2016 O Level Math 4048 Paper 1 SUGGESTED MS

General Certificate of Education Ordinary Level 2016 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 1 1 (a) 62x− B1 (b) 6 (1 3 )xy + B1 [2] 2 4 16 12 3 4 ( 4 ) 3 ( 4 ) ax ay by bx a x y b x y − + − = − − − M1 Grouping or M0 (4 3 )( 4 )a b x y= − − A1 or B1 per bracket [2] 3 (a) B1 (b) B1 [2] 4 ( ) 22 22 (5 1) (5 1) 25 10 1 25 10 1 nn n n n n + − − = + + − − + M1 Either expansion correct s.o.i. 2225 10 1 25 10 1 20 n n n n n = + + − + − = A1 Since 20 is a factor of the expression, it is a multiple of 20 for all positive integers n. AG Must conclude [2]  A B  A B

General Certificate of Education Ordinary Level 2016 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 2 5 Perimeter 22= cm Length + Breadth 11= cm (May be implied through guess and check, or other methods e.g. prime factorisation) M1 Find possible combinations of Lenth + Breadth s.o.i. 6 cm A1 [2] 6 1 (18.7)(12.8)sin 58.62 1465sin 2992 ABC ABC = = M1 Use of area of triangle s.o.i. 29.31685932 or 150.6831407 29.3 or 150.7 (1 d.p.) ABC =   =   A1 [2] 7 3 1 30 2 h = M1 Form equation 3311 30 23.81101578 23.8 (3 s.f.)30 2 2 h h=  =  = = A1 [2] 8 The widths of the flames are inconsistent / not constant. B1 Feature This may lead a reader to think that the gas bill in 2015 is twice the gas bill in 2012. B1 Comparison made [2] 9 2 2 31 ( 4) ( 4) 34 ( 4) xx x x +−− +−= − M1 Join fraction 2 1 ( 4) x x −= − A1 [2]

General Certificate of Education Ordinary Level 2016 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 3 10 (a) 3 or (175 2 )2 x x − B1 o.e. (b) If angle ADC 3 2 x= , 3 2 5 1802 x x+ + = If angle ADC (175 2 )x= −  , 3 2(175 2 )xx=− M1 Form equation with either angle 7 1752 50 x x = = 3 350 4 7 350 50 xx x x =− = = A1 [3] 11 (a) 229 ( 4) (4)x+ − − M1 Complete the square s.o.i. 27 ( 4)x=− + − A1 c.a.o. (b) (4, 7)− B1 FT from (a) [3] 12 In the US: Cost of diesel per litre 4.48 896 3.785 757== US Dollars M1 Cost per litre s.o.i. Cost of diesel per litre in SGD 896 1.242 1.47005582757=  = Singapore Dollars M1 Hawaii A1 M0A0 [3] 13 (a) 2520 m B1 (b) Let the heights of the planes be 1h and 2h . 1221 21 25120 120 hh − − − = M1 Difference s.o.i. 21 21 25120 120 3000 hh hh −=  − = A1 [3]

General Certificate of Education Ordinary Level 2016 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 4 14 (27, 22) B2 B1 for each correct coordinate [2] 15 (a) Angle AED 180 45 135= − =  (Adjacent angles on a straight line) M1 If any/all reason(s) is/are missing, or unacceptable short forms are used*, deduct only M1 from total in this part. A1 is still given if correct. Angle DAE 180 135 22 23= − − =  (Angles sum of triangle) M1 Angle ABC 180 23 34 123= − − =  (Interior angles, BC is parallel to AD) M1 Refle angle ABC 360 123 237= − =  (Angles at a point) A1 Do Not Accept the use of the following short forms: • ‘  ’ to denote ‘angle’ • ‘  ’ to denote ‘triangle’ • ‘//’ to denote ‘parallel’ *Unacceptable short forms to look out for (b) Angle ACD 180 45 38 97 90= − − =   (Angles sum of triangle) Must state reason Since angle ACD 90 , by right -angle in semicircle property, a semicircle with AD as its diameter does not pass through C B1 Must state circle property [5] 16 (a) Circumference 2π(10)= M1 Radius 10= cm s.o.i. 62.83185307 62.8 cm== (3 s.f.) A1 (b) By Pythagoras’ Theorem, Length of rectangle 2220 13 231= − = cm M1 Area 213 231 197.582894 198 cm= = = (3 s.f.) A1 [4] 17 (a) 4k = B1 5a= B1 (b) Gradient 100 4 4820 −= =−−− M1 48 4yx=− + A1 [4]

General Certificate of Education Ordinary Level 2016 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 5 18 (a) Use of ladder method or tree method to prime factorise M1 or B0 4227 A1 or B1 per factor (b) Since the powers of the prime factors are all even / are all multiples of 2 / are all divisible by 2, 784 is a perfect square. B1 (c) 7m= and 2n= B1 [4] 19 (a) Litres of petrol used 6.7 19629 1315.143100=  = M1 Total paid $2.42 1315.143 $3182.64606 $3182.65 (2 d.p.) = == A1 (b) Total distance travelled in 2024 100 19629135= M1 14540= km A1 [4] 20 (a) 1 : 2 500 000 1 cm : 25 km M1 Conversion to km s.o.i. Actual length 25 49.3 1232.5=  = km A1 (b) 1 cm : 25 km 1 cm2 : 625 km2 M1 Square map scale s.o.i. Area of Switzerland 241285 66.056 cm625== A1 c.a.o. [4] 21 (a) 200 84 200 − M1 84 s.o.i. 29 50= A1 o.e. (b) 90% 300 270 90% 300 330 = = M1 Find the bounds s.o.i. Number of students 100 64 36= − = A1 [4]

General Certificate of Education Ordinary Level 2016 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 6 22 (a) 312 B1 c.a.o. (b) Distance within shaded area on map 5.1= cm M1 Distance within shaded area in km 5.1 10 51=  = km M1 Time taken 51 35== 1 hour 27 minutes A1 [4] 23 (a) 17 units B1 (b) 4 16 20 30 DC OC OD OD OC DC =− =− −   =−         M1 Splitting of vectors s.o.i. or M0 D (20, –10) A1 or B1 for each coordinate (c) Trapezium B1 [4] 24 By Pythagoras’ Theorem, Radii of quadrant 2 2 2 22r r r r= + = = M1 s.o.i. Unshaded area ( ) 2 2 2 π2 π(2 )42 (1) (2) (3) r rr r r= + + − M3 M1 for triangles CGH and DEH (1) M1 for quadrant GEH (2) M1 for difference to find AEF and BGF (3) 23r= Fraction 22 22 3 3 3 (2 ) 4 4 rr rr= = = A1 c.a.o as the question asks for a fraction [5]

General Certificate of Education Ordinary Level 2016 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 7 25 (ai) 55AC=−ca  55CA=−ac  M1 s.o.i. 4 (5 5 ) 4 45 5 4 4 4 AP OP = − = − = + − = + c a c a a c a a c   1 (5 5 )5 54 CP OP = − = − = + − = + a c a c c a c a c   A1 (aii) 44PA=− ac  M1 s.o.i. 4 4 4 8 8 4PQ= − + + = +a c a c a c  A1 (b) Since 5 4CB PQ=  , PQ  is a scalar multiple of CB  and hence PQ is parallel to CB. B1 Find a relationship between PQ  and CB  (c) 5 : 4 B1 c.a.o. [6] END OF MARKING SCHEME

General Certificate of Education Ordinary Level 2016 Mathematics 4048/01 Syllabus 4048 Paper 1 SUGGESTED Solutions & Marking Scheme Mark Allocation Prepared by: Jordan Tew (@math_stuff_here) 8 BLANK PAGE