MARKING SCHEME

Higher 2 Tier GCSE Autumn 2008 Paper 115.52(4x – 1) – (2x + 7) = × 628x – 2 – 2x – 7 = 156x = 24x = 24/6 I.S.W. (= 4)16. (a) largest = 18 = (1·5)largest 1217. (a)middle = 15 = (1·5)middle 10smallest = 9 = (1·5)smallest 6(All same ratio, therefore similar triangles)(b) For example, QR = 1510 12(b)RQ = 12·5 (cm)3 2 and on the first branch5 52 7 and on the second branches992 7 ×5 9=144518. (a) (3x + 1)(7x – 1)x = -1/3 and 1/7(b)(i) (7x + 8)(7x – 8)(ii) 7x + 819. (Area) scale factor 16/100 (or equivalent)Length scale factor 4/10 or √16 / √1005 (cm)20. (a) 220 0(b) 52 0Explantion or calculation21. 5be – 3e = 7c – 10be ( 5b – 3) = 7c – 10be = (7c – 10b) / ( 5b – 3 )22. Use of the y values 9, 10 and their 9Use of trapezium rule or idea of sum areas23.523. (a) ½ LN - ON (= 2a -4b + 4a +16b)OR - ½ LN - OL (=-2a +4b + 8a +8b= 6a + 12b(b) Showing k = 2/3(c) Collinear or parallelPO is 2/3 x length OM or OM = 1.5 x POM1M1A1A14B2M1A14B1B1M1A14B2B1B2B16M1M1A13B1B1E13B1B1B13M1M1A13M1A1B1B1B15NotesMust clear fractions by a valid methodFor handling 2 of the 3 terms correctlyFor the Ms, 2x+ 7 is acceptable, but the first A mark is A0.For handling all 3 terms correctlyCollecting terms. F.T. until 2 nd error if at least M1 awardedUnsupported answer of x = 4 gets all 4 marks.B1 for finding any 2 corresponding ratiosB2 for ‘The triangles have been multiplied by 3/2 OR 2/3’OR‘the scale factor is 3/2 OR 2/3’, etcOR‘both cancel to 3:5:6’ – must be all 3 correct values.Any correct equation with only QR unknownC.A.O.C.A.O.C.A.O. Accept only on one branch provided the other branch isempty.F.T. their tree if probabilities are between 0 and 1 exclusive andNOT all ½.B1 for (3x 1)(7x 1)FT their pair of bracketsB1 for (7x….8)(7x….8)FT if possibleSC1 for relevant sight of 4 and 10 or √16 and √100CAOAlternate segment, cyclic quadrilateral and triangleCollect e termsFactoriseDivisionFT for equivalent stages and level of difficultyCAOIntention clear, e.g. ON + NPMust be simplified form10