Crest Sec 4NA Prelim MS Paper 2

2024 4NA Mathematics Prelim Paper 2 Marking Scheme Qns Workings Mark Allocation 1 1.673320053 = 1.67 B1 2 2 2 22 3 12 14 7 37 14 12 1 8 mn m n nn mn n n m n m  = = M1 A1 3a ( )( ) 216 49 4 7 4 7x x x− = − + B1 3b 12ab – 16bc + 3a – 4c = 4b (3a – 4c) + 3a – 4c = (3a – 4c) (4b + 1) M1 common factor A1 4a 5 ( 20) 25− − = B1 4b 25 8848 0.002825497288 = 0.00283 − =− − M1 A1 5a 2(10) 11(25) 9(35) 6(45) 3(65) 31 34.67741935 107534.7 / 31 + + + + = = 2 22 2 2 22(10) 11(25) 9(35) 6(45) 3(65) 1075 31 31 13.49644421 13.5 + + + + − = = B1 Mean M1 A1 S.D 5b She is correct as the corporate order of 60 madeleines is greater than the estimated mean, hence including this value in the calculation will increase the estimated mean. B1 6a 8, 4 B1 6b 28 4( 1) 28 4 4 32 4 n n n −− = − + =− B1

6c 32 4 42 4 32 42 4 74 74 4 118 2 n n n n n − =− =+ = = = Donovan is incorrect. As n is not an integer, – 42 is not a term in the sequence. M1 A1 7a 5 12 11 20 1 99 33= B1 7b 0 12 7 7 12 7 1 1 7 20 19 20 19 20 19 20 19 91 19         +  +  +                = M1 A1 7c 12 7 7 12 12 11 7 6 20 19 20 19 20 19 20 19 9 10         +  +  +                = M1 A1 8a Triangles ABE and DCE. Given AB parallel to CD, angle ABE = DCE (alternate angles), angle BAE = angle CDE (alternate angles), and angle AEB = angle DEC. Given that there are at least 2 pair of corresponding angles in both triangles with different corresponding length, triangles ABE and DCE are similar. B1 B1 for each pair of corresponding angles 8b 23 5 7.5 CE DE BE AE AE AE = = = 7.5 3 10.5AD= + = M1 corresponding ratios M1 value of EA A1 value of AD 9a Angle ABD = 180  – 123  = 57  Reason: angles in opposite segment B1 value B1 reason 9b Angle CBD = 90  – 57  = 33  Reason: right angled triangle in semi-circle B1 value B1 reason 9c Angle CAD = 33  Reason: angles in same segment B1 value B1 reason

11a ( )( ) ( )( ) 2 2 2 2 2 5 4 1 4 110 2 1 5 110 2 10 5 110 2 9 5 110 2 9 5 110 0 2 9 115 0 xx xx x x x xx xx xx + − − − = + − = − + − = − − = − − − = − − = M1 M1 expansion A1 11b 22 9 115 0xx− − = 2( 9) ( 9) 4(2)( 115) 2(2) 9 81 920 4 10.15964601 5.65964601( )      x x rorx ej − −  − − −= += =− 2 1 2(10.15964601)+1= 21.31929202=21.3x+= ( 5) 10.15964601 5 5.15964601=5.16x− = − = M1 quadratic formula M1 x values B1 B1 12a 3 6 10 3 610 1.8 r r r = = = M1 by similarity A1 12b Slant height of water = 223 1.8+ = 3.498571137 Curved surface area/surface area in contact with water (1.8)3.498571137 19.78395369 19.8 = = = M1 slant height/ ecf M1 A1 12c ( ) 2 10.17876021 1.8 33 = ( ) 21 6 12 10.17876023 133.821 134 − = = M1 vol of water M1 A1 13a 37 4000 1480100= 1480 12 17760= M1 A1

13b 2nd year/ age 26: 110 4000 4400100 3712 4400 19536100 =  = 3rd year/ age 27: 110 4400 4840100 3712 4840 21489.60100 =  = Total contribution for 3 years = 17760 19536 21489.60++ = 58, 785.60 M1 Calculate 10% annual increment M1 A1 13c 5 350000 17500 58785.60100 =  5 1500000 75000 58785.60100 =  58785.60 17500 41285.60−= Since 17500 58785.60 , Alex has sufficient CPF savings to cover the down payment for a 3-room resale flat only. Hence, 3 room resale flat is a better choice. M1 downpayment (ecf) A1 14a 180 55 125 360 125 235 − =  − =  M1 A1 14b 2 2 2 2 60 80 2(60)(80) cos 40 2645.973346 51.43902552 51.4 AB AB AB AB = + −  = = = M1 A1 14c 11(60)(80)sin 40 (55)(80)22 3742.690263 3740 + = = M1 Area of triangle A1 14d 1 15tan 55 15tan 55 15.2551187 15.3     − = = = = M1 A1 15a Household A: 350 – 90 = 260 Household B: 400 – 50 = 350 Household B has a higher range. B1 calculate range B1

15b 210 B1 15c 325 – 150 = 175 M1 A1 15d Household A is more consistent in the daily usage of water as the interquartile range is smaller than that of household B. B1 15e 65 100%200 32.5%  = M1 A1

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