[Tampines] [2023] 4E EM 4052 Prelims P1 MS

1 Mark Scheme Tampines Secondary School Mathematics Department Marking Scheme for 3E Math Preliminary Examination [√ means follow through] Total Marks : 90 No. Solutions Mark 1 𝐴 = 𝑃(1 + 𝑅 100)𝑛 𝐴 = 4500 (1 + 2.8 100)5 =5166.28 M1 A1 2 Diagram 4 B1 3 3 × 27𝑛 = 1 or 27𝑛 = 1 3 31 × 33𝑛 = 30 or 33𝑛 = 3−1 33𝑛+1 = 30 n = − 1 3 M1 A1 4 Listing or any method 8 numbers 13 on the left 14 14 14 14 14 15 …….. 8 numbers 16,17 on the right When x =1 , median is 14 When x =2, …. When x = 4 median is 14 When x = 5, median is 14.5 Range of x is 0 ≤ 𝑥 ≤ 4 or 0 ≤ 𝑥 < 5. M1 o.e A1 5a 9, 17 B2 5bi 1 3 B1 5bii 1 2 𝑥 + 1 2 𝑦 ≥ 17 2 ➔ 𝑥 + 𝑦 ≥ 17 = 1 12 M1 A1 or B2 6a 12 B1 6bi 12 37 B1 6bii − 35 37 B1

2 7 10𝑥 + 3𝑦 = 124 − − − − − − − −(1) 8𝑥 + 5𝑦 = 133 − − − − − − − −(2) Any method (1) × 5: 50𝑥 + 15𝑦 = 620 − − − − − − − −(2) (1) × 3: 24𝑥 + 15𝑦 = 399 − − − − − − − −(2) x = normal rate = $8.5 y = overtime rate = $13 M1 M1 any method A1 A1 8 Angle 𝑃𝐴𝑄 =.angle 𝐵𝐴𝐶 = angle ∠𝑆𝐴𝑈 = 𝜋 6 Area of triangle ABC = 1 2 (4𝑥)(4𝑥)𝑠𝑖𝑛 𝜋 6 = 4𝑥2 Area of sector APRQ/AUTS = 1 2 𝑟2𝜃 = 1 2 (8𝑥)(8𝑥) 𝜋 6 = 16𝜋 3 𝑥2 Area of triangle ASU / APQ = 1 2 (8𝑥)(8𝑥)𝑠𝑖𝑛 𝜋 6 = 16𝑥2 Area of segment = (2) 16𝜋 3 𝑥2 − (2)(16)𝑥2 (or at least 1 segment shown) Area of shaded region = 4𝑥2 + (2) 16𝜋 3 𝑥2 − (2)16𝑥2 = ( 32 3 𝜋 − 28)𝑥2 M1 M1 M1 M1 M1 A1 9 (6𝑛 + 1)2 − (6𝑛 − 1)2 = (6𝑛 + 1 − (6𝑛 − 1))(6𝑛 + 1 + 6𝑛 − 1) = 2(12n) = 24 n Or 36𝑛2 + 12 + 1 − (36𝑛2 − 12 + 1) = 24n M1 A1 10 (𝑛 − 2) × 180 = 720 3𝑥 + 135 + 115 + 164 + 90 = 720 x = 72 M1 A1

3 11a 7 − 8𝑥 + 𝑥2 = 𝑥2 − 8𝑥 + 7 = (𝑥 − 4)2 − 16 + 7 = (𝑥 − 4)2 − 9 = −9 + (𝑥 + (−4))2 M1 o.e A1 11b (4, -9) turning point indicated Correct cutting points at x-axis x = 1, 7 Correct cutting points at y-axis y = 7 B1 B1 B1 12a PQ // AC ∠𝐵𝑃𝑄 = ∠𝐵𝐴𝐶 (corresponding angles) ∠𝐵𝑄𝑃 = ∠𝐵𝐶𝐴 (corresponding angles) ∠𝑃𝐵𝑄 = ∠𝐴𝐵𝐶 (common angles) Since all corresponding angles of triangle 𝐴𝐵𝐶 and triangle PBQ are equal, hence triangle 𝐴𝐵𝐶 is similar to triangle PBQ. (AAA Similarity test or AA similarity test) M1 for any 1 correct equal angles shown A1 conclusion with 2nd angle shown 12b 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝐴𝐵𝐶 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝑃𝐵𝑄 = (3 2) 2 = 9 4 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝐴𝐵𝐶 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑎𝑝𝑒𝑧𝑖𝑢𝑚 𝐴𝑃𝑄𝐶 = 9 5 or 1.8 M1 A1

4 13a Mean = Σft Σt = 25×35+62×45+35×55+22×65+6×75 150 = 49.8 B1 13b Standard deviation = √Σf𝑡2 Σt − 𝑚𝑒𝑎𝑛(𝑡̅)2= √25 × 352 + 62 × 452 + 35 × 552 + 22 × 652 + 6 × 752 150 − 𝑚𝑒𝑎𝑛(𝑡̅)2 =10.565 = 10.6 minutes B1 13c His claim is wrong as the standard deviation measures consistency and how close the values are to one another, small standard deviation can mean most runners runs slower too. B1 14a = 4𝑥2 − 10𝑥𝑞 − 10𝑥𝑞 + 25𝑞2 by expansion or o.e = 4𝑥2 − 20𝑞𝑥 + 25𝑞2 M1 A1 or B2 14b 4𝑥2 − 20𝑞𝑥 + 25𝑞2 = 4𝑥2 + 40𝑥 + 100 −20𝑞 = +40 𝑞 = −2 25𝑞2 = 100 𝑞 = 2, 𝑞 = −2 Hence 𝑞 = −2 M2 A1 15 ( 3𝑥 4𝑦2) −2 = 1 ( 3𝑥 4𝑦2) 2 or ( 4𝑦2 3𝑥 ) 2 seen = 16𝑦4 9𝑥2 M1 A1

5 16a 1 : 65000 B1 16b 1 : 65000 1cm : 0.65 km 32 cm rep 20.8 km 32 cm M1 A1 16c 1 : 65000 1cm : 0.65 km Area scale: 1cm2 : 0.4225 km2 60 cm2 : 20.35 km2 20.35 km2 M1 A1 17a x = 4 B1 y = 2 B1 17b LCM = 24 × 33 × 5 × 7 B1 17c k = 5 B1 18a 12𝑛𝑚 − 3𝑛 − 4𝑚2 + 𝑚 = 3𝑛(4𝑚 − 1) − 𝑚(4𝑚 − 1) = (4𝑚 − 1)(3𝑛 − 𝑚) M1 A1 18b 8𝑥2 − 26𝑥 + 15 = 0 (2𝑥 − 5)(4𝑥 − 3) = 0 𝑥 = 5 2 𝑜𝑟 2.5 𝑥 = 3 4 𝑜𝑟 0.75 M1 A1 A1

6 19a 𝑃 = (4 2 3 6 0 3) B1 19b 𝑅 = (4 2 3 6 0 3) ( 12 2 25 −4 16 −3 ) =(146 −9 120 3 ) M1 for 2 values correct for the 2 by 2 matrix. 19c Store A $9 B1 B1 19d ($146 − 9)*0.9+($123)*0.95 = $240.15 M1 A1 20a 𝐵𝐶2 = 852 + 602 − 2(85)(60)𝑐𝑜𝑠115 𝐵𝐶2 = 15135.70627 𝐵𝐶2 = 123.0272 = 123 𝑚 M1 A1 20b tan 35 = 𝑇𝐴 85 𝑇𝐴 = 59.517 = 59.5 tan 𝜃 = 𝑇𝐴 𝑇𝐶 = 59.517 60 𝜃 = 44.768 = 44.8° M1 M1 A1 20c Area of triangle ABC = 1 2 𝑎𝑏𝑠𝑖𝑛𝑐 = 1 2 (85)(60) sin 115 =2311.084857 or 2550sin115 =2310 m2 1 2 𝑏ℎ = 2311.084857 𝑜𝑟 2550𝑠𝑖𝑛115 1 2 (123.02)ℎ = 2311.084857 𝑜𝑟 2550𝑠𝑖𝑛115 h = 37.572 = 37.6 M1 M1 o.e A1

7 21a 65 × 20 60 + 1 2 × 10 60 (𝑣 + 65) = 28.75 65 3 + 1 12 (𝑣 + 65) = 28.75 1 12 (𝑣 + 65) = 85 12 𝑣 + 65 = 85 v = 85 – 65 = 20 (shown) M1 M1 o.e A1 21b 𝑎 = 0−65 30 60 = -130 km/h Deceleration = 130 km/h M1 or o.e A1 21c Total distance travelled = 28.75 + 0.5(65) ( 30 60) = 45 𝑘𝑚 Average speed = 45 𝑘𝑚 1 ℎ = 45 km/h M1 A1 22a Distance of PR = √(−1 − 3)2 + (4 − 6)2 = √20 = 4.4721. . = 4.47 M1 A1 22b 𝑏 − 4 9 − (−1) = 𝑏 8 𝑏 − 4 10 = 𝑏 8 8𝑏 − 32 = 10𝑏 −32 = 2𝑏 b = – 16 𝑦 = − 16 8 𝑥 + 𝑐 4 = −2(−1) + 𝑐 4 = 2+c C = 2 𝑦 = −2𝑥 + 2 M1 M1 A1

8 22c 2𝑦 = 4𝑥 + 4 𝑦 = 2𝑥 + 2 Line k and line PQ are reflection of one another with respect to the y-axis. A1