2025 O Level Math 4052 Paper 1 SUGGESTED MS

General Certificate of Education Ordinary Level 2025 Mathematics 4052/01 Syllabus 4052 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 1 1 2.41 B1 c.a.o. [1] 2 (a) 20 B1 (bi) Number of scones with 700 ml of milk 700 18 37150 3=  = Number of scones with 900 g of flour 900 28 26275 11=  = M1 26 scones A1 (bii) Amount of butter 60 26 1958=  = A1 FT from (b)(i) [4] 3 Total interest earned 2.4% $5250 4 $504=   = M1 Calculate interest Total after 4 years $5250 $504 $5754+= A1 [2] 4 Calculation B because both the mass of gold and cost per gram of gold are rounded up. B1 Rounded up o.e. [1] 5 (a) 3 4 9 17 6 21 cc c − = + −= M1 Isolate c 7 2c=− A1 o.e. (b) 8 12 2 3 4 (2 3) (2 3) ab a b a b b + − − = + − − M1 Grouping o.e. (4 1)(2 3)ab= − + A1 (c) 22 (4 3 )(3 4 ) 12 9 16 12 x y x y x xy xy y +− = + − − M1 Expand 2212 7 12x xy y= − − A1 [6]

General Certificate of Education Ordinary Level 2025 Mathematics 4052/01 Syllabus 4052 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 2 6 (a) 43 2(5 3) 14 42x−  + −  43 2(5 3) 14 and 2(5 3) 14 42xx−  + − + −  Splitting 43 2(5 3) 14 29 2(5 3) 29 532 35 52 7 2 x x x x x −  + − −  + −  + − − 2(5 3) 14 42 2(5 3) 56 5 3 28 5 25 5 x x x x x + −  + +   M2 M1 for each side solved correctly o.e. 7 52 x−   A1 o.e. (b) B1 Correct inequality FT from (a) [4] 7 (a) Correct perpendicular bisector with arcs B1 (b) Correct angle bisector with arcs B1 (c) Correct shaded area B1 [3]

General Certificate of Education Ordinary Level 2025 Mathematics 4052/01 Syllabus 4052 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 3 8 (a) (–3, 9) B1 c.a.o. (b) 22( 7) 10PQ = − +  M1 12.20655562 units 12.2 units (3 s.f.) = = A1 [3] 9 Let one side of the cube be x. 2 2 6 125 125 125 66 x xx = =  = M1 Find one side of the cube s.o.i. Volume of cube 3 3 3 125 6 95.09072179 cm 95.1 cm (3 s.f.) = = = A1 [2] 10 (a) 2x= , 3y= B1 (b) 315k = B1 [2] 11 (a) The increase in cost per litre of fuel from the end of May to the end of June may not be linear in nature , where the cost might have increased suddenly close to the end of June. B1 The increase is non-linear o.e. (b) In between months, there may be drastic hikes in the cost of fuel per litre, which is not shown on the graph. B1 Increases may not be shown o.e. (ci) The vertical axis does not start from zero. B1 Does not start from zero o.e. (cii) A reader may think that the cost of fuel per litre at the end of June is about 8 times more of that at the end of January which is not true. B1 Any comparison made o.e. [4]

General Certificate of Education Ordinary Level 2025 Mathematics 4052/01 Syllabus 4052 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 4 12 8 4 7 6 6 8 4 7 2(3 4 2 ) 7 a b c c a b c a b − + = − + = − + = Since 2(3 4 2 )c a b−+ is always even for all integers a, b and c. However, if the result is 7, which is not even, then a, b and c cannot all be integers. B1 Explain by contradiction if a, b and c are all integers. [1] 13 Let angle CAD x= . Angle ACD 180 2 x−= . Also, angle ACB 42x= +  . 180 422 x x− =+ M1 Form equation 180 2 84 3 96 32 xx x x − = + = = A1 Angle ACD 32 42 74= + =  A1 [3] 14 Let the GNI in 2018 be x. GNI in 2019 1.0363x= GNI in 2020 0.97868172x= M1 Calculate GNI in 2020 s.o.i. GNI in 2021 1.0924x= Percentage increase 1.0924 0.97868172 100%0.97868172 xx x −= M1 o.e. 11.6195% 11.6% (3 s.f.) = = A1 [3] 15 (a) 2 2 ( 2)( 8) 2 8 16 10 16 y x x x x x xx = − − = − − + = − + M1 Form equation or M0 10b=− A1 or B1 16c= or B1 (b) 22 20 32y x x= − + B1 o.e. [3]

General Certificate of Education Ordinary Level 2025 Mathematics 4052/01 Syllabus 4052 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 5 16 (a) The tangent TA is perpendicular to the radius of the circle OA. Hence, angle OAT 90= . B1 Reason given (b) Angle AOT 180 90 42 48= − − =  (Angles sum of triangle) M1 With reason given Angle ACB 48 242 = =  (Angle at centre is twice the angle at centre) A1 With reason given (c) Angle AOC 180 2(23 ) 134= −  =  Reflex angle AOC 360 134 226= − =  M1 226 s.o.i. Shaded area 2226 π(8.7)360 =  M1 2 2 149.2774873 cm 149 cm (3 s.f.) = = A1 [6] 17 ( ) ( ) 453277 y− = M1 Change to base 7 12 1077 12 10 6 5 y y y −= =− =− A1 o.e. [2] 18 3 23 (3 )(2 3) a c d ad ad c d a +=− = + − M1 Remove fraction s.o.i. 6 2 9 3 6 9 3 ad ac ad c d ac ad c d = + − − + = + M1 Rearranging of terms o.e. (6 ) 9 3 93 6 a c d c d cda cd + = + += + A1 o.e. [3] 19 2 22 6 11 ( 3) (3) 11 y x x yx = + + = + − + M1 Attempt to complete the square o.e. 2( 3) 2yx= + + A1 Minimum point is (–3, 2) A1 [3]

General Certificate of Education Ordinary Level 2025 Mathematics 4052/01 Syllabus 4052 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 6 20 (a) Let angle APS x= Angle BPQ 180 90 90 xx= − − = − (Adjacent angles on a straight line) Angle BPQ 180 90 (90 ) xx= − − − = = Angle APS (Angles sum of triangle) M1 Prove angle APS and angle BPQ are equal o.e. PS QP= (Sides of a square are equal) Angle PAS = Angle QBP 90= (Interior angles of a square) M1 Hence, Triangles APS and BQP are congruent. (AAS Congruency Test) A1 Must state AAS (b) Let point X lie on BC and is vertically below point S. By Pythagoras’ Theorem, 2 2 2 25 25 1250QS = + = M1 Find QS o.e. 21250 31 17QX = − = cm M1 2 31 17 14 7 cm 24 cm PB PB AP = − = = = A1 A1 for both [6] 21 (a) 2 2 π (78) 13 π 20 r h = M1 Form equation o.e. 20 78 13 120 h h = = 120 cm A1 (b) 3 20 78 13 h = M1 Form equation o.e. 3 3 20 78 13 2078 13 h h = = M1 o.e. 90.04442251 90.0 (3 s.f.) h h = = A1 c.a.o. [5]

General Certificate of Education Ordinary Level 2025 Mathematics 4052/01 Syllabus 4052 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 7 22 Distance travelled by A 1 (20 )(60) 402 vv= + + M1 s.o.i. Distance travelled by B 100( 3)v=+ M1 s.o.i. 1100( 3) 75 ( 20)(60) 402v v v+ = + + + M1 Form equation 100( 3) 75 30( 20) 40 100 300 75 30 600 40 30 375 12.5 v v v v v v v v + = + + + + = + + + = = A1 [4] 23 25 34 3 5 2 (3 5) 5( 4) 3( 4)(3 5) x xx x x x xx +=+− − + + =+− M1 Join fraction 26 10 5 20 3( 4)(3 5)x x x x x− + + = + − M1 Remove fraction ( ) 226 5 20 3 3 12 5 20x x x x x− + = + − − M1 Expand ( 4)(3 5)xx+− 22 2 2 6 5 20 9 21 60 3 26 80 0 (26) (26) 4(3)( 80) 2(3) x x x x xx x − + = + − + − = −  − −= M1 Quadratic formula o.e. 2.407916139x= or 11.07458281x=− 2.41x= or 11.07x=− (2 d.p.) A1 Both seen [5] 24 33 4 12 44 4 12 16 81 81 16 ac ac −  =  M1 Flip fraction or M0 3927 8 ac= A1 c.a.o. or B1 for 27 8 or B1 for 39ac [2]

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