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 202