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 8i) 180 64 116BABN = − =  (int  s, //ABNN ) 360 127 116ABC = − −  (  s at a pt.) 117= 224.83 7.24 2(4.83)(7.24)cos117AC= + −  10.368= km (shown) M1 M1 A1 Must give reason Use cosine rule 8ii) sin117 sin 10.368 4.83 BCA= 24.52443BCA =  (5 d.p.) Bearing of A from C Reflex cN CB 360 24.52443 (180 127 )= − − −  282.47557= Bearing of A from C 282.5= (1 d.p.) M1 M1 A1 Use sine rule 8iii) Shortest distance 7.24 sin 24.52443=   3.00518= 3.01= km (to 3 s.f.) M1 A1 8iv) Distance of object to C 7.24 tan12=   1.54= km (to 3 s.f.) M1 A1

4E Prelim Math 2023 4 9 Let interior angle of polygon be x. 3 (5 2) 180x a b+ + = −   3 72 540x+ =  156x= 360 180 156n = −  15= M1 M1 A1 10ai) 2 5 b B1 10aii) DE DC CO OE= + + 2 5 b=− ab−+ 3 5ab=− + 23 35AD a b = − + 22 53ba=− M1 M1 A1 10b) AC AD DC=+ 2 3 a=− 2 3 OC=− therefor AC is parallel to OC . M1 A1 10ci) Let h be the common height of ADB and ACD from A to CB. 1 area of 2 1area of 2 DB hADB ACD CD h  =  3 2= B1 10cii) Since AC AD CD AO AE OE== ADC is similar to AOE . 2 area of 2 area of 5 ADB ACD  =  4 25= M1 A1 10ciii) Area area of : area of : area of ADB ACD AOE   6 : 4 : 25= area of 6 area of CDEO 25 4 ADB = − 2 7= M1 A1 M1 for finding the ratio of area of CDEO relative to area of ADB

4E Prelim Math 2023 5 11a) 0 B1 11b) Refer to graph • plot all points correctly • smooth curve B1 B1 11ci) Refer to graph B1 11cii) Gradient 0.8 ( 1) 03 −−= − 0.6 0.1=−  0.6 0.8yx=− + y-intercept 0.1 M1 A1 11d) 23 8 6 0xx− + = 63 8 0x x− + = 62 7 1xx x+ − =− + 1yx=− + Draw graph 1yx=− + . 23 8 6 0xx− + = has no solution because 627yx x= + − and 1yx=− + do not intersect for 0.5 7 x M1 M1 A1 M1 for either of these 2 working steps. M1 for drawing graph. 12a) 1500 x B1 12b) 1500 17 222.5 60 xx −= − 1500 1020 222.5xx −= − 21500( 22.5) 1020 2( 22.5 )x x x x− − = − 22 525 33750 0xx− + = (shown) M1 M1 A1 12c) 2( 525) ( 525) 4(2)(33750) 2(2)x − −  − −= 525 5625 4 = 150= or 112.5= M1 A1 A1 A1 for each solution 12d) Total time taken ( 22.5) 2 22.5x x x= + − = − minutes When 112.5x= Total time 2(112.5) 22.5=− 202.5= mins (reject since total time >4 hours) When 150x= Total time 2(150) 22.5=− 277.5= mins 4h 37.5min= M1 A1 Must reject with reason.

13a) Percentage difference 282 202 100%202 −= 39.60396= 39.6%= (to 3 s.f.) M1 A1 13b) The books can be stacked into columns. The configuration of the columns will fit into boxes. Configuration 1 (1 column) Max height 900 128 128 644= − − = Exceed max 400mm Max books 400 6 66=   books Total weight 66 75 4950=  = g Configuration 2 (2 columns) Max height 900 128 128 128 516= − − − = Exceed max 400mm Max books per column 2400 6 66 66 3=  =  books Max books 66 2 132=  = books Total weight 132 75 9900=  = g Configuration 3 (4 columns) Max height 900 128 128 128 128 388= − − − − = Max books per column 2388 6 64 64 3=  =  books Max books 64 4 256=  = books Total weight 256 75 19200=  = g M1 M1 M1 M1 These 3 M1 can be allocated to any of the 5 configurations, M1 for max height, M1 for max books, M1 for max weight. Configuration 1 Configuration 2 400 mm 128 mm 128 mm 400 mm 128 mm 128 mm 388 mm 128 mm 128 mm 128 mm 128 mm 128 mm Configuration 3

4E Prelim Math 2023 2 Configuration 4 (6 columns) Max height 900 128 128 128 128 128 260= − − − − − = Max books per column 1260 6 43 43 3=  = = books Max books 43 6 258=  = books Total weight 258 75 19350=  = g Configuration 5 (9 columns) Max height 900 128 6 132= −  = Max books per column 132 6 22=  = books Max books 22 9 198=  = books Total weight 198 75 14850=  = g Max books by weight 20000 75 266=   books Pick configuration 4 as most books. Total cost to manufacture and post to Australia 202 258 1.80= +  $666.40= Price per book 666.40 3 258 258 += $5.58294= $5.60= (to nearest ten cents) M1 A1