; PQ // RS OPQ /// OSR <;<;< 1 mark for 8x x x 4 + 5x 4 = 360 o Angles of quad / Hoeke van vierhoek = 360 o 18x = 360 o x = 20 o 6

7 x x 4 = 4(20 o ) (20 o ) 4 =80 o o 4 = 180 o Therefore / Dus ST // UV 8x x 4 = 8(20 o ) (20 o ) 4 =160 o o 4 = 180 o Therefore ST // UV Co-interior / Mede-binnehoek < =180 o Co-interior / Mede binnehoek < =180 o 4x x 4 1 mark for substitution / punt vir vervanging for 180 o Therefore / Dus ST//UV Co-interior / Mede binnehoek < =180 o 8x x 4 1 mark for substitution / punt vir vervanging 180 o Therefore / Dus ST//UV Co-interior / Mede binnehoek < =180 o mark answer / 1 punt vir antwoord For each side. 5 cm 5 cm 7

8 QUESTION 6 / VRAAG 6 Q/V ANSWER / ANTWOORD MARK ALLOCATION = x = x 2 x 2 = 12 x = 12 = 3, C = 2 πr = 2(3,14) 2 = 12,56 m 2 = 6,28 P = 6, , = 29,74 m 6.3 Area = (½ π 2 2 ) + (5 m 2 m) + (4 m 2 m) + (½ 3,46 m 2 m) = 6, ,46 = 27,74 m = x 2 x 2 = 12 x = or 3,46 2(3,14) 6,28 3, ,46 23,46 + 6,28 1 mark for answer/ punt vir antwoord (½ π 2 2 ) (5 m 2 m) (4 m 2 m) (½ 3,46 m 2 m) 6, ,46 2 marks for/ punte vir 27,74 m

GRAAD 12 SEPTEMBER 2012 WISKUNDE V3 MEMORANDUM

GRAAD 12 SEPTEMBER 2012 WISKUNDE V3 MEMORANDUM Province of the EASTERN CAPE EDUCATION NASIONALE SENIOR SERTIFIKAAT GRAAD SEPTEMBER 0 WISKUNDE V3 MEMORANDUM PUNTE: 00 Hierdie memorandum bestaan uit 3 bladsye. WISKUNDE V3 (SEPTEMBER 0) VRAAG. ; 4; 0;