2019 O Level Math 4048 Paper 1 SUGGESTED MS

General Certificate of Education Ordinary Level 2019 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 1 1 350 B1 [1] 2 The widths of the cards are not consistent, misleading the reader to th ink that the areas of the cards are used for comparison. B1 Relating to the use of areas to mislead reader o.e. [1] 3 (a) 4x B1 (b) 8a= B1 [2] 4 (a) 322 3 5 B1 (b) 3p= , 5q= B1 [2] 5 6 6 3200 1 3890100 3891 100 320 r r += += M1 Use of compound interest formula and isolate bracket term 6 389100 1 320 3.307837444 3.31 (3 s.f.) r =−  == A1 [2] 6 13 2 3 3 1 3 1 6 9 (2 3)(3 1) xx xx xx −−− − − += −− M1 Join fraction 38 (2 3)(3 1) x xx −+= −− A1 [2] 7 (a) 17 20 B1 o.e. (b) 8x= B1 [2]

General Certificate of Education Ordinary Level 2019 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 2 8 (3 5)( 2) 0pp− + = M1 Factorisation only 5 or 23pp= =− A2 A1 for each value M0A0 [3] 9 (a) 133% 3 B1 (b) 6 2 11 52 5 2 x x x + − − 2 11 19 28 4 x x x +   M1 Either inequality correctly solved o.e. 5 42 x−   A1 o.e. [3] 10 (a) 2 P kQ= (k is a constant) When 1 3Q= , 2P= . 2 12 54 3kk=  = M1 Find constant When 1 6Q= , 3 1154 64P ==  A1 o.e. (b) 1 8m= B1 o.e. [3] 11 (a) ( )( ) 22x y x y+− B1 c.a.o. (b) 6 2 3 1 2 (3 1) (3 1) ab b a b a a − − + = − − − M1 Grouping of terms or M0 (2 1)(3 1)ba= − − A1 or B1 per bracket [3]

General Certificate of Education Ordinary Level 2019 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 3 12 Refer to diagram below (a) Angle bisector with arcs B1 (b) Perpendicular bisector with arcs B1 (c) Correct shaded area B1 FT from (a) & (b) [3] 13 (a) A(8, –9) B1 (b) 22( 3) 7AB = − +  M1 7.615773106 7.62 (3 s.f.)== A1 [3]

General Certificate of Education Ordinary Level 2019 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 4 14 (a) 137 B1 (b) Bearing of T from D 137 55 082= − =  Bearing of D from T ( )360 180 82= − −  M1 180 82−  s.o.i. 262= A1 [3] 15 B3 B1 for correct shape of curve and relative position B1 for correct coordinates for x- and y-intercepts B1 for correct coordinate of turning point [3] ( 1,0)