Cambridge A Level Mathematics - Further 9231 — 2013 May/June Paper 2 · Variant 3

*3132110842* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/23 Paper 2 May/June 2013 3 hours Additional Materials: Answer Booklet/Paper Graph Paper List of Formulae (MF10) READ THESE INSTRUCTIONS FIRST If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet. Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. Where a numerical value is necessary, take the acceleration due to gravity to be 10 m s−2. The use of a calculator is expected, where appropriate. Results obtained solely from a graphic calculator, without supporting working or reasoning, will not receive credit. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 5 printed pages and 3 blank pages. JC13 06_9231_23/RP © UCLES 2013 [Turn over

2 1 A bullet of mass m kg is fired into a fixed vertical barrier. It enters the barrier horizontally with speed 280 m s−1 and emerges horizontally after 0.01 s with speed 30 m s−1. There is a constant horizontal resisting force of magnitude 1500 N. Find m. [4] 2 1 8a a A O B C P Q q A particle P travels on a smooth surface whose vertical cross-section is in the form of two arcs of circles. The first arc AB is a quarter of a circle of radius 1 8a and centre O. The second arc BC is a quarter of a circle of radius a and centre Q. The two arcs are smoothly joined at B. The point Q is vertically below O and the two arcs are in the same vertical plane. The particle P is projected vertically downwards from A with speed u. When P is on the arc BC, angle BQP is 1 (see diagram). Given that P loses contact with the surface when cos 1 = 5 6, find u in terms of a and g. [8] 3 Two uniform small smooth spheres A and B, of masses m and 2m respectively, and with equal radii, are at rest on a smooth horizontal surface. Sphere A is projected directly towards B with speed u, and collides with B. After this collision, sphere B collides directly with a fixed smooth vertical barrier. The total kinetic energy of the spheres after this second collision is equal to one-ninth of its value before the first collision. Given that the coefficient of restitution between B and the barrier is 0.5, find the coefficient of restitution between A and B. [9] 4 A particle P of mass m moves along part of a horizontal straight line AB. The mid-point of AB is O, and AB = 4a. At time t, AP = 2a + x. The particle P is acted on by two horizontal forces. One force has magnitude mg @2a + x 2a A−1 2 and acts towards A; the other force has magnitude mg @2a −x 2a A and acts towards B. Show that, provided x a remains small, P moves in approximate simple harmonic motion with centre O, and state the period of this motion. [5] At time t = 0, P is released from rest at the point where x = 1 20a. Show that the speed of P when x = 1 40a is 1 80 3ag, and find the value of t when P reaches this point for the first time. [4] © UCLES 2013 9231/23/M/J/13

3 5 A B C D E F G H O 4a 2a a 2a ABCD is a uniform rectangular lamina of mass m in which AB = 4a and BC = 2a. The lines AC and BD intersect at O. The mid-points of OA, OB, OC, OD are E, F, G, H respectively. The rectangle EFGH, in which EF = 2a and FG = a, is removed from ABCD (see diagram). The resulting lamina is free to rotate in a vertical plane about a fixed smooth horizontal axis through A and perpendicular to the plane of ABCD. Show that the moment of inertia of this lamina about the axis is 85 16ma2. [8] The lamina hangs in equilibrium under gravity with C vertically below A. The point C is now given a speed u. Given that the lamina performs complete revolutions, show that u2 > 192ï5 17 ag. 5 6 Six pairs of values of variables x and y are measured. Draw a sketch of a possible scatter diagram of the data for each of the following cases: (i) the product moment correlation coefficient is approximately zero; (ii) the product moment correlation coefficient is exactly −1. [2] On your diagram for part (i), sketch the regression line of y on x and the regression line of x on y, labelling each line. [2] On your diagram for part (ii), sketch the regression line of y on x and state its relationship to the regression line of x on y. [2] 7 Each of a random sample of 6 cyclists from a cycling club is timed over two different 10 km courses. Their times, in minutes, are recorded in the following table. Cyclist A B C D E F Course 1 18.5 17.8 19.2 22.3 16.5 20.0 Course 2 20.2 20.4 18.1 20.6 18.5 20.5 Assuming that differences in time over the two courses are normally distributed, test at the 10% significance level whether the mean times over the two courses are different. [8] © UCLES 2013 9231/23/M/J/13 [Turn over

4 8 The number, x, of a certain type of sea shell was counted at 60 randomly chosen sites, each one metre square, along the coastline in country A. The number, y, of the same type of shell was counted at 50 randomly chosen sites, each one metre square, along the coastline in country B. The results are summarised as follows. Σ x = 1752 Σ x2 = 55 500 Σ y = 1220 Σ y2 = 33 500 Find a 95% confidence interval for the difference between the mean number of sea shells, per square metre, on the coastlines in country A and in country B. [9] 9 A researcher records a random sample of n pairs of values of x, y, giving the following summarised data. Σ x = 24 Σ x2 = 160 Σ y = 34 Σ y2 = 324 Σ xy = 192 The gradient of the regression line of y on x is −3 4. Find (i) the value of n, [3] (ii) the equation of the regression line of x on y in the form x = Ay + B, where A and B are constants to be determined, [2] (iii) the product moment correlation coefficient. [2] Another researcher records the same data in the form x′, y′, where x′ = x k, y′ = y k and k is a constant. Without further calculation, state the equation of the regression line of x′ on y′. [2] 10 Jill and Kate are playing a game as a practice for a penalty shoot-out. They take alternate turns at kicking a football at a goal. The probability that Jill will score a goal with any kick is 1 3, independently of previous outcomes. The probability that Kate will score a goal with any kick is 1 4, independently of previous outcomes. Jill begins the game. If Jill is the first to score, then Kate is allowed one more kick. If Kate scores with this kick, then the game is a draw, but if she does not score then Jill wins the game. If Kate is the first to score, then she wins the game, and no further kicks are taken. (i) Show that the probability that Jill scores on her 5th kick is 1 48. [2] (ii) Show that the probability that Kate wins the game on her nth kick is 1 3 × 2n . [3] (iii) Find the probability that Jill wins the game. [3] (iv) Find the probability that the game is a draw. [3] © UCLES 2013 9231/23/M/J/13

5 11 Answer only one of the following two alternatives. EITHER A uniform rod AB rests in limiting equilibrium in a vertical plane with the end A on rough horizontal ground and the end B against a rough vertical wall that is perpendicular to the plane of the rod. The angle between the rod and the ground is 1. The coefficient of friction between the rod and the wall is -, and the coefficient of friction between the rod and the ground is 2-. Show that tan 1 = 1 −2-2 4- . [10] Given that 1 ≤45Å, find the set of possible values of -. [4] OR A researcher is investigating the relationship between the political allegiance of university students and their childhood environment. He chooses a random sample of 100 students and finds that 60 have political allegiance to the Alliance party. He also classifies their childhood environment as rural or urban, and finds that 45 had a rural childhood. The researcher carries out a test, at the 10% significance level, on this data and finds that political allegiance is independent of childhood environment. Given that A is the number of students in the sample who both support the Alliance party and have a rural childhood, find the greatest and least possible values of A. [10] A second random sample of size 100N, where N is an integer, is taken from the university student population. It is found that the proportions supporting the Alliance party from urban and rural childhoods are the same as in the first sample. Given that the value of A in the first sample was 29, find the greatest possible value of N that would lead to the same conclusion (that political allegiance is independent of childhood environment) from a test, at the 10% significance level, on this second set of data. [4] © UCLES 2013 9231/23/M/J/13

8 BLANK PAGE Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. 9231/23/M/J/13

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2013 series 9231 FURTHER MATHEMATICS 9231/23 Paper 2, maximum raw mark 100 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9231 23 © Cambridge International Examinations 2013 Mark Scheme Notes Marks are of the following three types: M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). B Mark for a correct result or statement independent of method marks. • When a part of a question has two or more "method" steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. • The symbol √ implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously "correct" answers or results obtained from incorrect working. • Note: B2 or A2 means that the candidate can earn 2 or 0. B2/1/0 means that the candidate can earn anything from 0 to 2. The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. • Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. • For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f., or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.81 instead of 10.

Page 3 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9231 23 © Cambridge International Examinations 2013 The following abbreviations may be used in a mark scheme or used on the scripts: AEF Any Equivalent Form (of answer is equally acceptable) AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) CAO Correct Answer Only (emphasising that no "follow through" from a previous error is allowed) CWO Correct Working Only – often written by a “fortuitous” answer ISW Ignore Subsequent Working MR Misread PA Premature Approximation (resulting in basically correct work that is insufficiently accurate) SOS See Other Solution (the candidate makes a better attempt at the same question) SR Special Ruling (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the light of a particular circumstance) Penalties MR –1 A penalty of MR –1 is deducted from A or B marks when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become "follow through " marks. MR is not applied when the candidate misreads his own figures – this is regarded as an error in accuracy. An MR–2 penalty may be applied in particular cases if agreed at the coordination meeting. PA –1 This is deducted from A or B marks in the case of premature approximation. The PA –1 penalty is usually discussed at the meeting.

Page 4 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9231 23 © Cambridge International Examinations 2013 Question Number Mark Scheme Details Part Mark Total 1 EITHER: Use I = Ft to find impulse I: Use I = m(v1 – v2) to find mass m: OR: Use v2 = v1 – at to find deceleration a: Use F = ma to find mass m: OR: Use s = ½ (v1 + v2)t to find distance s: Use Fs = ½ m(v1 2 – v2 2) to find mass m: S.R Taking v1 – v2 = 280 + 30 in any method: I = 1500 × 0.01 [= 15] M1 A1 m = I/250 = 0.06 M1 A1 a = 250/0⋅01 [=250000] (M1 A1) m = 1500/a = 0.06 (M1 A1) s = ½ × 310 × 0⋅01 [= 1⋅55] (M1 A1) m = 2 × 1500s/77500 = 0⋅06 (M1 A1) m = I/310 = 0⋅048[3] (max 2/4) (M1 A1) 4 4 2 State or imply reaction R is zero when contact lost M1 Use F = ma radially when contact lost: Use cos θ = 5/6 to find v2: Use conservation of energy at θ: If found by v2 = u2 + 2gh lose this A1 only) Combine to find u: mv2/a = mg cos θ [– R] M1 v2 = ag cos θ = 5ag/6 A1 ½mv2 – ½mu2 = mg{a/8 + a(1 – cos θ)} M1 A1 = 7mag/24 (A.E.F.) A1 u2 = 5ag/6 – 2 × 7ag/24 M1 u = √(¼ ag) or ½ √(ag) A1 8 8 3 Use conservation of momentum, e.g.: Use restitution (must be consistent with prev. eqn.): Find speed of B after striking barrier (ignore sign): Relate K.E. before and after collision: EITHER: Solve first two eqns for vA and vB (A.E.F): Substitute for vA, vB′ in KE eqn: Simplify and solve for e: OR: Use vA + 2vB = u in KE eqn to give e.g.: Solve for vA and vB: Find e from restitution eqn: mvA + 2mvB = mu B1 vA – vB = – eu B1 vB′ = ½ vB M1 (½mu2)/9 = ½mvA 2 + ½(2m)vB′2 M1 vA = ⅓(1 – 2e)u, vB = ⅓(1 + e)u M1 A1 u2/9 = (1 – 2e)2u2 /9 + ½(1 + e)2 u2 /9 A1 9e2 – 6e + 1 = 0, e = ⅓ M1 A1 81vA 2 – 18uvA + u2 = 0 or 81vB 2 – 72uvB + 16u2 = 0 (M1 A1) vA = u/9 and vB = 4u/9 (A1) e = (4u/9 – u/9)/u = ⅓ (M1 A1) 9 9

Page 5 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9231 23 © Cambridge International Examinations 2013 4 Find eqn. of motion at general point: (wrong sign loses this A1 only) Neglect (x/a)2 and higher powers: Find period from T = 2π/ω: m d2x/dt2 = mg(2a – x)/2a – mg{(2a + x)/2a}–½ M1 A1 d2x/dt2 = g(1 – x/2a) – g(1 – x/4a) = – gx/4a M1 A1 T = 2π√(4a/g) or 4π√(a/g) B1 5 Find v from v2 = ω2 (A2 – x2): Find t from x = a cos ωt (A.E.F.): v2 = (g/4a) {(a/20)2 – (a/40)2} M1 v = (1/80)√(3ag) A.G. A1 t = √(4a/g) cos–1 ½ = ⅔π√(a/g) M1 A1 4 9 5 Find MI of ABCD about O: Find MI of EFGH about O: EITHER: Find MI of final lamina about O: Find MI of final lamina about A: OR: Find MI of ABCD about A: Find MI of EFGH about A: Find MI of final lamina about A: IABCD = ⅓m{(2a)2 + a2} = 5ma2/3 B1 IEFGH = ⅓ (¼m){a2 + (½ a)2} = 5ma2/48 M1 A1 IO = IABCD – IEFGH ′ = 25ma2/16 M1 A1 IA = IO + ¾ m × 5a2 M1 A1 = 85ma2 /16 A.G A1 I′ABCD = IABCD + 5ma2 = 20ma2/3 (M1 A1) I′EFGH = IEFGH + ¼m × 5a2 = 65ma2/48 (M1 A1) IA = IEFGH – I′EFGH = 85ma2 /16 A.G (B1) 8 State or use u = (√20) aω Use energy when C above A to find ωmin: Hence find umin 2: B1 ½IA ωmin 2 = ¾ mg × (√20) a M1 A1 umin 2 = 20 (32/85)(¾√20) ag M1 = (192√5/17) ag A.G. A1 5 13 6 (i) (ii) Sketch 6 randomly scattered points (lose B1 here if not 6 points) B1 Sketch 6 (or more) points as if on negative gradient line B1 Show on (i) line labelled y on x between points and approx. horizontal B1 Show on (i) line labelled x on y between points and approx. vertical B1 Show on (ii) line labelled y on x passing through points B1 State or show on (ii) that x on y coincides with y on x B1 2 2 2 6

Page 6 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9231 23 © Cambridge International Examinations 2013 7 Consider differences e.g. course 2 – course 1: Calculate sample mean: Estimate population variance: (allow biased here: 23/9 or 2⋅556 or 1⋅5992) State hypotheses (A.E.F.), e.g.: Calculate value of t (to 3 s.f.): State or use correct tabular t value: (or can compare d with 1.44) Correct conclusion (AEF, on t, dep *B1): 1⋅7 2⋅6 –1⋅1 –1⋅7 2⋅0 0⋅5 M1 d = 4 / 6 = 0⋅6667 M1 s2 = (18 – 42/6) / 5 = 46/15 or 3⋅067 or 1⋅7512 M1 H0: µ1 – µ2 = 0, H1: µ1 – µ2 ≠ 0 B1 t = d/(s/√6) = 0⋅932[5] M1 A1 t5, 0.95 = 2⋅01[5] *B1 No difference between mean times B1 8 8 8 Find both sample means: Estimate both popn. variances (to 4 s.f.) (allow biased here: 72⋅36 and 74⋅64) EITHER: Estimate combined variance (3 s.f.): Use this s to find conf. interval: Evaluate: OR: Estimate common variance (to 3 s.f.): (note sx and sy not needed explicitly) Use this s to find conf. interval: Evaluate: x = 29⋅2, y = 24⋅4 B1 sx 2 = (55500 – 17522/60) / 59 and sy 2 = (33500 – 12202/50) / 49 M1 A1 sx 2 = 73⋅59 and sy 2 = 76⋅16 A1 s2 = sx 2/60 + sy 2/50 M1 = 2⋅750 or 1⋅6582 A1 ( x –y) ± 1⋅96 s M1 A1 4⋅8 ± 3⋅25 or [1⋅55, 8⋅05] A1 s2 = (59 sx 2 + 49 sy 2) / 108 or (55500 – 17522/60 + 33500 – 12202/50) / 108 = 74⋅8 or 8⋅652 or 3364/45 (M1 A1) ( x –y) ± 1⋅96 s √(1/60 + 1/50) (M1 A1) 4⋅8 ± 3⋅24[5] or [1⋅55[5], 8⋅04[5]] (A1) 9 9 9 (i) Find n using gradient: – ¾ = (192 – 24 × 34/n) /(160 – 242/n) M1 A1 4992/n = 1248, n = 4 A1 3 (ii) Find regression line x = Ay + B: M0 for y – 34/4 = –¾ ( x – 24/4) x – 24/4 = (–12/35) ( y – 34/4) M1 x = – 12y/35 + 312/35 or – 0⋅343y + 8⋅91 A1 2 (iii) Find coefficient r (ignore sign for M1): State regression line (B1 for each term on RHS): r = – 12/√(16 × 35) or – √(¾ × 12/35) M1 = –0⋅507 or –3√35/35 or –3/√35 A1 2 9

Page 7 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9231 23 © Cambridge International Examinations 2013 10 (i) Find prob. of Jill scoring on her 5th kick: (⅔)4 (¾)4 ⅓ = 1/48 A.G M1 A1 2 (ii) Find prob. of Kate winning on her nth kick: (⅔)n-1 (¾)n-1 ⅔ ¼ M1 A1 = 1/(3 × 2n) A.G. A1 3 (iii) Find prob. of Jill winning game: ⅓ ¾ {1 + (⅔ ¾) + (⅔ ¾)2 + … } M1 A1 = ¼ (1 + ½ + ¼ + … ) = ½ A1 3 (iv) Find prob that game is a draw ⅓ ¼ {1 + (⅔ ¾) + (⅔ ¾)2 + … } M1 A1 = ⅓ ¼ (1 + ½ + ¼ + … ) or 1 – P(Jill or Kate wins) = 1 – {½ + ⅓(½ + ¼ + … )} (M1 A1) = 1/6 A1 3 11 11 (a) Resolve horizontally and vertically: Relate limiting frictions and reactions at A and B: EITHER: Resolve vertically : Combine eqns to find RA: Take moments about B Rearrange and use FA = 2µRA Combine to find tan θ : OR: Resolve vertically: Combine eqns to find RB: Take moments about A: Rearrange and use FB = µRB Combine to find tan θ : OR: Take moments about centre of rod: Use first 3 eqns to eliminate 3 forces, e.g.: FA = RB B1 FA = 2µRA and FB = µRB B1 FB + RA = W B1 RA = W / (1 + 2µ2) M1 A1 2RA cos θ – 2FA sin θ = W cos θ M1 A1 RA – 2µRA tan θ = ½W RA = ½W / (1 – 2µ tan θ ) A1 tan θ = (1 – 2 µ2) / 4µ A.G M1 A1 FB + RA = W (B1) RB = 2µW / (1 + 2µ2) (M1 A1) 2RB sin θ + 2FB cos θ = W cos θ (M1 A1) RB tan θ + µRB = ½W RB = ½W / (tan θ + µ) (A1) tan θ = (1 – 2 µ2) / 4µ A.G. (M1 A1) (FA + RB) sin θ = (RA – FB) cos θ (M2 A1 A1) tan θ = (RB /2µ – µRB)/( RB + RB) (M1 A1 A1) = (1 – 2 µ2) / 4µ A.G. (A1) 10

Page 8 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9231 23 © Cambridge International Examinations 2013 Find positive value of µ when θ = 45°: Find positive value of µ when θ = 0°: State set of possible values of µ (A.E.F.): 2 µ2 + 4 µ – 1 = 0 µ = √(3/2) – 1 or 0⋅225 A.E.F M1 A1 µ = 1/√2 or 0⋅707 A.E.F B1 [√(3/2) – 1, 1/√2] or [0⋅225, 0⋅707] A1 4 14 (b) Tabulate observed values: Find corresponding expected values: Calculate value of χ2:] State or use correct tabular χ2 value: Use conclusion of independence to find eqn. for A: Find bounds for A (to 3 s.f.): (integer valued reqd for this A1) Relate new value χnew 2 to original χ2: Use conclusion of independence to find eqn. for A: Find Nmax with A = 29: (integer value reqd for this A1) Rural Urban Support A 60 – A Not support 45 – A A – 5 M1 A1 27 33 18 22 M1 A1 χ2 = (27 – A)2/27 + (A – 27)2/18 + (A – 27)2/33 + (27 – A)2/22 M1 = (50/297) (A – 27)2 or 0⋅168[35] (A – 27)2 A1 χtab 2 = χ1, 0.9 2 = 2⋅706 (to 3 s.f.) B1 (50/297) (A – 27)2 < χtab 2 M1 (A – 27)2 < 16⋅07 A1 Amin = 23 and Amax = 31 A1 χnew 2 = N × χ2 M1 0⋅168[35] N (A – 27)2 < χtab 2 M1 N < 2⋅706/(4 × 0⋅16835) = 4⋅02 A1 Nmax = 4 A1 10 4 14