MARKING SCHEME

Higher 2 Tier GCSE Autumn 2008 Paper 19. (a)(b) 33 , 3 , 3 , 7 , 710. (a) PlotsCurve3 3 × 7 2M1A1B1B14P1C1NotesFor a method that produces 2 prime factors fromthe set {3, 3, 3, 7, 7} before their second error. If their 2 nd primeand 2 nd error occurs at the same ‘level’ then allow M1.C.A.O. for the five correct factors. (Ignore 1s).F.T. their answer if at least one index form used with at least asquare. Ignore prime number requirement for this B mark.Use of brackets (3 3 )(7 2 ) OR dot 3 3 .7 2 gets the B1.The inclusion of any 1s as factors, for example, 3 3 ×7 2 ×1in their index form gets B0.F.T. their (a) if the M1 awarded.Allow B1 for 3969 OR 3 4 × 7 2P0 on 2 nd errorMust be a curve at least for x = −2 to x = 2Within ± ½ asmall square.11.(b) Line y = 6x-valuesAngle bisector of ∠ADC,Line parallel to AB distance 3cm from it.L1B14B1B1F.T. their graph if at least 2 readingsUse overlayLoci do not have to be accurate as long as the intention is clear.About 2 cm is required to identify the loci.Initially ignore extra horizontal lines or extra lines throughD. Then if first B2 awarded, extra lines get a penalty of −1.If a region is drawn this may remove the ambiguity and thelines offered will be identified and no penalty invoked.Correct region12. (a) Correct image(−4, −1) (0, −3) (1, 2)(b) Correct image(1, 2) (1,−1) (3, 1)13. (a) Least = 975Greatest = 1025(b) Use of 50 × "their greatest volume"= 50 × 1025= 51250 (cm 3 ) (= 51·25 litres)51·5 (litres) ≤ tank (≤ 52·5 litres)50 of the largest jugs will always fit into the tankbecause 51·25 is < minimum tank (51·5 litres).14. (a) 20c 9 d 5(b) 2a(3b – a)B13B2B24B1B1M1A1B1E27B2B24F.T. if their region is similar to the correct one, i.e. uses a linethrough the angle at D and a line parallel to AB.B1 for 2 correct vertices.B1 for correct reflection in y = x(0, 3) (−1, −2) (4, 1)B1 for 2 correct vertices.B1 for clockwise rotation of 90º about (−1, −2).(3, −1) (3, 2) (5, 1)B1 for anti-clockwise rotation of 90º about (−2, −1).(−5, –1) (–5, –4) (−7, –3)The last point will require extra grid lines.C.A.O.C.A.O.F.T. providing 1000 < "their greatest volume" ≤ 1100C.A.O. No need for upper bound.Note also the correct division arguments:51·5/1·025 gets M1, and evaluated as 50·2(4) gets the A1.The 51·5 gets the B1OR 51·5/50 gets M1 and 1·03 gets the A1.E1 for an explanation that only uses 52 litres for the tank andstates that it is always possible.B1 for 20c 9 d k OR 20c n d 5 OR kc 9 d 5Ignore any extra × signs.B1 for a(6b – 2a) OR 2(3ab – a 2 )9