2024 FMSS EMATH P2 PRELIM MS

1 2024 Sec 4E/5N Prelim Math Paper 2 Marking Scheme QN Solution Marks AO Level 1a(i) B1 N5 AO1 1a(ii) M1 A1 N5 AO1 1b M1 A1 N7 AO1 1c M1 (common denominator) A1 N7 AO1

2 QN Solution Marks AO Level 1d M1 M1 (Form Quadratic Eqn) M1 A1,A1 N7 AO2 2a Total Cost % profit B1 B1 N3 AO1 2b Total Amount Interest M1 A1 N10 AO1

3 QN Solution Marks AO Level 2c(i) B1 N1 AO1 2c(ii) M1 A1 N1 AO1 2d(i) Shenzhen Hotel (3N) B1 N3 A01 2d(ii) Shenzhen Hotel in SGD Hong Kong Hotel(2N) In SGD with conversion fees Total Cost M1 M1 A1 N3 A02

4 QN Solution Marks AO Level 3a(i) 136 cm B1 S1/AO1 3a(ii) LQ = 132 UQ = 142 IQR = 142 – 132 = 10 M1 for LQ/UQ A1 S1 AO1 3b No of girls = 2 B2 S1 AO1 3c 1. Casa Sec Sch girls are shorter as the median height is lesser than Landmark Sec Sch or vice versa 2. Casa Sec Sch has a smaller/larger spread of height as the interquartle range is lesser than Landmark Sec Sch or vice versa A1 Must state median A1 Must state spread. Accept more consistent if smaller spread is stated S1 AO3 3di Prob both had weekly allowance that is less than $25 B1 S2 AO1 3dii Prob one had at least $30 of weekly allowance and the other had less than $20 of weekly allowance M1 A1 or B2 S2 AO2 4a –1.5 (1dp) B1 N6/AO1 4b See page 5 P2 all points plotted correct P1 for 7 points plotted correct else P0 C1 N6/AO1 4c(i) See page 5 P1 L1 N6/AO1

5 4b and 4c(i) 0 10 20 30 40 50 60 70 80 90 100 0 1 2 3 4 5 6 7 8 x y

6 QN Solution Marks AO Level 4cii B2 for all B1 for 2 correct N6 AO1 4c(iii) A = 11, B = –6 M1 B1, B1 N7 AO2 5a M1 A1 A1 G7 AO1

7 QN Solution Marks AO Level 5b(i) M1 A1 G7 AO1 5(ii) M1 A1 A1 G7 AO2 5b(iii) Since vector OU and OS are scalar multiple of each other with the common point O, therefore O, S, and U lies on a straight line. M1 A1 –1M from whole qn if there is no vector notation B1 G7 AO3

8 QN Solution Marks AO Level 6(a)(i) angle BAD = 180º – 98º (angles in the opp segment) = 82º angle BAO = 82º – 30º = 52º angle BOA = 180º – 2(52º) (angle sum of isos triangle) = 76º angle OAE = 90º (tangent perpendicular to radius) angle OEA = 180º – 90º – 76º (angle sum of triangle) = 14º B1 B1 B1 minus 1 mark if no/wrong reason given G3 AO1 6(a)(ii) Since angle OAE = 90º (tangent perpendicular to radius), it formed a right angle in a semicircle therefore a circle with diameter OE will passes through A. AG1 G3 AO3 6(b) R: angle QTS = angle TRU (rt angle in semicircle) H: QS = TU (diameter of 2 equal circles) S: QT = TR (radii of 2 equal circles) By RHS, triangles STQ and URT are congruent OR A: angle QTS = angle TRU (rt angle in semicircle) S: QT = TR (radii of 2 equal circles) A: angle TQS = angle RTU (equilateral triangle TQR) By ASA, triangles STQ and URT are congruent OR A: angle QTS = angle TRU (rt angle in semicircle) A: angle TQS = angle RTU (equilateral triangle TQR) S: QS = TU (diameter of 2 equal circles) By AAS, triangles STQ and URT are congruent M2 for RHS –1 if no/wrong reason AG1 G2 AO3

9 QN Solution Marks AO Level 7a V olume of inner core Must show AG1 G5 AO3 7b V olume of cylinder M1 A1 G5 AO2 7c M1 A1 G5 AO2 7d Radius golf ball Surface Area of golf ball Number of golf balls B1 M1 M1 for area of plastic sheet A1 G5 AO2

10 QN Solution Marks AO Level 8a M1 M1 A1 G4 AO1 8b or Let DX be the shortest distance from the foot of D to AC. or M1 M1 A1 G4 AO2 8c Let ߠ be the angle of depression M1 A1 G4 AO2 8d Bearing of A from B or B1 G4 AO1