SCSS 2024 EM Prelim P2 MS

1 2024 4E/5N EM Prelim Paper 2 Marking Scheme (90 marks) 1a ( ) ( ) 3 1 2 5 34 4 3 1 3 2 5 12 4 6 15 27 2 2 27 xx xx xx x x +− +  − +  −   M1 A1 1bi ( ) ( ) ( ) 2 0.2 3 1.5 0.2 2 1.5 7 / 1.754 a a −−= +− =− − B1 1bii ( ) ( ) 23 2 2 2 3 2 2 3 2 3 2 2 3 2 32 2 32 2 bca bc a b c b c ab ac b c ab b c ac b a c ac c acb a c acb a −= + + = − + = − − =− − − =− − −−= − += − M1 M1 Either A1 1c ( ) ( ) ( )( ) ( ) 22 22 2 2 52 13 2 3 5 2 3 2 3 13 2 3 10 15 2 6 2 3 6 9 10 17 6 2 9 9 8 8 3 0 8 ( 8) 4(8)( 3) 2(8) 1.2906, 0.29057 1.291, 0.291 (3 ) x xx x x x xx x x x x x x x x x x xx x x x dp −=−− − − − =−− − − + = − − + − + = − + − − = − −  − − −= =− =− M1 – common denominator M1 – either expansion M1 A1A1

2 2a y = 7 B1 2b 7 points – B2 4 points – B1 Curve – B1 2ci 3.25 1.2x−   B1 2cii 2 2 2 120 2 12 1.5 0 1.52 2 1 1.5 1.5 1.65 0.25 xx xx xx y x or x + + = + + − = − + − =− =− =− =− M1 A1A1 2ciii ( ) 2 22 2 21 22 122 12 xx x x +−    = + − −       = + − Therefore, the minimum point is ( )1, 2−− M1 - either x or y value seen A1 – in the form ( )2x p q++

3 3a ( )3 3 2 2 2 32 32 3 2 2Vol of hemisphere = 33 18 13Vol of cone = 32 19 34 3 4 318 3 4 918 4 18 4 9 8 ( ) x x xy xy xy x x y x x y xy x y x shown        =    =    = = = = = M1 M1 A1 3b ( ) 2 2 22 22 2 2 3 2 38 2 964 4 265 4 2658.1394 / 2 3CSA of cone = 2 3 8.13942 38.356 l y x xx xx x xx xl xx x   =+  =+  =+ = =   =  = ( ) ( ) 2 2 2 22 2 3CSA of cylinder =2 3 2 28.274 CSA of hemisphere =2 3 56.549 3rim area= 3 2 21.206 xx x x x xx x     =  = −  = M1 4 correct – M3 3 correct – M2 2 correct – M1

4 3b 2 2 2 2 2 2 Total SA = 450 38.356 28.274 56.549 21.206 450 144.385 450 3.1167 1.7654 x x x x x x x + + + = = = = ( ) Total ht, = 3 3 8 3 3 14 14 1.7654 24.7156 24.7 (3 ) h y x x h x x x hx h h h sf ++ = + + = = = = M1 M1 A1 4a 1 25 B1 4b 16 3 5n nT n −= M1 - 3n− M1 – 5n A1 - Tn 4c ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 1 22 22 16 3 1 16 3 5 1 5 16 3 3 16 3 5 1 5 13 3 16 3 1 51 13 3 16 16 3 3 51 13 3 16 16 3 3 51 16 ()51 nn n nTT nn nn nn n n n n nn n n n n n nn n n n n n nn shownnn + −+ −− = − + − − −=− + − − − += + − − + − − = + − − − + += + −= + M1 M1 A1 4d ( ) ( ) 1 For 0 5 1 0 16 051 0nn n nn nn TT+  + − + − B1

5 5a ( ) ( ) 24grad of line 8 ( 4) 1 2 1 2 14, 4 into 2 144 2 2 1 22 l y x c subst y x c c c yx −−= −− =− =− + − =− + =− − + = =− + M1 A1 5b Line 6 18 3 1 32 m yx yx =− =− + The gradient of line l and line m are equal (gradient = 1 2− ) and the y- intercept not equal. Line l and line m are parallel. Therefore, line m does not intersect the line l. M1 – gradient value A1 – parallel 5c 1 22yx=− + -------(1) 2 3 4yx=− --------(2) ( ) ( ) 12 2 3 42 2 1 2 4 3 4 48 S xx x ubst i x to x x n − + = − − + = − = = ( ) ( ) 2 1 22 1 222 1 (2,1) 1Subst intox yx y y A = =− + =− + = M1 M1 – either x or y coordinate A1 5d 2 1 422 4 Area of AB un ts C i =  = M1 – BC = 4 units M1 – height of 2 units A1 -2 2 B C A (2, 1) l n

6 6a 90 (tan rad) 360 90 90 40 (sum of quad) 140 360 140 ( at a pt.) 220 220 ( at ctr = 2 at circum)2 =110 OPV ORV POR reflex POR PQR  = =  = − − − =  = −  =  =    M1 M1 A1 6b 180 140 (sum of isos )2 =20 ( in same seg) =43 43 20 23 180 140 ( on str. line) =40 (ext. of ) =23 40 63 OPR SPR STR OPU POU TUP OPU POU − =   =   = − =  = −   = +   + = M1 M1 A1 6c 2 ( at ctr = 2 at circum) =2 43 86 reflex 360 86 274 86area of minor sector = area of major sect or 274 =0.31387 area of major sector SOR STR SOR  =      =  = − =   No because the area of the minor sector is 0.314 of the area of major sector, which is less than 1 3 (0.333) M1 M1 A1

7 7ai 86 44 68 44 2 = 8 u v u v      +=     −           =−      −      −  B1 7aii ( ) ( )22 2 2 8 28 8.2462 8.25 units XY XY −= = − + = = M1 A1 7aiii Grad of = Grad of 48 1 6 2 4 47 32 1 32 ZY XY k k k OZ − =− − − − =−− = −= M1 A1 7bi ( ) , 84 4 , 8 4 33 , 8 4 4 63 ba b a ba ba ABD AB BD AD BD units BD units BE  + = =− −− −− =− M1 A1 7bii , 6 3 2 2 4 2 82 b a a b ba ba BEF BE EF BF BF BC BF  + = = − + − =− = =− M1 A1 7biii ( ) ( ) , 8 2 8 4 2 22 2 b a b a a a BCD BD DC BC DC BC BD AB AB DC  + = =− = − − − = = = ABCD is a trapezium because 2AB DC= , AB is parallel to DC 1 pair of opposite side parallel. M1 M1 - 2AB DC= , AB is parallel to DC A1 – awarded for ‘trapezium’ only if M1 awarded

8 8ai 50 80 40100= Median = $550 B1 8aii 25 80 20 $420100 75 80 60 $680100 680 420 $260 LQ UQ IQR  = == =  = == = =− = M1 A1 8b No of workers ( $540) 38 No of workers ( $540) 80 38 42 42% of workers ( $540) = 100%80 52.5% =  = − =  = M1 A1 8c No of workers ( $460) 26 No of workers ( $800) 73 No of workers btw $460 and $800 73 26 47 = = =− = M1 - either A1 8d Let the additional no of workers be x. No of workers now ( $650) 56 56 11 80 15 840 15 880 11 4 40 10 x x xx x x = + =+ + = + = = M1 M1 A1 9a Total amt 90 3 150 $540 = +  = B1 9b 2 Size of hall required (300 15) 1.65 7425 m =   = Hall C and Hall E (total area = 7600 m2) daily rental $8000 $5400 $13400 =+ = M1 A1

9 9c Cost for 1. hall rental = $13400 4 $53600 2. long tables 300 $2 4 $2400 6003. square tables $30 4100 $720 754. round tables $20 45 $1200 15005. chairs = $10 4 no of chairs needed = 300 4+30050 $1200 15  = =   = =   = =   =    == 00 6. security guard = 2 $10 12 3 $720 7. part-timers 4 $8 12 3 $1152 Total cost 53600 2400 720 1200 1200 720 1152 $60992 Total amt collected for 3 days 300 $540 $162000 Total entrance fee collected = 35    = =    = = + + + + + + = = = 000 6 =210000 Profit =162000+210000 60992 $311008 The profit of $311008 is more than the minimum target of $300000. The event organiser is correct.  − =  5 items – M3 4 items – M2 2 items – M1 M1 – either item 6 or 7 M1 M1 A1- comparison made