# Mathematics Grade 12 Paper 1 Western Cape Prelim 2024 Memo > Grade 12 Mathematics Paper 1 memorandum from the Western Cape Prelim Exam 2024, 17 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/wc-maths-p1-memo-sept-2024/d9c7151d-eb6e-4a27-9e05-8460f1c539aa - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 12 - Document type: Memorandum - Year: 2024 - Exam period: Western Cape Prelim Exam - Paper: 1 - Curriculum: CAPS - Pages: 17 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Mathematics](https://www.hlayiso.com/library/grade-12/mathematics) › [Mathematics Grade 12 Paper 1 Western Cape Prelim 2024 Memo](https://www.hlayiso.com/library/wc-maths-p1-memo-sept-2024/d9c7151d-eb6e-4a27-9e05-8460f1c539aa) ## About this document Mathematics · Grade 12 · Western Cape Prelim Exam · 2024. Memorandum, 17 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com + MATHEMATICS GR12 MARKING GUIDELINE-FINAL MATHEMATICS P1 SEPTEMBER 2024 MARKS: 150 TIME: 3 hours ✓ ONLY penalise for rounding in Q1.1.2 ✓ CALCULUS ONLY penalise for notation in Q7.1 NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • If a candidate has crossed out an attempt of a question and not redone the question, mark the crossed- out version. • Consistent accuracy applies in ALL aspects of the marking memorandum. Stop marking at the second calculation error. • Assuming answers/values in order to solve a problem is NOT acceptable. NOTA: • As 'n kandidaat 'n vraag TWEE KEER beantwoord, sien slegs die EERSTE poging na . • As 'n kandidaat 'n antwoord van 'n vraag doodtrek en nie oordoen nie, sien die doodgetrekte poging na. • Volgehoue akkuraatheid word in ALLE aspekte van die nasienriglyne toegepas. Hou op nasien by die tweede berekeningsfout. • Om antwoorde/waardes te aanvaar om 'n probleem op te los, word NIE toegelaat NIE. ### Page 2 Downloaded from hlayiso.com Mathematics P1 Marking guideline September 2024 QUESTION 1 1.1.1 (𝑥 − 1)(2𝑥 − 6) = 0 ✓ 𝑥=1 𝑥 = 1 or 𝑥 = 3 ✓ 𝑥=3 (2) 6 Accept 2 1.1.2 𝑥 2 − 7𝑥 − 7 = 0 −(−7) ± √(−7)2 − 4(1)(−7) ✓ Substitution 𝑥= 2(1) 7 ± √77 = 2 ✓ 7,89 = 7,89 or − 0,89 ✓ −0,89 −1 if rounding is incorrect. AO full marks (3) 1.1.3 2 6𝑥 + 7𝑥 > 5 6𝑥 2 + 7𝑥 − 5 > 0 ✓ Standard form (3𝑥 + 5)(2𝑥 − 1) > 0 ✓ Factors / formula 5 1 5 𝑥 < − or 𝑥 > ✓ 𝑥 < −3 3 2 1 ✓ 𝑥>2 AO 𝑓𝑢𝑙𝑙 𝑚𝑎𝑟𝑘𝑠 (4) • Learners must be penalised if they use the word 𝑎𝑛𝑑 (and not or) • They may use a ; • Loose one mark if include = sign 1.1.4 −6 1= + √𝑥 + 2 √𝑥 + 2 √𝑥 + 2 = −6 + 𝑥 + 2 or 𝑘 = −6 + 𝑘 2 ✓ Multiply by √𝑥 + 2 √𝑥 + 2 = 𝑥 − 4 𝑥 + 2 = 𝑥 2 − 8𝑥 + 16 ✓ square both sides 0 = 𝑥 2 − 9𝑥 + 14 ✓ standard form 0 = (𝑥 − 7)(𝑥 − 2) ✓ 𝑥=7 𝑥 = 7 or 𝑥 ≠ 2 ✓ 𝑥≠2 (5) 1 ### Page 3 Downloaded from hlayiso.com Mathematics P1 Marking guideline September 2024 1.2 6𝑦 + 2𝑥 = 4 𝑥 = 2 − 3𝑦 ✓ 𝑥 subject of formula 𝑥 2 + 𝑥𝑦 = 4 (2 − 3𝑦)2 + 𝑦(2 − 3𝑦) = 4 ✓ Subst. 4 − 12𝑦 + 9𝑦 2 + 2𝑦 − 3𝑦 2 = 4 6𝑦 2 − 10𝑦 = 0 ✓ standard form 2𝑦(3𝑦 − 5) = 0 ✓ factors / formula 5 −(−10) ± √(10)2 − 4(6)(0) 𝑦 = 0 or 𝑦 = ✓ both 𝑦 values 𝑦= 3 2(6) ✓ both 𝑥 values 𝑥 = 2 or 𝑥 = −3 (6) OR OR 2−𝑥 ✓ 𝑦 subject of formula 𝑦= 3 2−𝑥 ✓ Subst. 𝑥2 + 𝑥 ( )=4 3 3𝑥2 + 𝑥(2 − 𝑥) = 12 3𝑥2 + 2𝑥 − 𝑥2 = 12 2𝑥2 + 2𝑥 − 12 = 0 ✓ standard form 𝑥2 + 𝑥 − 6 = 0 (𝑥 − 2)(𝑥 + 3) = 0 ✓ factors / formula 𝑥 = 2 or 𝑥 = −3 ✓ both 𝑥 values −(1) ± √(1)2 − 4(1)(−6) 5 𝑥= 2(1) 𝑦 = 0 or 𝑦 = 3 ✓ both 𝑦 values (6) 1.3 4𝑥+1 √3. √48 − 22𝑥 22𝑥+2 ✓ 22𝑥+2 = √3. 4√3 − 2𝑥 2 = 12 − 4 ✓ 4 =8 ✓ Answer (3) OR OR 4𝑥+1 √3. √48 − 22𝑥 22𝑥+2 ✓ 22𝑥+2 = √144 − 2𝑥 2 = 12 − 4 ✓ 4 =8 ✓ Answer (3) 1.4.1 No, there will be no intersection between the graphs. ✓ answer Min value of 3(𝑥 − 1)2 + 5 is 5 ✓ reason (2) 2 ### Page 4 Downloaded from hlayiso.com Mathematics P1 Marking guideline September 2024 OR OR 2 3(𝑥 − 1) + 5 = 3 3(𝑥 − 1)2 = −2 −2 ✓ reason (𝑥 − 1)2 ≠ ✓ answer (2) 3 No, there will be no intersection between the graphs. OR OR 3(𝑥 − 1)2 + 5 = 3 3(𝑥 2 − 2𝑥 + 1) + 2 = 0 3𝑥 2 − 6𝑥 + 5 = 0 ∆= (−6)2 − 4(3)(5) ✓ reason ∆= −24 ∆< 0 ✓ answer No, there is no solution to the equation 𝑓(𝑥) = 𝑔(𝑥) (2) 1.4.2 3(𝑥 − 1)2 + 5 = 3 − 𝑘 3(𝑥 − 1)2 = −𝑘 − 2 3(𝑥 − 1)2 = −(𝑘 + 2)  Method ∴ 𝑘 + 2 < 0 for all real values of 𝑥  answer 𝑘 < −2 (4) OR OR 3𝑥 2 − 6𝑥 + 3 + 5 = 3 − 𝑘 3𝑥 2 − 6𝑥 + 3 + 5 + 𝑘 − 3 = 0 3𝑥 2 − 6𝑥 + 5 + 𝑘 = 0 ✓ sub ∆= (−6)2 − 4(3)(5 + 𝑘) ✓ 12𝑘 − 24 ∆= 36 − 60 − 12𝑘 ∆= −12𝑘 − 24 for real unequal roots −12𝑘 − 24 > 0 Method (∆> 0) −12𝑘>24 𝑘 < −2  answer (4) 3 Answer only full marks If AO 𝑘 ≤ −2 𝑚𝑎𝑟𝑘𝑠 4 [29] 3 ### Page 5 Downloaded from hlayiso.com Mathematics P1 Marking guideline September 2024 QUESTION 2 DAY 1 2 3 4 5 TIME 12 24 40 𝑎 𝑏 2.1.1 𝑎 = 60 ✓ 𝑎 = 60 𝑏 = 84 ✓ 𝑏 = 84 (2) 2.1.2 12 24 40 60 84 12 16 20 24 4 4 4 ✓ second difference can also be given at 2𝑎 = 4 2𝑎 = 4 𝑎=2 ✓ 𝑎 3(2) + 𝑏 = 12 𝑏=6 ✓ 𝑏 2 + 6 + 𝑐 = 12 𝑐=4 ✓ 𝑐 𝑇𝑛 = 2𝑛2 + 6𝑛 + 4 do not subtract marks if not given (4) 2.1.3 2 2𝑛 + 6𝑛 + 4 = 312 CA marking from 2.1.3 ✓ 𝑇𝑛 = 312 2 2𝑛 + 6𝑛 − 308 = 0 𝑛2 + 3𝑛 − 154 = 0 ✓ standard form (𝑛 − 11)(𝑛 + 14) = 0 𝑛 = 11 or 𝑛 ≠ −14 ✓ 𝑛 = 11 ✓ 𝑛 ≠ −14 (4) ∞ 2.2.1 4 4 ✓ First 3 terms ∑ 4𝑝1−𝑘 = 4 + + +⋯ 𝑝 𝑝2 𝑘=1 If only 2 terms, give the mark if r is correct in the next step 4 𝑟= ÷4 𝑝 1 ✓ 𝑟=𝑝 1 = 𝑝 4 𝑆∞ = =3 ✓ substitute into formula 1 1− 𝑝 Substitution must be 3 in the correct formula. 4=3− 𝑝 3 1=− 𝑝 𝑝 = −3 ✓ answer (4) 2.2.2 1 ✓ 𝑟 CA marking only if 𝑟=− 3 −1 < 𝑟 < 1 Series is convergent because −1 < 𝑟 < 1 ✓ reason 1 give full marks if learners only write −1 < 3 < 1 (2) 4 ### Page 6 Downloaded from hlayiso.com Mathematics P1 Marking guideline September 2024 2.3.1 3𝑥 − 4 −3𝑥 − 10 ✓ method +9= 2 2 3𝑥 − 4 + 18 = −3𝑥 − 10 ✓ Simplification 6𝑥 = −24 −24 𝑥= 6 𝑥 = −4 answer is given (2) OR OR 3𝑥 − 4 ✓ Method 𝑇3 = 2 Getting to equation (1) and (2) 3𝑥 − 4 𝑎 + 2𝑑 = 2 3𝑥 − 4 𝑎 + 2(3) = 2 3𝑥 − 4 𝑎+6= … … . . (1) 2 −3𝑥 − 10 𝑇6 = 2 −3𝑥 − 10 𝑎 + 5𝑑 = 2 −3𝑥 − 10 𝑎 + 5(3) = 2 −3𝑥 − 10 𝑎 + 15 = … … . . (2) 2 (2) − (1); −3𝑥 − 10 3𝑥 − 4 9= − 2 2 ✓ Simplification 18 = −3𝑥 − 10 − 3𝑥 + 4 For first step 6𝑥 = −24 −24 𝑥= 6 𝑥 = −4 (2) 2.3.2 3𝑥 − 4 𝑇3 = 2 ✓ Subt. 3(−4) − 4 𝑇3 = 2 ✓ −8 𝑇3 = −8 𝑑=3 𝑇1 = 𝑇3 − 3 − 3 𝑇1 = −8 − 3 − 3 𝑇1 = −14 ✓ −14 (3) OR OR 5 ### Page 7 Downloaded from hlayiso.com Mathematics P1 Marking guideline September 2024 𝑇3 = −8 ✓ 𝑎 + 2𝑑 = −8 𝑎 + 2𝑑 = −8 ✓ Substitution 𝑎 + 2(3) = −8 𝑎 = −14 ✓ −14 (3) 2.3.3 𝑆60 − 𝑆30 ✓ Method 60 30 ✓ 𝑆60 subst. or 4470 = [2(−14) + 59(3)] − [2(−14) + 29(3)] 2 2 ✓ 𝑆30 subs. or 885 = 4470 − 885 ✓ answer = 3585 (4) [25] 6 ### Page 8 Downloaded from hlayiso.com Mathematics P1 Marking guideline September 2024 QUESTION 3 3.1 𝑥 ∈ ℝ; 𝑥 ≠ 3 ✓ Answer (1) 3.2 2 𝑓(𝑥) = −1 𝑥−3 2 0= −1 ✓ 𝑦=0 𝑥−3 2 1= 𝑥−3 𝑥−3=2 ✓ Answer 𝑥=5 (2) 3.3 Hyperbola: Y  𝒙-int & 𝒚-int 𝒈  asymptotes  shape Line:  𝒙 and 𝒚 −int 𝟑 𝟓 X −𝟏  line X 𝒇 −𝟏 𝟐 𝟑 −𝟑 (5) 3.4 2 −1=𝑥−3 𝑥−3 2 =𝑥−2 𝑥−3 2 = 𝑥 2 − 5𝑥 + 6 2 −(−5) ± √(−5)2 − 4(1)(4) 𝑥 − 5𝑥 + 4 = 0 𝑥= ✓ standard form 2(1) (𝑥 − 4)(𝑥 − 1) = 0 ✓ factors / formula 𝑥 = 4 or 𝑥 = 1 3<𝑥≤4 ✓ CV  Inequality (4) [12] 7 ### Page 9 Downloaded from hlayiso.com Mathematics P1 Marking guideline September 2024 QUESTION 4 4.1 𝑥 = −3 ✓ Answer (1) 4.2 𝑦 = 𝑎(𝑥 + 5)(𝑥 + 1) ✓ substitute 𝑥-intercepts 5 = 𝑎(5)(1) ✓ substitute (0; 5) 𝑎=1 ✓𝑎 𝑦 = 1(𝑥 2 + 6𝑥 + 5) = 𝑥 2 + 6𝑥 + 5 ✓ equation (4) 4.3 2 𝑦 = 𝑥 + 6𝑥 + 5 CA from 4.2 𝑆𝑢𝑏𝑠𝑡. 𝑥 = −3: 𝑦 = (−3)2 + 6(−3) + 5 ✓ substitute 𝑥 = −3 𝑦 = 9 − 18 + 5 (1) 𝑦 = −4 answer given 4.4 −𝟑 < 𝑥 < −1 CA from 4.1 ✓ CV ✓ Inequality (2) 4.5 −𝑥 𝑔: 𝑦 = 𝑚 + k Subst. (−1; 0) 0 = 𝑚−(−1) − 4 𝑚=4 ✓ 𝑚=4 𝑔(𝑥) = 4−𝑥 − 4 ℎ(𝑥) = 4−𝑥 ✓ equation of ℎ ℎ−1 : 𝑥 = 4−𝑦 ✓ swop 𝑥 and 𝑦 do not have to show this step −𝑦 = log 4 𝑥 𝑦 = − log 4 𝑥 OR 𝑦 = log 1 𝑥 4 ✓ 𝑦 = − log 4 𝑥 or 𝑦 = log 1 𝑥 4 (4) [12] 8 ### Page 10 Downloaded from hlayiso.com Mathematics P1 Marking guideline September 2024 QUESTION 5 5.1 𝑦≥0 ✓ Answer (1) 5.2 √𝑥 = 3 ✓ 𝑦=3 𝑥=9 ✓ 𝑥=9 N(9; 3) (2) 5.3 𝑔(𝑥) = √𝑥 𝑔−1 : 𝑦 = √𝑥 where 𝑦 ≥ 0 𝑥 = √𝑦 where 𝑥 ≥ 0 ✓ swop 𝑥 and 𝑦 ✓ 𝑦 = 𝑥2 𝑦 = 𝑥2 where 𝑥 ≥ 0 ✓ 𝑥≥0 (3) [6] 9 ### Page 11 Downloaded from hlayiso.com Mathematics P1 Marking guideline September 2024 QUESTION 6 6.1 0,075 𝑛 ✓ Correct sub into formula 45 000 = 20 000 (1 + ) 12 9 161 𝑛 =( ) 4 160 9 ✓ Correct use of logs 𝑛 = log 161 ( ) 160 4 = 130,15 … = 131 months or 10 years and 10 /11 months ✓ Answer to the nearest month (3) 6.2.1 Original value of the equipment. ✓ Answer Accept other words if language Value at the start/ beginning barrier (1) 6.2.2 No, no matter how small the value, the next year it will be ✓ Explanation a fraction thereof. (1) OR OR 𝐸 = 𝑃(1 − 𝑖)𝑛 ✓ Explanation No, E represents an exponential function with an (1) asymptote at 𝐸 = 0. 6.2.3 48 033 = 106 239(1 − 𝑖)4 ✓ substitute into correct formula 4 48033 ✓ 4th root 𝑖 =1− √ 106239 𝑖 = 0,1799 … = 18% ✓ Answer Do not penalise for rounding (3) 6.3.1 0,102 −360 ✓ 𝑖 𝑥 [1 − (1 + 12 ) ] ✓ 𝑛 2 464 000 = ✓ Substitute into PV 0,102 12 𝑜𝑟0,0085 ✓ Answer 𝑥 = R21 988,40 Do not penalise for rounding (4) OR OR 0,102 360 0,102 360 𝑥 [(1 + ) − 1] 12 ✓ 𝑖 2 464 000 (1 + ) = 12 0,102 ✓ 𝑛 12 ✓ Substitute into FV 0,102 360 0,102 2 464 000 (1 + 12 ) ( 12 ) =𝑥 0,102 360 [(1 + ) − 1] 12 ✓ Answer 𝑥 = R21 988,40 (4) 10 ### Page 12 Downloaded from hlayiso.com Mathematics P1 Marking guideline September 2024 6.3.2 0,102 84 ✓ 2 464 000 (1 + 0,102 84 ) UB = 2 464 000 (1 + ) 12 12 0,102 84 0,102 84 ✓ 22 500[(1+ 12 ) −1] 22 500 [(1 + 12 ) − 1] 0,102 − 12 0,102 12  answer = R2 274 352,74 (3) OR OR 0,102 −𝑛 22 500 [1 − (1 + 12 ) ] 2 464 000 =  Sub into correct formula. 0,102 12 5236 0,102 −𝑛 = 1 − (1 + ) 5625 12 −𝑛 2017 389 ( ) = 2000 5625 𝑛 = 315,6158005 ∴ 231,6158 months left. 0,102 −231,6158  Months left. 22 500 [1 − (1 + 12 ) ] 𝑈𝐵 = 0,102 12 = R2 274 352,74 ✓ answer IF (3) 0,102 −276 22 500 [1 − (1 + 12 ) ] 𝑈𝐵 = 0,102 12 = 𝑅 2 391 073,84 ONLY 1 mark for answer [15] 11 ### Page 13 Downloaded from hlayiso.com Mathematics P1 Marking guideline September 2024 QUESTION 7 7.1 𝑓(𝑥) = −3𝑥 2 + 7 𝑓(𝑥 + ℎ) = −3(𝑥 + ℎ)2 + 7 = −3𝑥 2 − 6𝑥ℎ − 3ℎ2 + 7 ✓ 𝑓(𝑥 + ℎ) 𝑓(𝑥 + ℎ) − 𝑓(𝑥) = −3𝑥 2 − 6𝑥ℎ − 3ℎ2 + 7 + 3𝑥 2 − 7 = −6𝑥ℎ − 3ℎ2 ✓ 𝑓(𝑥 + ℎ) − 𝑓(𝑥) −6𝑥ℎ − 3ℎ2 ✓ substitute in formula ′ (𝑥) 𝑓 = lim ℎ→0 ℎ ℎ(−6𝑥 + 3ℎ) ✓ common factor/simplify = lim ℎ→0 ℎ = lim (−6𝑥 + 3ℎ) ℎ→0 ✓ answer = −6𝑥 Answer only – no marks. −1 for notation (5) 1 7.2.1 𝑦 = 8√𝑥 − √21 ✓ 8𝑥 2 1 = 8𝑥 2 − √21 1 𝑑𝑦 1 ✓ 4𝑥 −2 = 4𝑥 −2 𝑑𝑥 ✓ 0 [CA only if exponent is a fraction] (3) 7.2.2 𝑥 2 𝑦 = 2𝑥 3 + 4 ✓ 2𝑥 𝑦 = 2𝑥 + 4𝑥 −2 ✓ 4𝑥 −2 𝑑𝑦 = 2 − 8𝑥 −3 ✓ 2 𝑑𝑥 ✓ −8𝑥 −3 (4) 7.3 3 𝑦 = ℎ𝑥 + 𝑥 𝑑𝑦 ✓ Derivative = ℎ + 3𝑥 2 𝑑𝑥 ℎ + 3(−1)2 = 0 ✓ ℎ + 3(−1)2 = 0 ℎ = −3 ✓ answer (3) [15] 12 ### Page 14 Downloaded from hlayiso.com Mathematics P1 Marking guideline September 2024 QUESTION 8 8.1 −3𝑥 2 + 6𝑥 + 9 = 0 −(−2) ± √(−2)2 − 4(1)(3) ✓ 𝑓 ′ (𝑥) = 0 𝑥 2 − 2𝑥 − 3 = 0 𝑥= 2(1) (𝑥 − 3)(𝑥 + 1) = 0 ✓ Factors or formula 𝑥 = 3 or 𝑥 = −1 ✓ Answers D: 𝑥 = −1 and E: 𝑥 = 3 (3) 8.2 𝑓(𝑥) = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 𝑓 ′ (𝑥) = 3𝑎𝑥 2 + 2𝑏𝑥 + 𝑐 ✓ Derivative 3𝑎 = −3 𝑎 = −1 ✓ 𝑎 = −1 2𝑏 = 6 𝑏=3 ✓ 𝑏=3 𝑐=9 ✓ 𝑐=9 𝑓(𝑥) = −𝑥 3 + 3𝑥 2 + 9𝑥 − 𝟑𝟎 ✓ equation (5) 8.3 −1 ≤ 𝑥 ≤ 3 accept if you taught this ✓ CV OR ✓ Inequality −1 < 𝑥 < 3 according to me this is the correct answer (2) 8.4 Concave down: 𝑓 ′′ (𝑥) < 0 ✓ method −6𝑥 + 6 < 0 ✓ 𝑥=1 −6𝑥 < −6 ✓ answer ∴𝑥>1 (3) OR OR −1 + 3 𝑥= ✓ method 2 𝑥=1 ✓ 𝑥=1 ∴𝑥>1 ✓ answer (3) OR OR −(3) 𝑥 = 3(−1) 𝑥 = − 3𝑎 𝑏 ✓ method ✓ 𝑥=1 𝑥=1 ✓ answer ∴𝑥>1 (3) Answer only full marks [13] 13 ### Page 15 Downloaded from hlayiso.com Mathematics P1 Marking guideline September 2024 QUESTION 9 9.1 1 2 ✓ formula 𝑉= 𝜋𝑟 ℎ 3 ℎ2 + 𝑟 2 = 144 𝑟 2 = 144 − ℎ2 ✓ 𝑟 2 in terms of ℎ 1 𝑉= 𝜋ℎ(144 − ℎ2 ) ✓ substitute in volume formula 3 1 = 48𝜋ℎ − 𝜋ℎ3 answer given 3 (3) 9.2 𝑑𝑉 ✓ derivative = 48𝜋 − 𝜋ℎ2 𝑑ℎ ✓ =0 2 48𝜋 − 𝜋ℎ = 0 ℎ2 = 48 ℎ = 4√3 𝑜𝑟 6,93 ✓ answer Do not penalise for rounding (3) [6] 14 ### Page 16 Downloaded from hlayiso.com Mathematics P1 Marking guideline September 2024 QUESTION 10 10.1.1 (a) 90 ; (b) 120 ✓ a ; 𝑏 (2) 10.1.2 60 ✓ answer in any form P(Male and like B) = = 0,24 250 (1) 10.1.3 60 P(Male and like B) = = 0,24 250 130 P(𝐿𝑖𝑘𝑒 𝐵) = = 0,52 250 ✓ P(𝐿𝑖𝑘𝑒 𝐵) 150 𝑃(𝑀𝑎𝑙𝑒) = = 0,6 ✓ 𝑃(𝑀𝑎𝑙𝑒) 250 𝑃(𝐿𝑖𝑘𝑒 𝐵) × 𝑃(𝑀𝑎𝑙𝑒) = 0,52 × 0,6 = 0,312 ✓ 𝑃(𝐿𝑖𝑘𝑒 𝐵) × 𝑃(𝑀) = 0,312 ≠ 𝑃(𝑀𝑎𝑙𝑒 𝑎𝑛𝑑 𝑙𝑖𝑘𝑒 𝐵) ✓ Conclusion ∴Not independent (4) OR OR 150 𝑃(𝑀𝑎𝑙𝑒) = = 0,6 250 120 ✓ Both P(M) a & P(do not like B) 𝑃(𝑑𝑜 𝑛𝑜𝑡 𝑙𝑖𝑘𝑒 𝐵) = = 0,48 250 150 120 𝑃(𝑀𝑎𝑙𝑒) × 𝑃(𝑑𝑜 𝑛𝑜𝑡 𝑙𝑖𝑘𝑒 𝐵) = ( )( ) 250 250 = 0,288 ✓ Product 90 𝑃(𝑀 𝑎𝑛𝑑 𝑑𝑜 𝑛𝑜𝑡 𝑙𝑖𝑘𝑒 𝐵) = = 0,36 ✓ 𝑃(𝑀 𝑎𝑛𝑑 𝑑𝑜 𝑛𝑜𝑡 𝑙𝑖𝑘𝑒 𝐵) 250 P(M) × P(do not like B) ≠ P(M and do not like B) ∴ Not independent ✓ Conclusion (4) OR OR 70 P(Female and like B) = = 0,28 250 130 P(𝐿𝑖𝑘𝑒 𝐵) = = 0,52 ✓ P(𝐿𝑖𝑘𝑒 𝐵) 250 100 𝑃(𝐹𝑒𝑚𝑎𝑙𝑒) = = 0,4 ✓ 𝑃(𝐹𝑒𝑚𝑎𝑙𝑒) 250 𝑃(𝐿𝑖𝑘𝑒 𝐵) × 𝑃(𝐹𝑒𝑚𝑎𝑙𝑒) = 0,52 × 0,4 = 0,208 ✓ 𝑃(𝐿𝑖𝑘𝑒 𝐵) × 𝑃(𝑀) = 0,208 ≠ 𝑃(𝐹𝑒𝑎𝑙𝑒 𝑎𝑛𝑑 𝑙𝑖𝑘𝑒 𝐵) ∴Not independent ✓ Conclusion (4) 15 ### Page 17 Downloaded from hlayiso.com Mathematics P1 Marking guideline September 2024 10.2.1 8! = 40 320 ✓ 𝑎𝑛𝑠𝑤𝑒𝑟 Accept if only write 8! (1) 10.2.2 1 1 1 1 × × 𝑜𝑟 0,125 × 0,142857 × 0,166666 ✓ 8 8 7 6 1 1 −1 if not multiplying 1 ✓ 7×6 = 𝑜𝑟 0,002976 336 ✓ 𝑎𝑛𝑠𝑤𝑒𝑟 (3) OR OR 5×4×3×2×1 120 ✓ 5! 𝑜𝑟 8! 40320 ✓ Divide 8! 1 ✓ Answer = 336 (3) 10.3.1 5×5×9×9 5×5 = 2025 ✓9×9 Answer only full marks (2) 10.3.2 • 2 letters and 4 digits ( 2 letters en 4 syfers) 5×5×9×9×8×7=113 400 ✓ Method • 2 letters and 5 digits ( 2 letters en 5 syfers) 5 × 5 × 9 × 9 × 8 × 7 × 6 = 680 400 ✓ Method They would require 5 digits. 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