Linear Equations. 196 minutes. 191 marks. Page 1 of 50

Transcription

1 Linear Equations 196 minutes 191 marks Page 1 of 50

2 Q1. The perimeter of this L-shape is 56 cm. Not drawn accurately Set up and solve an equation to work out the value of x. x =... (Total 4 marks) Page 2 of 50

3 Q2. Work out the value of x. Not drawn accurately Answer... degrees (Total 4 marks) Q3. (a) Rearrange the formula to make w the subject of y = 3w + 8 Answer... (2) (b) Solve 5(x + 4) = 3x + 23 x =... (3) (Total 5 marks) Page 3 of 50

4 Q4. (a) Solve Answer x =... (3) (b) Rearrange this formula to make t the subject. s = 3t Answer t =... (2) (Total 5 marks) Page 4 of 50

5 Q5. The length of this rectangular tile is 6 times the width. Not drawn accurately Two tiles are put together to make this shape. Not drawn accurately The perimeter of the new shape is 24 cm. Work out the width of one tile. Answer... cm (Total 3 marks) Page 5 of 50

6 Q6. Here is a number machine. The output is three times the input. Work out the input x. x =... (Total 4 marks) Q7. (a) Solve 5x + 3 = 3(x + 2) Answer x =... (3) (b) 2(x + 16) + 4(x 5) simplifies to a(x + b) Work out the values of a and b Answer a =..., b =... (3) (Total 6 marks) Page 6 of 50

7 Q8. The triangle has lengths (4x 2) cm, (2x + 5) cm and (6x 9) cm. Find the value of x that makes this triangle equilateral. Answer... (Total 4 marks) Q9. (a) Solve the equation Answer x =... (1) Page 7 of 50

8 (b) Solve the equation Answer x =... (3) (Total 4 marks) Q10. (a) Solve the equation = 4 Answer x =... (1) (b) Solve the equation 8w 5 = 3w + 1 Answer w =... (3) (c) Simplify y + 2 y y Answer... (1) Page 8 of 50

9 (d) Factorise 15t + 25 Answer... (1) (e) Factorise z + 8z 2 Answer... (1) (Total 7 marks) Q11. (a) Explain why the sum of the angles in any quadrilateral is 360. (2) (b) A quadrilateral has one right angle. The other angles are 2x, 3x - 12 and x - 6 Not drawn accurately (i) Write down an equation in terms of x. Answer... (1) Page 9 of 50

10 (ii) Solve your equation and find the size of the largest angle in the quadrilateral Answer x =... degrees Largest angle =... degrees (3) (Total 6 marks) Q12. (a) Solve the equation Answer x =... (3) (b) Solve the inequality 3x + 8 < 29 Answer... (2) (Total 5 marks) Page 10 of 50

11 Q13. Solve 5x 9 = 3x + 11 x =... (Total 3 marks) Q14. Solve x =... (Total 4 marks) Q15. (a) Factorise fully 4x 2 6xy Answer... (2) Page 11 of 50

12 (b) Solve w =... (3) (Total 5 marks) Q16. Show that all sides of this quadrilateral could be equal. 7x 19 Not drawn accurately 6(x 2) 4x + 2 3(x + 3) (Total 5 marks) Page 12 of 50

13 Q17. Solve 8(x + 3) = 36 x =... (Total 3 marks) Q18. The diagram shows a square. Not drawn accurately Work out the perimeter of the square. Answer... cm (Total 5 marks) Page 13 of 50

14 Q19. The diagram shows a badge made from two similar pentagons. Not drawn accurately Work out the width of the badge. Answer... cm (Total 5 marks) Q20. (a) Factorise 3x 15 Answer... (1) (b) Multiply out 5(y + 4t 2) Answer... (2) Page 14 of 50

15 (c) Solve 3(w + 2) = 2w 1 w =... (3) (Total 6 marks) Q21. The perimeter of this triangle is 48 cm. Work out the value of x. x =... cm (Total 4 marks) Page 15 of 50

16 Q22. Solve 5x 2 = x + 16 x =... (Total 3 marks) Q23. AB is a straight line. Not drawn accurately Set up and solve an equation to work out the size of the obtuse angle. Answer... degrees (Total 4 marks) Page 16 of 50

17 Q24. Mr and Mrs Bell have twin daughters and a son. Mr Bell is four years older than Mrs Bell. Mrs Bell is three times older than their twin daughters. The twin daughters are seven years older than the son. The sum of the five ages is 150. Let x be the age of the twin daughters. Set up and solve an equation to work out the age of the twin daughters Answer x =... (Total 4 marks) Q25. (a) Multiply out 3(2c 1) Answer... (1) (b) Solve = 10 x =... (1) (c) Solve 3y + 6 = 30 7y y =... (3) (Total 5 marks) Page 17 of 50

18 Q26. Solve 3(x 2) = 5x Answer x =... (Total 3 marks) Q27. Grace wants to hire a taxi from home to the railway station. She normally uses Ace Taxis or Best Cars. Fixed charge Rate per kilometre Ace Taxis Best Cars Here is an advert for a new taxi firm, Cozycabs. Cozycabs No fixed charge 1.70 per kilometre The cost of this journey is the same using Ace Taxis and Best Cars. Let the distance from home to the railway station be x kilometres. Use this information to set up and solve an equation in x. Decide whether it is cheaper for Grace to hire a taxi from Cozycabs for the journey (Total 6 marks) Page 18 of 50

19 Q28. The diagram shows three angles on a straight line. Set up and solve an equation in x to help you work out the size of the smallest angle Answer... degrees (Total 4 marks) Page 19 of 50

20 Q29. The diagram shows a cuboid. The length is (5x + 1) cm. The width is (2x + 3) cm. The height is x cm. The length is 7 cm longer than the width. Work out the volume of the cuboid Answer... cm 3 (Total 5 marks) Q30. Solve 6x 5 = 2x + 13 x =... (Total 3 marks) Page 20 of 50

21 Q31. Solve 4(3x 7) = 20 x =... (Total 3 marks) Q32. The diagram shows a triangle ABC. AB = AC Not drawn accurately Show that the triangle is equilateral. (Total 4 marks) Page 21 of 50

22 Q33. Solve 9x 3 = 4x + 17 x =... (Total 3 marks) Q34. Solve x =... (Total 4 marks) Q35. Solve x =... (Total 4 marks) Page 22 of 50

23 . 3x or 2x seen for missing sides May be on diagram or in working B1 4x + 4x + 2x + 3x + 2x + x (= 56) oe 16x implies B1 their 16x = or or SC2 for or 5.09 or 5.1 SC2 for or 4.3 SC2 for 4 SC applies if method marks not awarded. ft [4] M2. x + 2x oe or states angles in quadrilateral = 360 Attempts to subtract from 360 x + 2x = 360 oe or or 132 seen dep x + 2x = oe or 3x = their 132 or their dep 44 [4] M3. (a) y 8 = 3w = w + Page 23 of 50

24 = w or = w SC1 or Do not ignore further work (b) 5x + 20 B1 5x 3x = or 2x = 3 their 5x 3x = 23 their oe ft [5] M4. (a) 12 x = 15 or 12 x = 5 3 oe 4 = 5 x = their or x = 12 their 15 or or 3 Page 24 of 50

25 (b) 3t = s 4 or = t + oe or or oe SC1 or [5] M5. 6x + 2x + 6x + 2x (=16x) Their 16x = 24 8x = 12 is M2 dep 1.5 (oe) or 9 after 1.5 seen oe SC1 14x = 24 leading to x = 24 / 14 oe Alternative method Guess a value and multiplies correctly by 16 x = 1 gives 16 x = 2 gives 32 Guesses a second value nearer to or brackets the correct answer and multiplies correctly by 16 dep 1.5 or 9 after 1.5 seen oe [3] M6. 3x or 5x 9 5x 9 = 3x oe = x Page 25 of 50

26 5x 3x = 9 or 2x = 9 oe dep on M2 dep 4.5 oe Alternative method 4.5 oe B3 correct trial using 4.5 but 4.5 not explicitly given as answer B2 correct trials using 4 and 5 (11 and 16) B1 any correct trial B4 [4] M7. (a) (5x + 3 =) 3x + 6 5x their 3x = their 6 3 or 2x = 3 oe B1 1.5 oe ft for linear equation if B0 scored ft (b) 2x + 32 or 4x 20 6x + 12 or 6(x + 2) a = 6 and b = 2 Accept ax + ab for ft from their 6x + 12 if earned SC2 a = 6 and b = 12 SC1 a = 6 ft [6] M8. 4x 2 = 2x + 5 or 4x 2 = 6x 9 or 6x 9 = 2x + 5 oe 4x 2 = 2x + 5 = 6x 9 Page 26 of 50

27 4x 2x = or 4x 6x = or 6x 2x = oe (x =) 3.5 oe = 12 and = 12 and = 12 Solving two of the equations and obtaining 3.5 for each solution Must show all sides = 12 or solving all three pairs of equations and getting 3.5 for each. T&I is M0 unless all 3 equations checked to be equal to 12, then award 3 marks out of 4. [4] M9. (a) 42 B1 (b) x + 1 = 6 9x x + 9x = 6 1 oe Allow one error 0.5 SC2 (x =) 1.25 SC1 4x = 5 [4] Page 27 of 50

28 0. (a) 80 (b) 8w 3w or Or better 5w = 6 B1 or 1.2 oe ft (c) y + 2y 2 or y(1 + 2y) Allow y + 2 y 2 B0 fw eg, y + 2y 2 = 3y 2 B1 (d) 5 (3t + 5) Allow 5 (3t + 5) or (3t + 5) 5 Condone missing final bracket B1 (e) z (z + 8) Allow z (z + 8) or (z + 8) z Condone missing final bracket B1 [7] 1. (a) Complete explanation eg, Quadrilateral can be divided into 2 triangles and Use of (n 2) 180 with n = 4 or Using (external angles) = 360 eg, (internal angles + external angles) = (internal angles) = B for partial explanation B0 for only B2 (b) (i) 3x 12 + x 6 + 2x + 90 = 360 or better eg, 6x + 72 = 360 B0 for 3x 12 + x 6 + 2x + 90 = 180 B1 Page 28 of 50

29 (ii) 6x = 288 or 6x = or x = (Their 288) 6 ft for 6x = 108 or 6x = or (Their 108) 6 x = 48 ft for x = (Their x) 12 for 35 x 63 SC1 48 with no working or using T & I SC2 (48 and)132 with no working or using T & I B1 ft [6] 2. (a) 23 2x = x = 3 gets allow one error = 2x 1.6 = 0.4x 4 f.t. if awarded. ft (b) 3x < 21 x < 7 3x = 21 gets iff recovered Must have inequality in answer. Accept.. [5] 3. 5x 3x or Implied by 2x or 20 2x = 20 Page 29 of 50

30 10 ft on one error only ft [3] 4. 2(2x + 3) 4(3x 3) or 4x x + 12 This mark is for the numerator of the LHS. Ignore any denominators. Three terms correct if expanded without brackets seen. 8x + 18 Their 8x + 18 = 16 This mark is for dealing with the denominators of the LHS and the value on the RHS NB 2(2x + 3) 4(3x 3) = 16 is M2 0.25, ¼, 2/8 oe ft on one error only. Do not accept 1/ 4 ft Alternative Method 1 (2x + 3) 2(3x 3) or 2x + 3 6x + 6 This mark is for the numerator of the LHS. Ignore any denominators. Three terms correct if expanded without brackets seen. 4x + 9 Their 4x + 9 = 8 This mark is for dealing with the denominators of the LHS and the value on the RHS NB (2x + 3) 2(3x 3) = 8 is M2 0.25, ¼, 2/8 oe ft on one error only. Do not accept 1/ 4 ft Page 30 of 50

31 Alternative Method 2 Three correct terms for x or x + = 2 or x + = , ¼, 2/8 oe ft on one error only. Do not accept 1/ 4 ft [4] 5. (a) 2x(2x 3y) B1 for correct partial factorisation eg 2(2x 2 3yx) or x(4x 6y) Do not accept further work B2 (b) 2w 1 = 8 4w or = 2 w Do not accept 8w 4 = 8 4w B1 2w + 4w = or + w = 2 + ft their 4 terms (w =) 1.5 oe ft [5] Page 31 of 50

32 6. Setting up a correct equation eg 7x 19 = 4x + 2 or 7x 19 = 6(x 2) B1 Collects their 4 terms eg 7x 4x = x = 7 Verifies that one side is equal to 30 or setting up another correct equation or substitutes their x into any expression and evaluates it correctly ft is only for their x = 7 B1ft Verifies that all sides are equal eg Solves A and B then: calculates 3 sides including C and D Solves A and B and A and C then: calculates 2 sides including D Solves A and B and C and D then: calculates one side of each pair e.g. A and C Solves any three pairs [5] 7. 8x + 24 (= 36) or x + 3 = oe 8x = 36 their 24 or x = their oe ft their equation if exactly one method mark awarded ft [3] Page 32 of 50

33 8. 2x 4 = x + 5 (P =) 2(2x 4) + 2(x + 5) or 6x + 2 oe B1 2x x = x + 2 = 4(x + 5) or 6x + 2 = 4(2x 4) x = 9 or side = 14 (Perimeter =) 4 their 14 Do not ft 4 their x or Strand (iii) Shows x = 9 (and each side is 14 (cm)) and perimeter is 56 (cm) 56 without working implies B1 Q1 [5] 9. oe Scale factor 3 or seen or implied 3x (x + 4) = 36x oe 36 3 (= 12) 3(x + 4) = 36 oe their 12 4 or 3x x = 36x Page 33 of 50

34 3x + 12 = 36 (x =) 8 or x + 4 = 12 or x = 8 or their 8 3 or 3x 2 24x = 0 or 3x 2 = 24x (3x =) [5] M20. (a) 3(x 5) (b) 5y + 20t 10 B1 for 2 correct terms. Penalise any incorrect further working. Eg 5y + 20t 10 = 25yt 10 is B1 5y + 20t 1 = 25yt 1 is B0 (error in expansion and incorrect further work) 5y + 20t 10 = 5(y + 4t 2) given as answer is B1 as shows a misunderstanding of expanding brackets. B1 B2 (c) 3w + 6 = 2w 1 w + 2 = w 3w 2w = 1 6 This mark is for rearranging their expansion correctly to get w terms one side and number terms on the other. w w = 2 (oe) 7 ft on one error ft [6] Page 34 of 50

35 M21. x x + 3x oe 48 9 x x + 3x = 48 oe 48 9 and 6 seen dep 6x = 48 9 or 6x = 39 oe their 39 6 dep 6.5 or SC3 for 13, 19.5 and 15.5 [4] M22. 5x x or 4x or or 18 oe 4x = oe ft their rearrangement with one error if awarded ft [3] M23. x x 23 = 180 oe 2x + 12 = 180 2x = or 2x = 168 (180 12) 2 or 84 Terms collected dep 119 Page 35 of 50

36 x = 84 and Strand (ii) an algebraic equation with both method marks awarded for correct algebra Q1 [4] M24. 3x + 4 (+) 3x (+) x (+) x (+) x 7 (= 150) oe 4 or 5 correct terms 3x x + x + x + x 7 = 150 oe ft their terms 9x 3 = 150 or 9x = oe ft their equation dep ft x = 17 SC3 for solution by trial and improvement [4] M25. (a) 6c 3 Mark final answer B1 (b) 200 B1 (c) 3y + 7y = 10y or 30 6 = 24 Allow rearrangement to get y terms on right ( 10y if correct) 10y = oe ft if M awarded and at most one error ft [5] Page 36 of 50

37 M26. 3x 6 (= 5x+ 8) 6 8 = 5x 3x or 3x 5x= ft for maximum of 2 marks if there is only one error this might be an error in the expansion or a sign error in rearranging the terms eg 3x 2 = (5x + 8) or 3x 2 = (5x + 8) 2 8 = 5x 3x 3x 5x = = 2x 2x = 10 x = 5 x = 5 Scores M0 ft must see 2nd or 3rd lines of working for 3x 6 = (5x + 8) or 3x 6 = (5x + 8) = 5x 3x 3x 5x = = 2x 2x = 2 x = 1 x = 1 Scores M0 ft must see 2nd or 3rd lines of working to enable ft 3x 6 = (5x + 8) or 3x 6 = (5x + 8) 6 8 = 5x + 3x 3x + 5x = = 8x 8x = 14 x = oe x = oe ft Scores M0 ft must see 2nd or 3rd lines of working to enable ft SC2 for x = 7 from 2x = 14 seen [3] M x or 4(.0) + 1.4x or x or x oe (an extra) ( )1.80 or 180p or 20p (per kilometre) seen x = 4(.0) + 1.4x or x = x or 1.6x 1.4x = 4(.0) 2.2 or 160x 140x = oe allow one error or 180p is equivalent to 20p per kilometre oe dep Page 37 of 50

38 (x =) 9 Journey is 9 kilometres their 9 or their 9 or 1.70 their 9 dep on second dep ( )16.6(0) and ( )15.3(0) ft their 9 Correct conclusion from their working with all steps shown Strand (iii) eg yes, it is cheaper ft Q1 [6] M28. 2x + 3x + 4x = 180 9x = 180 or x = ( ) or 180 seen and one trial worked out correctly eg = ( 2) or a different trial worked out correctly dep 40 Steps in setting up and solving equation clearly shown Strand (ii) Dependent on both method marks scored from an algebraic method Q1 [4] Page 38 of 50

39 M29. 5x + 1 = 2x oe 5x 2x = x = 9 or x = 3 oe Collecting terms from their linear equation using 5x + 1 and 2x + 3 B1 ft (5 their 3 + 1) (2 their 3 + 3) their 3 their 16 their 9 their 3 Their 3 must be positive to ft 432 Using x(10x 2 + 2x + 15x + 3) i.e. their (3 ( )) or their [5] M30. 6x 2x (= 4x) or (= 18) 4x = , etc. ft on one error incorrect cancelling after a correct fraction seen is not penalised ft [3] M31. 12x 28 (= 20) 3x 7 = x = x = x = + 7 This mark is for separating terms in their equation Page 39 of 50

40 4 ft if M0 or M0 ft [3] M32. Equates two sides 5w = 3w + 3 3w + 3 = w + 6 5w = w + 6 Collects like terms 5w 3w = 3 3w w = 6 3 5w w = 6 (w =) 1.5 dep Works out that all sides are 7.5 or solves another pair to get (w =) 1.5 Must have 3 rd side = 7.5 and one side using their equation = 7.5 as a minimum [4] M33. 9x 4x or x = 20 4 ft On one error ft [3] Page 40 of 50

41 M34. 3(10 x) or 30 3x Do not accept x = 3(10 x) Do not accept x = 30 3x x = 30 3x = 10 x dep 5x + 3x = Collecting their 4 terms (2 stages) oe = or dep on 3 rd ft [4] M35. 3(10 x) or 30 3x Do not accept x = 3(10 x) Do not accept x = 30 3x x = 30 3x = 10 x dep Page 41 of 50

42 5x + 3x = Collecting their 4 terms (2 stages) oe = or dep on 3 rd ft [4] Page 42 of 50

43 E1. Those students who used an algebraic approach rather than trial and improvement were generally more successful. Many students did not express the missing sides algebraically with 11x = 56 a common error. E2. Foundation Tier Many good attempts were seen to this question. A common error was to work out instead of Very few students set up an equation and some students used an incorrect value for the sum of the interior angles of a quadrilateral, usually 180, 280, 450 or 540. Some good signs of checking answers were seen on many scripts. Higher Tier Common errors were to divide 132 by 2 or 4. Some students assumed that opposite angles added up to 180, giving an answer of 60 for x. E3. Foundation Tier This question was generally found to be the most challenging on the paper. Responses to part (a), in particular, were poor. Part (a) had the highest proportion of students making no attempt with only a small minority making any progress. In part (b) many were able to expand the bracket but approximately half of those students made no further progress. A common error was to give 8x = 43. Higher Tier A majority gave the correct answer in part (a), but students continue to have difficulty with this topic. Part (b) was generally well answered. A small proportion of students collected terms incorrectly often leading to 8x = 43. E4. A few students attempted an algebraic rearrangement of the equation in part (a) and, of those that did so correctly, many could not progress beyond 12 x = 15. Common incorrect answers were 3, 4, 7 and 27. Some students attempted to use a flow chart method but could not set up the flow chart correctly nor do the correct calculations with the reverse operations. In part (b) many students simply swopped the letters s and t, so that t = 3s + 4 was a very common incorrect answer. Some students substituted various values for s and t to obtain a numerical answer. Page 43 of 50

44 E5. Foundation Tier This question was very poorly answered. Only a small number of students managed to give a completely correct solution. The vast majority were unable to access the question and make any progress. Many interpreted the length as 6 cm or as 6 more than x. Some students included the middle line in the diagram as part of the perimeter. Very few algebraic attempts were seen, with most students using trial and error. Higher Tier This question was well answered. There was some confusion with width, which was often given as the horizontal distance of 9 cm. The main errors were misinterpreting which diagram had a perimeter of 24 or, less often, taking the area as 24 cm². There were few trial and improvement methods seen with most students using an algebraic approach. E6. Foundation Tier Few students attempted an efficient algebraic method. Most managed just one correct trial. Some gave a trial for both 4 and 5 but did not go on to trial 4.5. Higher Tier Whilst there were a number of elegant algebraic solutions, the majority used a trial and improvement method to attempt to solve this problem. Many seemed unsure about how to test the 'three times' element or were doing it in their heads. Those who did use algebra usually had more success. E7. A few candidates managed to expand the brackets in part (a) and 3x + 6 was often seen. The most common error was to state that 5x + 3 = 8x. Other frequent errors were to eliminate the 3s to obtain 5x = + 2 or subtract the 2 to obtain 5x + 1 = 3x. There were hardly any correct answers in part (b) with many making no attempt. A response of a = 6 with b = 11 was common, which came from 2x + 4x = 6x and 16 5 = 11. Very few candidates progressed to 6x + 12, but then usually gave a = 6 and b = 12 as the answer. Of those candidates who expanded the brackets correctly, most tried to solve the equation 2x + 32 = 4x 20. E8. A majority found 3.5 but the most common method was trial and improvement. Providing that the value was checked (which by the nature of the method it generally was), this was awarded 3 of the 4 marks. Normally, trial and improvement is regarded as an acceptable method in problem solving but it was decided to only award a maximum of 3 as in this case there is an obvious algebraic approach. Solving any pair of equations to get x = 3.5 and then checking takes a few lines and little time, whereas trial and improvement tends to ramble all over the page and judging by the copious quantities of calculations in some cases must have taken considerable time. Candidates must learn to use an algebraic approach in problem solving. Algebra is still a weak area as candidates find it an abstract concept. If candidates can get success in using algebra in problem solving it will give it some relevance. Several candidates wrote 4x 2 = 2x + 5 = 6x 9 then added, for example, 2 to each side to obtain 4x = 2x + 7 = 6x 7. In some cases this was recovered but many then got lost, not knowing that only one pair was needed for a solution. Page 44 of 50

45 E9. The majority of candidates obtained the correct solution to the equation in part (a). About 60% of candidates failed to make any progress in solving this equation in part (b), with over 10% not making any attempt. Some made a mistake with rearrangement after obtaining x + 1 = 6 9x. Others scored up to two marks after making an error on their first step and attempting to solve x + 1 = 6 3x. Additional Examiner s Commentary Most of the remaining candidates in part (b) obtained a fully correct solution, typically proceeding as follows: Cross multiplying to get x + 1 = 3(2 3x) = 6 9x Isolating the x terms x + 9x = x = 5 Dividing through by 10 x = 5 10 = 0.5 E10. The majority of candidates answered part (a) of this question correctly although many gave their solution as 5. Nearly seventy percent obtained the correct solution, in part (b), nearly always from an algebraic approach although some used Trial and Improvement. Many of those who attempted a valid algebraic approach made slips usually with signs during the initial rearrangement. Obtaining 1.2 from 5w = 6 defeated some with 5/6 occurring fairly frequently. Nearly fifteen percent of candidates scored zero not encouraging for what is very nearly a banker topic. Only about a third of the candidates managed to obtain the mark for part (c) and not all of these obtained the simplest answer as y + 2 y 2 was also allowed. Clearly applying the correct hierarchy of operations to an algebraic expression is not well understood. A fairly common incorrect response was 2y 3. Candidates at Higher tier should expect routine questions similar to those in part (d) and about three-quarters of the candidates managed to answer it correctly. However, some gave their answer as 3t + 5 suggesting an incomplete understanding of the concept involved. Many candidates clearly did not appreciate the meaning of the term factorise. Part (e) saw a similar level of performance to that in part (c). Here incorrect responses included trying to factorise into two brackets and the completely misconceived 8z 3. Additional Examiner s Commentary (b) There was a mark here for combining 8b and -2b to obtain 6b There was a mark here for combining 3 and +7 to obtain 10 Candidates who are challenged to combine algebraic expressions should be encouraged to at least attempt the numbers, as half marks can be achieved without attempting any algebra. Page 45 of 50

46 E11. Intermediate Tier Part (a) was not well done with only about 20% of candidates gaining any marks. Most of these described the splitting into triangles method but some successfully remembered 180(n 2). There were many incorrect explanations with reference to angles on a straight line, opposite angles adding up to 180 and = 360 0, being fairly popular. Part (b)(i) was very badly done with many candidates confusing an equation with an expression. Only about 20% of candidates gained any marks for part (b)(ii), with less than half of these able to give a fully correct answer. The rare candidate who managed to start with the correct equation was often let down by poor rearrangement skills, with, for example, 6x 18 = 270often leading to 6x = 252. Most candidates who scored marks did so by division by 6 and/or following through accurately from their answer for x to the largest angle. Higher Tier In part (a) the descriptions were often incorrect, a common misconception being that the opposite angles of a quadrilateral add up to 180 or that a quadrilateral was simply a square that had been stretched a bit so the four 90 angles became a little bit less or more but it all balanced out in the end. The two triangles method was the most successful but many correctly quoted (n 2) and then substituted n = 4. Those candidates who tried to argue the exterior angle case often came to grief. 62% of candidates scored zero on this part. Part (b) was also disappointing. In (b)(i) an equation was required but many candidates only gave an expression. In part (b)(ii) the working was often very unclear and this led to errors in arithmetic, which might otherwise have been avoided. Credit was given for correct method and also follow through was allowed for the largest angle, but only for realistic answers (i.e., it had to be an obtuse angle). Only 43% of candidates scored all 3 marks in part (b)(ii). E12. Higher Tier Part (a) was well done by the majority, but many candidates are unable to deal with firstly the division by 5 and secondly the negative value for x. Answers of ±19 and 4 were common. Part (b) was very badly done. Too many candidates replace the inequality with an equals sign then fail to recover this in the answer. Others misinterpreted the question and gave answers of 6, 5, 4,. presumably assuming x is an integer. Many candidates introduced the sign. This was condoned this year but may not be in future. Intermediate Tier In (a) the correct answer was seen quite often but it seems to have been found mainly by trial and improvement. A correct algebraic solution was rarely seen. Of the candidates who tried to multiply both sides by 5, many made errors [usually sign errors] in the manipulation of the x term and/or the numbers Page 46 of 50

47 Part (b) was not well answered for a straightforward inequality. Many decided to convert to an equation and solved 3x + 8 = 29 to give an answer of x = 7 and consequently scored zero. Others just gave a list of integers which they thought fitted the inequality. E13. This question was well answered. The most common errors were in rearranging or poor arithmetic. E14. This question was not well answered. Combining the left hand side, either using the rules of algebraic fractions or finding a common denominator, was not carried out in the majority of cases. Many students who did successfully manage this did not then find the right hand side by cross multiplying by the appropriate denominator. The next major problem was in the expansion of the brackets with 12 and 6 commonly seen. E15. Responses to part (a) were very mixed with 2x(2x 3xy) a common incorrect answer. Many students did not attempt part (b). A large proportion of those who attempted this part either made an error in multiplying through by 4, or attempted to take terms in or out of the fraction incorrectly. A significant number attempted to use trial and error, often unsuccessfully. E16. Many students did not attempt this problem solving question and only a small minority gave a complete solution. The most common error was to add the sides instead of setting up an equation to solve. The complete solutions seen were often well presented and rigorous. E17. This question was well answered. It was common for students to arrive at and then incorrectly attempt to simplify or convert this to simplest form or decimals. It was rare to see division by 8 as the first step attempted. E18. Responses to this question were often poor, with many students working out an expression for the perimeter and equating this to zero. A significant minority obtained x = 9 but either made no further progress or went on to give 36 as the perimeter. Some students attempted to work out the area. E19. On this question students rarely scored part marks, giving fully correct or fully incorrect working, although a majority were successful. However a significant number of students gave x = 8 as the width of the badge. Examples of poor algebra were x + 4 = 4x and 3(x + 4) = 3x + 4. Many of the attempts by students at this question were unclear. E20. All parts of this question were well answered. Rearrangement errors were quite common in part (c). E21. This question was generally well answered with many giving an equation to solve and others using trial and improvement. Page 47 of 50

48 E22. The majority of students correctly manipulated the equation although some made errors. Poor arithmetic was the cause of some incorrect answers. E23. Most students were able to either obtain x = 84 or that the obtuse angle was 119 but many did not set up and solve an equation as required by the question. Some started with x + 35 = x 23. E24. There were a number of correct answers from using algebra and slightly more from using a trial and improvement method. However, there were also many unsuccessful attempts using trial and improvement usually due to arithmetical errors. The most common errors using algebra were omitting the twins or just using one of the twins. Some correctly found 55, 51, 17 and 10 by trial and improvement but when checking the total just added these four, forgetting the other twin and hence not realising they were correct. There were many algebraic expressions with mixed letters for the family members. The use of x + 7 rather than x 7 was a common error. E25. All parts were well answered. Part (c) had the most errors with rearrangement errors or poor arithmetic. E26. This question was less well answered than many similar ones on previous examinations. There were errors multiplying out the bracket, 3x 2 and 3x 5 being the most common ones. Few students successfully collected the terms on one side and the numbers on the other side. Many of those who reached 2x = 14 followed it with x = 7. E27. This question was generally poorly answered. Most students only went as far as writing one correct expression. Only able students formed an equation by equating the two expressions x and x. Others just tried experimenting with multiples of 1.60 or 1.40 or calculated and and then stopped. Although the question stated set up and solve an equation in x, many students did not form an equation but worked out the journey distance of 9km using multiples of 1.60 and 1.40, taking into account the fixed charges of 2.20 and 4. Only a small number of students compared the three costs and reached a valid conclusion. E28. This question was a good starter for almost all students. The question assessed quality of written communication and required students to follow an algebraic method. Although the majority of students complied, some used a numerical approach. A common error was to work out the value of x but not state the size of the smallest angle. Page 48 of 50

49 E29. Approximately half of the students gave a fully correct solution. Common errors in the linear equation included 5x + 1 2x + 3 = 7 or 5x + 1 = 7(2x + 3). Some students equated one of the sides to 7 giving 2x + 3 = 7 or 5x + 1 = 7. Students who obtained a value for usually continued to a correct volume although was quite common. E30. The majority of students knew to rearrange the equation but almost half made arithmetic or sign errors. Students often gained credit for continuing after their incorrect first step. However, division by 4 or the division of 18 by their coefficient of x was too challenging for many students. E31. This was very well answered by most students. The most common method used was to multiply out the bracket first. Setting out for this question was usually very clear. E32. Very few part marks were scored on this question as generally those students who set up an equation went on to solve it correctly and check that all three sides were equal. E33. This question was well answered. The main errors were in rearranging the variables or poor arithmetic. E34. A majority of students did not score marks on this question. Common errors included mistakes when multiplying through by 3, for example 30 x, transposition errors when collecting like terms on each side and conceptual errors, for example = 10 and x = 23. A significant proportion of students used trial and improvement but were often unsuccessful. E35. A majority of students did not score marks on this question. Common errors included mistakes when multiplying through by 3, for example 30 x, transposition errors when collecting like terms on each side and conceptual errors, for example = 10 and x = 23. A significant proportion of students used trial and improvement but were often unsuccessful. Page 49 of 50

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