2018 O Level Math 4048 Paper 1 SUGGESTED MS

General Certificate of Education Ordinary Level 2018 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 1 1 2 3 3 3, , 0.65,4 5 4   B1 c.a.o. [1] 2 3 6 2 8 y x   M1 Flip fraction or M0 9 12 y x= A1 or B1 for numerator and denominator [2] 3 (a) 3k = B1 (b) 2 3x= B1 [2] 4 Angle CBD 180 129 51= − =  (Adjacent angles on a straight line) Angle CBD 180 51 38 91= − − =  (Angles sum of triangle) M1 Since angle CBD 91 90=   , using the right -angle in semicircle property, BC is not a diameter of the circle. A1 AG [2] 5 2 2 (3 1) 2 9 6 1 2 n nn −+ = − + + M1 Expand ( ) 2 2 9 6 3 3 3 2 1 nn nn = − + = − + Since 3 is a factor of the expression, then it is a multiple of 3 for all integer values of n. A1 AG [2] 6 (a) The greatest number of marriages per 1000 males in 2014 and 2004 both occurred in the age group of 30-34 years. OR There are more marriages per 1000 males at a later age in 2014 than 2004. B1 Accept any reasonable comparison between both graphs (b) No because the mode age group for marriages per 1000 for both genders are within 30–34 years. B1 Accept any other comparison o.e. [2]

General Certificate of Education Ordinary Level 2018 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 2 7 (a) 'BA B1 (b) ( ) ( )''B A A B   B1 [2] 8 Total monthly payments $113 12 $1356=  = M1 Deposit $1651 $1356 $295= − = M1 Percentage $295 100% 20%$1475=  = A1 [3] 9 (a) Angle ABC 64= B1 (b) Angle AOC 128= B1 (c) Angle OAD 64= B1 FT from (a) [3] 10 Time taken 5408 7.500693481 7.50721= = = hours M1 Estimated time of arrival (Australian Time) is 1655 Singapore is Ahead by 1.50 hours A2 [3] 11 3 1 184 20 V = M1 Form equation 3 1 184 0.02320V =  = m3 A1 22.3 10 − m3 A1 [3]

General Certificate of Education Ordinary Level 2018 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 3 12 (a) 14 hours B1 (b) 2 kF d= , where k is a constant. Let 1 2 1 kF d= and 21 1.2dd= . ( ) 21 222 21 1 25 25 36 361.2 k k kFF dd d = = = =   M1 Finds a relation between 1F and 2F Percentage 11 1 25 536 100% 30 % 30.6%9 FF F − =  = = (3 s.f.) A1 Accept 3 s.f. [3] 13 (a) ( ) 22 4 3pq p q − B2 B1 for ‘2’ B1 for pq (b) 2 (3 1) (3 1)xy x x− + − M1 (2 1)(3 1)xy x= + − A1 [4] 14 Interior angle of C (6 2) 180 1206 −  = =  M1 Interior angle of B 360 120 90 150= − − =  M1 Exterior angle of B 180 150 30= − =  M1 Number of sides in B 360 1230 ==  A1 Alternatively (using sum of interior angles): ( 2) 180 150nn−  = M1 Form equation 180 360 150 30 360 12 nn n n −= = = A1 [4]

General Certificate of Education Ordinary Level 2018 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 4 15 Let x denote the number of males in the group. Number of females 4x=+ Total height of females 1624 Total height of females 162 648 x x =+  = + Total height of males 183 Total height of males 183 x x = = Total height of students 17124 Total height of females 342 684 x x =+  = + M2 M1 for any correct expression seen M1 for all correct 162 648 183 342 684 3 36 12 x x x x x + + = + = = M1 Find x Total number of students in the group 2(12) 4 28= + = A1 [4] 16 Perimeter of minor sector 2rr =+ Perimeter of major sector 2 (2 π)rr = + − M1 Either seen 2 (2 π ) 2(2 )r r r r + − = + M1 Form equation 22 π 4 4 32 π2 r r r r r r r r   + − = + =− M1 2π2 1.427728436 1.4283 −= = = rad (3 d.p.) A1 [4] 17 22 3x y ay x− = + M1 Remove fraction 22 3x y x ay− = + M1 Rearranging 2 2 ( 1) 3 3 1 x y ay ayx y − = + += − M1 3 1 ayx y += − A1 [4]

General Certificate of Education Ordinary Level 2018 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 5 18 (a) 22 2 5 109 2 12510 2 aa a += = M1 Form equation 2 25 4 25 5 42 a a = = = A1 o.e. (b) 48 37AB =  B1 (c) 11 5 16 48 13 1 12 37AC k −       = − =                 M1 Find AC  Since AC k AB  , the AC is not a scalar multiple of AB and hence C does not lie on AB. A1 [5] 19 (a) 128 grams B1 (b) 7x= B1 (c) 1. On average, the masses of the Apples are heavier than the Bananas as the median of 128 g is larger than the median of 125 g of the Bananas. B1 Must quote values 2. The masses of the Bananas are more consistent than the Apples as the range of 44 g is lower than the range of 58 g of the Apples. B1 Must quote values [4] 20 (a) 2126 2 3 7=   B1 (b) 14k = B1 (c) 3 7 11x=   M1 Any two factors 231x= A1 [4]

General Certificate of Education Ordinary Level 2018 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 6 21 When 2x=− , 11y= . 211 ( 2) ( 2) 3 4 2 8 (1) ab ab = − + − + − = −−− When 4x= , 1y=− . 21 (4) (4) 3 16 4 4 (2) ab ab − = + + + =− −−− M1 Substitution and either equation seen (1) 2 8 4 16 (3) (2) (3) 24 12 ab a  − = −−− + = M1 Elimination or substitution s.o.i. 1 2a= A1 o.e. 3b=− A1 [4] 22 Angle CBE = Angle BEC 35= Angle BCE 180 2(35 ) 110= −  =  M1 110 s.o.i. Area 211(8)(10)sin110 (8) sin11022= −  M3 M1 for ABCD M1 for BCE M1 for difference 45.1052458 45.1== cm2 (3 s.f.) A1 [5] 23 (a) 0.5x= B1 o.e. (b) ( ) 2 2 4 1 22 2 xx xx y − =−  − =−  =− M1 Find equation of line (no need to draw on grid, can be implied in working seen) 1.55x=− or 2.55x= A1 (ci) Drawing of tangent (Refer to graph on page 7) B1 (cii) Gradient 5 2 3 2 0 2 −= =−−− M1 s.o.i. 3 22yx=− + A1 o.e. [6]

General Certificate of Education Ordinary Level 2018 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 7 24 (a) Using sine rule, sin sin 29 4140 2392 4140sin 29sin 2392 DMS DMS  = = M1 o.e. Angle DMS 57.04452664 57.0= =  (1 d.p.) A1 c.a.o. (b) Bearing ( )360 180 (180 70 29 )= − − − +  M1 180 70−  s.o.i. 319= A1 [4] END OF MARKING SCHEME

General Certificate of Education Ordinary Level 2018 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 8 BLANK PAGE