2025 O Level Math 4052 Paper 2 SUGGESTED MS

General Certificate of Education Ordinary Level 2025 Mathematics 4052/02 Syllabus 4052 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 1 1 (a) 6 512 6 5 10 x x =− =− M1 Remove fraction 10 1 1 10 x x =− =− A1 o.e. (b) ( ) 2234c a b − M1 3 (2 )(2 )c a b a b= + − A1 (c) 3 4 1 (1) 5 6 8 (2) xy xy + = −−− − = −−− (1) 3 9 12 3 (3) (2) 2 10 12 16 (4) (3) (4) 19 19 xy xy x  + = −−−  − = −−− + = M1 Elimination or substitution s.o.i. 1x= A1 1 2y=− A1 o.e. (d) 3 5 2( 4) ( 4)( 3) xx xx + − + +− M1 Join fraction 3 5 2 8 ( 4)( 3) 3 ( 4)( 3) xx xx x xx + − −= +− −= +− M1 Simplify 1 4x= + A1 [10]

General Certificate of Education Ordinary Level 2025 Mathematics 4052/02 Syllabus 4052 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 2 2 (ai) (a) 26 minutes B1 (ai) (b) Upper Quartile 36= minutes Lower Quartile 14= minutes M1 Either one seen 22 minutes A1 (aii) Required probability 80 72 80 −= M1 72 s.o.i. 1 10= A1 o.e. (bi) 60 children B1 c.a.o. (ci) 54k = B1 (cii) 37.7 B1 [8]

General Certificate of Education Ordinary Level 2025 Mathematics 4052/02 Syllabus 4052 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 3 3 (a) $813.54 B1 c.a.o. (b) 35 cm : 5.6 km 1 cm : 0.16 km M1 Find 1 cm on map in any unit o.e. 1 cm : 160 m 1 cm2 : 25600 m2 M1 Square map scale 0.24 cm2 : 25600 m2 × 0.24 6144 m2 A1 (c) Let A : B : C 10 :15 :12x x x= M1 Equalize ratios o.e. 3 (12 ) 10 878 29 872 6 xx x x += = = M1 Finds 1 unit using any valid method o.e. 222 students A1 (di) ( )0.3 0.35 0.35 B1 c.a.o. (dii) ( )73.4 55.1 B1 o.e. (diii) Koh and Min obtained a final grade of 73.4 and 55.1 respectively. B1 o.e. [10]

General Certificate of Education Ordinary Level 2025 Mathematics 4052/02 Syllabus 4052 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 4 4 (ai) Angle ACB = Angle DAE = 90 (Given) Angle CAB = Angle DAE (Shared angle) M1 By AA Similarity test, the two triangles are geometrically similar. A1 Must state AA (aii) 1.25 1.25tan 30 2.165063509 cmtan 30 1.25 1.25sin 30 2.5 cm sin 30 sin 30 8sin 30 4 cm8 cos30 8cos30 6.29820323 cm8 DEDE AEAE AC AC BC BC =  = =  =  = =  =  = = =  = = M3 M1 for any 1 side M1 for any 3 sides M1 for all sides Perimeter ( ) ( )AC AE AB AD BD BC= − + − + + 17.34326674 cm 17.3 cm (3 s.f.) = = A1 (bi) Angle RST 180 2 x= −  (Base angles of an isos. triangle) M1 With reason given 180 2 135 3 45 15