GMSS 2024 4N Prelim EM P2 Marking Scheme

Marking Scheme GMS(S)/4N/EMath/Prelim P2/2024 Page 1 of 9 SECONDARY 4 NORMAL (ACADEMIC) MARKING SCHEME PRELIMINARY EXAMINATION 2024 ELEMENTARY MATHEMATICS PAPER 2 SECTION A Question Answer Marks Guidance 1(a)(i) 42.1 million = 4 21 000 00 = 42 100 thousand B1 1(a)(ii) 0.005 187 62 ≈ 0.0052 (4 decimal places) B1 1(b) 2 2 6.582 0.891 435.18 7 1 400 0.35    M1 A1 Total 4 Question Answer Marks Guidance 2(a) 5 2 5 7 49 7 7 7 7 2 5 3 a a a a         M1 A1 2(b) 3 4 4 3 4 4 3 4 4 4 3 16 16 2 8 x x x x                M1 A1 Total 4

Marking Scheme GMS(S)/4N/EMath/Prelim P2/2024 Page 2 of 9 Question Answer Marks Guidance 3(a) p = 40 – (3 + 7 + 12 + 8 + 5) p = 5 B1 3(b) 12 8 5p(more than 2) = 40 25 = 40 0.625    M1 A1 3(c) (0 3) (1 7) (2 5) (3 12) (4 8) (5 5)mean = 40 110 = 40 = 2.75            M1 A1 Total 5 Question Answer Marks Guidance 4(a) 15 units --- 180 1801 unit --- 15 = 12 2 units (exterior angle) ---- 12 2 = 24 360 24 = 15 n         M1 A1

Marking Scheme GMS(S)/4N/EMath/Prelim P2/2024 Page 3 of 9 4(b) 180 50angle   = 2 = 65 (isosceles trian gle) angle    = 360 230 = 130 (angles at a po int) angle = 180 130 20 = 30 (angles in sum of t CDE BAC ABC               riangle) angle 35 30 = 65 Conclusion: Since angle = 65 and angle C = 65 , by corresponding angles, line is parall el to line . ABD ABD DE AB CD         M1 M1 M1 M1 A1 Total 6 Question Answer Marks Guidance 5(a) For the first part of the journey, remaining distance = 300 - 180 = 120 km Time taken for remaining journey = 120 = 75 D s =1.6 h T otal DistanceAverage speed for entire journey = Total Time 300 = 1.8 1.6  88. 2 3529412 88.2 km/h (3 sig fig)   M1 M1 A1 Accept answer in fraction: 488 km/h17

Marking Scheme GMS(S)/4N/EMath/Prelim P2/2024 Page 4 of 9 5(b) 2 2 Total surface area = 2 (7.6)(2.5) 2 (7.6) = 482.2973042 482 cm (3 si g. fig)     M1 A1 Total 5 Question Answer Marks Guidance 6(a) 70% × $149 = $104.30 60% × $95 = $57 Total = $104.3 + $57 = $ 161.30 M1 M1 A1 6(b) Let x be the price of the 1st handbag paid after discount. x + (x −