2016 O Level Math 4048 Paper 2 SUGGESTED MS

General Certificate of Education Ordinary Level 2016 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 1 1 (ai) 10a= B1 (aii) 45 45 45 bca bc ab ac b c b ab ac c −= + + = − − = + M1 Rearranging of terms (4 ) 5 5 4 b a ac c ac cb a − = + += − A1 (b) 6 9 4 312 xx−+ = M1 Join fraction 10 9 36 10 45 9 2 x x x −= = = A1 o.e. (c) 4 3 18 (1) 6 2 1 (2) (1) 2 8 6 36 (3) (2) 3 18 6 3 (4) (3) (4) 26 39 xy xy xy xy x − = −−− + = −−−  − = −−−  + = −−− + = M1 Elimination or substitution s.o.i. 3 2x= A1 4y=− A1 (d) (3 2)(3 2) (3 2)( 4) xx xx +− +− M2 M1 for each factorisation 32 4 x x −= − A1 [11]

General Certificate of Education Ordinary Level 2016 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 2 2 (a) 96 60 24 48 = M B1 (b) 40 65 = N B1 (c) 96 60 40 7740 24 48 65 4080     ==         P B1 (d) It represents the total session fees collected from a 12- week block of sessions on weekdays and on weekends respectively. B1 (e) Total ( )12 19($40)(95%) 9($65)(95%)=  + M2 M1 for 95% o.e. M1 for 12 and no. of students $15333 A1 [7] 3 (a) AB CD= (Sides of a regular polygon) BC DE= (Sides of a regular polygon) Angle ABC = Angle CDE (Interior angle of a regular polygon) M1 All reasons given Triangles ABC and CDE are congruent. (SAS Congruency Test) A1 Must state SAS (bi) Exterior angle 180 160 20= − =  M1 s.o.i. 360 1820n== A1 (bii) Angle BCA 10= B1 (biii) Angle ACE 160 2(10 ) 140= −  =  Angle CAE 180 140 202 − = =  M1 Angle DEA 20 10 30= + =  A1 [7]

General Certificate of Education Ordinary Level 2016 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 3 4 (a) 5 53T = B1 (b) 2( 1) 3 2nT n n= + + + For an even value of n, 2( 1)n+ is odd and 3n + 2 is even. Hence, nT is odd for all even n. For an odd value of n, 2( 1)n+ is even and 3n + 2 is odd. Hence, nT is odd for all odd n. B1 Consider nth term for even and odd values of n (c) 2( 1) 3 2nT n n= + + + M1 2 2 1 3 2n n n= + + + + M1 Expand 2 53nn= + + (Shown) A1 (d) ( ) 1 22( 1) 5( 1) 3 5 3 ppTT p p p p + − = + + + + − + + M1 22 2 1 5 5 3 5 3p p p p p= + + + + + − − − M1 Expand 26p=+ A1 (e) If the difference is 4, 2 6 4 1pp+ =  =− . Since p cannot be negative, two consecutive terms of the sequence cannot have a difference of 4. B1 Alternatively: Since 0p , 20p then 2 6 6 4p+   . Thus, the difference is greater than 6, which cannot be 4. B1 [9]

General Certificate of Education Ordinary Level 2016 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 4 5 (a) 0.5p=− B1 (b) Refer to graph drawn on page 5 Correct scale drawn S1 Plotting of all 8 points correctly P1 Smooth curve passing through all points C1 (c) 3 3 582 5 2 6 62 x x x xy −= − − =  = M1 Since the line 6y= only intersects the curve once, then the equation only has one solution. A1 (d) Drawing of tangent M1 Green Line Gradient 4 ( 10) 3.522 −−== −− A1 (ei) Plotting of at least three relevant points P1 Line drawn for 14 x−   L1 Red Line (eii) 2.85x= B1 Accept ± 0.1 (eiii) 4A=− B1 12B=− B1 [13]

General Certificate of Education Ordinary Level 2016 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 5 3 522 xyx= − − 43yx=− Tangent for (d)

General Certificate of Education Ordinary Level 2016 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 6 6 (a) Angle DAO = Angle BAC (Shared angle) Angle ADO 90= (Tangent perpendicular to radius) Angle ABC 90= (Right angle in semicircle) M1 Either reason seen Triangles ABC and ADO are similar. (AA Similarity Test) A1 (b) Area of triangle ADO : Area of triangle ABC 221 : 2 1: 4== M1 Area of triangle BOC : Area of triangle ADO 4 1 1:1 2 :1 = − − = A1 (c) 33cos 65 cos 65OAAO=  =  M1 Shaded area 2 23π π(3)cos 65 =−   M2 M1 for each circle’s area 130.0311149 130== cm2 (3 s.f.) A1 [8]

General Certificate of Education Ordinary Level 2016 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 7 7 (a) Total number of mugs 15 1402= M1 1050= A1 (bi) 3600 x B1 (bii) 3600 50x− B1 (biii) 3600 3600 68 50 4xx += − M1 Form equation 2 3600 180000 3600 1750 xx xx −+ =− M1 Join fraction 2 2 7200 180000 17 850 17 8050 180000 0 (Shown) x x x xx + = − − + = A1 AG (biv) 2( 8050) ( 8050) 4(17)(180000) 2(17)x − −  − −= M1 Quadratic formula o.e. 400450 or 17xx== A2 (bv) Number of mugs 3600 450 50= − M1 9= A1 [12]

General Certificate of Education Ordinary Level 2016 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 8 8 (a) Using cosine rule, 2 2 26 3 7 2(3)(7) cos BAC= + −  M1 Use of cosine rule o.e. 42cos 22 11cos 21 BAC BAC = = M1 s.o.i. 1 11cos 21 58.41186449 58.4 (1 d.p.) (Shown) BAC − =  = = A1 AG (b) Total surface area 12 (3)(7)sin 58.41186449 10(6 3 7)2 =  + + + M2 M1 for triangle M1 for three rectangular sides 2177.8885438 178 cm (3 s.f.)== A1 (c) Let the vertical distance of C above AB be h. sin 58.41186449 3 3sin 48.1186449 h h = = M1 o.e. 2.55550626 2.56 cm (3 s.f.)== A1 (d) Let F denote the last unnamed vertex of the prism and X denote the point on FD vertically below E. By Pythagoras’ Theorem, 2 2 2 3 2.55550626 2.469387755FX = − = M1 o.e. 2 2 210 3 2.55550626 102.4693878AX = + − = M1 Let the angle of elevation be  . 2.55550626tan 102.4693878  = M1 14.16842544 14.2 = =  (1 d.p.) A1 [12]

General Certificate of Education Ordinary Level 2016 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 9 9 (ai) Percentage 11 100% 55%20=  = B1 (aii) Median 31.5= B1 (aiii) The mean might not be an appropriate average as there is an outlier of 69 minutes. B1 Mention ‘outlier’ o.e. (aiv) Standard deviation 10.9= minutes B1 (av) The times taken by the second group of students are more consistent than the first as the standard deviation of 8.64 minutes is lower than 10.9 minutes of the second group. B1 Must state values (bi) Legend: W – White ball R – Red ball B – Black ball B2 B1 for first ball’s probabilities B1 for second ball’s probabilities Deduct B1 for no legend or proper notation of W, R and B W R B W R B W R B W R B 9 16 5 16 2 16 8 15 2 15 5 15 9 15 2 15 4 1