2019 O Level Math 4048 Paper 2 SUGGESTED MS

General Certificate of Education Ordinary Level 2019 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 1 1 (a) 3 2 2 3 p r B1 (bi) 10a=− B1 (bii) 5 3 4 3 5 4 a ab b c b ab a c − = + + = − M1 Rearrange terms (3 ) 5 4 54 3 b a a c acb a + = − −= + A1 (ci) 22 77 922x   − − +       M1 Complete the square correctly 2 2 7 13 24 13 7 42 x x = − − =− + − + A1 In required form, accept in decimal form or mixed number o.e. (cii) 7 13,24  − B1 o.e. (d) 1 6( 3) 2( 3)( 1) xx xx − + − =−− M1 Join fraction ( ) 21 6 18 2 3 3x x x x x− + − = − − + M1 Expand 2 2 2 8 6 7 19 2 15 25 0 (2 5)( 5) 0 x x x xx xx − + = − − + = − − = M1 Factorise o.e. 5 or 52xx== A1 Both values o.e. [11]

General Certificate of Education Ordinary Level 2019 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 2 2 (a) Using cosine rule, 226.4 8.3 2(6.4)(8.3)cos 27CX = + −  M2 Deduct M1 for any wrong substitution 3.897366658 cm 3.90 cm (3 s.f.)== A1 (b) Using sine rule, sin sin112 6.4 7.5 XAB = M1 1 6.4sin112sin 7.5 52.29750374 52.3 (1 d.p.) XAB − =  = =  A1 (c) Angle ABC 180 112 52.29750374 27 42.70249626= − − + =  M1 Area of triangle ABC 1 (7.5)(8.3)sin 42.702496262= M1 2221.10871629 cm 21.1 cm (3 s.f.)== A1 [8]

General Certificate of Education Ordinary Level 2019 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 3 3 (a) Price of phone in UK 389$ $670.68965520.58== M1 Difference $670.6896552 $620=− $50.6896552 $50.69 (2 d.p.)== A1 (b) Price before sale 100 $78594= M1 $835.106383 $835 (nearest dollar)== A1 (ci) 59.2 10 B1 (cii) Percentage increase 76 6 1.20 10 7.85 10 100%7.85 10  − =  M1 52.86624204% 52.9% (3 s.f.)== A1 (ciii) Mean SMS per person per day in 2015 10 6 1.14 10 3655.54 10 =  M2 M1 for per person M1 for per day 5.637703378 5.64 (3 s.f.)== A1 [10]

General Certificate of Education Ordinary Level 2019 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 4 4 (a) 19 5p=− B1 o.e. (b) Refer to graph on page 5 Plotting of any 6 correct points P1 Plotting of all 9 points correctly P1 Smooth curve passing through all points C1 (c) 3.75x B1 Accept ±0.1 (di) Refer to graph on page 5 Plotting of any 3 relevant points within 44 x−   P1 Straight line passing through all points plotted for range L1 (dii) 3 5 1 105 x xx − + + = M1 Substitution s.o.i. 3 3 10 5 10 9 5 0 (Shown) x x x xx − + + = − − = A1 AG (diii) 2.65 or 0.55 or 3.25x=− − B2 B1 for any two correct B1 for all correct Accept ±0.1 [11]

General Certificate of Education Ordinary Level 2019 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 5 3 215 xyx= − + 5 10yx+=

General Certificate of Education Ordinary Level 2019 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 6 5 (ai) 2 18tt + B1 c.a.o. (aii) Product of middle two numbers 22 ( 6)( 12) 6 12 72 18 72 tt t t t t t = + + = + + + = + + M1 Difference ( ) 22 18 72 18 72 (Shown)t t t t= + + − + = A1 AG (aiii) 6 12 18 360t t t t+ + + + + + = M1 Form equation 4 324 81 t t = = M1 Largest number 81 18 99= + = A1 (bi) 8 12n+ B2 B1 for each term (bii) 8 12 4(2 3)nn+ = + Since 4 is a factor of the nth term, the terms of the sequence are all multiples of 4. B1 AG [9]

General Certificate of Education Ordinary Level 2019 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 7 6 (ai) Angle ABC 90= (right-angle in semicircle) B1 Must give reason Angle Angle 23CBD OAD= =  (angles in same segment) M1 Must give reason Angle ABD 90 23 67= − =  A1 Alternatively: Angle AOD 180 2(23 ) 134= −  =  (base angles of an isosceles triangle, angles sum of triangle) M2 M1 for each reason Angle ABD 134 2 67=  =  (angle at centre is twice the angle at circumference) A1 Must give reason (aii) Angle AED 180 67 113= − =  (angles in opposite segments) M1 Must give reason Angle EAD 180 113 42= − −  (angles sum of triangle) M1 Must give reason 25= A1 (bi) Minor arc length PQ 15.2 2(4) 7.2 cm= − = M1 Angle POQ 7.2 1.8 radians4== A1 (bii) Area of minor sector PQO 221 (4) (1.8) 14.4 cm2== M1 Area of major sector PQO 2π(4) 14.4 16π 14.4= − = − Area of major sector RSO 21 (6) (2 π 1.8) 36π 32.42= − = − M1 Either one seen Total shaded area 22 14.4 36 π 32.4 16π 14.4 59.23185307 cm 59.2 cm (3 s.f.) = + − − + == A1 [11]

General Certificate of Education Ordinary Level 2019 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 8 7 (ai) 9 4PQ −=  B1 (aii) 44 3 3 6 hhPR −     = − =     −       M1 s.o.i. or M0 4 6 1.5kk=  = A1 or B1 for each value 4 9(1.5) 9.5hh− =−  =− (bi) 5 2AC =− ba  B2 B1 for each term o.e. (bii) AB=−ba  M1 2 ()3XB=− ba  A1 o.e. (biii) 22 33 XY XY AC AC =  =    M1 Using similar triangles s.o.i. 2 5 5 2 3 2 3 3XY = − = − b a b a  A1 o.e. [9]

General Certificate of Education Ordinary Level 2019 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 9 8 (a) Volume 11(4 5)(6)(25) (4 6.5)(10)(25)22= + + + M2 M1 for area of trapezium M1 for volume 31987.5 m= A1 c.a.o. (b) By Pythagoras’ Theorem, 22 22 10 2.5 106.25 1 6 37 FE AG = + = = + = M1 M1 M1 for attempt to use Pythagoras’ Theorem M1 for either FE or AG found Total area ( )25 37 106.25=+ M1 Area of rectangle 22409.7631649 m 410 m (3 s.f.)== A1 (c) Let X be the point vertically below P on the rectangular base. By Pythagoras’ Theorem, 2210 25 725DX = + = M2 M1 for attempt to use Pythagoras’ Theorem M1 for DX Let the angle of elevation be  . 6.5tan 725  = M1 Attempt to use trigo ratio 1 6.5tan 725 13.57176784 13.6 (1 d.p.)  − =  = =  A1 [11]

General Certificate of Education Ordinary Level 2019 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 10 9 (ai) Median 71.5= B1 (aii) Range 43= B1 (aiii) Percentage 158 % or 58.3% (3 s.f.)3= B1 o.e. (aiv) There is a higher percentage / proportion of students who were awarded a merit in group B of 65%, than group A. B1 If percentages or figures are not quoted in either / both answers, deduct B1. There is a higher percentage / proportion of students who were awarded a distinction in group A of 520 %6 (or