Mathematics Grade 9 P1 Gauteng June 2023 Memo (Afrikaans)

Downloaded from hlayiso.com PROVINCIAL EXAMINATION PROVINSIALE EKSAMEN JUNE/JUNIE 2023 GRADE/GRAAD 9 MARKING GUIDELINES/ NASIENRIGLYNE MATHEMATICS/ WISKUNDE Keywords to consider/Sleutelwoorde om in ag te neem: M = Method/Metode CA = Consistent accuracy/Aaneenlopende akkuraatheid A = Answer/Antwoord 8 pages/bladsye

Downloaded from hlayiso.com MARKING GUIDELINES/ MATHEMATICS/WISKUNDE NASIENRIGLYNE GRADE/GRAAD 9 SECTION/AFDELING A QUESTION/VRAAG 1 1.1 1.2 1.3 1.4 1.5 B B D D C [5] SECTION/AFDELING B QUESTION/VRAAG 2 2.1 2.1.1 20A 1 mark for the HCF. 1 punt vir GGF. (1) 2.1.2 2 100A 1 mark for the LCM. 1 punt vir KGV. (1) 2.2 A rational numberA 1 mark for correct answer. ʼn Rasionale getalA 1 punt vir regte antwoord. (1) 2.3 2.3.1 73,33% A 1 mark for correct answer. 1 punt vir regte antwoord. OR/OF 73 13 % A (1) 2.3.2 27 A 1 mark for the answer. OR/OF 1 punt vir die antwoord. 27 30 A (1) 2.4 2.4.1 60 3  480 x M 1 mark for ratio/proportion. 1 punt vir verhouding/ 60x = 480 × 3 eweredigheid 1 mark for the final answer 1 440 1 punt vir finale antwoord. x OR/OF 60 2 marks if table or number line is used. x = 24 CA 2 punte as tabel of getallyn gebruik is OR/OF OR/OF 0 60 120 180 240 300 360 420 480 Consider alternative 0 3 6 9 12 15 18 21 24A mathematically correct responses which lead to the correct answer. Oorweeg alternatiewe wiskundig korrekte redenasies wat na die regte antwoord lei. (2) 2

Downloaded from hlayiso.com MARKING GUIDELINES/ MATHEMATICS/WISKUNDE NASIENRIGLYNE GRADE/GRAAD 9 2.4.2 Direct proportion CA 1 mark for correct answer. Direkte eweredigheidCA 1 punt vir regte antwoord. (1) 2.5  r  n 1 mark for correct A  P 1   substitution of p, r and n.  100  1 punt vir korrekte A = 3 030,05 1  100  A 4,5 5 substitusie van p, r en n. A = 3 030,05 104  M 1 mark for simplifying. ,5 5 100 1 punt vir vereenvoudiging. A = R3 775,993 … CA 1 mark for correct answer. A ≈ R3 776 CA 1 punt vir regte antwoord. OR/OF 1 mark for rounding-off correctly. A = P(1 + i)n where i = 100 4,5 = 0,045 1 punt vir korrekte A = R3 030,05(1 + 0,045)5 A afronding. A = R3 775,993 … CA A ≈ R3 776 CA (4) [12] QUESTION/VRAAG 3 3.1 –13 A 1 mark for answer. 1 punt vir antwoord. (1) 3.2 2 – (–1) A 1 mark for the expression./ =2+1 1 punt vir uitdrukking = 3 � 1 mark for the correct answer./ 1 punt vir regte antwoord. OR/OF 1 mark for answer only. 1 punt vir slegs antwoord. (2) 3.3 3.3.1  36 ( 1) 3  2 3 1 mark for −6 1 punt vir −6 = 168 M 1 mark for −1 1 punt vir −1 6 1 mark for answer. = 9 1 punt vir antwoord. = 23 C� 0 mark for answer only./ 0 punt vir slegs antwoord. (3) 3

Downloaded from hlayiso.com MARKING GUIDELINES/ MATHEMATICS/WISKUNDE NASIENRIGLYNE GRADE/GRAAD 9 3.3.2 3  6  ( 6  4)  5  7 1 mark for simplifying/ 1 punt vir vereenvoudiging. = 3  6  10  5  7 M 60 1 mark for dividing 5 . = 3 + 12 – 7 M 60 1 punt vir deling 5 . = 15 – 7 1 mark for answer. 1 punt vir antwoord. = 8 C� 0 mark for answer only. 0 punte vir slegs antwoord. (3) [9] QUESTION/VRAAG 4 4.1 9; 25; 92 ; 3 64; 1; 0,8 � 1 mark for 1st 3 correct numbers. 1 punt vir 1ste 3 korrekte getalle. 1 mark for last 3 correct numbers./1 punt vir laaste 3 korrekte getalle. (2) 4.2 3 64  64 � 1 mark for answer. 1 punt vir antwoord. (1) 4.3 4.3.1 5 1  511 � 1 mark for answer. 1 punt vir antwoord. OR/OF 5 1  15 � (1) 4.3.2 (pq)0 = 1 � 1 mark for answer. 1 punt vir antwoord. (1) 4

Downloaded from hlayiso.com MARKING GUIDELINES/ MATHEMATICS/WISKUNDE NASIENRIGLYNE GRADE/GRAAD 9 4.4 16a 2b 8 a 3b  1 mark for multiplying by 3a 3b 2 9a 4b the multiplicative inverse of the divisor./1 punt vir 16a 2b 9a 4b  M vermenigvuldiging met die 3a 3b 2 8a 3b vermenigvuldigingsinverse 6 2 van die deler. = (16(39)a 8)b 3 b  M 1 mark for product law 6 application./1 punt vir = 6ba C� toepassing van die produk- OR/OF wet. 16a 2b 8 a 3b 1 mark for quotient law  application./1 punt vir 3a 3b 2 9a 4b toepassing van die kwosient- 16�2 � 9�4 � wet. × M 3�−3 �2 8�3 � 1 mark for the answer 1 punt vir antwoord. = (2 × 3)a 2+4 – (–3) –3b1+1 –2 –1 M = 6a6b–1M 6 = 6ba C� (4) [9] QUESTION/VRAAG 5 5.1 216 � 1 mark for 216./1 punt vir 216. (1) 5.2.1 2; 6; 10 ……A 1 mark for all 3 terms. 1 punt vir al 3 terme. (1) 5.2.2 Tn = 4n – 2 CA 1 mark for 4n./1 punt vir 4n. 1 mark for –2./1 punt vir –2. (2) 5.2.3 Tn = 4n – 2 T15 = 4(15) – 2 M 1 mark for substitution./ T15 = 58 hearts/harte CA 1 punt vir vervanging. 1 mark for answer. 1 punt vir antwoord. (2) 5

Downloaded from hlayiso.com MARKING GUIDELINES/ MATHEMATICS/WISKUNDE NASIENRIGLYNE GRADE/GRAAD 9 5.2.4 39 × 2 = 78 hearts/harte A 1 mark for 78./1 punt vir 78. Tn = 4n – 2 1 mark substitution./ 78 = 4n – 2 M 1 punt vir vervanging. 80 = 4n 1 mark for answer. ∴ n = 20 CA 1 punt vir antwoord. OR/OF 1 mark for formula./ Tn = 2n – 1 M 1 punt vir formule. 39 = 2n – 1 M 1 mark substitution./ 40 = 2n 1 punt vir vervanging. n = 20 CA 1 mark for answer. 1 punt vir antwoord. (3) 5.3 12 cm A 2 marks for the answer. 2 punte vir antwoord. (2) [11] QUESTION/VRAAG 6 6.1 Variable/Veranderlike � 1 mark for answer. 1 punt vir antwoord. (1) 6.2 False/Vals � 1 mark for answer. 1 punt vir antwoord. (1) 6.3 (–12p + 12pq) – (2p+ pq) 1 mark for −pq./1 punt vir −�� = –12p + 12pq – 2p – pq� 1 mark for answer. = –14p + 11pq �� 1 punt vir antwoord. (2) 6.4 (ab + b) 2 2 1 mark for a b ./1 punt vir a b . 2 4 2 4 1 mark for 2ab3./1 punt vir 2ab3. = (ab2 + b) (ab2 + b) = a2b4 + 2ab3 + b2 � (2) 6.5 3x2 + 9 – 12x 1 mark for standardisation of the expression./1 punt vir = 3x2 – 12x + 9 � standaardvorm van uitdrukking. 1 mark for taking out the = 3(x2 – 4x + 3) � common factor./1 punt vir = 3(x – 3) (x – 1) � uithaal van gemene factor. 1 mark for each factor./1 punt vir elke faktor (x – 3) (x – 1). (4) 6

Downloaded from hlayiso.com MARKING GUIDELINES/ MATHEMATICS/WISKUNDE NASIENRIGLYNE GRADE/GRAAD 9 6.6 6.6.1 6p – 4 + 2(2p + 3) 1 mark for removing the brackets./1 punt vir verwydering = 6p –4 + 4p + 6� van hakies. = 10p + 2C� 1 mark for the answer. 1 punt vir antwoord. (2) 6.6.2 b 2  5b 2  3b b2 15b b 2  25 1 mark for (b + 5) (b – 5)./1 punt vir (b + 5) (b – 5). b(b  5) 3b(b  5) 1 mark for b(b + 5)./1 punt vir =  � (b  5)(b  5) 6b b (b – 5). 3b 2 1 mark for 3b(b – 5)./1 punt = vir 3b (b – 5). 6b b = �� 1 mark for the answer. 2 1 punt vir antwoord. (4) [16] QUESTION/VRAAG 7 7.1 4(2d + 6) = 32� 1 mark for answer. 1 punt vir antwoord. (1) 7.2 d(d – 2) = (d + 4) (d – 3) 1 mark for removing the brackets./1 punt vir verwydering d2 – 2d = d2 + d – 12 � van hakies. 1 mark for adding like terms. –3d = –12M 1 punt vir bymekaartel van gelyksoortige terme. d = 4 � 1 mark for the value of d. 1 punt vir waarde van d. Mary is correct/���� �� ��� � 1 mark for the final judgement. 1 punt vir finale slotsom. NB. correct judgement without calculations, NO Marks/ LW. korrekte slotsom sonder berekening verdien GEEN punte. (4) 7

Downloaded from hlayiso.com MARKING GUIDELINES/ MATHEMATICS/WISKUNDE NASIENRIGLYNE GRADE/GRAAD 9 x2 1 mark for multiplying all terms 7.3 x 2 3 by the LCD./1 punt vir 3x – (x – 2) = 6 � vermenigvuldiging met KGV. 3x – x + 2 = 6 � 2x = 4 1 mark simplification./1 punt vir x = 2 C� vereenvoudiging. 1 mark for answer. 1 punt vir antwoord. (3) 7.4 Let the number of first type of chocolate equal �./ 2 marks for setting up the Stel die 1ste soort sjokolade gelyk aan �. equation correctly./2 punte vir korrekte opstel van vergelyking. Number of first type of chocolate = x 1 mark for simplification./1 punt Number of second type of chocolate = 22 – x. vir vereenvoudiging. Aantal 1ste soort sjokolade= � 1 mark for answer. Aantal 2de soort sjokolade = 22 −� 1 punt vir antwoord. First kind is R8: Total amount for 1st kind = R8 × �. 1 mark for 13./1 punt vir 13. 2nd kind is R6: Total amount for 2nd kind is R6 (22 – x). 1ste soort kos R8: Totale bedrag vir 1ste soort = R8 × x. 2de soort kos R6: Totale bedrag vir 2de soort = R6 (22 – x) The total amount spent for both types of chocolates NB: is R150. Accept any variable (does not Die totale bedrag gespandeer op beide soorte is have to be �)./Aanvaar enige R150. veranderlike (hoef nie � te wees nie). ∴ 8x + 6 (22 – x) = 150 � Accept any reasonable and 8x + 132 – 6x = 150 � logical method to arrive at the 2x = 18 answer as correct./Aanvaar x = 9 � enige redelike en logiese metode ∴ there were 9 chocolates that cost R8/daar was 9 om die regte antwoord te verkry, sjokolade vir R8 and there were/en daar was as korrek. 22 – 9 = 13 �� chocolates that cost R6/ sjokolades vir R6. (5) [13] TOTAL/TOTAAL: 75 MARKS/PUNTE 8