2008 Autumn Paper 1 (Non calculator)Intermediate Tier14. (a)(b) 33 , 3 , 3 , 7 , 715. (a) PlotsCurve3 3 × 7 2MarksM1A1B1B14P1C1POST CONFERENCE MARK <strong>SCHEME</strong> (10/11/2008)Comments (Page 3)For 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 2– 1 on 2 nd error Within ± ½ asmall square.(b) Line y = 6x-values16. (a) Correct image(−4, −1) (0, −3) (1, 2)L1B14B2F.T. their graph if at least 2 readingsB1 for 2 correct vertices.B1 for correct reflection in y = x(0, 3) (−1, −2) (4, 1).(b) Correct image(1, 2) (1,−1) (3, 1)Angle bisector of ∠ADC,Line parallel to AB distance 3cm from it.B24B1B1B1 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.Use 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 through D.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 region18. (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).19. (a) 20c 9 d 5(b) 2a(3b – a)B13B1B1M1A1B1E27B2B24F.T. if their region is similar to the correct one, i.e. uses a linethrough the angle at D and a line parallel to ABC.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 )3
2008 Autumn Paper 1 (Non calculator)Intermediate Tier20. 4x ≥ – 22 OR equivalent21. 131222.x ≥ – 22/4 (I.S.W.) (–5·5)52(4x – 1) – (2x + 7) = × 628x – 2 – 2x – 7 = 156x = 24x = 24/6 I.S.W. (= 4)23. (a) largest = 18 = (1·5)largest 12middle = 15 = (1·5)middle 10smallest = 9 = (1·5)smallest 6(All same ratio, therefore similar triangles)(b) For example, QR = 1510 12RQ = 12·5 (cm)MarksB2B13B22M1M1A1A14B2M1A1POST CONFERENCE MARK <strong>SCHEME</strong> (10/11/2008)Comments (Page 4)B1 for each side provided there is a correct inequality sign.F.T. until 2 nd error.F.T. equivalent difficulty if B1 has been awarded.0 marks if they only use = sign. Condone use of the > sign.For all 4 correct.B1 for any 3 correct.Must 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.424. (a)3 2 and on the first branch5 5B1C.A.O.2 and 97 on the second branches9B1C.A.O. Accept only on one branch provided the other branch isempty.(b)2 7 ×5 9=1445M1A14F.T. their tree if probabilities are between 0 and 1 exclusive andNOT all ½.4