MARKING SCHEME

Paper 2 Higher Tier Autumn 2008 Mark Comments10. (a) Inter-quartile = reading at 225 – reading at 75= 37 to 41M1A1 Accept any reading in the range (71 to 73) – (32 to 34) andtherefore answers in the range 37 to 41.Allow consistent misread of the ‘Time’ scale for the M mark.For example, 66 – 26 = 40 OR 66 – 27 =39 get M1, A0However unsupported 39 or 40 get M1, A1.(b) 300 – reading at 90(300 – 265)= 3511. AD = 16 × sin56ºAD = 13·2(646) OR 13·3 (cm) OR 13M1A14M2A1Accept any reading for 265 in the range 260 – 270and therefore answers in the range 30 – 40 (whole numbersonly).All ranges are inclusive).Correct substituted sin ratio.M1 for sin56º = AD/16C.A.O.EC = (23 – 13·2(646)) = 9·735tan x = 9·735/15 (= 0·649026)∠CBF = 32·98 OR 3312. (a) Vol = 3x 2 (x+7)3x 3 +21x 2 = 3x 3 + 2x + 1(b) x = ( 2 + √{(-2) 2 – 4x21x-1}) / 2 x 21= (2+√88)/420.27 (and -0.18)(c) 1.6 (cm 3 )13. Overall strategy∏r 2 = 380r = 10.998…..72/360 x 2 x ∏ x rarc AB = 13.8(2… cm)14. (a) y α 1/x OR y = k/x3 = k/2y = 6/x(b)x -1 2 60y -6 3 0.115. x = 0.8232323… and 100x=82.323..815/99016. (a) P(www) = 4/10 x 3/9 x 2/8= 1/ 30 or equivalent(b) P(ssm) = 2/10 x 1/9 x 6/8 in any orderP(ssm)+P(sms)+P(ssm) OR 2/10 x 1/9 x 6/ 8 x3= 1/20 or equivalent17. (a) Method of finding an area of 1 bar4 correct areas AND intention to add all areas100(b) Group 20 to 25 identified9/30 or 21/30 considered20 + 9/30 x 5 OR 25 – 21/30 x 521.5B1M1A16M1A1M1A1A1B16B1M1A1M1A15B1M1A1B25M1A12M1A1B1M1A15M1M1A1M1M1M1A17F.T. ‘their AD’.Correct substituted tan ratio, based on their AD and/or ECF.T. ‘their AD’.PA−1 will apply if, e.g. , AD=13 is used (answer =33·67)Allow 1 slip in substitutionFT their x only if M1 in (b)Maybe in different sections of workingFT their rFT non linear onlyMaybe implied in part (b)FT their non linear expressionB1 for each value, do not accept 6/-1 for -6Or 10x and 1000x, or equivalent81.5/99 gains M1 onlyAreas are 10, 14, 17, 30, 14, 15CAO14