1. Q1. Prove that.................................................................................................................................. (Total 4 marks)

2. One fraction correct or n(n + 6) or (n + 5)(n – 4) M1 – Method mark n 2 + 6n and n 2 + 5n – 4n – 20 n 2 + 6n or n 2 + n – 20 Condone – n 2 + 5n – 4n – 20 Method mark n 2 + 6n – n 2 – 5n + 4n + 20 n 2 + 6n – n 2 – n + 20 A1 – Accuracy marks are awarded when following on from a correct method 5n + 20 and Answer given Must see all working for final mark Must see use of denominators for final mark A1 Accuracy marks are awarded when following on from a correct method [4]

3. Hardly any fully correct answers were seen. Few candidates knew that a common denominator was required. Many attempted to substitute numerical values into both sides as a test of validity. Common Mistakes – what did the examiners say?

4. The diagram shows the cross-section of a river bed, ABCDE. Not drawn accurately BD is the surface of the river. C is the midpoint of the river. A and E are points on the river bank that are 1 metre above the river surface. Taking the origin at B, metre as 1 unit on both axes and the river surface as the x-axis, the curve ABCDE can be represented by the equation (a) Show that the depth of the river at the mid-point C is 2 metres. (3) (b) Show that the distance AE is approximately 12.25 metres. (5) (Total 8 marks) (b) Show that the distance AE is approximately 12.25 metres (5) (Total 8 marks)

5. (a) 2x(x – 10) = 0 M1 – Method mark River ( x = ) 10 ( m wide) A1 – Accuracy marks are awarded when following on from a correct method When x = 5, y = (50 – 100) ÷ 25 = –2 A1 – Accuracy marks are awarded when following on from a correct method (b) 2x2 – 20x = 25 M1 – Method mark 2x2 – 20x – 25 = 0 M1 – Method mark Use quadratic formula, completing the square to solve equation Allow one error but not wrong formula M1- Method mark x = –1.123 or 11.123 A1 – Accuracy marks are awarded when following on from a correct method AE = 1.123 + 11.123 ≈ 12.25 m NB error with – b gives 1.123 and – 11.123 Allow last A1 A1 – Accuracy marks are awarded when following on from a correct method

6. This question was the second least successful question on the paper. There were very few fully correct answers. The majority of scripts were blank. Candidates had little idea of how to relate the x and y values. In part (a), few realised that y had to be put equal to zero. It was often put equal to 25. A few candidates started with y = –2 which was accepted as a method providing it led to x = 5 and that x = 10 was then tested to give y = 0. Very few who tried this method gained full marks. Several also began with y = 2 which led nowhere. In part (b), substituting = 12.25 was the most frequent error. The few that put the equation equal to 1 often obtained a quadratic expression but then failed to solve it correctly Common Mistakes – what did the examiners say?

7. (a) Simplify (9 + √7)(9 + √7) Give your answer in the form a + b√7 (2) (b) Prove that (Total 6 marks)

8. (a) 81 + 9 + 9 + or better 4 terms and any 3 correct M1 88 + 18 a = 88 b = 18 A1 (b) M1 A1 = 2 + 2 M1 = 2(1 + ) Strand (ii) Correct answer with a logical argument showing key steps Q1

9. Solve (Total 6 marks)

10. Sight of 10x or –3(2x – 1) or 3x(2x – 1) M1 – Method mark –6x + 3 or 6x 2 – 3x M1 – Method mark 6x 2 – 7x – 3 (= 0) A1 – Accuracy marks are awarded when following on from a correct method (2x – 3)(3x + 1) (= 0) M1 – Method mark x = 1.5 or – A1 – Accuracy marks are awarded when following on from a correct method Full answer with stages clearly shown Strand (ii) Q1 – Quality of written communication [6]

11. The value of a vintage car rises from £36 000 to £63 000. Work out the percentage increase in the price of the car. Answer............................................% (Total 3 marks)

12. M1 – method mark × 100 M2 complete and correct build up method. If any numerical errors calculations must be shown to give M2. M1dep – method mark 75 A1 Accuracy marks are awarded when following on from a correct method [3]

13. Many candidates did not know how to deal with the figures involved, with many dividing by 63 000. Those who did know what to do were sometimes held up by their arithmetic which was unnecessarily lengthy at times. Able candidates could virtually write down the answer. Less able candidates were often trying build up which rarely were of sufficient detail (once the accuracy had gone astray) to gain even a single mark. Common Mistakes – what did the examiners say?

14. Solve the equation Answer................................................. (Total 5 marks)

15. LHS x(x – 1) – 2(x + 1) Give M1 for x 2 – 3x + 2 if first line seen Allow invisible bracket if recovered. M1 LHS = x 2 – 3x – 2 Terms need not be collected. e.g.x 2 – x – 2x – 2 A1 (x – 1)(x + 1)(= x 2 – 1) On RHS or as denominator. x 2 – 1 can be written as x 2 – x + x – 1 M1 Their (x 2 – 3x – 2) = their (x 2 – 1) Dependent on first 2 M1’s DM1 – (= 0.33(3...)) Do not follow through. NB ‘cancelling’ x 2 on top and bottom of Gives correct answer. Give M1, A1, M1. M0, A0. A1 [5]

16. Too many candidates have no idea how to start this type of problem. Those candidates who had some idea could often make an attempt to find the numerator on the left-hand side. This was then often incorrectly evaluated as x2 – 3x + 2. Fewer knew to find the denominator or the right hand side as x2 – 1. A common error was to find the numerator as x2 – 3x ± 2 and the denominator as x2 – 1. The x2 were then cancelled out. This led to the correct answer but gained no credit as the answer came from wrong work. Common Mistakes – what did the examiners say?

17. Solve the equation Total 5 marks

18. (x – 2) + 5x(x + 1) = 3(x + 1)(x – 2) Allow 1 error M1 – Method mark 5x 2 + 6x – 2 = 3x 2 – 3x – 6 A1 - Accuracy marks are awarded when following on from a correct method 2x 2 + 9x + 4 = 0 M1 (2x + 1)(x + 4) = 0 A1 - Accuracy marks are awarded when following on from a correct method x = –1/2, –4 A1 - Accuracy marks are awarded when following on from a correct method [5]

19. Full marks on this question were rare. Many scored the first method mark for combining the numerator of the left hand side correctly but this was often followed by a failure to expand correctly, for example x – 2 + 5x2 + 5 was common. Many gained a method mark for getting the right hand side as 3(x + 1)(x – 2) but this was then expanded incorrectly. If they got this far only a few knew to rearrange into a quadratic of the form ax2 + bx + c = 0. At this stage earlier errors meant that this was difficult, if not impossible, to solve. This question inevitably was given up on at some stage. Common Mistakes – what did the examiners say?

20. Two gas supply companies have different ways of charging for the gas they supply. Alpha gasCO Fixed Charge £9.60 Price per kilowatt hour of gas First 5 kilowatt hours free then £1.30 for every kilowatt hour over 5. Beta gasCO Fixed Charge No fixed charge Price per kilowatt hour of gas £1.50 for every kilowatt hour. Find the number of kilowatt hours after which Alpha gasCo becomes cheaper than Beta gasCo. You might want to use some graph paper. You must show your method clearly. Answer................................... kilowatt hours (Total 4 marks)

21. 9.60 + (x – 5) × 1.30 Alt: M1 for graph of Alpha parcels M1 – Method mark = 1.50x M1 for graph of Beta M1 – Method mark 3.10 = 0.20x A1 - Accuracy marks are awarded when following on from a correct method x = 15.5 A1 answer. Accept 16 but not 15. T&I gets M1 iff taken as far as 15. A1 for both schemes at 15 A1 for both schemes at 16 A1 conclusion A1 - Accuracy marks are awarded when following on from a correct method [4]

22. Common Mistakes – what did the examiners say? This question was intended to give students a choice of methods and as such was considered a ‘using and applying mathematics’ question. Graph paper was supplied in case a graphical method was adopted. Pleasingly, many different methods were seen, although the main method was Trial and Improvement or writing out tables of costs for each kwh. Graphs when used were usually correct. Algebra was rarely seen but a few candidates used a logical method based on the difference in cost of each kwh and the difference in costs of 5 kwh for both schemes. This was essentially the algebraic method without letters. The main errors were careless arithmetic. About half of candidates scored full marks.

23. Make x the subject of the formula (Total 4 marks)

24. y(x – 3) = 3x + 4 M1 for cross-multiplying and expanding bracket M1 – Method mark yx – 3y = 3x + 4 A1 correct expansion A1 - Accuracy marks are awarded when following on from a correct method yx – 3x = 3y + 4 M1 – Method mark x(y – 3) = 3y + 4 M1 for clollecting terms and factorising A1 Accuracy marks are awarded when following on from a correct method x = (3y + 4)/(y – 3) A1 correct factorisation and division [4]

25. There was some good algebra demonstrated by candidates but many still have no idea how to start this Type of problem. There were many first lines of xy – 3 = 3x + 4. Candidates who managed the first step did not know to collect terms and factorise. Few candidates scored full marks. Common Mistakes – what did the examiners say?

26. A special packet of breakfast cereal contains 20% more than a normal packet. The special packet contains 600 g of cereal. How much cereal does the normal packet contain? Total 3 marks

27. 120% → 600 1.2 B1 – mark awarded independent of method 600 ÷ 120 × 100 600 ÷ 1.2 M1 – method mark 500 A1 - Accuracy marks are awarded when following on from a correct method [3]

28. Common Mistakes – what did the examiners say? Correct answers to this question were infrequent. Predictable incorrect responses were £480 or £720 obtained by subtracting or adding 20% of £600. Candidates who appreciated the link between 120% and £600 could not always use this fact to make progress and those who could and showed the full method sometimes made errors in their calculation. Use of the decimal multiplier, 1.2, was not seen. Some candidates arrived at £500 by an inspection method.

29. Solve the equation. 4 Marks.

30. 3(3x + 1) –2 (2x + 5) Could have 6 as denominator here Condone lack of brackets M1 – Method mark 9x + 3 – 4x – 10 A1 - Accuracy marks are awarded when following on from a correct method (their 5x – 7) = 6 M1 – Method mark x = 2.6 or A1 - Accuracy marks are awarded when following on from a correct method [4]

31. This question should not have been attempted by trial and improvement and those who did so were usually unsuccessful. Division by 2 and by 3 at the outset was a valid method but few who tried this succeeded in following through the resultant fractions correctly. Attempts at 3(3x + 1) – 2(2x + 5) = 6 almost always gained some marks, but there was frequent failure to achieve 9x + 3 – 4x – 10 at the next stage. Common Mistakes – what did the examiners say?

32. (a) n is a positive integer. (i) Explain why n(n + 1) must be an even number................................................................................................... (1) (ii) Explain why 2n + 1 must be an odd number................................................................................................... (1) (b) Expand and simplify (2n + 1) 2................................................................................................................................................................................................... Answer................................................. (2) (c) Prove that the square of any odd number is always 1 more than a multiple of 8.................................................................................................................................................................................................... (3) (Total 7 marks)

33. (a) (i) Even × odd, so even product or equivalent B1 (ii) 2 × n always even, so 2n + 1 is odd or equivalent B1 (b) 4n 2 + 2n + 2n + 1 3 or 4 terms correct M1 4n 2 + 4n + 1 Must simplify A1 (c) Odd 2 – 1 = (2n + 1) 2 –1 = 4n 2 + 4n = 4n(n + 1) Must factorise B1 = 4 × even Deduce ‘even’ connection B1 = multiple of 8 Concluding statement B1 [7]

34. In part (a) some candidates were confused by positive/negative and odd/even and could not offer any sensible reasoning. Whilst many good answers were seen (usually more for (ii) than (i)) there were also too many ‘partial’ solutions ie. only considering what happened when n was even and not continuing the argument for when n was odd. Such incomplete answers did not score the mark(s). Part (b) was well done by the vast majority of candidates; some made a mistake in one of the terms gaining just 1 of the 2 marks. Part (c) was meant to be a test of candidates’ ability to handle a ‘using and applying’ approach. The question did try to lead them into the algebraic approach that was necessary to gain any marks at all but even the most able candidates did not take the hint. There were some correct solutions but they probably number far less than 1% of the total entry. It is important that future candidates are shown the difference between ‘verify’ and ‘prove’. Common Mistakes – what did the examiners say?