Cambridge A Level Mathematics 9709 — 2017 Oct/Nov Paper 6 · Variant 1

2 The time taken by a car to accelerate from 0 to 30 metres per second was measured correct to the nearest second. The results from 48 cars are summarised in the following table. Time (seconds) 3 −5 6 −8 9 −11 12 −16 17 −25 Frequency 10 15 17 4 2 (i) On the grid, draw a cumulative frequency graph to represent this information. [3] (ii) 35 of these cars accelerated from 0 to 30 metres per second in a time more than t seconds. Estimate the value of t. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) Points (5.5,10), (8.5,25), (11.5,42), (16.5,46), (25.5,48) cf 50 40 30 20 10 0 5 10 15 20 25 time(sec) B1 B1 Axes labelled “cumulative frequency” (or cf) and “time [or t etc.] (in) seconds (or sec etc.)”. Linear scales – cf 0–48, time 2.5 – 25.5 (ignore <2.5 on time.) At least 3 values stated on each axis, but (0,0) can be implied without stating. B1 All points plotted accurately, (5, 10) etc. scores B0. Curve or line segments drawn starting at (5.5,10) and passing within ‘1 scale unit’ vertically and horizontally of plotted points 3 Question Answer Marks Guidance 2(ii) 48 – 35 = 13 t = 6.5 sec M1 Subt 35 (checked ±1 mm on graph) from 48 or 50, A1 6 ⩽ Ans ⩽ 7 2

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3 An experiment consists of throwing a biased die 30 times and noting the number of 4s obtained. This experiment was repeated many times and the average number of 4s obtained in 30 throws was found to be 6.21. (i) Estimate the probability of throwing a 4. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Hence (ii) find the variance of the number of 4s obtained in 30 throws, [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) find the probability that in 15 throws the number of 4s obtained is 2 or more. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) p = 0.207 B1 1 3(ii) Var = 30 × 0.207 × 0.793 = 4.92 B1 1 3(iii) P(⩾ 2) = 1 – P(0, 1) M1 = 1 – (0.793)15 – 15 1       (0.207)(0.793)14 M1 1 – P(0, 1) seen n =15 p = any prob = 0.848 A1 3

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4 The ages of a group of 12 people at an Art class have mean 48.7 years and standard deviation 7.65 years. The ages of a group of 7 people at another Art class have mean 38.1 years and standard deviation 4.2 years. (i) Find the mean age of all 19 people. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) The individual ages in years of people in the first Art class are denoted by x and those in the second Art class by y. By first finding Σx2 and Σy2, find the standard deviation of the ages of all 19 people. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) (48.7 12 38.1 7) 19 M1 Accept unsimplified (may be separate calculations) = 44.8 A1 2 4(ii) 7.652 = 2 2 48.7 12 Σ − x Σx2 = 29162.55 M1 Substitution in one correct variance formula 2 2 2 4.2 38.1 7 Σ = − y Σy2 = 10284.75 A1 One Σx2 or Σy2 correct (can be rounded to 4sf)) Combined var = (29162.55 10284..75) 19 + – 44.792 = 39447.3 19 – 44.792 M1 Using their Σx2 and Σy2 and their 4(i) in the variance formula Combined σ = 8.37 or 8.36 A1 4

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5 Over a period of time Julian finds that on long-distance flights he flies economy class on 82% of flights. On the rest of the flights he flies first class. When he flies economy class, the probability that he gets a good night’s sleep is x. When he flies first class, the probability that he gets a good night’s sleep is 0.9. (i) Draw a fully labelled tree diagram to illustrate this situation. [2] The probability that Julian gets a good night’s sleep on a randomly chosen flight is 0.285. (ii) Find the value of x. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Given that on a particular flight Julian does not get a good night’s sleep, find the probability that he is flying economy class. 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Mark scheme: 5(i) GNS x E 0.82 1 – x Not GNS GNS 0.9 0.18 F 0.1 Not GNS B1 B1 Shape, clear labels/annotation and all probs correct 2 5(ii) 0.82x + 0.18 × 0.9 = 0.285 M1 Eqn with x in , two 2-factors on one side x = 0.15 A1 2 5(iii) ( ) ( ) P E notGNS ( | ) P notGNS ∩ = P E notGNS M1 Attempt at P(E∩not GNS) seen as num or denom of fraction M1 Attempt at P(not GNS) seen anywhere = 0.82 0.85 1 0.285 × − = 0.975 A1 Correct answer 3

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6 (a) A village hall has seats for 40 people, consisting of 8 rows with 5 seats in each row. Mary, Ahmad, Wayne, Elsie and John are the first to arrive in the village hall and no seats are taken before they arrive. (i) How many possible arrangements are there of seating Mary, Ahmad, Wayne, Elsie and John assuming there are no restrictions? [2] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (ii) How many possible arrangements are there of seating Mary, Ahmad, Wayne, Elsie and John if Mary and Ahmad sit together in the front row and the other three sit together in one of the other rows? [4] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (b) In how many ways can a team of 4 people be chosen from 10 people if 2 of the people, Ross and Lionel, refuse to be in the team together? [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a)(i) M1 = 78 960 960 A1 2 6(a)(ii) not front row e.g. WEJ** in 3× 3! = 18 ways B1 3! seen mult by k⩾1 7 rows in 7 × 18= 126 ways B1 mult by 7 front row: e.g. *MA** in 4 × 2 = 8 ways M1 attempt at front row arrangements and multiplying by the 7 other rows arrangements, need not be correct Total 126×8 = 1008 A1 4 6(b) EITHER: e.g. *R** in 8C3 ways = 56 ways *L** in 8C3 = 56 ways (M1 Considering either R or L only in team **** in 8C4 = 70 ways M1* Considering neither in team DM1 summing 3 scenarios Total 182 ways A1) OR1: No restrictions 10C4 = 210 ways (M1 10C4 – , Considering no restrictions with subtraction *RL* = 8C2 = 28 M1* Considering both in team 210 – 28 DM1 subt = 182 ways A1) Question Answer Marks Guidance 6(b) OR2: R out in 9C4 = 126 ways L out in 9C4 = 126 ways (M1 Considering either R out or L out Both out in 8C4 = 70 M1* Considering both out DM1 Summing 2 scenarios and subtracting 1 scenario 126 + 126 – 70 = 182 ways. A1) 4

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