2020 O Level Math 4048 Paper 2 SUGGESTED MS

General Certificate of Education Ordinary Level 2020 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 1 1 (a) 2 1 5 4 23 6 3 10 8 xx xx +− +  − M1 Remove fraction 14 7 1 2 x x   A1 o.e. (b) 6 3 16 (1) 9 2 11 (2) (1) 2 :12 6 32 (3) (2) 3: 27 6 33 (4) xy xy xy xy − = −−− + = −−−  − = −−−  + = −−− (3) (4) 39 65x + = M1 Elimination or substitution s.o.i. 5 3x= A1 o.e. 2y=− A1 (c) 2 5(3 2 ) (3 2 ) xx x −− − M1 Join fraction 2 2 15 10 (3 2 ) 11 15 (3 2 ) xx x x x −+= − −= − A1 (d) 1 9 3 1527 a b   M1 Resolve negative index or M0 5 3 3b a= A1 or B1 for ‘3’ or B1 for 5 3 b a (e) ( ) 244 4( 2)( 2) (3 5)( 2) (3 5)( 2) x xx x x x x − +−=− + − + M2 M1 for each factorisation 4( 2) 35 x x −= − A1 o.e. [12]

General Certificate of Education Ordinary Level 2020 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 2 2 (a) 7 girls B1 (b) 30 26 24  = F B1 (c) 996 1316 = M B2 B1 for each correct element (d) The t otal fees collected by the holiday club for one morning is $996 and one afternoon is $1316. B1 (e) $11560 B1 (f) Number of children in P 1.5 (10 14) 36=  + = Number of children in Q 1.5 (12 16) 42=  + = Number of children in R 0.75 (16 20) 27=  + = M1 Total fees in Week 2 5 (36 $30 42 $26 27 $24) $14100=   +  +  = M1 Percentage change 14100 11560 100% 21.97231834%11560 −=  = 22.0% increase A1 o.e. [9]

General Certificate of Education Ordinary Level 2020 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 3 3 (a) 5.9 B1 (b) Refer to graph on page 4 Plotting of any 6 correct points P1 Plotting of all 9 points correctly P1 Smooth curve passing through all plotted points C1 (c) 0.7 or 4.3x= B2 Accept ±0.1 (di) Plotting of any 3 relevant points P1 Line drawn passing through plotted points L1 (dii) 0.9 or 5.1x= B2 Accept ±0.1 (diii) 63 2 9 2 3xx x  + − = − M1 Substitution seen o.e. 22 2 2 6 18 27 2 3 4 24 18 0 2 12 9 0 x x x x xx xx + − = − − + = − + = 12A=− A1 9B= A1 [13]

General Certificate of Education Ordinary Level 2020 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 4 3 2 3yx=− 629yx x= + −

General Certificate of Education Ordinary Level 2020 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 5 4 (a) Using cosine rule, 22660 950 2(660)(950)cos80AC = + −  M2 M1 for formula M1 for substitution 1058.463597 m 1060 m (3 s.f.) = = A1 AG (b) Using sine rule, sin sin 24 1058.463597 480 1058.463597sin 24sin 480 ADC ADC  = = M1 Apply sine rule formula Angle ADC 1 1058.463597sin 24180 sin 480 116.2454077 − = −  = M1 Bearing of D from C 360 (180 116.2454077 ) 296.2454078 296.2 (1 d.p.) = − −  = =  A1 (c) Time taken (600 950 1058.463597) 1000 9.5 + + = M1 16.85345429 mins 16 mins 50 secs (nearest 10 seconds) = = A2 [9]

General Certificate of Education Ordinary Level 2020 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 6 5 (a) By Pythagoras’s Theorem, Slant height 225.5 8 94.25= + = M1 Curved surface area ( )π(5.5) 94.25= M1 22167.7464128 cm 168 cm (3 s.f.)== A1 (bi) 3 Volume of water 6 27 75%Volume of glass 8 64 = =  B1 Hence, it is incorrect to say that the glass is filled to 75% of its capacity. AG (bii) Percentage 3 6 100%8 = M1 42.1875%= A1 c.a.o (biii) Volume of glass 21 242π(5.5) (8) π33== M1 Volume of water 242 1089π 42.1875% π3 32=  = M1 Let the radius of the cylindrical glass be r. 2 2 1089π (2.5) π 32 1089 80 r r = = M1 Make r the subject 1089 3.689512162875 3.6980r = = = cm (3 s.f.) A1 [10]

General Certificate of Education Ordinary Level 2020 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 7 6 (ai) BF CF= (tangents from external points are equal) DF EF= (tangents from external points are equal) Angle Angle BFD CFE= (vertically opposite angles) M2 M1 for any two correct statements M1 for all correct Hence, triangles BDF and CEF are congruent. (SAS Congruency Test) A1 Must state SAS (aii) (a) (90 ) x− B1 (aii) (b) 2x B1 (bi) Arc length 12(2π 1.8)=− M1 53.79822369 cm 53.8 cm (3 s.f.)== A1 (bii) Area of triangle KLM 21 (12) sin1.8 72sin1.8 2== M1 Area of major sector 21 (12) (2 π 1.8) 144π 129.62= − = − M1 Percentage of area that is shaded 2 2 π(12) 72sin1.8 144 129.6 100%π(12) − − += M1 13.14862333% 13.1% (3 s.f.)== A1 [11]

General Certificate of Education Ordinary Level 2020 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 8 7 (a) 1(10) 2(13) 3(9) 4(6) 5 6(2)Mean 2.68 50 q+ + + + +== M1 Knows to use mean formula s.o.i. 5 99 134 5 35 7 q qq += =  = A1 Total number of students 50= 7 40 50 3 p p + + = = A1 (b) SD 1.59= B1 c.a.o. (c) On average, the students watched more movies than the adults as the mean of 2.68 is higher than the adults’ of 2.04. B1 Must state values The number of movies watched by the students is more consistent than the adults as the standard deviation of 1.59 is higher than the adults’ of 1.92. B1 Must state values [6]

General Certificate of Education Ordinary Level 2020 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 9 8 (a) ( ) 2 27 ( 4) ( 1 5)AB= − − − + − − M1 6.708203932 units 6.71 units (3 s.f.)== A1 (b) Gradient of CD 5 ( 1) 24 ( 7) −−==− − − M1 8 2 8 2 8 4 4 2 5 3 BC OC OB OC =− −= − −     = + =     −         M1 Finds OC  o.e. 2 3 2(4) 5 y x c cc =+ = +  =− M1 25yx=− A1 (ci) 4 7 11 3 1 4AC −     = − =     −       M1 5.51 22XC AC ==   A1 o.e. (cii) 1.5 1OX −=  B1 o.e. [9]

General Certificate of Education Ordinary Level 2020 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 10 9 (a) B2 B1 for any two correct B1 for all correct Deduct B1 for any un-simplified fraction given (b) 21 22 1 18 nn nn −− = − M1 Form equation 2 2 21 22 462 1 8 n n n nn − − + =− M1 Expand 22 2 2 8 344 3696 7 343 3696 0 49 528 0 (Shown) n n n n nn nn − + = − − + = −+= A1 AG (c) 2( 49) ( 49) 4(1)(528) 2(1)n − −  − −= M1 Factorisation o.e. 16 or 33n= A2 A1 for each value (d) 16n= must be rejected as n must be at least 21. B1 o.e. (e) 21 12 233 32 M1 One case seen 21 44=