2018 O Level Math 4048 Paper 2 SUGGESTED MS

General Certificate of Education Ordinary Level 2018 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 1 1 (ai) 2 5 tv B1 o.e. (aii) 4( 3) 5(3 2 ) ( 3)(3 2 ) yy yy + − − +− M1 Join fraction 4 12 15 10 (3 2 )( 3) 14 3 (3 2 )( 3) yy yy y yy + − += −+ −= −+ A1 (b) (4 3)(4 3) (4 3)( 3) xx xx +− +− M2 M1 for each factorisation s.o.i. 43 3 x x −= − A1 (c) 2 20 (2 5)( 1) 2 5 2 5 20 xx x x x = + + + + + = M1 Expand 22 7 15 0 (2 3)( 5) 0 xx xx + − = − + = M1 3 2x= or 5x=− A1 o.e. [9]

General Certificate of Education Ordinary Level 2018 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 2 2 (a) Difference 12($3900) $44000 $2800= − = M1 s.o.i. Percentage increase $2800 4 100% 6 % 6.36%$44000 11=  = = A1 Accept 3 s.f. (b) Interest earned 6 2.35$6500 1 $6500100 = + −  M2 M1 for C.I. M1 for difference $972.0615149 $972.06== (2 d.p.) A1 (ci) 1£1 $1 £ £0.50761421321.9$1.97 7=  = = The cost per night for both hotels are the same in their own respective currencies. Since there is lesser pounds than euros per Singapore dollar, the hotel in London costs more per night than the hotel in Paris. B1 Show clear working with explanation (cii) Total cost for London hotel £145 3 £435== Total cost for Paris hotel €145 4 €580== M1 Total cost for any hotel s.o.i. In Singapore Dollars: Total for London £435 1.97 $856.95== Total for Paris €580 29000$ $878.7878...0.66 33= == M1 Conversion to SGD for any hotel s.o.i. Total to be paid with fee 29000101.8% $856.95 $ 33 =+  M1 101.8% o.e. $1766.981161 $1767== (nearest dollar) A1 [10]

General Certificate of Education Ordinary Level 2018 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 3 3 (a) 490 280 210 350 210 210 = P B1 c.a.o. (b) 0.75 1.50 2.25  = N B1 o.e. (c) 1260 1050 = T B2 B1 for each correct element (d) The total costs of selling Apple and Cherry pies from all small, medium and large sizes are $ 1260 and $1050 respectively. B1 Must state sizes and type of pie (e) Number of pies sold: Small 700= , Medium 408= , Large 348= M1 Number of pies Sales from pies sold: Small $735= , Medium $856.80= , Large $1096.20= M1 Total sales Profit $2688 ($1260 $1050)= − + M1 FT from (c) $378= A1 [9]

General Certificate of Education Ordinary Level 2018 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 4 4 (ai) 65n+ B2 B1 per term (aii) For any integer n, 3n is a multiple of 3. Since 5 is not a multiple of 3, the nth term is not a full multiple of 3. Thus, there is no term that is a multiple of 3. B1 o.e. (bi) 19 180 B1 (bii) 4 1 7 205 5 32 128 32 1435 35 k k kk − =− − = − M1 Remove fraction o.e. 163 1467k = M1 Isolate k 9k = A1 (biii) 41 1205 5 4 1 205 5 n n nn − − −  − M1 Remove fraction o.e. 9 206 822.888... or 22 9 n n   A1 Least value of 23n= A1 [10] 5 (a) 1 2OB −=  B1 (b) A(–7, 0) B1 D(–3, –3) B1 (c) Angle BAD 11 23tan tan64 −−   =+        M2 55.304846468 55.3 (1 d.p.)= =  A1 (d) ( ) 2 23 ( 1) ( 3 2)BD= − − − + − − M2 M1 for each correct difference 5.3851648 units 5.39 units (3 s.f.)== A1 [9]

General Certificate of Education Ordinary Level 2018 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 5 6 (a) 3500 x B1 (b) 3500 10x− B1 (c) 3500 3500 2110xx −=− M1 Form equation 2 3500 3500 35000 2110 xx xx −+ =− M1 Join fraction ( ) 2 2 2 2 35000 21 210 21 210 35000 0 7 3 30 5000 0 3 30 5000 0 (Shown) xx xx xx xx =− − − = − − = − − = A1 AG (d) 2( 30) ( 30) 4(3)( 5000) 2(3)x − −  − − −= M2 M1 for correct discriminant M1 for formula 46.1298756 or 36.1298756 46.13 or 36.13 (2 d.p.) x=− =− A1 A1 Deduct A1 if not given to 2 d.p. (e) Time taken 3500 2(46.1298756) 10= − M1 Must use at least 5 s.f. o.e. 42 minutes 30 seconds (nearest 10 seconds)= A1 [11]

General Certificate of Education Ordinary Level 2018 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 6 7 (a) Angle Angle 90OCA ODB= =  (Tangent perpendicular to radius) M1 (Radii of circle) Angle Angle (Vert. opp. angles) OC OD AOC BOD = = M1 Both reasons given Hence, triangles OAC and OBD are congruent. (AAS Congruency Test) A1 Must state AAS (bi) cos30 7 cos307 AC AC=  =  M1 Find AC or OC using trigo s.o.i. Area of triangle OAC 1 (7)(7 cos30 )sin 302=   M1 Area of triangle formula s.o.i. 2210.608811196 cm 10.6 cm (3 s.f.)== A1 Alternatively: cos30 7 cos307 AC AC=  =  M1 sin 30 7sin 30 3.5 cm7 OC OC=  = = M1 Area of triangle OAC 22 1 (3.5)(7 cos30 )2 10.608811196 cm 10.6 cm (3 s.f.) = == A1 (bii) Sector area 260 49 π(3.5) π360 24 =  = M1 s.o.i. Shaded area 22 49π(7) π(3.5) 2(10.608811196) 2 π 24 = − − +  M1 2107.0640761015 107 cm (3 s.f.)== A1 [9]

General Certificate of Education Ordinary Level 2018 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 7 8 (a) Let M be the centre of the square base ABCD. 2 2 2 22 15 2252 225 2 AM AM AM AM += =  = M1 Find AM o.e. 22 22520 512.52AE = + = M1 Find 2AE o.e. 512.5 22.63846284 22.6 cm (3 s.f.)AE= = = (Shown) A1 AG (b) Let F be the midpoint of AB. 7.5 cmAF = M1 s.o.i. 7.5cos 512.5 BAE= M1 1 7.5cos 512.5 70.65263004 70.7 (1 d.p.) BAE − =  = =  A1 (c) Total surface area ( ) 2 115 4 (15) 512.5 sin 70.652630042 = +   M2 M1 for area of triangle M1 for all 5 sides 2865.80028090249 866 cm (3 s.f.)== A1 (d) 2 Area of base 16 16 Area of base 20 25 PQRS ABCD == M1 Area of base PQRS 2216 15 144 cm25=  = A1 [11]

General Certificate of Education Ordinary Level 2018 Mathematics 4048/02 Syllabus 4048 Paper 2 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 8 9 (ai) 3.0 3.5m 3.5 4.0m 4.0 4.5m 19 12 4 B1 (aii) Mean 3.26= B1 (aiii) SD 0.524= B1 (aiv) The mid-values of the intervals are used in calculation, and not the individual data values. B1 Mid-value o.e. (av) Number of babies 10% 50 5=  = From the graph, there are 5 babies with mass less than or equal to 2.575 kg. M1 2.575 kg s.o.i. Thus, there are 90% of babies with mass greater than 2.575 kg and the data from the hospital does not support the claim. A1 Alternatively: From the graph, there are 41 newborn babies who have a mass greater than 2.8 kg. M1 Percentage 41 100% 82% 90%50=  =  Thus the data from the hospital does not support the claim. A1 (bi) 2 5 B1 c.a.o. (bii) (a) 13 12 25 24 M1 13 50= A1 o.e. (bii) (b) 96 225 24 M1 M1 for one case s.o.i. 9 50= A1 [11]

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