2023 Springfield 3G3 Mathematics (4052) (P2-MS)

22 Marking Scheme 3Exp P2 1a 50 8 4 5  = M1 A1 AO1 1b 31 3 1 0.65 10 10 0.65 10 10 6500 a a a − −  =  = = M1 A1 AO1 2a 52 5 2 0 7 7 1 77 5 2 0 5 2 k k k k − −+ = = − + = = M1 A1 AO2 2b 3 5 15 2 22 3 5 15 2 , 15 2 22 5 20 , 7 3 74 , 3 7 43 x x x x x x x xx xx x −  −  + −  − −  +  −   − −   M1 M1 A1 AO2 2c B1 3a Based on the perimeter, the various combination for Length and Breadth are: 1. 4+4+9+9 2. 5+5+8+8 3. 6+6+7+7 6320 2 5 320 2 2 2 2 2 2 5 8 5 = =       from the prime factorisation of 320, the perimeter combination should be 5+5+8+8. Thus the height of the cuboid will be 8 cm M1 M1 A1 AO3 -3 -2 1 3 4 5 -1 0 2 7 3−

23 [Turn Over 3b n = 20 B1 AO2 4a 16 oC B1 AO1 4b 16 8850 m 88501 16 1 553.125 m 5 553.125 2765.625 m C C C → → → = M1 A1 AO2 5a Let x be the length and y be the breadth ( )( ) area new area 1.1 0.82 new area 0.902 0.902% change 100% % change 9.8% xy xy xy xy xy xy = = = −= =− M1 M1 A1 AO2 5b amt in Yen 32400 6200 amt in Yen 38600 total amt in Yen 1.05 38600 total amt in Yen 40530 40530total amt in SGD 106.80 total amt in SGD $379.49 =+ = = = = = M1 M1 A1 AO2 6a 3000 x B1 AO1 6b 3000 6x− B1 AO1 6c 2 2 3000 3000 50 69 3000 3000( 6) 50 ( 6) 9 18000 50 ( 6) 9 162000 50 300 6 3240 0 xx xx xx xx xx xx −=− −− =− =− =− − − = M1 M1 A1 AO2

24 6d ( ) 2 2 6 3240 0 ( 6) 6 4(1)( 3240) 2(1) 60 or 54( ) 3000no. of kg of vegetable 60 6 no. of kg of vegetable 55.6 kg xx x x x rej − − = − −  − − −= = =− = − = M1 M1 A1 AO2 7a 2 2 3 31.5 1.5 4.25 yx x a a =+ =+ = B1 AO1 7b Correct plotting of points – 2 mark [correct 6 out of 8 points – 1 mark] Smooth curve – 1 mark AO2 7c Drawing of line y = 9 2.8 0.05x= M1 A1 AO1

25 [Turn Over 7d Drawing of tangent Grad = 5.67 0.1 M1 A1 AO1 7e Drawing of tangent 3.9 0.1y= M1 A1 AO3 8a 17 12 5 5cos 17 72.895 72.9 OB AOB AOB AOB = − = =  =   =  M1 A1 AO1 8b reflex 360 72.9 2 reflex 214.2 214.2major arc length 2 (17)360 major arc length 63.554 major arc length 63.6 cm AOC AOC   = −   =  =  = = M1 M1 A1 AO2 8c 2 2 145.8area of sector (17) 360 area of sector 367.71 area of sector 368 cm OAEC OAEC OAEC =  = = M1 A1 AO1 8d 2 1area of minor segment 367.71 (17)(17)sin( 72.9 2)2 area of minor segment 286.49 area of minor segment 286 cm ABEC ABEC ABEC = −  = = M1 A1 AO2 9a AO2 B O A D

26 35cos30 40.415 40.4 cm AD AD AD = = = M1 A1 9b 40.415 12 28.415 cm BG BG =− = M1 A1 AO1 9c 2 2 2 40 28.415 49.065 49.1 cm AG AG AG =+ = = M1 A1 AO2 9d sin 30 28.415 14.208 14.2 cm HG HG HG = = = M1 A1 AO2 9e 2 1area of triangle 2 1area of triangle