Cambridge A Level Thinking Skills 9694 — 2012 Oct/Nov Paper 3 · Variant 3

This document consists of 7 printed pages and 1 blank page. IB12 11_9694_33/4RP © UCLES 2012 [Turn over *4512179406* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level THINKING SKILLS 9694/33 Paper 3 Problem Analysis and Solution October/November 2012 1 hour 30 minutes Additional Materials: Answer Booklet/Paper Electronic Calculator 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. Do not use staples, paper clips, highlighters, glue or correction fluid. DO NOT WRITE ON ANY BARCODES. Calculators should be used where appropriate. Answer all the questions. Start each question on a new answer sheet. 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.

2 © UCLES 2012 9694/33/O/N/12 1 Study the information below and answer the questions. Show your working. New members of a company are invited to play a simple strategy game as part of their induction day. Each game consists of 12 rounds, played between two players. Both players choose a sequence of 12 letters made up of As, Bs and Cs. They then compare their lists letter by letter. Points are awarded as follows. Letters played 2 points for… 0 points for… A and B A B B and C B C C and A C A When players play the same letter, they both get one point. Three of the players have decided not to change their sequences from game to game. Luke plays B A B A C A C A A A C A. Maggie repeats the sequence A B C over and over again. Nigel plays A B A B B B C A C A B C. (a) What will Maggie’s and Nigel’s scores be when they each play a game against Luke? [2] (b) In some games, one player may have scored enough points to ensure a win before all rounds have been completed. In which of the three games between Luke, Maggie and Nigel does this occur, and after which round? [2] A fourth player, Ophelia, takes a different approach, which involves always starting with A, and subsequently playing whatever letter her opponent played in the previous round. (c) When Luke, Maggie and Nigel each play against Ophelia, which of the three scores the least and what is that score? [2] (d) Pedro is going to play against Ophelia. Give a sequence of letters that he could play which would ensure that she scored nothing. [1] (e) Quentin joins in and plays against Luke, Maggie and Nigel. Suggest a sequence of 12 letters which would produce the maximum total score for Quentin over the three games. [3]

3 © UCLES 2012 9694/33/O/N/12 [Turn over 2 Study the information below and answer the questions. Show your working. In order to check that they have recorded credit card numbers correctly, businesses can do the following: • Change the digits in the odd positions (counting from the left: first, third, etc.) according to the table below. Original 0 1 2 3 4 5 6 7 8 9 Changed 0 2 4 6 8 1 3 5 7 9 • Add up all sixteen digits in the modified number. • Check that this total is a multiple of ten. This works because one of the digits in the credit card number is chosen to ensure that a multiple of ten results from this procedure. This digit is called the ‘check digit’. (a) What is the missing digit (#) in 4920 1641 01#4 1711? [2] A common mistake made when copying down numbers is to switch two adjacent digits (e.g. 24 miscopied as 42). The check digit system will detect single errors of this kind, except for one pair of digits. (b) Which pair of digits in a card number will not be detected if they are switched? [1] Card numbers are often read out as pairs of two-digit numbers; in the above example, “forty-nine, twenty, sixteen” etc. Some of these can be misheard, and in English this is particularly common for fourteen two-digit numbers: 13 heard as 30, or 30 heard as 13, and similarly 14 can be confused with 40, 15 with 50, 16 with 60, 17 with 70, 18 with 80, and 19 with 90. (c) How many of these fourteen mistakes will be detected, if just one of them occurs when the sixteen-digit number of a card is read? [2] (d) In how many of the sixteen positions could the check digit be located? (Assume nothing other than the information given.) [1] (e) Four digits of a card number were blanked out “for security reasons”. 4579 3991 #### 2607 How many possible valid card numbers could this represent? [2] (f) A national identity register using sixteen-digit numbers for individuals was proposed, and it was suggested that the numbers should be allocated so that they could not be confused with a credit card number. Specify a simple way to achieve this (without restricting the range of available credit card numbers). [2]

4 © UCLES 2012 9694/33/O/N/12 3 Study the information below and answer the questions. Show your working. A games designer is trying to create rectangular mazes made up of identically-sized square rooms, and needs to quantify how complicated they are. In order to do this, he defines the longest path of a maze as the number of rooms that a participant would visit in completing the maze, if she were to take a wrong door whenever faced with a choice. Rooms visited more than once are counted for each visit. The rooms all consist of four walls with closed doors in one or more of them. As a result the participant cannot know whether she has reached the exit until she opens the ‘exit’ door. The designer assumes that the participant marks each door that she tries, and will not repeat any pathways that she has already fully explored. Because the maze designer is considering mazes of different sizes, he decides to use the complexity coefficient of the maze as the measure of its difficulty, which he defines as the longest path divided by the size of the maze (the number of rooms it consists of). (a) Show that the complexity coefficient of the following maze is 1.44 (rounded to 2 decimal places), and complete the path that the participant takes: 1, 2, 1, 4… [2] 7 8 9 4 5 6 1 2 3 exit entrance Mazes must not have loops in them. A loop is where a pathway leads a participant back to a room already visited without retracing her steps. The entrance and the exit must be in different rooms. A participant must be able to reach all rooms in the maze. (b) Draw a different maze involving a 3 x 3 grid of rooms, which also has a complexity coefficient of 1.44 and which has its entrance and exit in the same places as the maze above. [1] (c) (i) The maximum complexity coefficient for a 3 x 3 maze is 1.78. Draw an example. (The entrance and exit may be in different positions to those in the example above.) [2] (ii) Find the maximum complexity coefficient for a 5 x 5 maze. [2] The designer also wants to consider the shortest way out of a maze. He defines the simplicity coefficient as the shortest path from the entrance to the exit divided by the size of the maze. In considering this, the designer restricts his investigation to rectangular mazes whose entrances and exits are in the rooms at diagonally opposite corners of the maze. (d) Consider rectangular (including square) mazes made up from no more than 18 rooms. What is the lowest simplicity coefficient for such mazes? Give an example of a maze with this simplicity coefficient. [3]

5 © UCLES 2012 9694/33/O/N/12 [Turn over The designer thinks it will be useful to consider the difference between the longest and shortest paths for particular mazes. When considering this, he continues to restrict himself to mazes in which the entrance and exit are in rooms at diagonally opposite corners of the maze. (e) (i) For a maze made up from no more than 18 rooms, what is the maximum difference between the longest and shortest paths? Draw an example. [2] (ii) For a 16 x 21 maze, what is the maximum difference between the longest and shortest paths? You are not required to draw an example. [3]

6 © UCLES 2012 9694/33/O/N/12 4 Study the information below and answer the questions. Show your working. The Rodstein Musical Society is staging its production of North Atlantic at the Carousel Theatre each night this week from Monday to Saturday, with an extra performance tomorrow afternoon (Saturday). Tickets are priced at $15 and $12, except Monday when all tickets were $10. All the tickets for tonight (Friday) and tomorrow night have been sold already, but there are still some available for tomorrow afternoon. This is the theatre’s seating plan for tomorrow afternoon’s performance. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 A 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 B 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 C 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 D 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 E 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 1 2 3 4 5 6 F G H I J K L M N O P Q R S T F G H I J K L M N O P Q R S T 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 STAGE Rows A – E: $15 Rows F – T: $12 indicates ticket sold

7 © UCLES 2012 9694/33/O/N/12 Two $12 tickets have just been sold to a customer, who chose two seats next to each other, as far forward as possible. They are about to be marked off as sold on the seating plan. (a) (i) In which row are the two seats that have just been sold? [1] (ii) How many $15 tickets have been sold so far for tomorrow afternoon’s performance? [1] (iii) How many $12 tickets have now been sold for tomorrow afternoon’s performance? [1] This is a summary of the sales of the tickets and programme booklets from the performances earlier this week. Number of tickets sold $15 $12 $10 Total Ticket Sales Programme Sales Monday – – 344 $3440 $279 Tuesday 102 273 – $384 Wednesday 115 – $5433 $351 Thursday 136 368 – $6456 $438 Two figures have been left out. (b) (i) What was the income from total ticket sales for Tuesday’s performance? [1] (ii) How many $12 tickets were sold for Wednesday’s performance? [2] The Society’s funds will receive all the profit from this week’s performances of North Atlantic. Its profit is obtained by subtracting the hire of the theatre and all other costs (except those related to programme booklets) from the income from total ticket sales. The selling of programme booklets at the performances is organised and run as a separate venture, with all the profit to be donated to a local youth club. The cost of hiring the theatre is $1500 per day, plus $2 per ticket sold. (c) (i) If no more tickets are sold for tomorrow afternoon’s performance, what will be the total cost of hiring the theatre this week? [3] (ii) What is the minimum profit the Society will make from this week’s production when the other costs are $4360? [3] Sales of programme booklets, at $3 each, have been very disappointing. 3000 were printed at a cost of $1200. $2470 was raised for the youth club last year, and this year’s target is to beat this figure. At an emergency meeting of the Programme Committee this morning, it was acknowledged that if sales do not improve, the amount raised will fall some way short of the target. It was decided to reduce the price, and that the new price should be the minimum multiple of $0.10 that will achieve the target if at least half of the remaining programme booklets are sold. (d) What price did the Programme Committee set? [3]

8 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. © UCLES 2012 9694/33/O/N/12 BLANK PAGE

CAMBRIDGE INTERNATIONAL E×AMINATIONS GCE Advanced Level MARK SCHEME for the October/November 2012 series 9694 THINKING SKILLS 9694/33 Paper 3 (Problem Analysis and Solution), maximum raw mark 50 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 October/November 2012 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9694 33 © Cambridge International E×aminations 2012 1 (a) What will Maggie’s and Nigel’s scores be when they each play a game against Luke? [2] Nigel 15 points, Maggie 14 points. 1 mark for each correct score. (b) In some games, one player may have scored enough points to ensure a win before all rounds have been completed. In which of the three games between Luke, Maggie and Nigel does this occur, and after which round? [2] Luke versus Nigel. It is decided in the eleventh round. 1 mark for the correct players and 1 mark for the correct round. (c) When Luke, Maggie and Nigel each play against Ophelia, which of the three scores the least and what is that score? [2] Maggie loses 1 – 23 against Ophelia. The following table shows the results, with the player on the left’s score shown first. M N O L 10 – 14 9 – 15 12 – 12 M 11 – 13 1 – 23 N 7 – 17 1 mark for Maggie, 1 mark for her score (1). 1 mark for one player’s score for each of the three games calculated correctly. (d) Pedro is going to play against Ophelia. Give a sequence of letters that he could play which would ensure that she scored nothing. [1] C B A repeated Ophelia A C B A C B A C B A C B Pedro C B A C B A C B A C B A

Page 3 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9694 33 © Cambridge International E×aminations 2012 (e) Quentin joins in and plays against Luke, Maggie and Nigel. Suggest a sequence of 12 letters which would produce the maximum total score for Quentin over the three games. [3] Quentin’s options are shown in the table below, where * can be any letter. Luke B A B A C A C A A A C A Maggie A B C A B C A B C A B C Nigel A B A B B B C A C A B C Quentin C/A A * C/A A/B * B/C C/A B/C C A/B B/C 3 marks for a correct string of letters. If 3 marks cannot be awarded, award 2 marks for a string of letters with one mistake. 1 mark for a string with two mistakes. 2 (a) What is the missing digit (#) in 4920 1641 01#4 1711 ? [2] 4 9 2 0 1 6 4 1 0 1 # 4 1 7 1 1 8+9+4+0 + 2+6+8+1 + 0+1+ +4 + 2+7+2+1 = 55. 8+9+4+0 + 2+6+8+1 + 0+1+5+4 + 2+7+2+1 = 60. To make multiple of 10, we must add 5. As the missing digit is in an odd position, we need whatever turns into 5, namely 7. 2 marks for “7” (working essential). 1 mark for getting 5, or for 1 (by using transformation rather than inverse transformation). 1 mark for one consistently applied mistake [e.g. converting all the numbers according to the table: 8+9+4+0+2+3+8+2+0+2+8+2+5+2+2 = 57; so missing digit is 6. (b) Which pair of digits will not be detected if they are switched? [1] It is easiest to consider how much the odd digits are changed (mod 10): 0 1 2 3 4 5 6 7 8 9 0 2 4 6 8 1 3 5 7 9 +0 +1 +2 +3 +4 +6 +7 +8 +9 +0 The only adjacent digits that can be swapped over are 0 and 9.

Page 4 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9694 33 © Cambridge International E×aminations 2012 (c) How many of these fourteen mistakes will be detected, if just one of them occurs when the sixteen-digit number of a card is read? [2] All of them (2 marks – working essential). Either check them all: 13 30 14 40 15 50 16 60 17 70 18 80 19 90 23 60 24 80 25 10 26 30 27 50 28 70 29 90 5 6 6 8 7 1 8 3 9 5 0 7 1 9 Or observe that 12 and 20 would clash (2+2 4+0) so other 1x x0 would not. 1 mark for checking at least two of the pairs (e.g. 13 & 30, and 14 & 40) (d) In how many of the sixteen positions could the check digit be located? (Assume nothing other than the information given.) [1] Any position could be the necessary check digit because the mapping of the odd positions is invertible (accept 16). (e) Four digits of a card number were blanked out “for security reasons”. 4579 3991 #### 2607 How many possible valid card numbers could this represent? [2] 1 mark for appreciating that three digits could be anything, and then the fourth is fixed. 2 marks for 10 × 10 × 10 = 1000. SC: 1 mark for 10 000 (f) A national identity register using sixteen-digit numbers for individuals was proposed, and it was suggested that the numbers should be allocated so that they could not be confused with a credit card number. Specify a simple way to achieve this (without restricting the range of available credit card numbers). [2] To guarantee no clashes, it suffices to have the remainder mod 10 as something fixed but non-zero: 2 Marks. Any clear system that would distinguish the national identity card system from the credit card system but would restrict the available credit card numbers, e.g. “ID numbers start with 0”, scores 1 mark. OR 1 mark for recommending using the algorithm, and then altering the details of the check digit process which produces numbers which will work in some cases.

Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9694 33 © Cambridge International E×aminations 2012 3 (a) Show that the complexity coefficient of the following maze is 1.44 (rounded to 2 decimal places), and complete the path that the participant takes: 1, 2, 1, 4… [2] 1 mark for correct completion of the sequence: 1, 2, 1, 4, 5, 6, 3, 6, 5, 4, 7, 8, 9 1 mark for seeing the calculation: 13 ÷ 9 = 1.44 (b) Draw a different maze involving a 3 × 3 grid of rooms, which also has a complexity coefficient of 1.44 and which has its entrance and exit in the same place as the maze above. [1] 7 8 9 4 5 6 1 2 3 1 mark for a different maze with the same entrance and exit, which involves a longest path of 13. Allow entrance and exit in the other walls of square 1 and 9, or entrance & exit reversed. (c) (i) The maximum complexity coefficient for a 3 × 3 maze is 1.78. Draw an example. [2] 2 marks for an example which includes a longest path of 16; this requires each room to be visited to twice, except the end of the ‘deadend pathway’ and the exit. The exit and entrance must therefore be on adjacent squares. For example: 7 8 9 4 5 6 1 2 3 7 8 9 4 5 6 1 2 3 7 8 9 4 5 6 1 2 3 1 mark for a maze with complexity coefficient which is greater than 13/9, or a maze in which the entrance and exit are in adjacent rooms, but the interior walls are incomplete/confused. (ii) Find the maximum complexity coefficient for a 5 × 5 maze. [2] Maximum complexity coefficient = ((23 × 2) + 2)/(5 × 5) = 48/25 = 1.92 [2 marks] If 2 marks cannot be awarded, award 1 mark EITHER for seeing the number 48, OR for evidence that the candidate appreciates that all but one or two of the rooms in the maze must be visited (at least) twice OR a correctly drawn 5 × 5 maze (with adjacent entrance and exit) OR a correctly generalized case from their (c)(i).

Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9694 33 © Cambridge International E×aminations 2012 (d) Consider rectangular (including square) mazes made up from no more than 18 rooms. What is the lowest simplicity coefficient for such mazes? Give an example of a maze with this simplicity coefficient. [3] Simplicity coefficient of any 4 × 4 maze with a shortest path of 7 is 7/16 = 0.4375. This minimises (L + W – 1) ÷ (LW). An example of such a maze is given below: 14 13 9 5 15 16 10 11 12 6 7 8 1 2 3 4 Award 1 mark for an example of a rectangular maze (of any dimensions permissible given the size) which includes a ‘shortest path’ (e.g. a corridor around the outside – or of the same length, but down the middle). Award 1 mark for the 4 × 4 case, or appropriate variants, as shown above. Award 1 mark for an appropriately calculated simplicity coefficient that is not equal to 1. (e) (i) For a maze made up from no more than 18 rooms, what is the maximum difference between the longest and shortest paths? Draw an example. [2] Maximum difference = 20, achievable in a 6 × 3 maze, as shown below. Longest path for a maze = 28; shortest path for this maze = 8. 16 15 9 3 17 18 10 11 12 4 5 6 14 13 7 1 8 2 2 marks for the answer (20) and an appropriate maze. If 2 marks cannot be awarded, award 1 mark for an appropriate maze with no clear calculation of the difference, or a maze and difference correctly calculated for the 4 × 4 solution (whose difference is 25 – 7 = 18), or a different rectangular maze (not 3 × 6 or 4 × 4), or “28 – 8 = 20” seen with an incomplete/empty maze.

Page 7 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9694 33 © Cambridge International E×aminations 2012 (ii) For a 16 × 21 maze, what is the maximum difference between the longest and shortest paths? You are not required to draw an example. [3] Maximum difference = 2 × (15 × 20) = 600 [3 marks] If 3 marks cannot be awarded, award 1 mark for the number 36 [the shortest path] and 1 mark for an attempt to calculate length of the longest path in the 15 × 20 subsection of the maze. Alternatively, 2 marks for 300 OR any number between 598 to 602 seen. 4 (a) (i) In which row are the two seats that have just been sold? [1] Answer: (Row) L (ii) How many $15 tickets have been sold so far for tomorrow afternoon’s performance? [1] Answer: 162 (iii) How many $12 tickets have now been sold for tomorrow afternoon’s performance? [1] Answer: 376 + 2 = 378 (b) (i) What was the income from total ticket sales for Tuesday’s performance? [1] Answer: $4806 (ii) How many $12 tickets were sold for Wednesday’s performance? [2] Answer: 309 If 2 marks cannot be awarded, award 1 mark for subtracting $1725 (115 × $15) from $5433 or an answer of ($)3708. (c) (i) If no more tickets are sold for tomorrow afternoon’s performance, what will be the total cost of hiring the theatre this week? [3] Answer: $16 014 Accept a correctly calculated answer that carries forward one or more incorrect answers from (a) and (b). Strict follow-through marks awarded. If 3 marks cannot be awarded, award 1 mark each for inclusion of the following (maximum 2 marks): • $9000 for 6 days’ hire. • 660 tickets sold for Friday and Saturday nights. • 540/(162 + 378) tickets sold for Saturday afternoon. • Calculated total number of tickets sold multiplied by $2.

Page 8 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 CODE # © Cambridge International E×aminations 2012 (ii) What is the minimum profit the Society will make from this week’s production when the other costs are $4360? [3] Answer: $23 647 This is $20 135 (Monday – Thursday) + $8460 (Friday) + $8460 (Saturday night) + $6966 (Saturday afternoon) – $16 014 (theatre hire) – $4360 (other expenses) Accept a correctly calculated amount that carries forward one or more incorrect answers from (a), (b) and (c). If 3 marks cannot be awarded, award 2 marks for (total ticket sales of) $44 021. OR Award 1 mark for evidence of each of the following (maximum 2 marks): • $20 135 ticket sales Monday to Thursday (e.g. $3440 + $4806 + $5433 + $6456 seen within calculation). • $8460 ticket sales Friday and/or Saturday night (e.g. $8460 or $16 920 seen within calculation). • $6966 ticket sales Saturday afternoon (e.g. $6966 seen within calculation) – or their FT ticket sales. (Where appropriate, accept amounts correctly calculated using carried forward figures.) (d) What price did the Programme Committee set? [3] Answer: $1.80 Total sales so far = $1452, so profit is $252 at present, $2218 short of the target. Number of programmes sold so far = 484 (1452/3), so there are 2516 left, of which the aim is to sell 1258. For 3 marks to be awarded, appropriate working must be sighted, as well as the answer $1.80. If 3 marks cannot be awarded, award 2 marks for evidence of $2218/1258 or an answer of $1.76 or $1.80. OR Award 1 mark for evidence of each of the following: • (They aim to raise another) $2218 • (They aim to sell a further) 1258 programmes (or 2516 programmes)