ADSS Prelim 2025 P2 [Answers]

Marking Scheme for Sec 4E/5N EMath Prelim P2 (2025)1a22225205(4)55105(2)(2)(2)(2)(1)21xxxxxxxxxxxx M1A11b232324122174721129abaccabccaabc M1A11c11239193123427273abbaba M1A11d 2222(1)181()33522133221530260(23)(2)0322yyyyyyyyyyyyyyor M1M1A1A12aThe range of time taken = 64 – 21 = 43 minA12bThe median time taken 333433.52 minA12cThe mean time taken 21212223242527303334373842464751596418

= 35.777 = 35.8 minB12dLet time taken that one of the students took to complete the survey be x(35.777)(18)1.435.835.82010064422036.301241.012min41minxxxxx (to nearest minutes)M1M1A13a450x A13b45020x A13c 2245045040206045045090002(20)3270002(20)24027000020135000()xxxxxxxxxxxxshown M1M1A13d (to 2 dp)B1B1A1A14aiA = 371719153325323945254533391532194517393732 28922702 A1

4aiiA represents the total selling prices for all chocolate cakes, strawberry cakes and marble cakes of Store M and Store N respectively.A14aiiiMatrix B = 371719153325403426 = 25122300 A14aivC289225121127022300 38011$782402 A14avC represents the total profit of selling the three cakes by both shops. A14bia 332PSSRPQQRbSRaPSSRabbSRab→→→→→→→→ A14bib1222SMSRabSM→→→ A14bic22222PMPSSMabPMbaPMb→→→→→ A14bid 2522521455PTPMaPTbPTab→→→→ A14bii44455554532PQQTPTQTPTPQaQTbabaQTbaTSSMTMTSTMSMaa→→→→→→→→→→→→→→→

B1B1A15a 1st chosen pupil 2nd chosen pupil AnnAnn 11xy 1xy Other pupils1yxyx Other pupils1xyxy A1A1A1 5bP(Ann is the second candidate) 1111xyxyxyxy M1 21xy A15cGiven11xy=151Therefore, x + y + 1 = 15 x + y = 14M1

Given x = 6, 6 + y = 14y = 14 ‒ 6 = 8The total number of girls = 1+8 = 9. M1A16aiVol of cylinder 22(12)(35)15833.62697rh cm3B1Vol of cone 2211(12)150.796447433rhhhcm315833.62697 + 150.7964474 h = 18850h = 20 cmB1Surface area of cylinder22rrh2(12)2(12)(35) = 3091.327 cm2 M1Surface area of cone 22(12)(1220879.2868305rl cm2B1Total surface area of the container = 3091.327 + 879.2868 cm2 = 3970 cm2 (3 sig fig)A16aiiTotal volume of water in the container when water reaches 30 cm from the vertex of the cone23(1885015833.62697)(12)(3020)7540.266451cm M1B1Time taken 7540.26645130.16130.2250 sA16biI disagree that the area of triangle EBA to the area of triangle ECA is 16 : 25.Because 0.50.5areaofEBABAhareaofECACAh 15. M1A16bii15areaofEBAareaofECA 15areaofECAcm2 B1Since ACDE is a parallelogram, area of ACDE = area of ∆ECA x 2 = 15 x