2023 O Level Math 4052 Paper 1 SUGGESTED MS

General Certificate of Education Ordinary Level 2023 Mathematics 4052/01 Syllabus 4052 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 1 1 Sima $270 B1 Ken $210 B1 [2] 2 (a) 68.5x= B1 111.5x= B1 (b) 77.9 B1 [3] 3 (a) 968cd B1 (b) ( ) 22 2 16 25 9 25 5 25 16 9 5 k k  +  = += M1 Factorise out 225 s.o.i. 365 25 5 5 6 kk k =  = = A1 [3] 4 (a) Since the indices of M are all even or are all factors of 2 or are all divisible by 2, M is a perfect square. B1 (bi) 227k = B1 (bii) LCM 20 10 152 7 11=   B1 [3] 5 (a) The rate 0.25% should be multiplied to the respective balance of each month. Taking 3% of $4500 is taking simple interest per annum, not compound interest. B1 Describe how simple/compound interest works (b) Total value 12 0.25$4500 1 100 =+  M1 $4636.871806 $4636.87 (2 d.p.) = = A1 [3]

General Certificate of Education Ordinary Level 2023 Mathematics 4052/01 Syllabus 4052 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 2 6 (a) 2 2 14 ( 7) 49 x x b xb −+ = − − + 2 22 ( ) 25 2 25 xa x ax a +− = + + − M1 Expand or complete the square s.o.i. or M0 7a=− A1 or B1 24b=− or B1 (b) 7x= B1 FT from (a) [3] 7 (a) New Total 118% 550= M1 o.e. 649= A1 (b) Total remainder 100 81 18045=  = M1 Calculate remainder s.o.i. Total postcards 100 18072= M1 o.e. 250= A1 [5] 8 (ai) 114 cm B1 (aii) Upper Quartile 106= minutes Lower Quartile 125= minutes M1 Either one seen 19 minutes A1 (b) 121h= B1 (ci) 115.75 cm B1 c.a.o. (cii) 12.1 cm B1 [6] 9 In France: 780€ €488.93625021.5953$780 == M1 Conversion In Singapore: €488.9362502 €1 €0.6281233977 $1 €0.62 (2 d.p.) $1 780 +== = A1 Accept 4 d.p. [2]

General Certificate of Education Ordinary Level 2023 Mathematics 4052/01 Syllabus 4052 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 3 10 Gradient 3 ( 2) q p −= −− M1 o.e. 31 23 3 9 2 3 11 q p qp pq − =+ − = + =− A1 [2] 11 3 (3 1) 2(4 1) (4 1)(3 1) a a a aa + − − −+ M1 Common denominator 2 2 9 3 8 2 (4 1)(3 1) 9 5 2 (4 1)(3 1) a a a aa aa aa + − += −+ −+= −+ M1 [2] 10 (ai) ( ) 22xy x y+ B1 (aii) 215 10 12 8 5 (3 2 ) 4 (3 2 ) cd ce d de c d e d d e − + − = − + − M1 Grouping or M0 (5 4 )(3 2 )c d d e= + − A1 or B1B1 (b) 2210 15 4 6x ax ax a+ − − M1 2210 11 6x ax a= + − A1 [5] 13 73 25 5 n n + =+ M1 Form equation 35 5 75 3 2 40 20 nn n n + = + = = A1 [2]

General Certificate of Education Ordinary Level 2023 Mathematics 4052/01 Syllabus 4052 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 4 14 Since D is equidistant from A, B and C, AD BD CD== . Let angle CBD x= . Angle BCD x= (base angles of an isos. triangle) Angle BDC 180 2 x= − (angles sum of triangle) M1 Angle ADB 2x= (adj. angles on a straight line) Angle ABD 90 2 x= − (base angles of an isos. triangle) A1 Hence, angle ABC 90 xx= − + (shown) AG Alternatively: Since D is equidistant from A, B and C, AD, BD and CD can be taken as three radii of a circle, centre D, where AD BD CD== . M1 Attempt to link points to a circle Since AD CD= , AC is the diameter of the circle . Also, B lies on the circumference of the circle. Thus, by right angle in semicircle property, angle ABC is a right angle. A1 AG Mention diameter or semicircle with reason given [2] 15 ( ) 2 22 (2 1) 2( 5)(2 1) 4 4 1 2 2 10 5 n n n n n n n n − − − − = − + − − − + M2 M1 for each correct expansion 224 4 1 4 22 10 18 19 3(6 3) n n n n nn = − + − + − = − = − A1 Factorise out 3 o.e. Since 3 is a factor of the expression, it is a multiple of 3 for all integer values of n. AG Must conclude [3] 16 Since the ratio of the volumes are 331 :3 1: 27= , it is misleading as it is supposed to be tripled. B1 Compare volumes o.e. [1] 17 (a) B1 (b) ( ) ( ) 'C D C D   B1 o.e. [2] A B 

General Certificate of Education Ordinary Level 2023 Mathematics 4052/01 Syllabus 4052 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 5 18 (a) By similar triangles, 220 10 20 20 2 (Shown) hr hr d h r =  = = − = − B1 AG (b) 2 2 21 1 1π(10) (20) π π3 3 3 r h r h−= M1 Form equation 2 3 3 2000π 2π (2 ) 500 500 7.93700526 rr r r = = == M1 Find r o.e. 320 2 500 4.12598948 4.13 (3 s.f.)d = − = = A1 [4] 19 B2 B1 for coordinates on x and y-axis B1 for shape and scale drawn [2] 20 (3 4)( 4) (2 1 3)(2 1 3) xx x x x x +− + + + + − − M2 M1 for each factorisation s.o.i. (3 4)( 4) (3 4)( 2) 4 2 xx xx x x +−= +− −= − A1 [3] O (–7, 0) (3, 0) (0, 21) x y ( 3)( 7)y x x=− − +

General Certificate of Education Ordinary Level 2023 Mathematics 4052/01 Syllabus 4052 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 6 21 (a) 103.2 3 402.3 77 129 2 383 88.9 x x += + T B2 B1 for any 3 correct elements B1 for all correct (b) 103.2 3 129 2 5.6xx+ = + + M1 Form equation 31.4x= A1 (c) 1.60 2.80 = B B1 c.a.o. [5] 22 (a) 22 3 (4 ) 2 (3 2 ) 12 3 6 4 x x y x x y x xy x xy − = + − = + M1 Expand 276 6 7 xy x xy = = A1 (b) 2(3 4 ) 16 2(2 3 2 ) 14 2 16 10 4 6 4 16 x x y x x y x y x y yx + − + = + + − + = + −= M1 Find equation o.e. 66 4 167 x x −= M1 Substitute 8 167 14 x x = = A1 [5]

General Certificate of Education Ordinary Level 2023 Mathematics 4052/01 Syllabus 4052 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 7 23 (a) 304 B1 (b) Using cosine rule, 2263 95 2(63)(95)cos 64AC = + −  M2 Deduct M1 for any wrong value used 88.01532465 88.0 m (3 s.f.)AC== A1 c.a.o. (c) Using sine rule, 1 sin sin 64 95 88.01532465 95sin 64sin 75.9584626888.01532465 BAC BAC −  =  = =   M1 Bearing ( )360 180 60 75.95846268= − −  −  M1 180 60−  s.o.i. 164.0415373 164.0 (1 d.p.)= =  A1 c.a.o. [7] 24 (a) 2 kI D= (k is a constant) When 150D= , 1370I = . 7 21370 3.0825 10150 k k=  =  M1 Find constant When 228D= , 7 2 3.0825 10 228 592.9709141 593 (3 s.f.) I = == A1 (b) The Sun’s radiation would be quartered. B1 ‘Quartered’ o.e. [3] 25 (a) 3, 8, 13 B2 B1 for any two B1 for all correct (b) 52n− A1 [3]

General Certificate of Education Ordinary Level 2023 Mathematics 4052/01 Syllabus 4052 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 8 26 By Pythagoras’ Theorem, 228 6 10 cmAH = + = M1 s.o.i. 2 2 2 8 12 208AF = + = M1 2208 6 244 15.62049935 cmAG= + = = M1 Shortest length 17 15.62049935 1.379500648 1.38 cm (3 s.f.) =− == A1 [4] 27 (a) NM =−ab  2 ()5NP=− ab  MN =−ba  3 ()5MP=− ba  M1 Find NP  or MP  22 55 23 55 1 (2 3 ) (Shown)5 OP OP = + − =+ =+ b a b ab ab   33 55 23 55 1 (2 3 ) (Shown)5 OP OP = + − =+ =+ a b a ab ab   A1 AG (b) 19 55OQ= +