Cambridge A Level Thinking Skills 9694 — 2018 May/June Paper 3 · Variant 3

1 Archie found the drawing below on the wall of a temple ruin. He has labelled the four dice A, B, C and D. A D B C Archie initially assumes that the four dice in the drawing are identical copies of a standard die, in which the numbers of spots on opposite faces always add up to 7. (a) (i) How many spots are on the face of die D which is touching die C? [1] (ii) Explain how you can deduce that there are three spots on the face of die B which is touching die C. [2] Archie knows that the total number of spots on the four rectangular sides of an arrangement of four dice was thought to have a special significance. (b) What are the two possible totals for the number of spots on the sides of the arrangement in the drawing that Archie has found? [3] Archie showed the drawing to his friend George, who explained that these drawings always showed two identical standard dice and a pair of other dice, known today as Sicherman dice. In a pair of Sicherman dice, one has faces with 1, 2, 2, 3, 3 and 4 spots, while the other has faces with 1, 3, 4, 5, 6 and 8 spots. On each die the numbers of spots on opposite faces always add up to a fixed number for that die. (c) What are the two fixed numbers for Sicherman dice? [1] (d) Assuming that George is correct, (i) Which two of the dice in the drawing must be the Sicherman dice? Explain your answer. [2] (ii) What is the smallest possible total number of spots that there could be on the sides of the arrangement in the drawing? [1]

Mark scheme: Question Answer Marks 1(a)(i) 3 1 1(a)(ii) 1 mark for use of opposite faces adding to 7 anywhere in explanation. 2 1 mark for reference to matching the orientation seen on die C or deduced about D. 1(b) The total excluding the spots on die A is 1+1+5+2+4+4=17. 3 The two faces of die A that are included in the total are either 6–3 or 1–4 as it must match the orientation of numbers that is true for die B, so these add to either 5 or 9. The totals are 22 and 26 2 marks for one or other, without or without a second number OR max 2 of 1 mark for 17 B+C+D 1 mark for either 6–3 or 1–4 for die A 1 mark for difference of 4 (min 19 max 29) ft from 4 in (a)(i) 16 BCD and 21 and 25 1(c) 5 and 9 1 1(d)(i) A must be the Sicherman die with 1,2,2,3,3,4 (is the only die without a 5 or 2 6) B cannot be a Sicherman because it has a 2 and a 6. C cannot be a Sicherman because the 4 and 5 would have to be opposite. Since neither B nor C for above reasons, A and D must be the Sicherman pair. 2 marks for correct pair with reasons for excluding B and excluding C. 1 mark for any of three deductions above, or answer A and D without justification. 1(d)(ii) The hidden face of die B is still a 4. 1 The hidden face of die D could be 3 or a 6, so choose 3. The two back faces of die A could be 1 and 2 – total 3. The smallest possible total is 19.

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2 Trains run from Athos to Ethos, stopping at Banta, Chanta and Danter on the way. A researcher has noticed that there is a pattern in the number of people who get on and off at the different stations. • At Banta, twice as many people get on as get off. • At Chanta, one third of the people on the train get off, and then 30 people get on. • At Danter, 10 more people get off than get on. (Where the number of people getting on or off is variable, that number can be 0 and the pattern still holds.) The researcher assumes that this pattern holds for every train that leaves Athos bound for Ethos. At each stop, those people who are getting off the train do so before anyone gets on. All trains are empty before people get on at Athos. On one occasion, 120 people got on the train at Athos and 24 people got on at Banta. (a) How many people were on the train when it arrived at Ethos? [2] (b) (i) What is the greatest number of people who could have made the whole journey from Athos to Ethos? [2] (ii) What is the least number of people who could have made the whole journey from Athos to Ethos? [1] On another occasion, 150 people got on the train at Athos and 150 people arrived at Ethos. (c) How many people got on the train at Banta? [3] On a third occasion, the number of people on the train between Banta and Chanta was equal to the number of people on the train between Chanta and Danter. (d) How many people were on this train when it arrived at Ethos? [2]

Mark scheme: 2(a) 120 + (24 – 12) + (30 – 44) – 10 = 108 2 1 mark for correct processing at Banta (132), or ft from error at Banta, or 118. 2(b)(i) 12 of original people get off at Banta, leaving 108A (+ 24B). 2 At Chanta, (44 get off: 24B +) 20A, leaving 88A (+ 0B + 30C) For greatest number of A to arrive at Ethos, none will get off at Danter, so greatest number who travel Athos to Ethos is 88. 1 mark for 88(A) at Danter with subsequent wrong working OR 1 mark for identifying that the limit is when the maximum number of non-As get off at Chanta. ft clear Banta number from (a)–20 2(b)(ii) All the people who started at Athos who are still on the train at Danter can 1 get off at Danter Least number who travel Athos to Ethos is 0 2(c) 130 people on train before people got on at Chanta (150 + 10 – 30) 3 So 130 = two thirds of people arriving at Chanta So 195 people leave Banta and 195 – 150 = 45 is the reduction at Banta. 45 × 2 = 90 people got on the train at Banta 2 marks for 195 seen or 1 mark for 130 seen OR 3/2x to determine arriving at Chanta OR –150 and double to determine arriving at Banta OR 2  x  2 marks for an algebraic statement: 150 + + 30 − 10 = 150 (oe)   3  2  2 OR 1 mark for ( attempt ) + attempt = 150 3 2(d) Number between Chanta and Danter must be equal to (two thirds of this 2 number) + 30, so number = 90 [1] Number at Ethos = (number between Chant and Danta) −10 = 80

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3 A local store has started to make and sell cakes alongside its other products. The cakes each require 150 g flour, 200 g sugar and 2 eggs. Flour and sugar each cost 60¢ per kg and eggs cost 20¢ each. The total cost of making an individual cake is the cost of the ingredients plus $3, which covers all the other standing costs. The store owner expected that he would be able to sell 40 cakes each week. At the end of each week, any unsold cakes are thrown away. (a) Show that the total cost to make 40 cakes is $144.40. [2] In the first week he made 40 cakes, which were priced at $5 each. Only 10 of the cakes were sold. When asked, the customers said that this was because the prices were too high. In the second week 40 cakes were made, but the price was reduced to $4.50. (b) What is the smallest number of cakes that would need to be sold so that the total money received for the cakes was higher than in the first week? [2] In fact 20 cakes were sold in the second week. As a result, the store owner assumes that for every 5¢ by which he reduces the price of a cake, an extra customer will buy a cake. The price must always be a multiple of 5¢. (c) If the store owner’s assumption is correct, what is the maximum profit possible if he makes 30 cakes? [2] (d) What number of cakes should the store owner make if he wants to achieve the maximum profit? [4] In the second week (when 20 cakes were sold at $4.50 each) no customer bought more than one cake, but a number of customers commented that they would have bought a second cake if there were a discount available. Based on these comments, the store owner is considering offering a 10% discount to customers who buy two cakes. This discount would apply to both of the cakes. However, no customer would be allowed to buy more than 2 cakes. (e) (i) If all 20 customers in the second week had bought 2 cakes, what profit would the store owner have made? [2] (ii) If this offer had been in place, how many of the customers would have had to have bought a second cake in order for the shopkeeper to make a profit? [3] [Question 4 begins on the next page]

Mark scheme: 3(a) Price of ingredients for a cake is 61¢ soi [1] 2 350 g of sugar or flour is 0.35 × $0.60 = $0.21 2 eggs cost $8 ÷ 20 = $0.40 Total is $0.61, so the total cost of making a cake is $3.61. 40 cakes therefore cost $144.40 to make. AG If 2 marks cannot be awarded, award 1 mark for calculating that the cost to make one cake is $3.61 OR at least one of 3.6 4.8 8.4 or 24.40 for 40. 3(b) In the first week the sales amounted to a total of 10 × $5 = $50 [1] 2 At $4.50 each, 11 would amount to $49.50, so 12 is the smallest number to make more money than in the first week. 1 mark for “greater than 11”. 3(c) The cakes would need to be sold at $4 each [1] 2 which means that there would be a profit of $0.39 on each one. 30 × $0.39 = $11.70 ft cost from (a) 3(d) 19 4 Making one extra cake increases the costs by $3.61, but 5¢ is lost from the profit from all of the other cakes that would have been sold. Therefore, for any given number of cakes, the cost of producing one extra can be thought of as $3.61 + $0.05 × the number of cakes. Number of Total cost for Selling price for cakes one extra extra cake 30 $5.11 $3.95 20 $4.61 $4.45 19 $4.56 $4.50 18 $4.51 $4.55 So it is worth making one extra cake when only 18 are made, but not when 19 are made. 19 cakes gives the best profit of $17.86. If four marks cannot be awarded, award one mark for: • Calculating the profit for a particular number of cakes made. • Calculating the profit for a second case. • Calculating another case which improves on the profit from the first case. Alternatively, solutions involving calculus should be rewarded thus: Finding the price (p) as a function of the number of cakes (x): p = 550 – 5x [1] Finding profit π = (189 – 5x)x oe [1] Differentiating and equating to zero or determining mirror line [1] 19 cakes ft their 3.61 from (a) 3(e)(i) The cakes will all be sold at $4.05 2 Profit of 44¢ per cake [1] 40 × $0.44 = $17.60 Alternatively, $4.05 × 40 soi [1] $162 – $144.40 = $17.60 ft their 3.61 from (a) 3(e)(ii) Producing the 40 cakes cost $144.40. 20 cakes at $4.50 will bring in $90, so 3 an extra $54.40 is required. [1] Since both cakes get the 10% discount, the second cakes effectively cost 80% of $4.50, or $3.60. [1] 15 such cakes would produce $54, so 16 customers would need to take advantage of the offer. OR 1 mark for correct calculation of profit when a second cake is bought by a specified number of the 20 customers

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4 Linker is an electronic game in which the player tries to score as many points as possible in three minutes by linking from one letter to another successfully. This is the device used to play Linker. From SCORE A B C D E F A FROM TIME B C To TO D A B C E F D E F SET When the SET button is pressed, digits from 0 to 5 appear in the grid, 0 appears in the SCORE display, 3:00 appears in the TIME display and a letter (A, B, C, D, E or F) appears to the right of the grid in the FROM display. As soon as the player presses one of the TO buttons the game begins and a link is made. The time starts to count down towards 0:00 and the relevant number of points are added to the score. The letter on the button that was pressed now appears in the FROM display ready for the next link to be made. This continues until the TIME display shows 0:00. However, if the sequence of letters entered by the player repeats the same three letters in a row, the game is over. Because of the limited time available to make as many links as possible and the threat of causing the game to end prematurely, players often do not try to score the maximum amount available from every link. For example, suppose a player were to be given the following grid and the letter D in the FROM display: From A B C D E F A 1 2 0 4 5 3 B 0 5 1 3 2 4 C 2 3 4 5 0 1 To D 4 0 2 1 3 5 E 3 1 5 0 4 2 F 5 4 3 2 1 0 If the player were to enter, in order, F D B B F C B B B, the links made and points scored would be: From To (D) F = 2 points F D = 5 points D B = 3 points B B = 5 points B F = 4 points F C = 1 point C B = 1 point B B = 5 points B B = 5 points If the player’s next entry were F, they would score another 4 points, but the game would be over because B B F would have occurred twice in the sequence of the player’s entries. Their final score would be 35 points. If the player’s next entry were B, they would score another 5 points, but the game would be over because B B B would have occurred twice in the sequence of the player’s entries. Their final score would be 36 points. (a) Liam is playing a game of Linker. This is his grid. From A B C D E F A 5 3 2 0 4 1 B 0 4 1 5 3 2 C 3 5 0 2 1 4 To D 1 2 3 4 0 5 E 4 0 5 1 2 3 F 2 1 4 3 5 0 His starting letter in the FROM display was B. His third entry was D, which took his score to 14. Since then he has entered, in order, F E F C. (i) What, in order, were his first two entries? [2] (ii) What is his total score at present? [2] [Question 4 continues on the next page] (b) Inga is about to start a game of Linker. This is her grid. From A B C D E F A 2 4 3 1 0 5 B 3 5 0 4 2 1 C 4 2 1 3 5 0 To D 1 0 5 2 4 3 E 5 3 4 0 1 2 F 0 1 2 5 3 4 Her starting letter in the FROM display is A. Her intended strategy is to score 5 points as often as possible, only scoring less when a 5-point link would end the game. Show how Inga can score a total of 47 points for her first ten links, without causing the game to be over. [2] For each game, the grid that appears on the screen is the following grid, or a rotation of it, with each letter from P to U replaced by a different digit from 0 to 5: From A B C D E F A P T R U S Q B R S P Q T U C T Q U S R P To D U R T P Q S E Q P S R U T F S U Q T P R Each version of the grid that appears on the screen has its own grid code, which is shown in a small display on the side of the device. This code consists of the letter P, Q, R or S, indicating the orientation of the grid, followed by the numerical values of P, Q, R, S, T and U, in order. For example, the grid that Liam is playing with has the grid code Q251043. This means that Q is in the top-left corner and P = 2, Q = 5, R = 1, S = 0, T = 4, U = 3. (c) (i) All the possible grid codes are programmed into the device. How many are there? [1] (ii) Write, in order from left to right, the digits of the top row of the grid with the grid code R403215. [2] (iii) What is the grid code for the game that Inga is about to start? [2] (d) When Kerry plays Linker, he always starts by entering A A A D D D very quickly. This often gives him a reasonable score in a very short time before he settles down to give more thought to the rest of the game. On one occasion, however, he did this and scored a total of 1 point only for his first six links. Draw a grid for which entering A A A D D D will result in a total score of 1 point only. State the code for your grid. [4]

Mark scheme: 4(a)(i) C F 2 From B, entries C F D score 5 + 4 + 5 = 14. Award 1 mark for any of the following: • B C (from B, entries B C A score 4 + 5 + 5) • C E (from B, entries C E A score 5 + 5 + 4) • E F (from A, entries E F D score 4 + 5 + 5) • F D (from E, entries F D D score 5 + 5 + 4) 4(a)(ii) 29 2 14 + 3 (D to F) + 3 (F to E) + 5 (E to F) + 4 (F to C) Award 1 mark for either of the following answers: • 15, which fails to include the points already scored • 26, which fails to include the link from D to F 4(b) (She can score 47 points by entering) 2 E C D F A E C E C A OR E C D F A E C A E D Award 2 marks for either answer. E C D F A E C is the limit for scoring consecutive 5s (total 35), then either A E D scores 3 + 5 + 4 or E C A scores 4 + 5 + 3. If 2 marks cannot be awarded, award 1 mark for appreciation that (A) E C D F A E must be the first six entries. 4(c)(i) 2880 (4 × 6 × 5 × 4 × 3 × 2 × 1) 1 4(c)(ii) 3, 4, 1, 0, 5, 2 2 If 2 marks cannot be awarded, award 1 mark for one the following: • 4,1,3,5,2,0 (the top row of grid P403215) • 0,5,4,2,1,3 (the top row of grid Q403215) • 2,0,5,1,3,4 (the top row of grid S403215) • 3,1,2,4,5,0 (right-hand column read upwards or R downwards) 4(c)(iii) S540213 2 Consideration of, for instance, the relative positions of the 2s (the digit in the top left corner), the code letter must be S. The top row of 243105 therefore corresponds to SQUTRP. If 2 marks cannot be awarded, award 1 mark for any code beginning with S that contains the digits 0 to 5 once. OR Award 1 mark for one of the following, which only matches the top row of Inga’s grid with the master grid: • P253041 (with 243105 corresponding to PTRUSQ) • Q325104 (with 243105 corresponding to QUPSTR) • R412530 (with 243105 corresponding to RPTQUS) 4(d) The 1 point must have been scored for the link from A to D, so A to A and D 4 to D must score 0. This must be a grid with a grid code of the form P01 or S01. Award 4 marks for any correct grid with a P01 or S01grid code. e.g. A B C D E F A 0 2 3 1 4 5 B 3 4 0 5 2 1 C 2 5 1 4 3 0 D 1 3 2 0 5 4 E 5 0 4 3 1 2 F 4 1 5 2 0 3 If 4 marks cannot be awarded, award 3 marks for any grid with all the 0s and 1s consistent with a P01 or S01grid code (given) and no digit appearing more than once in any column. e.g. A B C D E F A 0 2 3 1 4 5 B 2 4 0 5 3 1 C 3 5 1 4 2 0 D 1 3 2 0 5 4 E 4 0 5 2 1 3 F 5 1 4 3 0 2 If 3 marks cannot be awarded, award 2 marks for any grid with all the 0s and 1s consistent with a P01 or S01grid code, even if code not stated. i.e. (pto) 4(d) A B C D E F A 0 1 B 0 1 C 1 0 D 1 0 E 0 1 F 1 0 or A B C D E F A 0 1 B 0 1 C 0 1 0 D 1 0 E 1 0 F 1 0 If 2 marks cannot be awarded, award 1 mark for valid code seen OR any grid that contains the following: A B C D E F A 0 B C D 1 0 E F

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