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