MARKING SCHEME

MATHEMATICS - 3 TIERPAPER 2 - INTERMEDIATE WITH COURSEWORK2008 Autumn Paper 2 (Calculator allowed)Intermediate Tier1. (a) (£)24·10 − (£)5·20 = (£)18·9(0)(£)18·9(0) ÷ (£)3·15(0)= 67 days(b) 26 × 791002. Correct image(Allow ± 2mm)3. (a) x = 180 − 65 − 65= 504.= 20·54 I.S.W.(b) y = 360 – 90 – 72 – 116= 82(a) 3–11(b) (x =) 36(c) 6a – 10b(d) –55. (a) Euros = 800 × 1·49= (€) 1192(b) Present = 96·85/1·49= (£) 656. (a) 25 (%) OR ¼(b) You cannot tell from the pie charts because youwould need to know the number of boys andnumber of girls.ORYou can tell from the pie chartsif the number of boys and girls were equalOR1/3 of the boys was more than about 2/3 of thegirls.7. (a) x( x – 2)(b) 6x + 18 I.S.W.(c) n + 5 OR (n+5)/7 OR n+5/77(d) –195MarksM1A1B1M1A15B22M1A1M1A14B1B1B1B2B27M1A1M1A14B1B2ORB23B1B1B2B1B16POST CONFERENCE MARK SCHEME (15/11/2008)Comments (Page 1)For the complete method that leads to the number of additionaldays.F.T. ‘their 6’+1. Allow ‘6 additional days’Must be a whole number of days, and rounded up if needed.Allow M1 for methods such as 10% = 7·920% = 15·8, 5% = 3·95 etc as long as there isC.A.O. enough correct work to convince that theyAllow 20·54% understand percentages and are finding 26% bya proper method of partitioning.If not all correct, then B1 for three correct vertices.Use of a different scale factor should be marked then MR−1C.A.O.C.A.O.Do not penalise extra =x or x= or =n or n= in this question.F.T. ‘their 3’ – 14 provided their answer is negative.C.A.O. Allow embedded solutions, 36/4 = 9B1 for either term.If B2 awarded, then − 1 once only for any inappropriate extraalgebra, e.g. 6a – 10b = −4ab gets B2 then –1.B1 for the 16 OR the –2116d – 21e gets B0C.A.O.C.A.O.Along these linesORAlong these linesC.A.O.C.A.O.C.A.O.B1 for n + 5 ÷ 7SC1 for 2 2 – 5 AND 10 2 – 5 OR Equivalent5