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 of triangle 1 the area of parellogram 6 ABF ABD ABD ABCD ABF ABCD = == = 7a 1− c From the graph, there is only one intersection point between the line yk= and the curve for some values of k hence there is only one solution for some values of k. di 3 3 3 9 4 0 9 1044 12 1 244 1 24 xx xx x xx yx − − = − − = − + = + =+ ii 0.45 or 2.75x=− − 8a 5 (5 7) 24 59T =  + = b 2 2 ( 2) 4 4 244 64 n n n n n n nn + + + = + + + = + + c 2 1 2 2 22 2 ( 1) 6( 1) 4 2 1 6 6 4 41 ( 4 1) ( 6 4) 2 10 3 pT p p p p p pp p p p p pp − = − + − + = − + + − + = + − + − + + + = + + d 2 2 2 10 3 303 2 10 300 0 10 or 15 ( ) pp pp p NA + + = + − = =− 9a 115 5500 4000 $9615 ++ =

2024 Sec 4/5 Prelims EM Marking Scheme b 32 3 470 80 20 80 70( 32) 83 5600 20 150 2240 83 5600 20 150 2240 23240 140 Total distance 140 140 32 312 312Amount of fuel used 8.8 27.456100 Cost (27.456 1.70) 350 90 486.6752 486.6752 1.20 $584.0102100 xx xx x x x ++= ++ = + = += = = + + = =  = =  + + = = 4 Jim is incorrect as the tips included is less than 20% of the amount he should be paying.