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 (Shown) xx x x − = + = = A1 AG (bii) 12 sides B1 (biii) Sum of interior angles of PQRST (5 2) 180 540= −  =  M1 s.o.i. Angle PTS 540 3(150 ) 452 − = =  M1 s.o.i. Angle RTP 45 15 30= − =  A1 [12]

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) 5 5 (a) 671 100 2 2 (2) M1 Numerator or denominator s.o.i. 5702= A1 (b) Initial value $1280= At time tT= , value 1.5 $1280 $1920=  = M1 s.o.i. Tangent drawn on graph M1 Gradient 2400 1000 6.6 2 304.3478261 304.35 (2 d.p.) −= − == A1 To 2 d.p. Accept ± 5 304.35 dollars/year B1 Correct units (a) When 0x= , 1 4y= . 011 44 ka k=  = M1 o.e. When 2x= , 9y= . 2119 46aa−=  = M1 When 3x= , 3 1 1 1 4 6 864y ==  . A1 c.a.o. [9]

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) 6 6 (ai) Area ( )1 (9)(12)sin 1.1 0.82=+ M1 o.e. 2251.10020474 cm 51.1 cm (3 s.f.)== A1 (aii) Using sine rule, ( ) 9 sin1.5 sin π 1.5 1.1 AC = −− M1 ( ) 9sin1.5 17.41499707 cmsin π 1.5 1.1AC == −− A1 Using cosine rule, 2217.41488707 12 2(17.41488707)(12)cos 0.8CD= + − M1 12.49346372 cmCD= A1 3 12.493463725 7.496078231 cm 7.50 cm (3 s.f.) CM = == A1 c.a.o. (bi) By Pythagoras Theorem, 2216 16 512 1 5122 tan 70 1 512 tan 70 31.08415863 cm2 PR PY XY PY XY = + = = =  = = M1 Find height of pyramid Volume of pyramid 231 (16) (31.08415863) 2652.51487 cm3== M1 Number of cubes 3 2652.51487 61.86... 613.5= = = M1 Maximum number of cubes s.o.i Remaining volume 332652.51487 61(3.5) 37.13987 cm= − = M1 3 3 4 π 31.139873 3 37.139874π r r = = M1 2.069746976 2.07 (3 s.f.)== A1 [13]

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) 7 7 (a) 1. Disagree; The elements {1, 2, 3, 4}  . B1 o.e. 2. Agree; The factors of 24 are all not multiples of 5. B1 o.e. (bi) B1 c.a.o. (bii) B1 FT from (b)(i) (ci) 83 220 B1 c.a.o. (cii) 18 17 16 12 11 10 30 29 28 30 29 28      +         M2 M1 for each case 37 145= A1 c.a.o. [8]  P Q  P Q

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) 8 8 (ai) When xp= , 21yp=+ . 1 15(1 2 1)( )24 pp+ + = M1 Form equation 2 2 15( 1) 4 4 4 15 4 4 15 0 (Shown) pp pp pp += += + − = A1 AG (aii) (2 3)(2 5) 0pp− + = M1 3 2p= or 5 2p=− (Reject) A2 A1 for reject o.e. (aiii) 2 230 (1 4)2AC = − + − M1 o.e. 3.354101966 units 3.35 units (3 s.f.)== A1 (bi) 2x= B1 (bii) Gradient 5 ( 2) 3 1 ( 4) 5 − − −= =−−− M1 o.e. 3 5 360 (2) 55 y x c cc =− + =− +  = M1 Find y-intercept 36 55 5 3 6 3 5 6 yx yx xy =− + =− + += A1 c.a.o. [11]

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) 9 9 (a) 3 hour 31 minutes B1 c.a.o. (b) 336 km B1 c.a.o. (c) Total distance travelled 95 260 135 317 807= + + + = km Time taken to drive 807 90== 8 hours 58 minutes M1 * Manchester: Range 420 95 260 65 km = − − = Range to be charged 336 65 271 km= − = York: Range 336 135 201 km =− = Range to be charged 336 201 135 km= − = M1 FT from * Calculate either range to be charged s.o.i. Time taken to charge 271 30193 42.124 mins = = Time taken to charge 135 3064 63.28125 mins = = M1 FT from drive time Calculate either time taken to charge s.o.i. Cost to charge 45$ 271 $121.95100=  = Cost to charge 25.1$ 135 $33.885100=  = M1 FT from time taken Calculate either cost to charge s.o.i. Total time taken 10= hours 43.405 minutes Total Cost $155.835= M1 Both correct Erin’s calculations are correct. A1 M0A0 (d) As a larger p ercentage of the car battery is charged, the charging rate decreases, which will length the total time of the journey. B1 [9] END OF MARKING SCHEME

General Certificate of Education Ordinary Level 2025 Mat