Greendale Sec 4048_02 Answer Scheme

1 2022 4E5A Preliminary Examination Paper 2 Mark Scheme 1a 3 12 1 34 12 4 6 3 12 6 3 4 116 xx xx xx x −+< −< + − <+ < M1 A1 1b ( ) ( ) ( ) ( ) ( ) 2 2 2 2 72 4343 7 24 3 43 7 86 43 13 8 43 x xx xx x xx x x x − −− −−= − −+= − −= − M1 A1 1c 1 10 2 6 1 63 2 10 5 49 49 7 m n nn mm −   = =  M1 A1 1d ( ) ( )( ) ( )( ) ( )( ) ( ) 2 2 2 50 32 5 11 12 2 25 16 54 3 25 4 5 4 54 3 25 4 3 x xx x xx xx xx x x − −− − = +− +−= +− −= − M1 M1 A1 1e 4 5 16xy−= ------ (1) 6 3 10xy+= ------ (2) (1) 3: 12 15 48 (3) (2) 5: 30 15 50 (4) (3) (4): 42 98 123 12 in(2),3 16 2 3 103 14 3 10 113 xy xy x x Substitute x y y y × −= → × +=→ += = =  += += =− M1 A1 A1

2 2a 8 15 10 15 6 8 1[ 20 30 20 35 13 14 ]3 6 5 12 6 9 5 24 38 1 53 793 27 16 2128 3 2117 2633 91 5 3 A A A        = ++           =     =   M1 A1 2b The number of bus trips per day made by each type of bus of the 2 companies. B1 2ci 3BA exists because B is a 1x3 matrix and A is a 3x1 matrix B1 2cii The elements in this 1x2 matrix represents the total number of bus trips made by Company A and B respectively. B1 2di ( ) 8 15 112 25 50 20 30 165     B1 for ( )12 25 50 B1 for 1 1   2dii [(8 12 ) (20 25 ) (6 50 )] $22 896 22 $19712 trips passengers trips passengers trips passengers ×+ ×+ ×× = × = M1 A1

3 3a 9.3 B1 3b P1-plot all points P1-join up points P1-smooth curve 3c 83 15 83 15 0 83 10 5 x x x x x x += +−= +− = Draw y = 5, From the graph, x = 0.6 or x = 4.4 B1, B1 3di Line 21yx= + drawn. x 0 3 6 y 0.5 2 3.5 P1 P1 3dii From the graph, x = 1 or x = 3.15 B1, B1 3diii 22 2 82 3 10 1 166 20 1 6 16 20 5 21 16 0 21, 16 xx x xx x x xx x xx BC  +− =+ +−= + +− =+ − += ∴= − = M1 M1 A1

4 4ai Bearing of from 360 (180 126) 306 AD =−− = ° B1 4aii 270 242 28 480 sin28 sin36 383.3821 383.4 (1 d.p) ABD AD AD m ∠= − = ° =°° = = M1 M1 A1 (b) Let be the angle of depression 4800tan 630 82.5226 82.5 x x x = = = ° M1 A1 (c) 180 36 36 108 383.3821 sin36 sin108 236.94317 237 (3 s.f) APD PD PD m ∠ = −− = ° =°° = = M1 M1 A1

5 5a 22 2 3 Let be the height of the cone that was r emoved. By similar triangles, 1 65 56 46 1.5 Volume of paint bottle 11= (5 )(20) ( 5 7.5) ( 1 1.5)33 1765.5755 1765.6 (shown) x x x xx x x cm ππ π =+ = + = = + ×× − ×× = = M1 A1 M1 (volume of cylinder), ecf M1 (volume of frustrum), ecf A1 (answer shown correctly) (b) 1765.6 1500 1001765.6 15.043 15.0% (3 s.f) − × = = B1 (c) Total mass 30 1500(3) 4530 4.53 g kg = + = = M1 A1 (d) 3 2 3 2 3 2 2 3 1 3 1 3 1 Base area of larger bottle 3= (5)1 163.3694... 163 (3 s.f) L S L S L S V