XMSS E Math Prelim P2 Mark Scheme

Solution/ Mark Scheme 1 (a) ( ) ( ) 2 7 7 025 2 7 7 25 5 2 7 2 7 --- [M1] 10 35 14 2 12 21 37 1 (also accept ) --- [A1]44 xx xx xx xx x x +− −= +− = + = − + = − =− =− − Alternative 2 7 7 0 25 5(2 7) 2(7 ) 0 --- [M1] 10 35 14 2 0 12 21 37 1 (also acce pt ) --- [A1]44 xx xx xx x x +− −= + − − = + − + = =− =− − (b) (c) 22 2 2 2 15 3 28 15 8 23 20 --- [B1] x x y y x y x y y  = = 2 2 2 (2 3)( 4) 2 2 11 12 2 2 11 10 0 ---[M1] ( 11) ( 11) 4(2)(10) ---[M1]2(2) 11 41 4 1.15 or 4.35 --- [A1, A1] xx xx xx x x x − − = − + = − + = − −  − −= = =

(d) ( ) 22 22 3 2 6 6 3 2( 3 ) 6 ( 3 )( 2) --- [M1 for numerator; M1 for denominator]( 3 )( 2 ) 2 --- [A1]2 ax bx a b a ab b x a b a b a ab b a b x a b a b x ab + − − +− + − += +− +−= +− −= −

Solution/ Mark Scheme 2 (a) Each compound increases the principal amount, which in turn leads to greater interest being generated. Cody should invest in account A as it provides more frequent compounding than account B, hence, generating greater interest. OR Cody should invest in account A because the interest is calculated 12 times each year as compared to only once a year for Account B. Each time, the interest is calculated on a larger amount of money. Therefore, account A will generate greater interest. *Do not accept if students provide a calculated example and arrive at a conclusion just based on calculation. Key idea of more frequent compounding in account A, which will lead to more interest generated, should be featured in students’ explanation. (b) 3 3 3 3 20000 1 22823.32 --- [M1]100 22823.32 1 100 20000 22823.32 1 --- [M1]100 20000 22823.32 1100 20000 4.4999 (5 s x x x x x += += += =− = .f.) 4.5 (1 d.p.) --- [A1] (c)(i) 40 $156000 36 $2800 --- [M1]100 $163200 --- [A1]  +  =

(c)(ii) 3 $156000(0.92) --- [M1] $121475.328 (3 d.p.)= Method 1 $121475.328 100% 74.433% --- [M1]$163200 100% 74.433% 25.6% (3 s.f.) --- [A1] = −= Method 2 loss incurred: $163200 $121475.328 $41724. 672 $41724.672% loss: 100% --- [M1]$163200 25.566% 25.6% (3 s.f.) ---[A1] −=  = =

(d) [A1] for either Amount of S$ spent in America: $368 2 1.35 $993.60 [M1 for finding cost of hotel in US in SGD] 1Amount of S$ spent in Canada: $250 3 $735. 2941 (4 d.p.) 1.02   =   = [M1 for finding cost of hotel in Canada in SGD] 1.5Total amount spent: $993.60 $735.2941 ($99 3.60 $735.2941) $1754.82 100 OR 101.5 ($993.60 $735.2941) $1754.82 100 + +  + = +  =

Solution/ Mark Scheme 3 (a)(i) Median mark = 35 (a)(ii) 38 30 --- [M1] 8 --- [A1] − = (a)(iii) 60th percentile: 60 80 48100= From graph, 60th percentile: 36 --- [B1] (b) 85 80 68 --- [M1]100 80 68 12 12 students scored less than marks. Fro m graph: 25 ---[A1]xx = −= = (c)(i) Chemistry Test Median mark: 32 IQR: 40 25 15−= Physics Test Median mark: 35 IQR: 8 1. The students perform better for the Physics test due to a higher median mark of 35, as compared to the Chemistry test, with a lower median mark of 32. --- [B1] 2. The performance of the students were more consistent for the Physics test due to a lower interquartile range of 8, as compared to the Chemistry test, which has a higher interquartile range of 15. --- [B1] *Note: Students need to draw reference to the values of median/ IQR in their explanation to be awarded the mark. (c)(ii) The entire box-and-whisker plot would shift to the right by one unit/ one mark. --- [B1]

Solution/ Mark Scheme 4 {2,3, 4,...15} {4,8,12} {2,3, 4,6,8,12} A B  = = = (a)(i) Elements in B’: 5,7,9,10, 11,13,14,15 --- [B1] [Accept if students write: {5,7,9,10,11,13,14,15} ] (a)(ii) B *Note: A is a proper subset of B. 2 A 3 4 8 6 12 (a)(iii) Answer: 0 --- [B1] Answer: 2,3,6 --- [B1]

(b)(i) Physics Not Physics Literature 13 5 ---[B1] Not Literature 7 15 2 Let represent number of students who to ok both Physics and Lit. 11 --- [M1]40 39 10 10 ( 1) 40 39 156 0 ( 13)( 12) 0 --- [M1] 13 or 12 (rej) x xx xx xx xx xx −= − =  − − = − + = = =− Note: Award full credit if students managed to obtain the first column correctly by trial and error, with relevant workings provided. i.e. 13 7 20 By trial and error, 13 12 1 40 39 10 Therefore, there are 13 students taking Physics and Literature. 7 students take Physics but not Literature. += = (b)(ii) 18 22 2 --- [M1, allow ecf]40 39 33= --- [A1]65  [A1] for both correct

Solution/ Mark Scheme 5 (a) 32 32 32 3 2 21π(2 ) π(2 ) --- [M1, M1]33 21π(8 ) π(4 )33 16 4 16 4 4 --- [A1] r r h r r h r r h rh r hr = = = = = (b) 3 3 3 3 2 2 π(2 ) 450 --- [M1]3 16 π 4503 450 3 16π 450 3 16π 2.9947 (5 s.f.) --- [A1] Total surface area: 2π( ) π( ) , where 2 r r r r r R R l R r = = = = = += 22 22 2 2 (2 ) (4 ) --- [M1] 20 20(2.9947) =13.392 l R h rr r =+ =+ = = Alternatively 22 (radius) 2 2.9947 5.9894 (height) 4 2.9947 11.9788 5.9894 11.9788 --- [M1] 13.392 l = = =+ = 2 3 3 Total Surface area: 2π(2 2.9947) π(2 2.9947)(13.392) --- [M1] 477.38 cm ( 5 s.f.) 477 cm --- [A1]  +  =  M1: applying formula to find vol of hemisphere correctly. M1: applying formula to find vol of cone correctly.

Solution/ Mark Scheme 6 (a) (angles in the same segment) (base angles, isos. tr t iangle) = = is triangles an t d are congruenA the common side. By AS est, PQS SRP OSP OPS QSP QSO OSP R S PO OPS RPS PS SQP PR  =  =  = +  +  --- [A1]. *max 1 mark is deduced directly from the question for any wrong reason given. (b) 64 2 ( at centre = 2 at circumference) 128 90 (tangent perpendicular to radius) 360 90 90 128 --- [M1] 52 64 52 116 ( 180 ) By property of an POS OPT OST PTS PTS PRS  =    =  = =   = − − −  =  + = + =    gles in opposite segment, since 180 , we cannot draw a circle passing through P, R, S and T. PTS PRS +   M1 for any two correct; M2 for all three correct [A1]