MARKING SCHEME

MATHEMATICS - 2 TIERPAPER 1 - HIGHER WITH COURSEWORKHigher 2 Tier GCSE Autumn 2008 Paper 11. 300 / 10 (=30)(£) 270 and (£) 302. 14 x ¾3. 360 / 6= 10.54. (a) 4x28x = 28/4 I.S.W. (=7)(b) 3x – 21 = 273x = 48x = 48/3 I.S.W. (=16)(c) 2x = 6 x 35. ∠DCB = 180 – 80= 100°x= 18/2 ISW (=9)11 bottlesM1A12M1A1A13M1= 60 (0) A12B1B1B1B1B1B1B1B18M1A1NotesOr equivalent strategy that leads to 10.5 or 11Need not be stated if answer of 11 is givenAccept an answer of 11 for all 3 marksCollecting the x termsAllow implicit solutions such as :6×7 – 11 = 17 + 2×7Clearing bracket. B2 for x – 7 = 9Collecting termsAnswers only get full marks.Allow implicit solns. e.g.3(16 – 7) = 27C.A.O. Allow on the diagram.F.T. until 2 nd errorF.T. until 2 nd error∠DCE = (360° − 100° − 135°)= 125 (°)x = 125 (°)6. (a) 302 + 2°(b) Their 145° bearing from ATheir 243° bearing from BTown C.1657. (a) × 100300= 55 (%)(b)14 3= 14×=1 723= 6 (mph)8. (a) Line of best fit by eye.Line through (61,59)(b) From their lineB1B14B1M1M1A14M1A1M1B1A15M1A1B13F.T. their angles, but not 80C.A.O. Do not accept 058°.+ 2° Use overlay. Watch for an unambiguous point on one orboth of the correct bearings and award the mark(s).F.T. if at least M1 and 2 intersecting lines.If the correct point C is unambiguously indicated even withoutthe bearing lines then award M1, M1, A1.C.A.O. SC1 for 55/100For substituted distance/time. Accept 2·3(3) or 2·2 for this markFor dealing with time correctly. 14/140 gets M1, B1.C.A.O.42/7 gets M1,B1,A0. 0·1 (miles per minute) gets M1, B1, A0.Must have positive gradient, look fit for purpose and have atleast 3 points above their line and 3 points below their line.If point not plotted, then the line must pass through the 2mmsquare 60-62/58-60 inclusive.F.T. if their line has a positive gradient. Reading should beexact, if the point is on the grid lines, else it should be read toeither side of the 2mm square, if the point is inside a square.Axes interchange is marked as correct then MR−18