Canberra EM 2024 Prelims Paper 2 MS

2024 Sec 4/5 Prelims EM Marking Scheme 1a 4 6 58 (1) 3 5 46 (2) 12 18 174 (3) 12 20 184 (4) (3) (4) 2 10 5 7 mn mn mn mn n n m + = −−−−− + = −−−−− + = −−− + = −−− − − =− = = b 2 31 2 50 5 31 2( 5)( 5) 5 3 2( 5) 2( 5)( 5) 10 2( 5)( 5) x xx x x x x xx xx x xx −−− =− − + − −+= −+ −= −+ ci 5− ii 3 3 3 3 25 72 25 72 (7 2) 25 (7 2) 25 yx z yx z x z y xzy = + = + += += d 2 2 23 20(3 1)( 1) 23 2(3 1)( 1) 2 3 2(3 1)( 1) 2 3 6 4 2 6 6 5 0 1.54 or 0.541 x xx x xx x x x x x x xx x + −=+− + =+− + = + − + = − − − − = =− 2ai (a) 40.5 (b) 35.5 (c) 42 33 9 − =

2024 Sec 4/5 Prelims EM Marking Scheme ii 90 69 100%100 21% −  = bi 5 9 9 5 20 19 20 19 9 38     +        = ii 6 14 13 14 6 13 14 13 6 20 19 18 20 19 18 20 19 18 91 190        +   +             = 3a 20000 4 80000 80000 10040 $200000 =  = b 3 (55000)(0.97)(3) 100 $1600.50 101600.50 55000100 $7100.50 3.7855000 1 100 $61475.72916 61475.72916 55000 $6475.73 I A I = = + = =+  = =− = Wayne should choose package A as it will generate a higher interest. c (150 7) (380 6) 3330AUD 3330 1.020.91 3732.527 $3733  +  =   = 

2024 Sec 4/5 Prelims EM Marking Scheme 4a Let 180 2 180 (180 2 ) Since , QEC QEC ECD AEF QEC CED EDC ECD EDC EC ED        =  =  =  = =  = −  = − − − =  = = b 90 90 (90 ) Triangle and are similar. (AA) AFE QFC ECF FCQ AEF FCQ AEF QCF     =   = −  = − − =  = = c 2 2 360 2(61) 180 58 58Area of sector (14) 360 99.204514 68 1Area of triangle (14) sin 582 83.1087 1342 Area of segment =99.2045 CED ECD ECD CD   = − − = = = = = 1468 83.10871342 16.0958 16.1 − =  5a 192 180 12 137 12 125 (shown) ABC −=  = − = b 2 2 2 2 35 68 2(35)(68) cos125 8579.223037 92.62409966 92.6 AC AC AC = + − = =  c 125 85 40 180 40 70 70 68 sin 40 sin 70 46.5147 46.5 DBC BDC DC DC  = − =  = − − = = =  (Alternate angles are equal) (Angle bisector) (Sum of angles on a straight line) (Sum of angles in a triangle) (Vertically opposite angles are equal) (Angle in a semi-circle) (Tangent perpendicular to radius)

2024 Sec 4/5 Prelims EM Marking Scheme d 21Area of triangle (68) sin 402 1486.124954 11486.124954 46.51472 63.89915247 63.9 h h = = =   =  e tan 40.5 68 58.07748661 58.07748661tan 63.89915247 42.3 h h   = = = = 6a AD DB DA AB DE = =+ = = = 6p -6p + 6p + 4q 4q 2q b AE AD DE AG =+ = = 6p + 2q 3p + q c BG BA AG=+ =− = -6p 4q + 3p + q -3p - 3q d 3 3 GF GA AF BG GF =+ =− = = - p q + 2p -p - q BG is parallel to GF and there is a common point G; therefore B, G and F lie on a straight line.

2024 Sec 4/5 Prelims EM Marking Scheme e the area of triangle 1 the area of triangle 3 the area of triangle 1 3 the area of parellogram 2 6 the area