ANDSS 4E2024EM_Prelim P2_MS_10th September

4E EMath Prelim P2 2024 Mark Scheme 1a) Since area 1536= , 21536 (24) (24) l  =+ 40l = cm 22height 40 24 32= − = cm Volume of cone 21 (24) 323 =    6144= cm3 (Shown) M1 M1 A1 1bi) 346144 3 r=   3 6144(3)= 4r =16.641 =16.6 cm (to 3 s.f.) M1 A1 1bii) 2Surface area 4 (16.641)= 3479.91555= 3480= cm2 (to 3 s.f.) M1 A1 2a) 24 20000 1 21337.05100 r+= 24 21337.051 100 20000 r+= 24 21337.051 100 20000 r+= 1.002700002= 0.270r = (to 3 s.f.) M1 A1 2b) loan amount $1774.40 84= $149049.60= 100% 2.78% 7+ 119.46%= Loan amount without interest 100%$149049.60 119.46%= $124769.4626= Price of Car 100%$124769.4626 70%= $178242= (nearest dollar) M1 M1 A1 2c) April : SGD980 JPY112210= June : JPY112210=SGD949.32318 Percentage loss 949.32318 980 100%980 −= 3.13%=− (to 3 s.f.) M1 M1 A1 Must circle loss.

4E Prelim Math 2023 2 3) Let the number of couples needed be x . 16 0.437 16 x xx + =+ + + 16 0.4(53 2 )xx+ = + 16 21.2 0.8xx+ = + 26x= M1 M1 A1 4i) 0a= 5b= B1 B1 4ii) 46 B1 4iii) mean 45.1= Standard Deviation 9.58= B1 B1 4iv) The group of 20 students have higher marks since the mean is higher, and they have more consistent marks as their standard deviation is lower. B1 B1 5a) Let y be the number of 50 cent coins and x be the number of 20 cent coins. 73yx+= --------(1) ( )0.5 ( )0.2 28.10yx += ------(2) From (2) ( ) 56.2 0.4( )yx=− -----------(3) Sub (3) into (1) 56.2 0.4 73xx− + = 0.6 16.8x= 28x= 73 28 45y= − = They have 45 50-cent coins and 28 20-cent coins. M1 M1 M1 A1 A1 Forming 1st equation Forming 2nd equation For performing substitution/elimination A1 for 45 and A1 for 28 5b) 22 41 (2 ) (5) (2 5) x xx −−− 41 (2 5)(2 5) (2 5) x x x x=− − + − 4 (2 5) (2 5)(2 5) xx xx −+= −+ 4 2 5 (2 5)(2 5) xx xx −−= −+ 25 (2 5)(2 5) x xx −= −+ 1 25x= + M1 M1 A1 • Factorise first denominator • Combine into single fraction

4E Prelim Math 2023 3 6a) BA BO OA=+ 19 12 3 −−   =+       10 15 −= B1 6b) ( ) ( ) 22 10 15BA = − + 18.0= units (to 3 s.f.) M1 A1 7i) Let m be the mid point of AB. MB 1 (65.32)2= (property of chord) 32.66= mm 1 32.66sin 45MOB − =  0.81216= rad 2AOB MOB =  1.624328= 1.6243= rad (shown) M1 A1 M1 for using sine 7ii) 22(45 ) (45) 1389.29x + − = 22(45 ) (45) 445.08953x+ − = 4.699x= 4.70= mm (to 3 s.f.) M1 A1