2017 O Level Math 4048 Paper 1 SUGGESTED MS

General Certificate of Education Ordinary Level 2017 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 1 1 4k =− B1 [1] 2 3 6.5x−   B1 o.e. [1] 3 (a) 37 grams B1 (b) 210 grams B1 [2] 4 B2 B1 for correct x and y values B1 for correct shape and relative position [2] 5 (a) 16.5v= B1 o.e. (b) 33 0.48568 = m/s2 (3 s.f.) B1 Accept exact FT from (a) [2] 6 (a) {2, 10} A B1 (b) 3 B B1 (c) BC = B1 [3] O y x –3 8 24 ( 8)( 3)y x x=− − +

General Certificate of Education Ordinary Level 2017 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 2 7 Let x be the number of adults who joins the club. 58 7 95 10 x x + + M1 Form inequality 580 10 665 7 13 85 28 3 xx xx +  +    A1 Solve inequality o.e. Smallest number is 29 adults A1 [3] 8 Angle FDE 360 96 108 156= − − =  M1 Exterior angle 180 156 24= − =  M1 Number of sides 360 1524== A1 Alternatively (using sum of interior angles): ( 2) 180 156nn−  = M1 Form equation 180 360 156 24 360 15 nn n n −= = = A1 [3] 9 23 2ππ 3 2 3 r h r hr = = M1 Expression for h in terms of r Total surface area 2 22π 2π 3r r r =+  M1 210 π3 r= A1 [3]

General Certificate of Education Ordinary Level 2017 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 3 10 Refer to diagram below (a) Perpendicular bisector with arcs B1 (b) Angle bisector with arcs B1 (c) Correct shaded area B1 FT from (a) & (b) [3] 11 3(3 4) 4 16 xx−− = M1 Join fraction 9 12 4 6xx− − = M1 Remove fraction 5 18 18 5 x x = = A1 o.e. [3]

General Certificate of Education Ordinary Level 2017 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 4 12 5 1 6 1 8 9BC − − −     = − =            M1 Find CB  o.e. 22( 6) 9BC = − +  M1 10.81665383 10.8== units (3 s.f.) A1 [3] 13 4 2.51 674.79100PP+ − = M1 Find total amount s.o.i. 4 4 2.51 1 674.79100 674.79 2.511 100 P P + − = = +− M1 $6500.982 6 $65 08 00.P== (2 d.p.) A1 [3] 14 (a) (2 3)( 4)xx+− B2 B1 Each bracket (b) ( )( )2(2 3) 3 2 3 4yy− − − − M1 Use result from (a) (4 6 3)(2 7) (4 3)(2 7) yy yy = − + − = − − A1 M0A0 [4] 15 Angle ABC 90= (Right-angle in semicircle) B1 8.1tan 13.8ACB= M1 1 8.1tan 30.4110812713.8ACB − 