Outram Sec 4E5N Maths P2 Prelim 2022 Solutions

1 1a 3500 12 38000percentage increase = 100%38000 10.5% −  = 1b A = 101.85000(1 )100+ =5976.51 Interest earned = 5976.51 – 5000 =$976.51 (nearest cent) 1c Cost of hotel in Switzerland = 140 4 101.8 0.7 100   =$814.40 Cost of hotel in London = 101.8120 3 1.73 100   =$634.01 Total cost = $814.40 + $634.01 =$1448 (nearest dollar) 2a 43 132 x x+ − 4 3 2 6 10 1 1 10 xx x x +  − − − 2b 3 2 3 1 10 18 12 233 24 10 37 3 15 10 15 4 39 52 4 3 1 2 xy xy xy xy x x y + =− −− − =− −− − =− −− + =− −− =− =− = 2c 22 22 2 36 25 2 3 3 (2 3) 6( 5) 2( 5)(2 3) ( 5)(2 3) 3 (2 3) 6( 5) 2( 5)(2 3) 6 9 6 30 2(2 10 3 15) 6 15 30 4 14 30 2 29 0 (2 29) 0 290or 2 x xx x x x x x x x x x x x x x x x x x x x x x x xx xx xx +=−+ +− +=− + − + + + − = − + + + − = − + − + − = − − += += = =− M1 A1 A1 M1 M1 M1 M1 M1 A1 A1 M1 A1 A1 M1 M1 M1 A1 each

2 2d 500 5(100) 100 300 3(100) 100 200 2(100) 100 200 300 500 2 2 32 3 3 27 4 4 16 4 3 2 == == ==  2e 24 6 8 12 12 8 x y x y y x − − −   = = 3a 4.5 3c 53 12 53 4 8 8 x x x x y += + − = = 0.40 3.60x or= 3dii 1.10 2.80x or= 3diii 22 2 3 4 5 342 103 4 6 8 3 4 6 10 8 3 12 10 0 12; 10 x x x xx x x x x x xx AB + = + − + = + − + = + − − + = =− = 4a In 1 min, larger pipe fills 1 x of tank. In 1 min, smaller pipe fills 1 10x+ of tank. In 1 min, both pipes fill 11 10xx+ + of tank. 2 2 1 1 1 10 20 ( 10) 1 ( 10) 20 20(2 10) ( 10) 40 200 10 30 200 0( ) xx xx xx x x x x x x x x shown += + ++ =+ + = + + = + − − = M1 A1 M1 A1 M1 A1 each M1 M1 M1 M1 A1 A1 each

3 4b 2( 30) ( 30) 4(1)( 200) 2(1) 30 1700 2 35.6 5.62( ) x or rej − −  − − −= = =− 4c 1 2045.61 43.9%  = 5ai sin 27.5 15.3 7.06 AD AD = = 5aii 2215.3 7.06 13.6 AC=− = 5aiii 13.6tan 35.6 13.6 tan 35.6 19.0 AB AB = =  = 5b Angle of depression = angle of elevation of D from B 7.06tan 19.0 20.4 AOE AOE =  =  5c 2213.6 19.0 23.37 11 23.37 13.6 19.022 11.1m BC AP AP = + =   =   = 5d 22 2 2 2 7.06 19.0 20.27 20.27 23.37 15.3 2(23.37)(15.3)cos 58.9 BD BCD BCD = + = = + −   =  6a TD= s TC TD DC=+ = s + 4r 6bi Since 3TC PQ= , so TC//PQ. () () TCD PQD corr s CTD QPD corr s  =   =  By AA property, triangles TCD and PQD are similar. M1 A1 each M1 A1 M1 M1 M1 M1 A1 M1 A1 M1 M1 M1 A1 M1 A1 M1 M1 M1

4 6bii 2 3 9 1 6 : : log 1 : 9 1 : 12 1 : 9 : 108 Area TCD PQ Area PQD PQ Area TCD TD Area CAD AD Area PQD Area TCD Area of paralle ram == == 1 : 108 7ai ( ) ( ) 37 ( ) 180 37 37 106 106 2 53 at centre = 2 at circumference OAB isos AOB sumof sin a ADB  =    = − − =     =  =    7aii 180 22 37 37 53 31 (sum of s in a ) OBD = − − − − =    7aiii 180 37 31 112 (opp angles of a cyclic quad) AED = − − = 7b Let the radius be r. 2 2 2 22 2 Area of sector 106 360 106Area of sector 360 1Area of AOB = sin1062 106 122.5