Cambridge A Level Thinking Skills 9694 — 2017 Oct/Nov Paper 3 · Variant 2

1 Fred earns his living as a taxi-driver, offering lifts between his home town of Honeytrees and the airport, a distance of 30 km. He charges a basic rate of $32 for a single journey. The journey in either direction takes at least 40 minutes and at most 70 minutes, depending on the traffic. Fred takes a minimum break of 10 minutes between each single journey. He works a shift of at most 7 hours each day. Fred’s first journey is always from Honeytrees to the airport and his final journey must arrive in Honeytrees before the end of his shift. There are always enough passengers wishing to travel from Honeytrees to the airport. Often he is able to pick up a customer for the return journey. If not, he drives back after his 10 minute break, without a passenger. (a) What is the greatest amount of money that Fred is able to take in one shift? [2] Fred decides to change the price that he charges. For each single journey, there will be a basic charge of $28 for the first 40 minutes, plus $0.60 for every further 3 minutes, or part thereof. (This means that as soon as the timer has moved beyond any multiple of 3 minutes, he adds the next $0.60.) (b) Assuming that there are always sufficient customers at Honeytrees and at the airport, find the greatest amount of money that Fred is certain to take in one shift. [3] The Transport Agency has announced that there will be roadworks between Honeytrees and the airport in both directions for the foreseeable future. This means that journey times will be increased by 20%. (c) Find Fred’s maximum charge for a single journey. [1] Fred wants to move into a new house and he needs to increase his income. He decides to work a longer shift of up to 9 hours each day. He still has a 10 minute break between each single journey, but after 4 single journeys he has a longer break of 30 minutes. The roadworks are still in place. There are always sufficient customers at Honeytrees, but not necessarily at the airport. Last Wednesday, Fred made a number of journeys. Two of these journeys took the maximum length of time and the remaining journeys each took 65 minutes. (d) (i) How long after the beginning of his shift did Fred complete his last journey? [2] (ii) Find the difference between the greatest and the least amounts of money that Fred could have taken. [2]

Mark scheme: Question Answer Marks 1(a) The final journey could be 70 minutes long either way, and so will not be 2 embarked on after (7 hours – 150 minutes =) 4.5 hours (or 270 minutes) [1 mark] In which case only 6 paid journeys possible (40+10+40 + 10+40+10+40 + 10+40+10+40 = 290) 6 × $32 = $192 Condone $256 (8 journeys soi [1 mark] at $32) 1 mark for max journeys for their time limit. 1(b) 270 minutes last time to set out; 4 journeys each over 60 minutes (and three 3 rests) would exceed this, so total fare cannot be less than 4 × ($28 + (7 × 0.6)) = 4 × $32.20 = $128.80 If 3 marks not awarded, award 1 mark each for the following (max 2): • Time threshold 270 mins oe (may be seen in (a)) • Acknowledgement that minimum paid time is what’s certain • Correctly deducing number of returns for their threshold • Pays in both directions (dependent on previous mark) SC: 2 marks for 61 or over 60 minutes seen OR 1 mark for 51 minutes seen OR 2 marks for $224 or $131.20 or $136 or $168 OR 1 mark for $68 or $112 1(c) Maximum journey time = 84 minutes (20% greater than 70 minutes) 1 charge = $28 plus (15 × $0.60) = $37 1(d)(i) 2 journeys of 84 minutes plus 4 journeys of 65 minutes. 2 Time = 84 + 10 + 84 + 10 + 65 +10 + 65 + 30 = 358 minutes plus (65 + 10 + 65) = 498 minutes = 8 hours 18 minutes. 1 mark for 6 journeys (supported) OR arriving at 358 minutes for the section of his shift up to his first extended break soi. SC: 1 mark for 508 – from wrongly including an extra 10 minutes at end 1(d)(ii) Greatest = $37 × 2 + 4 × $33.40 = $207.60 2 Least = 3 × 65 minute journeys =$100.20 Difference = $107.40 Award 1 mark for $207.60 or $100.20

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2 Moses has a storage room, in the shape of a cuboid, enclosing a space measuring 220 cm horizontally, 240 cm vertically and 600 cm back. 240 cm 600 cm 220 cm He has 130 identical boxes measuring 50 cm by 60 cm by 80 cm. They can be stacked in any orientation. Moses stacks boxes into the storage room, all with the same orientation, with the 50 cm edge vertical. (a) What is the maximum number of boxes he can fit in this way? [2] (b) What is the largest number of boxes that can be stored if they can be placed in any orientation, but all the boxes must be placed in the same orientation as each other? [2] Moses’ wife, Leillah, claims that there is a way of placing the 130 boxes into the storage room, but not all the boxes would be in the same orientation. (c) Show that Leillah is correct. [3] Moses now exchanges his boxes for larger boxes measuring 50 cm by 60 cm by 160 cm, which hold twice as much and are better value. The boxes do not have to be placed in the same orientation as each other. (d) Moses’ daughter, Janet, claims that 63 of these larger boxes can be fitted into the storage room. Is she correct? [3]

Mark scheme: 2(a) He can fit in 2 × 4 × 10 = 80 or 3 × 4 × 7 = 84 boxes. 2 1 mark for either calculation. 2(b) If he places the 50 cm side horizontal and 80 cm side vertical he can fit in 2 4 × 3 × 10 = 120 boxes. 1 mark for 12 × 3 × 3 = 108 boxes OR 7 × 4 × 4 = 112 boxes OR 2 × 12 × 4 = 96 boxes seen. 2(c) For example, 96 boxes can be placed as 2 × 4 × 12, leaving a space 60 × 3 240 × 600, into which a further 36 boxes can be placed, with the 60 cm side horizontal and the 80 cm side vertical. This makes a total of 132 boxes (so 130 is obviously possible). 3 marks for any demonstration of ⩾130 boxes If 3 marks not awarded, award 1 mark each for the following (max 2): • Volume calculation to show sufficient • Splitting the 220 into a combination of 50s, 60s and 80s • Covering a side with 220 end 2(d) For example, with 160 horizontally, he can fit 1 × 4 × 12; this leaves room for 3 1 × 1 × 12 with 60 horizontally. On top of this second set can be fitted another 1 × 1 × 3, making a total of 63. AG 1 mark for filling an entire face OR for room volume = 66 boxes (if not given in (c)) OR 2 marks for obtaining at least 60 (with supporting working) OR 3 marks for (at least) 63 (with supporting working)

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3 Tickets for next year’s Glastonbourne Music Festival are soon to be released to the general public. They can only be bought by calling the box office, which opens at 8 am on the release date. There are discounted tickets available for the first 100 callers. Sally is keen to buy discounted tickets, and decides to call before 8 am in the hope of being at the front of the queue when the box office opens. She calls at 6 am precisely and is told by an automated message that she is 560th in the queue. She stays on the line, and one minute later she is told that she is 540th. (a) Using this information alone, state when Sally should expect to reach the front of the queue. [1] When Sally reaches the front of the queue, another message informs her that the box office has not opened yet, and that she must try again later. She decides to call again immediately. She assumes that no new people will call the box office, and that everyone else who has already called will also call again immediately after reaching the front of the queue. (b) If this continues until after the box office opens, at what time will Sally be able to buy her tickets? [1] However, when Sally calls for the second time, she is told that she is 588th in the queue. She realises that in fact some new people have called the box office. She assumes that these new callers have joined the queue at a constant rate since her first call. Sally wants to predict, using this new information, when she will be able to buy tickets. She wants to keep her calculations as simple as possible, so she decides to place all her calls precisely at the start of a minute. For example, if she were to arrive at the front of the queue at 7 seconds past 6.49 am, she would place her next call at 6.50 am. If however she were to arrive at 6.49 am exactly, she would place the call immediately, i.e. at 6.49 am. On this basis, Sally predicts that she will be able to buy tickets at 27 seconds past 8.01 am. (c) Show how Sally reached her prediction. [3] Sally wonders if she could avoid waiting unnecessarily in the queue. She uses the information she has so far to work out the best time to call in order to arrive at the front of the queue as soon after 8 am as possible. She will only consider placing the call precisely at the start of a minute. (d) (i) Calculate at what time she should call, and the precise time she will reach the front of the queue. [2] (ii) What is the latest time that she could call and still expect to buy discounted tickets? Show at what time she will arrive at the front of the queue. [2] Sally realises that her assumption that all callers will re-join the queue as soon as they reach the front is unlikely to be realistic. She now assumes that half of the callers will give up and not call again after reaching the front of the queue. So, each time she calls and is told how many people are in the queue, only half of those people will be in the queue the next time she calls. Because of this extra complexity, Sally decides it will be easier for her to rejoin the queue again each time she reaches the front. (e) (i) How many new callers does Sally now calculate are joining the queue every minute? [2] (ii) Will Sally be able to buy discounted tickets? Show what time she should expect to get through to the box office. [4] [Question 4 begins on the next page]

Mark scheme: 3(a) 20 people in one minute: 560/20 = 28 minutes. She will reach the front at 1 06:28 am. 3(b) 06:28 – 06:56 – 07:24 – 07:52 – 08:20 1 08:20am 3(c) 3 Time joined Number of people Minutes wait 06:00:00 560 560/20 = 28 06:28:00 588 588/20 = 29.4 06:58:00 618 618/20 = 30.9 07:29:00 649 649/20 = 32.45 She reaches the operator at 08:01:27 AG Award 1 mark for converting 588 into minutes wait or time joined Award 1 mark for 618 seen Award 1 mark for accurate application: implied by sight of 07:29 and 649/20 oe 3(d)(i) Award 2 marks for the correct time (07:28) and the correct time that the front 2 of the queue is reached (08:00:24). If 2 marks cannot be awarded, award 1 mark for either the following: • at 07:28 the queue has 648 people in it, which means a 32.4 minute wait; • at 07:27 the queue has 647 people in it, which means a 32.35 minute wait, arriving at the front at 07:59:21. 3(d)(ii) 100 discounts at 20 per minute suggests that the last ones will go to the 2 caller getting through just before 8.05 am. Joining at 07.33, there will be 653 people in the queue. 653/20 = 32 m 39 s, leading to 08:05:39 seconds: too late. Therefore 07:32 is the latest time, from which she gets to the front at 08:04:36. Award 2 marks for the correct time (07:32) and the correct time that the front of the queue is reached (08:04:36). If 2 marks cannot be awarded, award 1 mark for either of the following: • at 07.32 am the queue has 652 people in it, which means a 32.6 minute wait; • at 07.33 am the queue has 653 people in it, which means a 32.65 minute wait, arriving at the front at 08:05:39. 3(e)(i) Of the 560 callers, only 280 re-join. 2 So 588 – 280 = 308 have joined in the meantime. 308 over 28 minutes: 11 per minute 1 mark for 308 soi 3(e)(ii) 08:03:12 (07:30:00 + 33.2 minutes wait), so Yes 4 If 4 marks cannot be awarded: Award 1 mark for clearly calculating the three components for any case: minutes wait (queue/20), number re-starting (queue/2) and number arriving (minute × (their) 11). Award 1 mark for correctly applying the process iteratively once (implied by 06:28:00 as second time of starting). Award 1 mark for a second iteration applied correctly (implied by either 06:58 or 07:30 AND the number of people in the queue at those times).

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4 Strategy is a TV quiz show. Four contestants take part in each edition. Throughout the show every question is presented simultaneously to the contestants, together with four possible answers, A, B, C and D, only one of which is correct. Each contestant has an electronic keypad which allows them to select one of the options. The first part of the show is called Piggyback. It consists of five rounds of 10 questions on specific subjects. Every correct answer selected by a contestant adds $5 to their own prize pot. Before the show begins, the contestants are informed of the five question categories and each of them has to make two strategic decisions: • One category must be nominated “double”. The contestant adds $10 to their prize pot, instead of $5, for every correct answer they select in this round. • A second category must be nominated “piggyback”. The contestant does not take part in this round, but has the same amount of money added to their prize pot for the round as the contestant who adds the most to their prize pot as a result of answering the questions. None of the contestants knows in advance what categories the others have nominated “double” and “piggyback”. This means that a “piggyback” could be worth up to $100 if one or more of the others have nominated the same category “double”. Occasionally all four contestants nominate the same category “piggyback”, in which case the round becomes void and all the piggybacks are wasted. The two contestants with the highest prize pots at the end of Piggyback progress to the second part of the show, called Freeze, taking their prize pots with them. Freeze consists of 10 general knowledge questions. For each question, the first to select an answer freezes the other out of that answer. The other contestant can then choose to pass or select a different answer. During Freeze, every correct answer adds $25 to the contestant’s prize pot, but $10 is deducted for every incorrect answer. A contestant who passes because they have been frozen out of the answer they want to select is not penalized, but if neither contestant selects an answer to a question within 10 seconds, $40 is deducted from both prize pots. The contestant with the higher prize pot at the end of Freeze is the show’s winner and takes their prize pot through to Multiplier, the final part of the show. This consists of 20 general knowledge questions, but with a time limit of only 5 seconds per question. During Multiplier, $20 is deducted from the prize pot for an incorrect answer and $40 is deducted if no answer is selected within 5 seconds. The amount of prize money that the winner takes home is their final prize pot multiplied by the number of correct Multiplier answers. Ben, Jodie, Olivia and Toby were today’s Strategy contestants. The following table details the prize pots they acquired during the Piggyback rounds: Round 1 Round 2 Round 3 Round 4 Round 5 Total History Music Sport Geography Science prize pot Ben $40 $90 $45 $80 $40 $295 Jodie $35 $45 $40 $25 $45 $190 Olivia $80 $40 $35 $80 $45 $280 Toby $35 $90 $45 $35 $35 $240 In Freeze, Olivia was first to select an answer to all but the last of the 10 general knowledge questions. She appeared to make no effort at all to answer the last question. The performances of Ben and Olivia during Freeze are summarized below. Question 1 2 3 4 5 6 7 8 9 10 Ben pass correct pass incorrect pass pass pass pass correct pass Olivia correct incorrect incorrect correct incorrect correct correct correct incorrect pass Olivia was today’s winner. She went on to answer 12 of the Multiplier questions correctly, but answered 7 incorrectly and failed to answer the other one within the 5-second time limit. (a) What is the largest amount of money that anyone could possibly win on an edition of Strategy? [2] (b) (i) How much was in Ben’s prize pot and how much was in Olivia’s prize pot after today’s ninth Freeze question? [2] (ii) Suggest why Olivia did not attempt to answer the last question. [1] (c) (i) In total, how many questions did Olivia answer correctly today? [2] (ii) How much did Olivia win? [2] (iii) What is the maximum extra amount Olivia could have won if she had selected an answer to the last Freeze question? [2] (d) Deduce which categories were each of today’s four contestants’ “double” and “piggyback” choices. [4]

Mark scheme: 4(a) Maximum for Piggyback is 2 × $100 (“double” & “piggyback”) + 3 × $50 = 2 $350 [1 mark] Maximum for Freeze is 10 × $25 = $250 $600 × 20 = $12 000 4(b)(i) Ben’s prize pot: $335 ($295 + 2 × $25 – $10) [1 mark] 2 Olivia’s prize pot: $365 ($280 + 5 × $25 – 4 × $10) [1 mark] 4(b)(ii) If Olivia had selected an incorrect answer and Ben had selected the correct 1 answer, Ben would have won (by $5), whereas not attempting the question was the only way she was guaranteed to win by at least $5. 4(c)(i) In Piggyback, History and Geography must have been her “double” and 2 “piggyback” categories (to be able to add more than $50 to her prize pot). She must have answered 8 questions correctly in one of the two categories and not taken part in the other. Her total number of correct answers in Piggyback was therefore 8 + 8 + 7 + 9 = 32. 32 (in Piggyback) + 5 (in Freeze) + 12 (in Multiplier) = 49 If 2 marks cannot be awarded, award 1 mark for evidence of appreciation that she answered 32 questions correctly in Piggyback OR for an answer of 57 (counting piggyback answers as her own) 4(c)(ii) Her prize pot after the ninth Freeze question was $365, then $325 after she 2 did not attempt the last question. This was reduced by $180 during Multiplier (7 × $20 + 1 × $40) before multiplication by 12. 12 × $145 = $1740 If 2 marks cannot be awarded, award 1 mark for evidence of appreciation that her prize pot was reduced by $180 during Multiplier. 4(c)(iii) If she had answered the last question correctly, $25 would have been added 2 to her prize pot instead of $40 deducted. $65 × 12 = $780 If 2 marks cannot be awarded, award 1 mark for an answer of $300 (which fails to reinstate the deducted $40) OR for sight of $2520 4(d) Any amount over $50 must be due to a “double” or a “piggyback”. Only 4 Olivia recorded over $50 in round 1, so: Olivia: history/round 1 “double”; geography/round 4 “piggyback” [1 mark] Olivia must have piggybacked Ben’s $80 in round 4, so: Ben: geography/round 4 “double”; music/round 2 “piggyback” [1 mark] Ben must have piggybacked Toby’s $90 in round 2, and, of the other rounds, only in round 3 is Toby’s amount equal to the greatest amount, so: Toby: music/round 2 “double”; sport/round 3 “piggyback” [1 mark] “Double” must result in a multiple of $10, and only in round 5 is Jodie’s amount equal to the greatest amount, so: Jodie: sport/round 3 “double”; science/round 5 “piggyback” [1 mark]

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