[FMS] [2023] 4E EM 4052 Prelims P2 MS

1 2023 Sec 4E/5N Prelim Math Paper 2 Marking Scheme QN Solution Marks AO Level 1a(i) or 0.16 B1 N5 AO1 1a(ii) M1 (x multiply) A1 N5 AO1 1b M1 (common denominator A1 N5 AO1 1c (1)x3, (1)x2 M1 (for subst/elim method) A1, A1 N7 AO1

2 QN Solution Marks AO Level 1d M1 (Form Quadratic Eqn) A1, A1 N7 AO1 2a CPF funds Ordinary Account B1 B1 N3 AO1 2b Amount Petrol used Petro Cost B1/M1 B1/A1 N4 AO1 2c Return after 1st yr Return after 2nd yr Let x % be rate of interest for 2nd year M1 A1 N3 AO2

3 QN Solution Marks AO Level 2d Option B Option A because the total charge for Option B is higher than Option A M1 (converting to SGD) M1 (1.5% fee) A1 (only if Option B is calculated correctly N10 AO3 3a(i) 150 min B1 S1/AO1 3a(ii) LQ = 138 UQ = 160 IQR = 160 – 138 = 22 M1 for LQ/UQ A1 S1 AO1 3b B1 S1 AO1 3c Between 140 mins 160 mins Airlines A has 90 – 34 = 56 movies Airlines B has 106 – 14 = 92movies Airlines B because there are more movies for Ali to choose. M1 for finding no of movies A1 S1 AO3 3d Prob of first movie is still playing B1 S2 AO2 3e Prob of choosing at least 2 comedy movies M1 A1 S2 AO1 4a 5.8 (1dp) Graph_ B1 P2 all points plotted correct P1 for 7 points plotted correct else P0 C1 N6/AO1

4 4a x y

5 QN Solution Marks AO Level 4b B1 N6 AO2 4c(i) B1, B1 N6 AO2 4c(ii) Refer to graph M1 for drawing A2 for all correct A1 for 1 or 2 correct N6 AO2 5a(i) B1 G7 AO1 5a(ii) B1 G7 AO1

6 QN Solution Marks AO Level 5a(iii) M1 A1 G7 AO2 5b(i) M1 A1 –1M from whole qn if there is no vector notation G7 AO1 5b(ii) M1 A1 G7 AO1

7 QN Solution Marks AO Level 5b(iii) Since vector AB is a scalar multiple of vector DC, AB is parallel to DC. ABCD is trapezium because AB is parallel to DC. M1 M1 A1 Must explain scalar multiple therefore AB is // DC G7 AO3 6(a) (i)(a) angle EBC = 64º (angles in the same segment) B1 G3 AO1 6(a) (i)(b) angle EOC = 2 × 64º = 128º (angles at centre = 2 × angles at circumference) reflex angle EOC = 360º-128º = 232º (angles at a point) B1 G3 AO1 6(a) (i)(c) angle EBC = 64º (angles in the same segment) angle EDC = 180º – 64º = 116º (angles in opposite segments) angle EDG = 180º – 116º – 46º =18º (adj angles on a str line) M1 A1 For Qn 6a(i) minus 1 mark if no/wrong reason given G3 AO1 6(a) (ii) CH = DH (tangent from ext point) angle CHD = 180º–2(46º) = 88º Since angle CHD does not form a right angle in a semi circle, therefor a semicircle with CD as diameter will not pass through H. M1 A1 G3 AO3

8 QN Solution Marks AO Level 6b(i) A: angle ORS = angle PRQ (common angle) Angle OSR = xº (base angle of isosceles triangle) Angle OSP = 180º – xº (adj angles on a str line) Angle PQR = xº (angles in opposite segments) A: Angle PQR = angle OSR = xº By AA, triangles ORS and PRQ are similar. M1 showing both corr angles No mark if no/wrong reason A1 G2 AO3 6b(ii) Since xº = 60º, triangle ORS is equilateral OR : QR 1 : 2 triangle ORS : triangle PRQ 1 : 4 area of triangle ORS and quadrilateral PQOS. 1 : 3 B1 B1 or B2 G2 AO1 7a M1 M1 simplification AG1 N7 AO3 7b M1 A1, A1 N7 AO1

9 7c Area of paper Cost of paper M1 for Area M1 for conversion to m2 A1 G5 AO2

10 8a Angle NBC = 180º – (360º – 290º) =110º Bearing of A from B = 118.4º + 110º =228.4º (1dp) OR Bearing of A from B = 360º – (70º + 68.041º) =222.0º (1dp) M1 M1 A1 B1 OR M1 A1 M1 A1 G4 AO2 N