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)− (5,0) (2, 9) (0, 5) y x 2( 2) 9yx=− − + O

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) 5 16 (a) OT is shared OA OB= (radii of a circle) M1 Must state both Angle Angle 90 OAT OBT= =  (tangent is perpendicular to radius) M1 Triangles OAT and OBT are congruent. (RHS Congruency Test) A1 Must state RHS Alternative Method 1: OT is shared OA OB= (radii of a circle) M1 Must state both TA TB= (tangents from external points are equal) M1 Triangles OAT and OBT are congruent. (SSS Congruency Test) A1 Must state SSS Alternative Method 2: OA OB= (radii of a circle) Angle Angle 90 OAT OBT= =  (tangent is perpendicular to radius) M1 TA TB= (tangents from external points are equal) M1 Triangles OAT and OBT are congruent. (SAS Congruency Test) A1 Must state SAS (b) Refer to diagram below Two points plotted along the straight line passing through T and O B1 [4]

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) 6 17 (a) 3.6 km/h B1 (bi) 12.24 pm B1 Accept 1224 (bii) 8 km B1 (d) 22 m/s9 B1 o.e. [4] 18 (ai) {5, 6, 10, 15} B1 (aii) {5} B1 (b) RQ = B1 Deduct B1 for each extra answer 16 P B1 [4] 19 Let the original ratio be 7x : 5x . 7 60 3 5 60 2 x x − =− M1 Form equation 14 120 15 180xx− = − M1 Remove fraction 60x= M1 Finds one unit Ann’s current amount 7($60) $60 $360= − = A1 [4] 20 (a) Let h be the height of the smaller bottle. 4 30 10 4 3010 h h = = M1 12 cm A1 (b) Let d be the diameter of the base of this bottle. 3 500 4 125 d = M1 Form equation 34 4 6.349604208 6.35 cm (3 s.f.)d =  = = 6.35 cm A1 [4]

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) 7 21 (a) 24 14 29 20 15 x  = Q B1 (b) 800 822 290 40 45 232 800 835 xx  = + + T B2 B1 for any 3 correct elements B1 for all elements (c) 11x= B1 [4] 22 Gradient of AB 5 1 2 7 ( 2) 3 −−= =−−− M1 Gradient of PQ 3 2= M1 o.e. 3 2 374 (5)22 y x c cc =+ = +  =− M1 Find y-intercept o.e. 37 22yx=− A1 o.e. [4] 23 (a) Measured area 26 5 30 cm== M1 22 22 30 cm :19.2 m 1 cm : 0.64 m 1 cm : 0.8 m 0.8n= A1 (b) Total varnish needed 19.2 1.2 litres16== M1 Least amount 2($42.50) $24=+ M1 $109= A1 [5]

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) 8 24 (ai) 180 cm to 200 cm B1 Must state units (aii) 2178 cm3 or 179 cm (3 s.f.) B1 (bi) 24 B1 (bii) 48 cm B1 (c) The heights of the second set of trees are more consistent than the first as the interquartile range of 36 cm is lower than 48 cm, of the first. B1 Must state values of interquartile range [5] 25 (ai) 2(2) (2) 11 8 4 2 19 (Shown) pq pq + = + += B1 AG (aii) 8 4(8 ) 2 19 pq qq =− − + = M1 Substitution or elimination s.o.i. 32 4 2 19 2 13 6.5 qq qq − + = − =−  = A1 o.e. 8 6.5 1.5p= − = A1 o.e. (b) Sum of first 9 terms 23(9) 7(9) 306= + = Sum of first 10 terms 23(10) 7(10) 370= + = M1 Either one calculated 10th term of sequence 370 306 64=−= A1 [6] END OF MARKING SCHEME