2017 O Level Math 4048 Paper 2 SUGGESTED MS

General Certificate of Education Ordinary Level 2017 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 1 1 (a) 42 53 3 12 10 5 xx xx +−  +− M1 Remove fraction 82 1 4 x x − − A1 o.e. (b) 2 2 (5 2 ) 3 (5 2 ) y y y y −− − M1 Join fraction 2 2 2 2 10 4 3 (5 2 ) 74 (5 2 ) y y y y yy y −−= − −= − A1 (c) 3 3 2 35 18 10 53 h j j k h k M1 Flip fraction or M0 5 24 12 j hk= A1 or B1 for 12 or B1 for 5 24 j hk (d) 1 12 4 816 v t   M1 Flip fraction or M0 3 22 v t= A1 or B1 for 1 2 o.e. or B1 for 3 2 v t (e) 2 2( 4) 3( 2) 1( 2)( 4) 2 8 3 6 2 4 8 xx xx x x x x x + + − =−+ + + − = − + − M1 Remove fraction 2 3 10 0 ( 2)( 5) 0 xx xx − − = + − = M1 Factorise o.e. 2x=− or 5x= A1 A1 for both [11]

General Certificate of Education Ordinary Level 2017 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 2 2 (ai) Mean 75.41 10 12 = M1 664.5083333 10 4.51 10=  =  (3 s.f.) 4.51 million A1 (aii) Percentage 6 7 5.09 10 100%5.41 10 =  M1 9.408502773% 9.41%== (3 s.f.) A1 (b) Number of people 61.591 10 101.6%=   M1 o.e. 61.616456 10= A1 c.a.o. (c) Number of visitors 7100 1.509 1097=   M1 771.555670103 10 1.56 10=  =  (3 s.f.) A1 o.e. [8] 3 (a) 50 bacteria B1 (b) 70.7p= (3 s.f.) B1 c.a.o. (c) Refer to graph on page 3 Correct scale used to draw graph S1 Plotting of all 8 points correctly P1 Smooth curve passing through all points C1 (d) 3.325 hours B1 Accept ± 0.05 (ei) Drawing of reasonable tangent at (2, 200) M1 Gradient 450 20 138.7096774 139 (3 s.f.)3.8 0.7 −= = =− A1 Accept ± 5 (eii) It represents the rate of growth of bacteria on the 2nd hour to be 139 bacteria per hour. B1 FT from (e)(i) [9]

General Certificate of Education Ordinary Level 2017 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 3 t 50 2 ty=

General Certificate of Education Ordinary Level 2017 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 4 4 (a) ( ) 22( 4 4) 10 ( 2)AB= − − + − − M1 14.4222051 14.4== units (3 s.f.) A1 (b) Gradient 10 ( 2) 3 4 4 2 −−= =−−− M1 3 2 310 ( 4) 42 y x c cc =− + =− − +  = 3 42yx=− + A1 o.e. (ci) 3 2 5 2 3 5 35 (1)22 xy yx yx += =− + =− + −−− M1 Express in the form y mx c=+ or find gradient Since the gradients of line p and AB are equal and they have distinct y-intercepts, both lines are parallel and won’t intersect. A1 Conclusion (cii) 6 4 37 (2)yx= − −−− Substitute (1) into (2) 356 4 3722xx− + = − M1 Substitution or elimination s.o.i. 9 15 4 37 13 52 74, 2 xx x xy − + = − = = =− A1 74, 2  − A1 o.e. [9]

General Certificate of Education Ordinary Level 2017 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 5 5 (a) Angle ACB 122 64 58= − =  M1 s.o.i. Using cosine rule, 22104 135 2(104)(135)cos58AB= + −  M1 118.9994414 119== m (3 s.f.) A1 (b) Using sine rule, sin sin 58 135 118.9994414 ABC = M1 Use of sine rule 47.82987621ABC =  Bearing ( )360 180 122 47.82987621= − −  +  M1 180 122−  s.o.i. 349.8298762 349.8= =  (1 d.p.) A1 (c) sin 58 104 104sin 58 TA TA = = M1 88.197002 88.2== m (3 s.f.) A1 (d) tan 27 135 135 tan 27 CD CD = = M1 68.78593568 68.8== m (3 s.f.) A1 [10]

General Certificate of Education Ordinary Level 2017 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 6 6 (a) 2RC=+pq  B2 B1 for each term (bi) Since 2RC ST=  , then RC and ST are parallel. B1 Must state Angle RDC = Angle SDT (Common angle) Angle DRC = Angle DST (Corresponding Angles, RT is parallel to ST) Angle DCR = Angle DTS (Corresponding Angles, RT is parallel to ST) M1 Any two reasons given Triangles RCD and STD are similar. (AA Similarity Test) A1 Must state AA (bii) Let h denote the height of the parallogram Area of triangle STD : area of parallelogram ABCD 1 1 1(1) : 2 (6)( )2 2 2 hh=  M1 1: 24= A1 (biii) 2QA=− −pq  M1 or M0 2 2 2QR=− − + =− +p q q p q  A1 or B1 per term [9]

General Certificate of Education Ordinary Level 2017 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 7 7 (a) Height ( 5)x=− cm B1 Width 2( 5) (2 10)xx= − = − cm B1 FT from height (b) ( )(2 10) ( 5) 2(2 10) 2 48x x x x x− + − − + = M1 Form equation 2 2 22 2 10 ( 5)(4 20 2 ) 48 2 10 ( 5)(6 20) 48 2 10 6 30 20 100 48 x x x x x x x x x x x x x x − + − − + = − + − − = − + − − + = M1 Expand s.o.i. ( ) 2 2 2 8 60 52 0 4 2 15 13 0 2 15 13 0 (Shown) xx xx xx − + = − + = − + = A1 AG (c) (2 13)( 1) 0xx− − = M1 Factorise o.e. 13 2x= or 1x= A2 A1 for each value (d) 1x= is rejected as the height of the cuboid is negative. B1 o.e. (e) Volume ( )6.5(6.5 5) 2(6.5) 10= − − M1 29.25= cm3 A1 c.a.o. [11]

General Certificate of Education Ordinary Level 2017 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 8 8 (a) 1 20cos 30 2cos 3 AOM AOM − = =  M1 1 Angle 2 22cos 3 96.37937021 96.4 (3 s.f.) AOB AOM − = =  = =  A1 AG (b) Shaded area 2296.37937021 1 π(30) (30) sin 96.37937021360 2 =  −   M3 M1 for sector M1 for triangle M1 for difference 309.748208 310== cm2 (3 s.f.) A1 (ci) Volume of water 309.748208 150= M1 FT from (b) 46462.2312 46500== cm3 (3 s.f.) A1 (cii) Volume of water needed 2π(30) 150 46462.23122=  − M1 FT from (c)(i) 165595.2729= cm3 A1 165.5952729 170== litres (nearest 10 litres) A1 [11]

General Certificate of Education Ordinary Level 2017 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 9 9 (ai) (a) 42 km/h B1 (ai) (b) Lower quartile 37= Upper quartile 46= M1 Either one seen IQR 46 37 9= − = km/h A1 (aii) Number of people who exceed speed limit 80 73 7=−= M1 s.o.i. Percentage 7 100% 8.75%80=  = A1 (aiii) On average, the cars passing the checkpoint in the morning drove at a faster speed than those in the afternoon as the median speed of 42 km/h is higher than 40 km/h in the afternoon. B1 Must state figures The cars passing the checkpoint in the afternoon drove at a more consistent speed than those in the morning as the interquartile range of 8 km/h is lower than 9 km/h of those in the morning. B1 Must state figures (bi) 17 40 B1 (bii) (a) 54 80 79 M1 1 316= A1 (bii) (b) 16 51 280 79 M1 One case s.o.i. 102 395= A1 [12]

General Certificate of Education Ordinary Level 2017 Mathematics 4048/02 Syllabus 4048