ACSBR 2024 4E5N Prelim MATH P2 solutions

Mathematics Paper 2 Marking Scheme Secondary 4 Express / 5 Normal Academic Preliminary Exams 2024 Anglo-Chinese School (Barker Road) Qn Steps/Answer Remarks 1 (a) USD = 10000 𝑥 (b) 10000 𝑥 − 0.030 − 10000 𝑥 = 166 10000𝑥 − 10000(𝑥 − 0.030) = 166𝑥(𝑥 − 0.030) 166𝑥2 − 4.98𝑥 − 300 = 0 83𝑥2 − 2.49𝑥 − 150 = 0 (c) 𝑥 = 2.49 ± √(−2.49)2 − 4(83)(−150) 2(83) 𝑥 = 1.359 or 𝑥 = −1.329 (d) 20000 1.359 − 0.030 = USD 15 049 Accept 15 048 2 (ai) 𝑠 = [5 − 3(−3)2] ÷ (−3)2 𝑠 = −2 4 9 or 𝑠 = − 22 9 Cannot accept 3 sf (aii) 𝑠𝑞2 = 𝑟 − 3𝑞2 𝑠𝑞2 + 3𝑞2 = 𝑟 𝑞2(𝑠 + 3) = 𝑟 𝑞 = ±√ 𝑟 (𝑠 + 3) (b)(i) (7𝑛 − 1)2 − (𝑛 − 1)2 = [(7𝑛 − 1) + (𝑛 − 1)][(7𝑛 − 1) − (𝑛 − 1)] = 6𝑛(8𝑛 − 2) = 12𝑛(4𝑛 − 1) = 12(4𝑛2 − 𝑛) is also accepted (b)(ii) 𝑝 − 3𝑛 + 12𝑛2 − 4𝑛𝑝 (7𝑛 + 1)2 − (𝑛 − 1)2 = 𝑝(1 − 4𝑛) − 3𝑛(1 − 4𝑛) 12𝑛(4𝑛 − 1) = (3𝑛 − 𝑝)(4𝑛 − 1) 12𝑛(4𝑛 − 1) = 3𝑛 − 𝑝 12𝑛

Mathematics Paper 2 Marking Scheme Secondary 4 Express / 5 Normal Academic Preliminary Exams 2024 Anglo-Chinese School (Barker Road) 3 (a) One possible answer • ET = AT (tangents from external points are equal) • TO is common • OE = OA (radii of circle) Triangle TOA is congruent to triangle TOE (SSS test) Another possible answer • angle TAO = angle TEO = 90° (tangent perpendicular to radius) • TO is common • OE = OA (radii of circle) Triangle TOA is congruent to triangle TOE (RHS test) (bi) angle AOT = (180 – 90 – 32)° = 58° (tangent perpendicular to radius) angle ABF = 1 2angle AOT 1 2 (58°) = 29° (angle at centre is twice angle at circumference) angle OFG = angle ABF = 29° (alternate angles, OF //BA) (bii) angle ACF = angle ABF =29° (angles in the same segment) angle CAE = 180 – 90 – 29 = 61° (OT is perpendicular bisector of chord AE) angle CDE = 180 – 61 = 119° (angles in opposite segments) (c) As angle OET and angle OAT are right-angles, by the property of right angle in a semicircle, OT is a diameter and points E and A will lie on the circumference. OETA are thus four points on the circumference of this circle. Or Angle AOE + angle ATE = (58 × 2) + (32 × 2) = 180° Angle OET + angle OAT = 180°. By the property of angles in opposite segments, OETA are thus four points on the circumference of this circle. 4 (a)(i) 𝐴𝐵⃗⃗⃗⃗⃗ = 2𝑏 − 𝑎 (a)(ii) 𝐵𝐶⃗⃗⃗⃗⃗ = − 3 4 𝐵𝐴⃗⃗⃗⃗⃗ = − 3 4 (2𝑏 − 𝑎) 𝑂𝐶⃗⃗⃗⃗⃗ = 𝑂𝐵⃗⃗⃗⃗⃗ + 𝐵𝐶⃗⃗⃗⃗⃗ = 2𝑏 − 3 4 (2𝑏 − 𝑎) = 3 4 𝑎 + 1 2 𝑏 (b)(i) 𝑂𝑃⃗⃗⃗⃗⃗ = 𝑂𝐵⃗⃗⃗⃗⃗ + 𝐵𝑃⃗⃗⃗⃗⃗ = 2𝑏 + 3𝑎 (b)(ii) 𝑂𝑃⃗⃗⃗⃗⃗ = 3𝑎 + 2𝑏. 𝑂𝐶⃗⃗⃗⃗⃗ = 1 4 (3𝑎 + 2𝑏) As 𝑂𝐶⃗⃗⃗⃗⃗ = 1 4 𝑂𝑃⃗⃗⃗⃗⃗ , and O is a common point, O, C and P lie on a straight line.

Mathematics Paper 2 Marking Scheme Secondary 4 Express / 5 Normal Academic Preliminary Exams 2024 Anglo-Chinese School (Barker Road) (ci) 3:1 (cii) OAC : OAB : OAD 1 : 4 2 : 1 4 : 2 Therefore, OAC : OAD is 1 : 2. 5 (a) 𝑝 = −4 (b) (c) The maximum point of the curve is 93m. ±1m (d) 2.9 – 0.3 = 2.6 ±0.2s (e) Tangent drawn correctly –24 (±4) m/s

Mathematics Paper 2 Marking Scheme Secondary 4 Express / 5 Normal Academic Preliminary Exams 2024 Anglo-Chinese School (Barker Road) 6 (a) volume of water = 𝜋(102)(80) + ( 2 3)(𝜋)(103) = 8666 2 3 𝜋 (b) Capacity of one conical cup = ( 1 3)(𝜋)(32)(5.3) = 50 cm3 Alternative (c) Volume of water remaining after dispensing 250 cups = 8666 2 3 𝜋 − (250 × 1 3 𝜋(32)(5.3) Volume of water dispensed for 250 cups = 250 × 15.9𝜋 = 4691 2 3 𝜋 or 14739 cm2. Volume of water in cylinder = 4691 2 3 𝜋 − 2 3 𝜋(103) Height of water dispensed for 250 cups = 250×15.9𝜋 100𝜋 = 4025𝜋 or 12645 cm2. Height of water in cylindrical section = 4025𝜋 𝜋(102) or 12645 𝜋(102) Height of water remaining in dispenser = 90 − 125 𝜋 = 40.25 or 40.250 Height of water remaining in dispenser = 40.25+10 or 40.250+10 = 50.25 cm or 50.3 cm = 50.2 (3sf) (d) Slant height of cup = √32 + 5. 32 = 6.0902 Curved surface area of cup = 𝜋(3)(6.0902) = 57.399 cm2 250 cups will cost 57.399 × 250 × 0.003 = 43 cents Accept 44 cents

Mathematics Paper 2 Marking Scheme Secondary 4 Express / 5 Normal Academic Preliminary Exams 2024 Anglo-Chinese School (Barker Road) 7 (a) 18.5 × 7500 ÷ 100000 M1 1.3875 km A1 c.a.o. (b) 62 = 72 + 5. 52 − 2(7)(5.5) 𝑐𝑜𝑠 ∠ 𝑃𝑄𝑅 M1 Or equivalent method leading to the correct bearing 𝑐𝑜𝑠 ∠ 𝑃𝑄𝑅 = −43.25 −77 M1 𝑎𝑛𝑔𝑙𝑒 𝑃𝑄𝑅 = 55.827° M1 Bearing of Q from R is (180+55.827) = 235.8° A1 (c) Area of QPR = 1 2 (7 × 75)(5.5 × 75) 𝑠𝑖𝑛 5 5.827° √M2 = 89 585 m2 = 89 600 m2 A1 (d) Let shortest distance from R to PQ be X. 𝑠𝑖𝑛 ∠ 𝑃𝑄𝑅 = 𝑅𝑋 𝑄𝑅 𝑠𝑖𝑛 5 5.827° = 𝑅𝑋 412.5 M1 Or equivalent method 𝑅𝑋 = 341.28m M1 Let greatest angle of elevation be y. 𝑡𝑎𝑛 𝑦 = 75 𝑡ℎ𝑒𝑖𝑟 𝑅𝑋 M1 y =12.4° A1

Mathematics Paper 2 Marking Scheme Secondary 4 Express / 5 Normal Academic Preliminary Exams 2024 Anglo-Chinese School (Barker Road) 8 (ai) 15 marks (aii) 18 – 12 6 marks (b) 20.5 marks (c) Marks (x) 4 ≤ 𝑥 < 10 10 ≤ 𝑥 < 15 15 ≤ 𝑥 < 20 20 ≤ 𝑥 < 24 Number of students 6 14 14 6 (ci) (6×7)+(14×12.5)+(14×17.5)+(6×22) 40 =14.85 (cii) 4.62 Alternative answer (c) Marks (x) 4 ≤ 𝑥 < 10 10 ≤ 𝑥 < 15 15 ≤ 𝑥 < 20 20 ≤ 𝑥 < 24 Number of students 7 13 14 6 (ci) (6×7)+(14×12.5)+(14×17.5)+(6×22) 40 =14.7125 (cii) 4.76 (d) 15 (e) The students performed better in Mathematics as the median score (15 marks) was higher than Chemistry’s (14 marks) The students performed more consistently in Chemistry as the interquartile range (3 marks) is lower than Mathematics (6 marks)

Mathematics Paper 2 Marking Scheme Secondary 4 Express / 5 Normal Academic Preliminary Exams 2024 Anglo-Chinese School (Barker Road) 9 (a) $4100 (1 + 4 100) 5 = $4988 = $5000 (b) Justification: Choose the highest COE price recorded in the past 12 months to cover the worst-case scenario EV 60kWh Cost price of car $(100,000+105,000 – 30,000 =$175,000 Minimum downpayment 40% × $175,000 = $70,000 1998cc petrol car Cost price of car $(110,000+150,000) = $260,000 Minimum downpayment 30% × $260,000 = $78,000 6 months of Lee’s salary = 6 × $5000 = $30,000 Savings – downpayment EV 60kWh: $(105,000 – 70,000) = $35,000 1998cc petrol car: $(105,000 – 78,000) = $27,000 conclusion Lee can only afford the downpayment for the EV 60kWh. ($35,000 – $30,000 = $5000) (c) To determine is Lee can afford the car in Jan 2024. i) Road Tax [250 + 3.75(60 − 30)] × 0.7826 ÷ 6 = $47.246 ii) Loan amount 175000 – 70000 = $105,000 Interest 105,000×2.78×7 100 = $20,433 (iii) Monthly Instalment (105,000 + 20,433) ÷ 84 =$1493.25 iv) Other costs (monthly basis) 4700 12 + 600 = $991.667 Total monthly expense $2532.1625 Spare cash for the month 5000 – 2700 = $2300 v) Affordability Spare cash – total monthly expenses 2300 – 2532.1625 = – $232.1625 Lee would not be able to afford the car. However, if we take into consideration that Lee still has $5,000 remaining after making the downpayment in part (b), with $30,000 already set aside, he would be able to afford the car