Mathematics Grade 10 Paper 1 North West November 2024 Memo

PROVINCIAL ASSESSMENT PROVINSIALE ASSESSERING GRADE/GRAAD 10 G MATHEMATICS P1/WISKUNDE V1 NOVEMBER 2024 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 100 TIME/TYD: 2 hours/uur These marking guidelines consists of 7 pages. Hierdie nasienriglyne bestaan uit 7 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P1/Wiskunde V1 2 NW/November 2024 Grade 10/Graad 10 – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 1 1.1.1 𝑥 2 − 16𝑥 + 15 = (𝑥 − 1)(𝑥 − 15)  answer/antwoord (1) 1.1.2 𝑎(6𝑎 + 2𝑏) − 3(3𝑎 + 𝑏)  common factor/ = 2𝑎(3𝑎 + 𝑏) − 3(3𝑎 + 𝑏) gemeenskaplike faktor = (3𝑎 + 𝑏)(2𝑎 − 3)  answer/antwoord (2) 2 3 3 1.2.1 (33 ) / √(33 )2 3  √(33 )2 = 32 =9  9 OR / OF 3 3 ( √27)2  ( √27)2 = 32 =9  9 (2) 1.2.2 𝑥 + 3 𝑥 3 − 27 𝑥−3 × 2 × 2 𝑥−3 𝑥 − 3 𝑥 − 9 𝑥 + 3𝑥 + 9  × 𝑥 2 +3𝑥+9 𝑥 + 3 (𝑥 − 3)(𝑥 2 + 3𝑥 + 9) 𝑥−3  (𝑥 − 3)(𝑥 2 + 3𝑥 + 9) = × × 2 𝑥−3 (𝑥 − 3)(𝑥 + 3) 𝑥 + 3𝑥 + 9  (𝑥 − 3)(𝑥 + 3) =1  answer/antwoord (4) 3𝑥+1 (1−3) 3𝑥 (3−32 )  common factor numerator/ 1.2.3 / 3𝑥 (32 +1) 3𝑥 (32 +1) gemeenskaplike faktor teller 31 (−2)  common factor denominator = 10 gemeenskaplike faktor (−6) noemer =  simplifying/vereenvoudig 10 3  answer/antwoord =− 5 OR/OF 3𝑥 .3 − 3𝑥 .32  common factor numerator/ 3𝑥 .32 + 3𝑥 gemeenskaplike faktor teller 3𝑥 (3−32 )  common factor denominator =3𝑥 (32 +1) gemeenskaplike faktor −6 noemer =  simplifying/vereenvoudig 10 3  answer/antwoord = −5 (4) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P1/Wiskunde V1 3 NW/November 2024 Grade 10/Graad 10 – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 2.1.1 2. 32𝑥 = 2. 32  factorising/faktoriseer 32𝑥 = 32  simplifying/vereenvoudig 2𝑥 = 2  simplifying/vereenvoudig 𝑥=1  answer/antwoord (4) 2.1.2 2𝑥 − 1 ≥ 3(2𝑥 + 1)  simplifying /vereenvoudig 2𝑥 − 1 ≥ 6𝑥 + 3  multiplication brackets −4𝑥 ≥ 4  simplifying/vereenvoudig 𝑥 ≤ −1  answer/antwoord (4) 2.1.3 2𝑥 2 − 𝑥 − 10 = 0 (2𝑥 − 5)(𝑥 + 2) = 0  factorising/faktoriseer 2𝑥 – 5 = 0 of 𝑥 + 2 = 0  equal to 0/stel gelyk aan 0 5 1 𝑥 = 2 /2 2 / 2,5 or / of 𝑥 = −2 answers/antwoorde (4) 2.2 (𝑥 + 5)(𝑥 − 5) = 𝑥 2 − 25  𝑥 2 − 25  𝑥2 − 25 (𝑥 + 1)(𝑥 − 25) = 𝑥 2 − 24𝑥 − 25  −24𝑥 −24𝑥 must be added at (𝑥 + 5)(𝑥 − 5) −24𝑥 moet by (𝑥 + 5)(𝑥 − 5) bygetel word  answer/antwoord (4) 2.3 𝑦 = 3𝑥 − 2………………...................... (1)  3𝑥 − 2 7𝑥 – 2𝑦 = 8 ………………………………. (2) 7𝑥 – 2(3𝑥 – 2) = 8  substitution/vervanging 7𝑥 – 6𝑥 + 4 = 8 𝑥 = 4  𝑥 = 4 𝑦 = 3(4) − 2 𝑦 = 10  𝑦 = 10 (4) [20] QUESTION/VRAAG 3 3.1.1 𝑇𝑛 = 3𝑛  3n (1) 3.1.2 𝑇300 = 3(300)  substitution/vervanging 𝑇300 = 900  answer/antwoord CA from/van 3.1.1 (2) 3.1.3 1500 = 3𝑛  substitution/vervanging 𝑛 = 500  answer/antwoord CA from /van 3.1.1 (2) 3.2 TERMSTERMS..repeat after the fifth letter.  see pattern/sien patroon Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P1/Wiskunde V1 4 NW/November 2024 Grade 10/Graad 10 – Marking Guidelines/Nasienriglyne TERMSTERMS..herhaal na die vyfde letter. 383 = 76 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 3  ÷5 5  answer/antwoord The 383rd letter will be a R / Die 383ste letter sal ‘n R wees. (3) 3.3.1 36 ; 43 answers/antwoorde (2) 3.3.2 Add 7 to the previous term / Tel 7 by die vorige term. answer/antwoord (2) [12] QUESTION/VRAAG 4 4.1 3 x $ 450 = $ 1 350  3 x $ 450 = $ 1 350 $ 1 350 x R 9,10  $ 1 350 x R 9,10 = R 12 285  answer/antwoord (3) 4.2.1 Balance/Balance = R20 000 – R3000 – R2000  20 000 – 3000 – 2000 = R15 000  answer/antwoord Total installments/Totale paaiemente = 24 x R900  24 x R900 = R21600  Total installments/Totale Total interest paid = Total installments – Balance paaiemente Totale rente betaal= Totale paaiemente - Balans = R21 600 – R15 000  R21 600 – R15 000 = R6 600  answer/antwoord (6) 4.2.2 𝐴 = 𝑃(1 + 𝑛𝑖)  formula/formule 21 600 = 15 000(1 + 2𝑖)  substitution/invervanging  simplification/vereenvoudig 21 600 − 1 = 2𝑖 15 000 0,22 = i  answer/antwoord r = 22 % (4) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P1/Wiskunde V1 5 NW/November 2024 Grade 10/Graad 10 – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 5 5.1  𝑓(𝑥): intercepts/afsnitte  𝑓(𝑥): shape & turning point vorm en draaipunt  𝑔(𝑥): intercept/afsnit  𝑔(𝑥): shape/vorm  𝑔(𝑥): asymptote/asimptote (5) 5.2 TP (0;4)  answer/antwoord (1) 5.3 The graph of 𝑓 is reflected in the 𝑥 – axis /  reflected/refleksie  𝑥-axis/as OR 𝑓 is reflected in the line 𝑦 = 0.  reflected/refleksie  𝑦=0 (2) [8] QUESTION/VRAAG 6 6.1 𝑐 = −4  𝑐 = −4 −8 𝑚= =2  𝑚=2 −4 OR /OF Substitute/Vervang A (4;4) 𝑔(𝑥) = 𝑚𝑥 − 4  substitution/vervanging 4 = 𝑚(4) − 4 8 = 4𝑚 𝑚=2  𝑚=2 OR / OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P1/Wiskunde V1 6 NW/November 2024 Grade 10/Graad 10 – Marking Guidelines/Nasienriglyne Δ𝑦 𝑚= Δ𝑥 4 − (−4) =  substitution/vervanging 4−0 8 =4 =2  𝑚=2 (2) 6.2 𝑦 ∈ 𝑅; 𝑦 ≠ −2  𝑦∈𝑅  𝑦 ≠ −2 (2) 6.3 4 4 − 2 = 2𝑥 − 4  − 2 = 2𝑥 − 4 𝑥 𝑥 4 − 2𝑥 = 2𝑥 2 − 4𝑥 0 = 2𝑥 2 − 2𝑥 − 4  0 = 2𝑥 2 − 2𝑥 − 4 B (2; 0) E (−1; −6)  B (2; 0)  E (−1; −6) (4) 6.4 −1 ≤ 𝑥 < 0 or 𝑥 ≥ 2  −1 ≤ 𝑥 < 0  𝑥≥2 CA from/van 6.3 (2) 6.5 4 4 𝑓(4) = − 2  𝑓(4) = 4 − 2 4 = −1  −1 𝐴𝐶 = 4 + 1  𝐴𝐶 = 4 + 1 = 5 units/eenhede =5 (3) [13] QUESTION/VRAAG 7 7.1 𝑦 = −2  𝑦 = −2 (1) 7.2 𝑎=1  𝑎=1 𝑞 = −2  𝑞 = −2 0 = (1)𝑏1 − 2 ∴𝑏=2  𝑏=2 ∴ ℎ(𝑥) = 2𝑥 − 2  ℎ(𝑥) = 2𝑥 − 2 CA from/van 7.1 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P1/Wiskunde V1 7 NW/November 2024 Grade 10/Graad 10 – Marking Guidelines/Nasienriglyne 7.3 𝑦 = −1  𝑦 = −1 OR / OF 20 − 2 = −1 (1) [6] QUESTION/VRAAG 8 8.1 𝑃(𝐴) + 𝑃(𝐵) = 0,43 + 0,37  0,43 + 0,37 = 0,80 Not complementary events /  conclusion/gevolgtrekking Nie komplimentêre gebeurtenisse. 𝑃(𝐴) + 𝑃(𝐵) ≠ 1 (2) 8.2 𝑃(𝐴 𝑎𝑛𝑑 𝐵) ≠ 0  𝑃(𝐴 𝑎𝑛𝑑 𝐵) ≠ 0 Not mutually exclusive events/  conclusion/gevolgtrekking Nie onderling uitsluitende gebeurtenisse. (2) 8.3  0,2709 𝑩  0,1591 𝑨  0,2109  0,3591 0,2709 0,1591 0,2109 0,3591 (4) [8] QUESTION/VRAAG 9 9.1 𝑃(𝑁𝑜𝑡 𝐴) = 1 − 𝑃(𝐴) / 𝑃(𝑁𝑖𝑒 𝐴) = 1 − 𝑃(𝐴) 2 =1− 2 3  1−3 1 1 =  3 3 (2) 1 9.2 𝑃(𝐵) = 1 − 𝑃(𝐵 ) 3 =1− 3 5  1−5 2 2 =  5 5 (2) 9.3 𝑃(𝐴 𝑜𝑟/𝑜𝑓 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 and/𝑒𝑛 𝐵)  formula/formule 4 2 2 = 3 + 5 − 𝑃(𝐴 and/𝑒𝑛 𝐵)  substitution/substitusie 7 52  answer/antwoord 𝑃(𝐴 and/𝑒𝑛 𝐵) = 105 (3) [7] TOTAL/TOTAAL: 100 Copyright reserved