2020 O Level Math 4048 Paper 1 SUGGESTED MS

General Certificate of Education Ordinary Level 2020 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 1 1 –1.46 B1 [1] 2 Diagram 3 B1 [1] 3 Angle BOA 66= (alternate angles, CB is parallel to OA) Angle OBA 90= (tangent perpendicular to radius) M1 s.o.i. Angle OAB 180 90 66 24= − − =  A1 [2] 4 (a) The vertical axis does not start from zero. B1 Do not accept the use of ‘y-axis’ (b) It may seem that the newspaper circulation has decreased significantly by 5 times from 2013 to 2016. B1 o.e. [2] 5 (a) 102 B1 (b) 24 21 3 10 a b ab M1 Flip fraction 2 14 5b= A1 o.e. [3] 6 Let one side of the hexagon be x. 2 2 1 ( ) sin 60 6Area of hexagon 2 1Area of triangle (4 ) sin 602 x x  =  M1 Some attempt to use area of triangle formula 3 : 8 A1 [2] 7 (a) 83 minutes B1 (b) 108 minutes and 69 minutes B1 [2] 8 (a) 0.25 B1 o.e. (b) Total 9 0.15= M1 60= A1 [3]

General Certificate of Education Ordinary Level 2020 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 2 9 4 10 15 320 xx−+ =− M1 Join fraction 6 15 60x− + =− M1 6 75 12.5 x x − =− = A1 o.e. [3] 10 (a) 22 99 422x   + − −       M1 Completing the square 2 9 97 24x= + − A1 o.e. (b) 9 97,24 −− B1 o.e. [3] 11 (a) 92n+ B1 (b) 9 2 335n+= M1 Equate 9 333 37 n n = = A1 [3] 12 Kim $171 B1 Pat $69 B1 Xin $45 B1 [3] 13 (a) 0y= B1 (b) 24 16xy=− M1 Rearrange 2 16 4 16 4 yx yx −= −= A1 o.e. [3]

General Certificate of Education Ordinary Level 2020 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 3 14 Total amount of water 100 378 450 litres84=  = M1 100 84 s.o.i. Capacity 100 45060= M1 100 60 s.o.i. 750 litres= A1 [3] 15 12 850 1 1120100 r+= M1 Correct use of formula 12 1121 100 85 r+= M1 Remove brackets 12 112100 1 85 2.325354541 2.33 (3 s.f.) r =−  == A1 [3] 16 (ai) {s, a, r, e} B1 (aii)  or { } B1 o.e. (b) ( ) 'PQ B1 [3] 17 (a) 1 : 200000 1 cm : 20 km M1 Express map in the form cm : km Distance in cm 950 47.5 cm20== A1 (b) 22 1 cm : 20 km 1 cm : 400 km M1 Area in 22 330803cm 827.0075 cm400== A1 c.a.o. [4]

General Certificate of Education Ordinary Level 2020 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 4 18 45tan 36 BC= M1 45 61.93718642 mtan 36BC ==  M1 45tan 61.93718642 22ADB= + M1 Angle of elevation 1 45tan 61.93718642 22 28.19643473 28.2 (1 d.p.) − =  + = =  A1 [4] 19 (a) 9 6 5 15x y x y+ − + M1 Expand 4 21xy=+ A1 (b) 3 (4 3 ) 2 (4 3 )a b x y b x− − − M1 Grouping or M0 (4 3 )(3 2 )b x a y= − − A1 or B1 per bracket [4] 20 (a) Appropriate method to prime factorise M1 or M0 232 3 11 A1 or B1 for any two correct factors and B1 for all correct (b) 2p= B2 B1 for any two correct values B1 for all correct 3q= 1r = (c) HCF 66= B1 c.a.o. [5]

General Certificate of Education Ordinary Level 2020 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 5 21 (ai) Angle EDC 102= B1 because of interior angles where AB is parallel to CD. B1 Must state parallel sides AB and CD (aii) Angle BCD 96= B1 because of the sum of interior angles in a pentagon. B1 Must mention pentagon (b) Angle BAC 18= (Base angles of an isosceles triangle) M1 Must state reason Since angle CAE + angle AED 120 78 18 180= + − =  , they sum up to interior angles where AC is parallel to ED. Thus, as ACDE has two pairs of parallel sides, it is a parallelogram. A1 [6] 22 (a) By similar triangles, 36 36 24 BD = 36 36 54 m (Shown)24BD=  = B1 AG (b) 1 ( 30)(37 24) 15862 CF+ + = M1 22 mCF = M1 Total area 1 1 1(24)(36) (36)(54) (22)(54) 15862 2 2= + + + M1 23177 m= A1 [5] 23 (a) 13.5 hours B1 (bi) 40% 140 56= M1 s.o.i. 14.75n= A1 M0A0 (bii) 60 12 140 − M1 60 or 12 s.o.i. 12 35= A1 [5]

General Certificate of Education Ordinary Level 2020 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 6 24 (a) (1 )OP m m= + −ac  B1 o.e. (bi) ( )4 (1 )OB m m= − −ac  M1 Factorise 4 s.o.i. 4OB OP=  Thus, OB  is a scalar multiple of OP  , w ith common point O, then O, P and B lie on a straight line. A1 AG Must state common point in conclusion (bii) CB OB OC=−  M1 s.o.i. in any step 3 4m= A1 o.e. 3k = A1 (c) 3 : 4 B1 [7] END OF MARKING SCHEME