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 −  = =   M1 30.4x= (3 s.f.) (Angles in same segment) A1 With reason [4] 16 (a) 39.55 minutes B1 c.a.o. (b) 9.42 minutes B1 (c) Number of adults 72 4499 =  M1 Either angle s.o.i. 32= A1 [4]

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) 5 17 Angle OBA 2x= (Base angles of an isos. triangle) M1 s.o.i. Angle OBC 5x= (Base angles of an isos. triangle) M1 s.o.i. 5 5 2 180x x x+ + = M1 Form equation 12 180 15 x x = = A1 [4] 18 (a) 2 2 28 2 7 63 3 7 = = M1 Either one seen 2 2 228 63 2 3 7 =   M1 Since the powers of 28 63 are all even / multiples of 2 / divisible by 2, it is a perfect square. A1 (b) 227k = M1 98k = A1 [5] 19 (a) 28 170 95 176 89x = P B1 (b) 1314 3 1236x = + R B2 B1 for each correct element (c) The total amount collected in Week 1 and 2 by the Tennis club are $1314 and $(3 1236)x+ respectively. B1 (d) 26x= B1 [5] 20 (a) 32, 25, 18p q r= = = B2 B1 for any two B1 for all correct (b) 46 7 n− B2 B1 for each term (c) 546 7 246 41 7nn− =−  = Since n is not a postive integer, –246 is not a term of the sequence. B1 [5]

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) 6 21 (a) For values of k less than the minimum value of the curve 22 3 7y x x= − − , the li ne yk= does not intersect the curve and thus the equation has no solutions. B1 (b) 22 12 0xx− − = B1 c.a.o. (ci) 34yx=− B1 (cii) Drawing of line in (c)(i) M1 FT from (c)(i) 0.45x=− or 3.4x= A2 [6]

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) 7 22 (a) 12.17(1.615 2) 5 1.615A += − M1 Substitution 12.99691285 13.00A== (2 d.p.) A1 c.a.o. (b) ( 2) (5 ) ( 2)5 bcA A c b cc +=  − = +− M1 Remove fraction 52A Ac bc b− = + M1 Expand 52 ( ) 5 2 Ac bc A b c A b A b + = − + = − M1 Factorise s.o.i. 52Abc Ab −= + A1 [6] 23 (a) Let y be the cost of the shampoo and x be its quantity. Thus y kx= , where k is a constant. When 30.80x= , 400y= . 400 1000 30.80 77k == When 19.25x= , 250y= . 250 1000 19.25 77k == When 5.95x= , 75y= . 75 1500 5.95 119k == M1 Any two distinct values of k calculated Since k is not a constant, then the cost of the shampoo is not directly proportional to its quantity. A1 Must conclude (b) Let h be the height of the 75 ml bottle. 3 75 18.4 250 h = M1 Form equation 33 75 75 18.418.4 250 250 h h=  =  M1 12.31756628 cm 12.3= cm (3 s.f.) A1 [5] END OF MARKING SCHEME

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) 8 BLANK PAGE