Develop a model — Paper 3 — A Level Thinking Skills 9694 topical questions

Question 1

Square Deal is a game for two players, played over a number of rounds. In each round both players have a 4 × 4 grid onto which numbered til…

Accepted Answer

4(a) 249 (points) (12 + 72 + 62 + 22 and 72 + 62 + 72 + 52) 3 If 3 marks cannot be awarded, award 1 mark for each of the following: • Identification of all three lines (and no others) that add up to square number totals: (4,6,3,3) (1,7,6,2) and (7,6,7,5). • Correct calculation of the value of at least one of the three lines: 70, 90 and 159 points respectively. 4(b) 2 and 4 with justification 3 Award 1 mark the correct pair chosen Award 1 mark each for recognition of the following: • 0 would have allowed him to make a Square Deal. • 7 would have produced a column value of 90 points (or “greater than 84”) 4(c)(i) 352 (82 + 82 + 72 + 22 and 82 + 72 + 72 + 32) 3 Award 2 marks for an answer of 340 or more. (340 fails to appreciate that one of the 8s can be used in a row and a column, and is made up of 82 + 82 + 72 + 22 and 72 + 72 + 62 + 52.) Award 1 mark for sight of 181 or identification of (8,8,7,2) as the best possible line OR for a row and a column that each add up to 25. 4(c)(ii) Award 1 mark for any complete grid that does not contain any numbers that 2 would not be allowed (e.g. three 0s, four 1s, one or more 9s etc.). Award 1 mark for any grid (even if incomplete) that would produce 352 or the candidate’s answer to (c)(i) – provided it is more than 158. 4(d) 1 placed in third row of Russell’s grid and 5 placed in bottom row of 4 Gordon’s grid + explanation Award 1 mark for stating (or indicating clearly in some other way) that the 1 should be placed in the third row down (or other suitable description) of Russell’s grid and the 5 should be placed in the bottom row of Gordon’s grid. Award 1 mark each (up to a maximum of 3) for any of the following observations: • The four tiles left in the bag are 3, 4, 4 and 6. OR There is still a 3 in the bag, but no 2 or 8. • (So) placing 5 in Gordon’s bottom row guarantees that his score for the round will be 0. • 1 on Russell’s grid (in third row/second column) guarantees a score for the round (02 + 62 + 12 + 22 = 41). • Unless Gordon takes the 3 from the bag, he will have to place a number in Russell’s last square (4 or 6) that will create a row that scores points (82 + 62 + 72 + 42 = 165) OR a column that is better than the one already in place (52 + 62 + 12 + 42 = 78). SC1: if no other marks can be awarded, award 1 mark for stating that tile 1 should be placed in the third row of Russell’s grid, AND tile 5 can go in either of Gordon’s empty spaces.

Question 2

Richard owns a company which produces a range of different chocolates. He owns a shop where he sells standard boxes of the chocolates. He a…

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3(a) 3 minutes for each of the boxes, plus a total of 28 minutes for the 56 2 chocolates. 34 minutes in total. Allow 15 minutes and 19 minutes both given (the times for each box). 1 mark for either 28 minutes for the 56 chocolates or the total time for either box calculated (15 or 19 minutes). 3(b) The total is 4 × 48 + 18 = 210 chocolates. 3 The largest boxes (48 chocolates) take 27 minutes to pack. [1 mark] There are two hours available, so 4 boxes can be packed, (with an extra 12 minutes left). [1 mark] Award 1 mark for an attempt to find a set of at most 5 boxes that can be packed in the 2 hours available. SC1: 234 chocolates in one box. 3(c) To reach the correct proportions requires packing 10 additional medium 3 boxes and 4 additional large boxes, (which will take a total of 120 minutes). [1 mark] Packing 1 small, 2 medium and 1 large box will take a total of 30 = (4 +8 + 8 + 10) minutes. [1 mark] There will be enough time to pack 12 additional sets of 1 small, 2 medium and 1 large, so in total Richard should pack 12 small, 34 medium and 16 large boxes. [1 mark] 3(d) 84 boxes 3 The assistant is considerably slower packing small and large boxes compared to Richard, but only slightly slower packing medium boxes, so the assistant should be assigned to pack medium boxes. (In 3 hours, 20 boxes can be packed.) [1 mark] It will take Richard 140 minutes to pack 10 boxes each of small and large, leaving 340 minutes more packing time. [1 mark] In 340 minutes, 11 sets in the ratio 1 : 2 : 1 can be packed, so in total 20 + 20 + 11 × 4 = 84 boxes. [1 mark] SC1: Has both pack in the ratio 1 : 2 : 1, obtains total of 80 boxes. OR SC1: Assistant only takes 40 minutes, so can only complete 4 sets/16 boxes. 3(e) With Richard working on his own the number of boxes in stock has reduced 4 by 11 small, 18 medium and 10 large boxes over the week. [1 mark] It would be most efficient for Richard to package the small and large boxes and the assistant to package the medium boxes, plus some of the medium boxes that Richard had been packaging. Richard needs 11 × 4 + 10 × 10 = 144 minutes to package the additional small and large boxes. [1 mark] This means that Richard will be able to package 144 ÷ 8 = 18 fewer medium boxes, which will need to be packaged by the assistant (in addition to the 18 other medium boxes that are required). [1 mark] OR (SC1) If assistant only for 11S 18M 10L, 77 + 162 + 150 = 389 minutes, (rounding up to 7 hours). A total of 36 medium boxes will require 36 × 9 = 324 minutes, so the assistant should be employed for 6 hours each week since it must be a whole number of hours. [1 mark for rounding up their answer]

Question 3

Richard owns a company which produces a range of different chocolates. He owns a shop where he sells standard boxes of the chocolates. He a…

Accepted Answer

3(a) 3 minutes for each of the boxes, plus a total of 28 minutes for the 56 2 chocolates. 34 minutes in total. Allow 15 minutes and 19 minutes both given (the times for each box). 1 mark for either 28 minutes for the 56 chocolates or the total time for either box calculated (15 or 19 minutes). 3(b) The total is 4 × 48 + 18 = 210 chocolates. 3 The largest boxes (48 chocolates) take 27 minutes to pack. [1 mark] There are two hours available, so 4 boxes can be packed, (with an extra 12 minutes left). [1 mark] Award 1 mark for an attempt to find a set of at most 5 boxes that can be packed in the 2 hours available. SC1: 234 chocolates in one box. 3(c) To reach the correct proportions requires packing 10 additional medium 3 boxes and 4 additional large boxes, (which will take a total of 120 minutes). [1 mark] Packing 1 small, 2 medium and 1 large box will take a total of 30 = (4 +8 + 8 + 10) minutes. [1 mark] There will be enough time to pack 12 additional sets of 1 small, 2 medium and 1 large, so in total Richard should pack 12 small, 34 medium and 16 large boxes. [1 mark] 3(d) 84 boxes 3 The assistant is considerably slower packing small and large boxes compared to Richard, but only slightly slower packing medium boxes, so the assistant should be assigned to pack medium boxes. (In 3 hours, 20 boxes can be packed.) [1 mark] It will take Richard 140 minutes to pack 10 boxes each of small and large, leaving 340 minutes more packing time. [1 mark] In 340 minutes, 11 sets in the ratio 1 : 2 : 1 can be packed, so in total 20 + 20 + 11 × 4 = 84 boxes. [1 mark] SC1: Has both pack in the ratio 1 : 2 : 1, obtains total of 80 boxes. OR SC1: Assistant only takes 40 minutes, so can only complete 4 sets/16 boxes. 3(e) With Richard working on his own the number of boxes in stock has reduced 4 by 11 small, 18 medium and 10 large boxes over the week. [1 mark] It would be most efficient for Richard to package the small and large boxes and the assistant to package the medium boxes, plus some of the medium boxes that Richard had been packaging. Richard needs 11 × 4 + 10 × 10 = 144 minutes to package the additional small and large boxes. [1 mark] This means that Richard will be able to package 144 ÷ 8 = 18 fewer medium boxes, which will need to be packaged by the assistant (in addition to the 18 other medium boxes that are required). [1 mark] OR (SC1) If assistant only for 11S 18M 10L, 77 + 162 + 150 = 389 minutes, (rounding up to 7 hours). A total of 36 medium boxes will require 36 × 9 = 324 minutes, so the assistant should be employed for 6 hours each week since it must be a whole number of hours. [1 mark for rounding up their answer]

Question 4

Five days every week, Faridah sells ice cream at the beach from a box on the back of her bicycle. Each morning she cycles from her home to …

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Question Answer Marks 1(a) The scenic route loses 20 × 25 = 500 g, whereas the direct route loses 2 60 × 10 = 600 g. Award 1 mark for either of these masses, or for ‘scenic by 100 g’. No marks for unsupported answer. 1(b) With the better-insulated box, the direct route loses 10 × 20 = 200 g, 1 whereas the scenic route loses 5 × 50 = 250 g. So Faridah could save 500 g – 200 g = 300 g. 1(c) It would take 2 days, each day saving 300 g of ice cream, to recoup $1, so it 2 would take 30 days to recoup the $15, which corresponds to 6 weeks. Award 1 mark for 50¢ per day or equivalent. 1(d) In terms of ice-cream saving, the $0.80 bus fare corresponds to 2 600 × 0.8 g = 480 g of ice cream. Currently, Faridah is losing 500 g of ice cream a day, so the bus journey would be an improvement if she lost less than 20 g of ice cream on the journey, which means that the journey would have to take less than 20 minutes. Award 1 mark for comparing the net cost of taking a bus journey (for an arbitrary number of minutes) and the net cost of not; this includes consideration of 0 minutes, which reduces to the equivalence between the bus fare and 480 g of ice cream OR for an algebraic representation: (t/600) + 0.8 = 500/600. 1(e) The total quantity needed is 2300 g 3 1 mark for any comparison of profit from n normal ice-creams on scenic route (500 g lost, $1 for 600 g) with profit from n luxury ice creams on direct route (800 g lost, $1 for 500 g). e.g.: 800 g of luxury gives no money, whereas 800 g of normal yields 50 cents. 1 further mark for any improved comparison of quantities, or a comparison of rates (e.g. every additional 300 g of ice cream would be sold for 10¢ more if it is the better quality ice cream, so 1500 g is needed to compensate for the 50¢ loss). Alternatively: (q – 500)/600 = (q – 800)/500 [2 marks; 1 mark for either side correct]

Question 5

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 …

Accepted Answer

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).

Question 6

Strategy is a TV quiz show. Four contestants take part in each edition. Throughout the show every question is presented simultaneously to t…

Accepted Answer

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]

Question 7

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 …

Accepted Answer

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).

Question 8

Strategy is a TV quiz show. Four contestants take part in each edition. Throughout the show every question is presented simultaneously to t…

Accepted Answer

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]

Question 9

Leon is an architect and he uses coloured plastic building blocks to help him model new housing developments. Each building block measures …

Accepted Answer

3(a) Suitable diagram with suitable calculation. 1 3(b) 8 for base × 5 for height, so 40 1 3(c) Type A house: 60 Red + 8 Green; Type B house: 40 Blue + 5 Green 3 Number of blocks required = 1200 Red, 800 Blue and 260 Green Cost = (2 × $55 + $22) + (4 × $22) + ($22 + 2 × $6) = $254 ft 1 mark for each brick colour, correctly calculated (3 marks, only awarded if correctly summed): Green bricks cost $34 Red bricks cost $132 Blue bricks cost $88 If no marks awarded for brick costs, award 1 mark for calculating the best cost for a specified number of bricks, greater than 500 (Type A = $156 and Type B = $122). 3(d) $200 = 3 × $55 + 1 × $22 + 2 × $6 + $1 2 OR 2 × $55 + 4 × $22 + $2 OR 1 × $55 +6 × $22 + 2 × $6 + $1 OR 9 × $22 + $2 [1] Number of blocks = 1800 so number of Type A houses = 1800/60 = 30 Award 1 mark for 1800 seen without supporting working 3(e) In the model, 2 Area of Type A plot is 8 × 4 + 16 = 48 cm2 (for 192 m2) Area of Type B plot = 5 × 5 + 10 = 35 cm2 (for 140 m2) Area for 20 of each type = 20 × 83 = 1660 cm2 Actual area = 1660 × 4 = 6640 m2 1 mark for 48 or 35 or 192 or 140 or 1660 soi SC: 1 mark for 26560 m2 3(f)(i) Area of 10 houses = 5 × their 48 + 4 × their 35 + 100 = 480 (cm2) [1] 4 Actual area = 4 × their 480 = 1920 (m2) OR available area = 30 000/4 = 7500 (cm2) [1] Divide 30 000 by their 1920 OR 7500 by their 480 (= 15.625) [1] Number of houses must be a whole number, so 15 × 5 = 75 Type A 15 × 4 = 60 Type B 15 × 1 = 15 Type C ft [1] Award final ft if ratio is calculated using the integer part of 30 000/their area for plot of 10 houses. 3(f)(ii) 75 Type A, 60 Type B and 15 Type C have area 1920 × 15 = 28 800 ft [1] 2 Area left for car parking is 30 000 – 28 800 = 1200 square metres ft [1] OR (0.625/15.625) = proportion unused ft [1] × 30 000 = 1200 square metres ft [1]

Question 10

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 nu…

Accepted Answer

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

Question 11

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. …

Accepted Answer

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

Question 12

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 a…

Accepted Answer

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

Question 13

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 nu…

Accepted Answer

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

Question 14

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. …

Accepted Answer

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

Question 15

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 a…

Accepted Answer

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

Question 16

Birdnest village school educates children for 5 years. All the children from the village of Birdnest attend the school; they are known as N…

Accepted Answer

2(a) An increase of 10 each year would result in a total of 35 + 5 × 10 = 85. 1 2(b)(i) An extra 20 Cuckoos resulted in an extra 10 cars, so 50%. 1 2(b)(ii) If all 35 cars in year zero are from Cuckoos, there are 70 Cuckoos before 1 the expansion. Thus there would be 5 × 20 – 70 = 30. 2(c) Stopping fourth year children resulted in 55 – 52 = 3 fewer cars than with no 3 change, so 3 fifth year Nesters no longer coming by car. [1] This means there were originally 15 Nesters coming by car and 20 outsiders. [1] Hence Cuckoos were 40 of the children. The remaining 60 would be Nesters. [1] 2(d) Half of Cuckoos and quarter of Nesters come by car, but only those not in 2 the last year. The expansion of Cuckoos hasn’t reached the final year yet, so 32 + 60 Cuckoos not in last year. 12 + 46 = 58 1 mark for first step of method: number of Nesters (48) OR Cuckoos (92) not in final year OR 1 mark for method to find number of cars used by those in their final year (N/4 + C/2) SC: 1 mark for 62 (= 12 + 50), ignoring proportion of Cuckoos in last school year Alternatively: There would be 10 extra cars, but 4 final year Cuckoos no longer drive, so 52 + 6 = 58 2(e) There would be 68 cars in both the fourth and fifth year (since the 20 2 Cuckoos from the first year would now reach their last year.) But, if any Cuckoos could come by car, half of the 140, i.e. 70 would, so this is not fewer. 1 mark for 68 or 70 seen. Year of 0 1 2 3 4 5 expansion Nesters 60 60 60 60 60 60 Cuckoos 40 60 80 100 120 140 Cuckoos not in 32 52 72 92 112 112 final year Nesters by car 15 15 12 12 12 12 Cuckoos by 20 30 40 46 56 56 car Total cars 35 45 52 58 68 68

Question 17

Jaspreet owns a business making suits to order – he only makes the suits once a customer has ordered them. Customers can order any number o…

Accepted Answer

3(a) Trousers bought with jacket: $115 1 Additional pair of trousers: $40 Total price = $155 3(b) The quickest way to complete the order is for one tailor to make the jacket 1 (20 hours) and one tailor to make the trousers (20 hours in total). Therefore 20 hours are needed in total. 16 hours of work will be completed on Tuesday and Wednesday, so the items will be ready on Thursday. 3(c) Four working days is a total of 32 hours, so each tailor can make either 3 3 pairs of trousers ($120 each, so $240) 1 pair of trousers and 1 jacket ($125 – discount $10 each, so $230) 2 waistcoats ($100 each, so $200) The maximum total price would be $240. If 3 marks cannot be awarded, award 1 mark for (max 2): calculating the income per hour for two of the three items ($4, $4.25, $3.33) OR correctly calculating one of the three options above (120/240, 125/250, 100/200) correctly applying the discount (115/230) SC: 2 marks for an answer of $250 for 1 trousers and 1 jacket (forgetting the discount) OR an answer of $120 (forgetting there are two tailors) 3(d) The total time for the order is 5 × 10 + 7 × 20 + 3 × 15 = 235 hours. [1] 3 This means that the shortest time is 120 hours. [1] One way to achieve this would be for Harry to do 6 jackets and Joe to do 5 trousers, 1 jacket and 3 waistcoats Alternative: Harry does 1 trouser, 4 jackets and 2 waistcoats, and Joe does 4 trouser, 3 jacket and 1 waistcoat [1] 3(e) The total time needed for completing the orders is 7 hours for the order that 3 is still in progress, 10 hours, 25 hours and 45 hours for the other three orders, making a total of 87 hours for all of the work. [1] There is a total of 2 × 5 × 8 = 80 hours available if no extra hours are worked, so 7 extra hours will be needed. [1] If Harry is given 40 hours from the three orders (so that he has 4 extra hours), this will leave Joe with 3 extra hours. For example, Harry could be allocated two pairs of trousers followed by the jacket, and Joe the two waistcoats and a pair of trousers. [1] Alternatively, for solutions using scheduling: A schedule which allows for the non-urgent orders to be completed on time. [1] A schedule in which both tailors are occupied for the full 40 hours of the normal week. [1] Answer of 7 hours. [1] 3(f) If Harry can’t work any extra hours then he needs to have work allocated 1 that gets him as close as possible to his 40 hours. After he has completed the 4 hours to finish the waistcoat he can be allocated 35 hours of work to make a total of 39 in the week. 8 extra hours will be needed. 3(g) Since there are 7 hours more work needed than are available by the end of 3 the week, one of the orders must be completed on Monday. Delaying any one order to be finished on Monday will allow the others to be completed on time. [1] The order that is due on Wednesday would need a 40% reduction. The discount would be $16. The order that is due on Friday would need a 20% reduction. The discount would be $18. [1 for the value of either discount] The best option involves losing $16.

Question 18

David is trying to work out the bonuses that he will pay to his employees for their work over the past six months. The city in which they w…

Accepted Answer

4(a) North: 80 + 99 = 179 1 East: 95 + 92 = 187 South: 36 + 53 = 89 West: 86 + 85 = 171 The most sales took place in East zone 4(b) Carol had the highest sales in Mar 2 and the second highest sales in Jan and Apr Total bonuses were 2 × $50 + $150 = $250 1 mark for an answer showing an incorrect judgement for ONE of Carol’s monthly bonuses: e.g. 50 + 150 = $200 or 50 + 50 + 50 + 150 = $300. 4(c) Tanya works in the West zone, so the first 10 sales are worth 10 points in 2 total. [1] The remaining 10 sales are worth 2 points each, so the total is 30 SC: 1 mark for 40 or ‘2 each’ 4(d) Neither North zone employee will receive any bonus 4 Neither East zone employee will receive any bonus In the South zone all sales were worth 2 points, so Rachel will have a bonus of $600 and John will not get a bonus. In the West zone, both employees will receive bonuses. Bonuses will be awarded to Rachel, Oliver and Tanya [1] (dependent on no others identified) Rachel had a bonus of $600 [1] Oliver had a bonus of $1200 [1] Tanya had a bonus of $1100 [1] SC: 1 mark for identification that North and East zone employees do not receive bonuses; may be implied by correct points totals for A, C, F and M seen. 4(e) Martin would have received bonuses for most sales in 2 of the months and 3 second highest in 1 of the months, which would have been $350. He would not have received any bonuses from the points. Therefore his total bonus in the old system was $350. [1] Under the new system, Martin would have earned 47 points in the first three months and then 52 points in the second three months, so would receive no bonus for the first three months and $200 in the second three months. [1] Martin’s total bonus would be $150 less. 4(f) A bonus of $1000 requires a total of 60 points for the three month period. 3 Since Oliver is in the West zone he would achieve 30 points from 10 sales every month and would only need an additional 5 sales per month (at 2 points each) to reach 60 points. Oliver’s minimum target would be 15 sales per month. [1] Since John is in the South zone he can achieve 60 points by making 10 sales per month (at 2 points each). [1] Oliver would need to make 5 sales more per month than John. SC: 2marks for 15 difference in total sales (rather than number per month)

Question 19

The Järnvägar Railway Company wants to build a level line between Vänster and Höger. In some places the land is higher than the required le…

Question 20

Birdnest village school educates children for 5 years. All the children from the village of Birdnest attend the school; they are known as N…

Accepted Answer

2(a) An increase of 10 each year would result in a total of 35 + 5 × 10 = 85. 1 2(b)(i) An extra 20 Cuckoos resulted in an extra 10 cars, so 50%. 1 2(b)(ii) If all 35 cars in year zero are from Cuckoos, there are 70 Cuckoos before 1 the expansion. Thus there would be 5 × 20 – 70 = 30. 2(c) Stopping fourth year children resulted in 55 – 52 = 3 fewer cars than with no 3 change, so 3 fifth year Nesters no longer coming by car. [1] This means there were originally 15 Nesters coming by car and 20 outsiders. [1] Hence Cuckoos were 40 of the children. The remaining 60 would be Nesters. [1] 2(d) Half of Cuckoos and quarter of Nesters come by car, but only those not in 2 the last year. The expansion of Cuckoos hasn’t reached the final year yet, so 32 + 60 Cuckoos not in last year. 12 + 46 = 58 1 mark for first step of method: number of Nesters (48) OR Cuckoos (92) not in final year OR 1 mark for method to find number of cars used by those in their final year (N/4 + C/2) SC: 1 mark for 62 (= 12 + 50), ignoring proportion of Cuckoos in last school year Alternatively: There would be 10 extra cars, but 4 final year Cuckoos no longer drive, so 52 + 6 = 58 2(e) There would be 68 cars in both the fourth and fifth year (since the 20 2 Cuckoos from the first year would now reach their last year.) But, if any Cuckoos could come by car, half of the 140, i.e. 70 would, so this is not fewer. 1 mark for 68 or 70 seen. Year of 0 1 2 3 4 5 expansion Nesters 60 60 60 60 60 60 Cuckoos 40 60 80 100 120 140 Cuckoos not in 32 52 72 92 112 112 final year Nesters by car 15 15 12 12 12 12 Cuckoos by 20 30 40 46 56 56 car Total cars 35 45 52 58 68 68

Question 21

Jaspreet owns a business making suits to order – he only makes the suits once a customer has ordered them. Customers can order any number o…

Accepted Answer

3(a) Trousers bought with jacket: $115 1 Additional pair of trousers: $40 Total price = $155 3(b) The quickest way to complete the order is for one tailor to make the jacket 1 (20 hours) and one tailor to make the trousers (20 hours in total). Therefore 20 hours are needed in total. 16 hours of work will be completed on Tuesday and Wednesday, so the items will be ready on Thursday. 3(c) Four working days is a total of 32 hours, so each tailor can make either 3 3 pairs of trousers ($120 each, so $240) 1 pair of trousers and 1 jacket ($125 – discount $10 each, so $230) 2 waistcoats ($100 each, so $200) The maximum total price would be $240. If 3 marks cannot be awarded, award 1 mark for (max 2): calculating the income per hour for two of the three items ($4, $4.25, $3.33) OR correctly calculating one of the three options above (120/240, 125/250, 100/200) correctly applying the discount (115/230) SC: 2 marks for an answer of $250 for 1 trousers and 1 jacket (forgetting the discount) OR an answer of $120 (forgetting there are two tailors) 3(d) The total time for the order is 5 × 10 + 7 × 20 + 3 × 15 = 235 hours. [1] 3 This means that the shortest time is 120 hours. [1] One way to achieve this would be for Harry to do 6 jackets and Joe to do 5 trousers, 1 jacket and 3 waistcoats Alternative: Harry does 1 trouser, 4 jackets and 2 waistcoats, and Joe does 4 trouser, 3 jacket and 1 waistcoat [1] 3(e) The total time needed for completing the orders is 7 hours for the order that 3 is still in progress, 10 hours, 25 hours and 45 hours for the other three orders, making a total of 87 hours for all of the work. [1] There is a total of 2 × 5 × 8 = 80 hours available if no extra hours are worked, so 7 extra hours will be needed. [1] If Harry is given 40 hours from the three orders (so that he has 4 extra hours), this will leave Joe with 3 extra hours. For example, Harry could be allocated two pairs of trousers followed by the jacket, and Joe the two waistcoats and a pair of trousers. [1] Alternatively, for solutions using scheduling: A schedule which allows for the non-urgent orders to be completed on time. [1] A schedule in which both tailors are occupied for the full 40 hours of the normal week. [1] Answer of 7 hours. [1] 3(f) If Harry can’t work any extra hours then he needs to have work allocated 1 that gets him as close as possible to his 40 hours. After he has completed the 4 hours to finish the waistcoat he can be allocated 35 hours of work to make a total of 39 in the week. 8 extra hours will be needed. 3(g) Since there are 7 hours more work needed than are available by the end of 3 the week, one of the orders must be completed on Monday. Delaying any one order to be finished on Monday will allow the others to be completed on time. [1] The order that is due on Wednesday would need a 40% reduction. The discount would be $16. The order that is due on Friday would need a 20% reduction. The discount would be $18. [1 for the value of either discount] The best option involves losing $16.

Question 22

David is trying to work out the bonuses that he will pay to his employees for their work over the past six months. The city in which they w…

Accepted Answer

4(a) North: 80 + 99 = 179 1 East: 95 + 92 = 187 South: 36 + 53 = 89 West: 86 + 85 = 171 The most sales took place in East zone 4(b) Carol had the highest sales in Mar 2 and the second highest sales in Jan and Apr Total bonuses were 2 × $50 + $150 = $250 1 mark for an answer showing an incorrect judgement for ONE of Carol’s monthly bonuses: e.g. 50 + 150 = $200 or 50 + 50 + 50 + 150 = $300. 4(c) Tanya works in the West zone, so the first 10 sales are worth 10 points in 2 total. [1] The remaining 10 sales are worth 2 points each, so the total is 30 SC: 1 mark for 40 or ‘2 each’ 4(d) Neither North zone employee will receive any bonus 4 Neither East zone employee will receive any bonus In the South zone all sales were worth 2 points, so Rachel will have a bonus of $600 and John will not get a bonus. In the West zone, both employees will receive bonuses. Bonuses will be awarded to Rachel, Oliver and Tanya [1] (dependent on no others identified) Rachel had a bonus of $600 [1] Oliver had a bonus of $1200 [1] Tanya had a bonus of $1100 [1] SC: 1 mark for identification that North and East zone employees do not receive bonuses; may be implied by correct points totals for A, C, F and M seen. 4(e) Martin would have received bonuses for most sales in 2 of the months and 3 second highest in 1 of the months, which would have been $350. He would not have received any bonuses from the points. Therefore his total bonus in the old system was $350. [1] Under the new system, Martin would have earned 47 points in the first three months and then 52 points in the second three months, so would receive no bonus for the first three months and $200 in the second three months. [1] Martin’s total bonus would be $150 less. 4(f) A bonus of $1000 requires a total of 60 points for the three month period. 3 Since Oliver is in the West zone he would achieve 30 points from 10 sales every month and would only need an additional 5 sales per month (at 2 points each) to reach 60 points. Oliver’s minimum target would be 15 sales per month. [1] Since John is in the South zone he can achieve 60 points by making 10 sales per month (at 2 points each). [1] Oliver would need to make 5 sales more per month than John. SC: 2marks for 15 difference in total sales (rather than number per month)

Question 23

Joshua is a taxi driver who is considering working only on Fridays, doing a 12-hour shift from 08:00 to 20:00. He is permitted to collect p…

Accepted Answer

2(a) Income = ($2 + 10 × $1 =) $12. Cost = 10 × 2 × $0.2 = $4. 1 Difference = $8 AG 2(b) A profit of $14 would require a journey distance of (14 – 2)/0.6 = 20 km 2 1 mark for $0.60 soi Trial & Improvement approach: a correct calculation for a distance greater than 10km, and an improvement attempt [1] 2(c) The average number of jobs per hour will be 40/(2 × 10) = 2, so he 2 estimates that he will do 12 × 2 = 24 during his shift. So he will make a profit of 24 × 8 = $192. Award 1 for 24 journeys OR for a correct hourly profit of $16. 2(d) He will be able to complete 12 trips at $14 profit each, so will only make 1 $168. This is less than $192. OR His hourly profit will drop from $16 to $14. OR Longer journeys means fewer fixed fares oe 2(e) A total profit per shift of $240 corresponds to 30 journeys earning $8 per 2 journey. This is 600 km, so he will need to travel at 50 km/h. OR A total profit per shift of $240 corresponds to an hourly profit of $20. To achieve this, an average speed of 20/16 × 40 = 50 km/h will be needed. 1 mark for either 30 journeys, or 5/4 oe 2(f) For each 10 km journey, his fuel cost will increase by 20 × $0.05 = $1. [1] 2 Therefore, he will need to charge $1 ÷ 10 more per kilometre, so $1.10. If neither of the above marks can be awarded, award 1 mark for a correct algebraic expression for the profit : Hourly : 2(2 + 10r – 5) = $16 oe OR Shift profit: ($2 x 24 + 240r) – $0.25×480 (= $192) oe OR Journey Profit: (2+10r)–(0.25×2×10) = $8 oe

Question 24

An international cycling competition is held every year in Pelatonia. Countries are invited to send a squad of cyclists to take part in the…

Accepted Answer

Question Answer Marks 1(a) 8 1 1(b)(i) 28 – 6 = 22 in the first 4 events; 2 events each, so 11 + 6 = 17 1 1(b)(ii) 2 Event Number Maximum Allocation of cyclists number of in each teams allowed team per squad Individual trial 1 4 I J K G Manhattan 2 4 AB CD EF GH Chase 2 3 AB CD EF Derby 4 1 I J K H Road race 6 1 LMNOPQ 1 mark for any solution with six different letters uniquely in the Road Race. 1(c) 3 + 6 + 6 + 4 + 6 = 25 medals 1 1(d)(i) (7 golds can be awarded as:) 1 (6 in the) Road Race and (1 in the) Individual trial or (4 in the) Derby, (2 in the) Manhattan and (1 in the) Individual trial or (4 in the) Derby, (2 in the) Chase and (1 in the) Individual trial 1(d)(ii) (7 golds can be awarded as:) 2 (6 in the) Road race + (1 in the) Individual trial: Possible silvers: 1, 2, 2, 4, 0 in the 5 events, a total of 9 or (4 in the) Derby, (2 in the) Manhattan/Chase and (1 in the) Individual trial: Possible silvers: 1, 2, 2, 0, 6 in the 5 events, a total of 11 Greatest number of silver = 11 1 mark for 9 seen 1(e) (7 golds can be awarded as:) 2 (6 in the) Road race + (1 in the) Individual Trial: Other medals: 2, 4, 4, 4, 0 = 14 or (4 in the) Derby, (2 in the) Manhattan or Chase and (1 in the) Individual Trial: Other medals: 2, 4, 4, 0, 6 = 16 Smallest possible total is 14 + 7 = 21 www Award 1 mark for 14, 16 or 23

Question 25

An international cycling competition is held every year in Pelatonia. Countries are invited to send a squad of cyclists to take part in the…

Accepted Answer

Question Answer Marks 1(a) 8 1 1(b)(i) 28 – 6 = 22 in the first 4 events; 2 events each, so 11 + 6 = 17 1 1(b)(ii) 2 Event Number Maximum Allocation of cyclists number of in each teams allowed team per squad Individual trial 1 4 I J K G Manhattan 2 4 AB CD EF GH Chase 2 3 AB CD EF Derby 4 1 I J K H Road race 6 1 LMNOPQ 1 mark for any solution with six different letters uniquely in the Road Race. 1(c) 3 + 6 + 6 + 4 + 6 = 25 medals 1 1(d)(i) (7 golds can be awarded as:) 1 (6 in the) Road Race and (1 in the) Individual trial or (4 in the) Derby, (2 in the) Manhattan and (1 in the) Individual trial or (4 in the) Derby, (2 in the) Chase and (1 in the) Individual trial 1(d)(ii) (7 golds can be awarded as:) 2 (6 in the) Road race + (1 in the) Individual trial: Possible silvers: 1, 2, 2, 4, 0 in the 5 events, a total of 9 or (4 in the) Derby, (2 in the) Manhattan/Chase and (1 in the) Individual trial: Possible silvers: 1, 2, 2, 0, 6 in the 5 events, a total of 11 Greatest number of silver = 11 1 mark for 9 seen 1(e) (7 golds can be awarded as:) 2 (6 in the) Road race + (1 in the) Individual Trial: Other medals: 2, 4, 4, 4, 0 = 14 or (4 in the) Derby, (2 in the) Manhattan or Chase and (1 in the) Individual Trial: Other medals: 2, 4, 4, 0, 6 = 16 Smallest possible total is 14 + 7 = 21 www Award 1 mark for 14, 16 or 23

Question 26

An online toy store sells all of its toys in boxes that are wrapped and then tied with coloured string. All of the boxes have dimensions of…

Accepted Answer

Question Answer Marks 1(a) (20 + 12 + 10 + 12) × 2 + 16 = 124 cm AG 1 1(b)(i) 10 + 10 + 10 + 10 + 4 = 44 cm 1 1(b)(ii) 2 + 10 + 10 + 10 + 10 = 42 cm 1 1(b)(iii) 40 cm [1 mark] 4 Box starts Red Blue White 1 0 44 40 40 2 124 44 40 40 3 248 42 42 40 4 372 40 44 40 5 496 40 44 40 6 620 40 40 44 7 744 40 40 44 8 868 42 40 42 Boxes 4, 5, 6, 7 [3 marks] Award of marks for incorrect final answers : Correct 1 2 3 4 1 0 2 0 1 Given 34 00 11 22 3 5 0 0 1 2 6 0 0 0 1 OR For the 'box selection' marks, award 1 mark each for the following points seen in working (max 2), if the correct four boxes have not been given: • the correct amount of red for at least two boxes (4 onwards) • box ribbon starting lengths for at least three ribbons beyond 124 (either relative to 0 or to the red sections) 1(c)(i) For one large box, height 3 × 12 = 36 cm, 2 string used = (30 + 2 × 36) × 2 + 16 = 220 cm Difference = (3 × 124) – 220 = 152 cm 1 mark for a solution which omits the 16 cm for the bow, but is otherwise correct, giving 168 cm OR Saving on string will always be 2 bows + the length used on the faces that are together = 2 × 16 + 4 × (20 + 10) = 32 + 120 = 152 cm 1 mark for evidence of recognising that 2 × (20 + 10) is saved between two boxes. 1(c)(ii) The saving on string will always be 2 bows + the length used on the faces 1 that are together. These faces do not include any lengths of string equivalent to the height of the box (nor do the bows). Clear explanation required for the mark.

Question 27

Electronic passports (e-passports) are only useful if their origin can be checked. To make things simpler, individual countries use ‘trust …

Accepted Answer

2(a)(i) There must be 4. 1 2(a)(ii) All 12 distinct ordered pairs. 1 2(b) Both C and E need to trust others, and there are solutions with just 2: 2 EA and CA (or CB or CD or CE), or CA and EC (or EB or ED) 2 marks for any complete and correct diagram (condone omission of given 5 arrows) 1 mark for any diagram in which all trust relationships present, but also superfluous ones but with a maximum of 4 extra connections. SC: 1 mark for links identified but not shown in diagram. 2(c)(i) One country has to trust all others, and all others have to trust it: a six- 2 pointed star. 2 × 6 = 12 policies are required. 1 mark for any directed graph with desired property but which is not minimal SC: 1 mark for correct structure with 6 nodes instead of 7. 2(c)(ii) The departure of the central country would remove all trusted links for 1 everyone. 2(d) Rotational symmetry means that we only have to look at one case. 1 Longest chain was 3, is now 5, so difference of 2 Must be supported 2(e) 2 All countries are the same (by symmetry). The minimum steps others are: Skip 1: 1,1,2,2,3,3; Skip 2: 1,2,1,2,3,2 So Skip 2 is better by 1 step in 6 passport checks. 1 mark for Skip 2 with supporting evidence of decision. 1 mark for satisfactory quantification of the difference: allow any clear, precise explanation, e.g. average number of checks, total checks for all chains for 1 particular country, etc.

Question 28

Peregrine is spending a week walking along a famous mountain path. He is planning to walk from hostel to hostel, staying each night at a di…

Accepted Answer

3(a) Peregrine has been walking 07:00 until 11:00 = 4 hours = 12 km 2 other walker could have been walking from 06:00 until 11:00: 5 hours = 15 km 1 mark for either of 12 km or 15 km seen 9 × 3 = 27 km 3(b) Effectively 8 am start for other walker : 9 km / 3 hour walk 1 So Peregrine will reach his destination at 11am + 3 hours = 14:00 3(c) If he does not meet someone by 12:00 [1] he should turn back. 2 The person he meets may have been walking since 06:00 (18 km travelled) and he may therefore have that far to go [1]. 3(d) Furthest out and back would be 06:00 + 18 km = 12:00. 2 If P left at 9 am he will have walked 9 km by then They walk towards each other and meet at 13:30 [2] Award 1 mark for considering the distance or time walked by someone starting at any time between 06:00 and 09:00 3(e) At 10:20 the first walkers will have walked for 4 hours 20 minutes = 13 km [1] 4 He has walked 1 hour = 3 km. So 16 km [1] between hostels, so halfway point is 8 km in. This will take Peregrine 2 hours 40 minutes to accomplish: 12:00 [1] Picnickers leave at 10:20 in order to reach halfway point at 13:00 They will meet at 12:30 3(f)(i) The family must be at the hostel by nightfall: 1 14:00 to 18:00 = 4 hours = 8 km = 2 hours 40 minutes for Peregrine 3(f)(ii) If the family sets off at 6 am, and walks without stopping, they walk ‘2t’ km in 1 the hours after 6 am and he walks 3(t – 3) km. These are equal when t = 9 : at 15:00 3(g) Maximum distance to next hostel = 7 hours = 21 km 2 For the runner this is 21/4 [1] hours at 4 km/h. He must have set off no later than 12:45

Question 29

This map shows the ferry routes on Lake Veronica, together with the time taken to sail from each stop to the next. Carleton 31 mins 15 mins…

Accepted Answer

4(a) Via Detlie, earliest arrival time is 09:41 + 31 mins = 10:12. 2 Via Toth, earliest is 09:43 + 21 mins = 10:04, so earliest is 10:04. 1 mark for either time correct AND an appropriate conclusion from their times if two are given. SC: 1 mark for 10:18 or 10:10 (+6 minutes, uses departure times) SC: 1 mark for 10:06 or 09:58 (–6 minutes, uses sum of travel times from Munro) SC :1 mark for 10:25 or 10:30 (using the departure times from Carleton) 4(b) Earliest arrival times are 16:30 + 21 mins = 16:51 aboard Frances and 16:01 2 (from Ockelman Island) + 12 mins = 16:13 aboard Marie, so 38 mins. 1 mark for either time correct OR 44 minutes (6 minutes on Ockelman Island omitted) 4(c) By 13:00, Marie had stopped five times at the island. According to the 2 record, 23 + 28 + 37 + 45 + 34 = 167 visitors had arrived and (0 +) 7 + 11 + 18 + 26 = 62 had departed, so there were 105 visitors on the island at that time. 1 mark for recognition that the first five rows of figures (and only the first five) are involved in the calculation (indicated by ‘167 arrived’ or ‘62 departed’), or for calculating the number on the island after the 3rd (70), or 4th (97) or 6th (109) ferries have left. 4(d) Cost of Day Roamer tickets = (2 × $15) + (2 × $9) = $48 3 Extraction of ticket prices $5.10 and $3.00 (Munro – Toth) and $6.30 and $3.80 (Ockelman Is.) 2 × ($5.10 + $3.00 + $6.30 + $3.80) = $36.40 Difference $11.60 1 mark each for any of the following (max 2): $48 [whole family day roamer] $18.20/$36.20 [half family/whole family ticket by ticket] $3.60/$7.20 [adult(s) difference] $2.20/$4.40[child(ren) difference] SC: 2 marks for final answer of $5.80 (uses one-way tickets OR only buys tickets for 1 × adult + child) 4(e) 52 × $15 + 31 × $9 = $1059 2 Number of adult tickets = 52 Search method: The criteria for the search are tickets = 83 and income = $1059 1 mark for an initial search that meets either one of the criteria AND for an adjustment that gets closer to the solution Algebraic method (1 mark for parsing algebraically) a + c = 83 AND 15a + 9c = 1059 OR 15x + 9(83 – x) = 1059 4(f) Detlie first [CDMTC] 4 Depart C 10:25, depart D 13:27, arrive M 13:53 Depart M 17:20 3 hours 27 minutes / 207 minutes at Munro Toth first [CTMDC] Depart C 10:30, depart T 12:57, arrive M 13:14 Depart M 16:24 3 hours 10 minutes / 190 minutes at Munro She should therefore visit the towns in the order Detlie, Munro, Toth. 4 marks for 3 hours 27 minutes / 207 minutes and 3 hours 10 minutes / 190 minutes AND statement of the order Detlie, Munro, Toth. 3 marks for 3 hours 27 minutes / 207 minutes OR 3 hours 10 minutes / 190 minutes. 2 marks for three or more correct arrival/departure times in Munro. 1 mark for two correct arrival/departure times in Munro (allowing the 6 minute discrepancy) SC: using the departure times from the table rather than the journey times (i.e. 6 minutes later): 1 mark for arrival and departure times from Munro [13:59 and 17:20, 13:20 and 16:24] OR the time spent on Munro for either route [3 h 21 m or 3 h 04 m] 2 marks for 3 h 21 m AND 3 h 04 m

Question 30

An online toy store sells all of its toys in boxes that are wrapped and then tied with coloured string. All of the boxes have dimensions of…

Accepted Answer

Question Answer Marks 1(a) (20 + 12 + 10 + 12) × 2 + 16 = 124 cm AG 1 1(b)(i) 10 + 10 + 10 + 10 + 4 = 44 cm 1 1(b)(ii) 2 + 10 + 10 + 10 + 10 = 42 cm 1 1(b)(iii) 40 cm [1 mark] 4 Box starts Red Blue White 1 0 44 40 40 2 124 44 40 40 3 248 42 42 40 4 372 40 44 40 5 496 40 44 40 6 620 40 40 44 7 744 40 40 44 8 868 42 40 42 Boxes 4, 5, 6, 7 [3 marks] Award of marks for incorrect final answers : Correct 1 2 3 4 1 0 2 0 1 Given 34 00 11 22 3 5 0 0 1 2 6 0 0 0 1 OR For the 'box selection' marks, award 1 mark each for the following points seen in working (max 2), if the correct four boxes have not been given: • the correct amount of red for at least two boxes (4 onwards) • box ribbon starting lengths for at least three ribbons beyond 124 (either relative to 0 or to the red sections) 1(c)(i) For one large box, height 3 × 12 = 36 cm, 2 string used = (30 + 2 × 36) × 2 + 16 = 220 cm Difference = (3 × 124) – 220 = 152 cm 1 mark for a solution which omits the 16 cm for the bow, but is otherwise correct, giving 168 cm OR Saving on string will always be 2 bows + the length used on the faces that are together = 2 × 16 + 4 × (20 + 10) = 32 + 120 = 152 cm 1 mark for evidence of recognising that 2 × (20 + 10) is saved between two boxes. 1(c)(ii) The saving on string will always be 2 bows + the length used on the faces 1 that are together. These faces do not include any lengths of string equivalent to the height of the box (nor do the bows). Clear explanation required for the mark.

Question 31

Electronic passports (e-passports) are only useful if their origin can be checked. To make things simpler, individual countries use ‘trust …

Accepted Answer

2(a)(i) There must be 4. 1 2(a)(ii) All 12 distinct ordered pairs. 1 2(b) Both C and E need to trust others, and there are solutions with just 2: 2 EA and CA (or CB or CD or CE), or CA and EC (or EB or ED) 2 marks for any complete and correct diagram (condone omission of given 5 arrows) 1 mark for any diagram in which all trust relationships present, but also superfluous ones but with a maximum of 4 extra connections. SC: 1 mark for links identified but not shown in diagram. 2(c)(i) One country has to trust all others, and all others have to trust it: a six- 2 pointed star. 2 × 6 = 12 policies are required. 1 mark for any directed graph with desired property but which is not minimal SC: 1 mark for correct structure with 6 nodes instead of 7. 2(c)(ii) The departure of the central country would remove all trusted links for 1 everyone. 2(d) Rotational symmetry means that we only have to look at one case. 1 Longest chain was 3, is now 5, so difference of 2 Must be supported 2(e) 2 All countries are the same (by symmetry). The minimum steps others are: Skip 1: 1,1,2,2,3,3; Skip 2: 1,2,1,2,3,2 So Skip 2 is better by 1 step in 6 passport checks. 1 mark for Skip 2 with supporting evidence of decision. 1 mark for satisfactory quantification of the difference: allow any clear, precise explanation, e.g. average number of checks, total checks for all chains for 1 particular country, etc.

Question 32

Peregrine is spending a week walking along a famous mountain path. He is planning to walk from hostel to hostel, staying each night at a di…

Accepted Answer

3(a) Peregrine has been walking 07:00 until 11:00 = 4 hours = 12 km 2 other walker could have been walking from 06:00 until 11:00: 5 hours = 15 km 1 mark for either of 12 km or 15 km seen 9 × 3 = 27 km 3(b) Effectively 8 am start for other walker : 9 km / 3 hour walk 1 So Peregrine will reach his destination at 11am + 3 hours = 14:00 3(c) If he does not meet someone by 12:00 [1] he should turn back. 2 The person he meets may have been walking since 06:00 (18 km travelled) and he may therefore have that far to go [1]. 3(d) Furthest out and back would be 06:00 + 18 km = 12:00. 2 If P left at 9 am he will have walked 9 km by then They walk towards each other and meet at 13:30 [2] Award 1 mark for considering the distance or time walked by someone starting at any time between 06:00 and 09:00 3(e) At 10:20 the first walkers will have walked for 4 hours 20 minutes = 13 km [1] 4 He has walked 1 hour = 3 km. So 16 km [1] between hostels, so halfway point is 8 km in. This will take Peregrine 2 hours 40 minutes to accomplish: 12:00 [1] Picnickers leave at 10:20 in order to reach halfway point at 13:00 They will meet at 12:30 3(f)(i) The family must be at the hostel by nightfall: 1 14:00 to 18:00 = 4 hours = 8 km = 2 hours 40 minutes for Peregrine 3(f)(ii) If the family sets off at 6 am, and walks without stopping, they walk ‘2t’ km in 1 the hours after 6 am and he walks 3(t – 3) km. These are equal when t = 9 : at 15:00 3(g) Maximum distance to next hostel = 7 hours = 21 km 2 For the runner this is 21/4 [1] hours at 4 km/h. He must have set off no later than 12:45

Question 33

This map shows the ferry routes on Lake Veronica, together with the time taken to sail from each stop to the next. Carleton 31 mins 15 mins…

Accepted Answer

4(a) Via Detlie, earliest arrival time is 09:41 + 31 mins = 10:12. 2 Via Toth, earliest is 09:43 + 21 mins = 10:04, so earliest is 10:04. 1 mark for either time correct AND an appropriate conclusion from their times if two are given. SC: 1 mark for 10:18 or 10:10 (+6 minutes, uses departure times) SC: 1 mark for 10:06 or 09:58 (–6 minutes, uses sum of travel times from Munro) SC :1 mark for 10:25 or 10:30 (using the departure times from Carleton) 4(b) Earliest arrival times are 16:30 + 21 mins = 16:51 aboard Frances and 16:01 2 (from Ockelman Island) + 12 mins = 16:13 aboard Marie, so 38 mins. 1 mark for either time correct OR 44 minutes (6 minutes on Ockelman Island omitted) 4(c) By 13:00, Marie had stopped five times at the island. According to the 2 record, 23 + 28 + 37 + 45 + 34 = 167 visitors had arrived and (0 +) 7 + 11 + 18 + 26 = 62 had departed, so there were 105 visitors on the island at that time. 1 mark for recognition that the first five rows of figures (and only the first five) are involved in the calculation (indicated by ‘167 arrived’ or ‘62 departed’), or for calculating the number on the island after the 3rd (70), or 4th (97) or 6th (109) ferries have left. 4(d) Cost of Day Roamer tickets = (2 × $15) + (2 × $9) = $48 3 Extraction of ticket prices $5.10 and $3.00 (Munro – Toth) and $6.30 and $3.80 (Ockelman Is.) 2 × ($5.10 + $3.00 + $6.30 + $3.80) = $36.40 Difference $11.60 1 mark each for any of the following (max 2): $48 [whole family day roamer] $18.20/$36.20 [half family/whole family ticket by ticket] $3.60/$7.20 [adult(s) difference] $2.20/$4.40[child(ren) difference] SC: 2 marks for final answer of $5.80 (uses one-way tickets OR only buys tickets for 1 × adult + child) 4(e) 52 × $15 + 31 × $9 = $1059 2 Number of adult tickets = 52 Search method: The criteria for the search are tickets = 83 and income = $1059 1 mark for an initial search that meets either one of the criteria AND for an adjustment that gets closer to the solution Algebraic method (1 mark for parsing algebraically) a + c = 83 AND 15a + 9c = 1059 OR 15x + 9(83 – x) = 1059 4(f) Detlie first [CDMTC] 4 Depart C 10:25, depart D 13:27, arrive M 13:53 Depart M 17:20 3 hours 27 minutes / 207 minutes at Munro Toth first [CTMDC] Depart C 10:30, depart T 12:57, arrive M 13:14 Depart M 16:24 3 hours 10 minutes / 190 minutes at Munro She should therefore visit the towns in the order Detlie, Munro, Toth. 4 marks for 3 hours 27 minutes / 207 minutes and 3 hours 10 minutes / 190 minutes AND statement of the order Detlie, Munro, Toth. 3 marks for 3 hours 27 minutes / 207 minutes OR 3 hours 10 minutes / 190 minutes. 2 marks for three or more correct arrival/departure times in Munro. 1 mark for two correct arrival/departure times in Munro (allowing the 6 minute discrepancy) SC: using the departure times from the table rather than the journey times (i.e. 6 minutes later): 1 mark for arrival and departure times from Munro [13:59 and 17:20, 13:20 and 16:24] OR the time spent on Munro for either route [3 h 21 m or 3 h 04 m] 2 marks for 3 h 21 m AND 3 h 04 m

Question 34

An online toy store sells all of its toys in boxes that are wrapped and then tied with coloured string. All of the boxes have dimensions of…

Accepted Answer

Question Answer Marks 1(a) (20 + 12 + 10 + 12) × 2 + 16 = 124 cm AG 1 1(b)(i) 10 + 10 + 10 + 10 + 4 = 44 cm 1 1(b)(ii) 2 + 10 + 10 + 10 + 10 = 42 cm 1 1(b)(iii) 40 cm [1 mark] 4 Box starts Red Blue White 1 0 44 40 40 2 124 44 40 40 3 248 42 42 40 4 372 40 44 40 5 496 40 44 40 6 620 40 40 44 7 744 40 40 44 8 868 42 40 42 Boxes 4, 5, 6, 7 [3 marks] Award of marks for incorrect final answers : Correct 1 2 3 4 1 0 2 0 1 Given 34 00 11 22 3 5 0 0 1 2 6 0 0 0 1 OR For the 'box selection' marks, award 1 mark each for the following points seen in working (max 2), if the correct four boxes have not been given: • the correct amount of red for at least two boxes (4 onwards) • box ribbon starting lengths for at least three ribbons beyond 124 (either relative to 0 or to the red sections) 1(c)(i) For one large box, height 3 × 12 = 36 cm, 2 string used = (30 + 2 × 36) × 2 + 16 = 220 cm Difference = (3 × 124) – 220 = 152 cm 1 mark for a solution which omits the 16 cm for the bow, but is otherwise correct, giving 168 cm OR Saving on string will always be 2 bows + the length used on the faces that are together = 2 × 16 + 4 × (20 + 10) = 32 + 120 = 152 cm 1 mark for evidence of recognising that 2 × (20 + 10) is saved between two boxes. 1(c)(ii) The saving on string will always be 2 bows + the length used on the faces 1 that are together. These faces do not include any lengths of string equivalent to the height of the box (nor do the bows). Clear explanation required for the mark.

Question 35

Electronic passports (e-passports) are only useful if their origin can be checked. To make things simpler, individual countries use ‘trust …

Accepted Answer

2(a)(i) There must be 4. 1 2(a)(ii) All 12 distinct ordered pairs. 1 2(b) Both C and E need to trust others, and there are solutions with just 2: 2 EA and CA (or CB or CD or CE), or CA and EC (or EB or ED) 2 marks for any complete and correct diagram (condone omission of given 5 arrows) 1 mark for any diagram in which all trust relationships present, but also superfluous ones but with a maximum of 4 extra connections. SC: 1 mark for links identified but not shown in diagram. 2(c)(i) One country has to trust all others, and all others have to trust it: a six- 2 pointed star. 2 × 6 = 12 policies are required. 1 mark for any directed graph with desired property but which is not minimal SC: 1 mark for correct structure with 6 nodes instead of 7. 2(c)(ii) The departure of the central country would remove all trusted links for 1 everyone. 2(d) Rotational symmetry means that we only have to look at one case. 1 Longest chain was 3, is now 5, so difference of 2 Must be supported 2(e) 2 All countries are the same (by symmetry). The minimum steps others are: Skip 1: 1,1,2,2,3,3; Skip 2: 1,2,1,2,3,2 So Skip 2 is better by 1 step in 6 passport checks. 1 mark for Skip 2 with supporting evidence of decision. 1 mark for satisfactory quantification of the difference: allow any clear, precise explanation, e.g. average number of checks, total checks for all chains for 1 particular country, etc.

Question 36

Peregrine is spending a week walking along a famous mountain path. He is planning to walk from hostel to hostel, staying each night at a di…

Accepted Answer

3(a) Peregrine has been walking 07:00 until 11:00 = 4 hours = 12 km 2 other walker could have been walking from 06:00 until 11:00: 5 hours = 15 km 1 mark for either of 12 km or 15 km seen 9 × 3 = 27 km 3(b) Effectively 8 am start for other walker : 9 km / 3 hour walk 1 So Peregrine will reach his destination at 11am + 3 hours = 14:00 3(c) If he does not meet someone by 12:00 [1] he should turn back. 2 The person he meets may have been walking since 06:00 (18 km travelled) and he may therefore have that far to go [1]. 3(d) Furthest out and back would be 06:00 + 18 km = 12:00. 2 If P left at 9 am he will have walked 9 km by then They walk towards each other and meet at 13:30 [2] Award 1 mark for considering the distance or time walked by someone starting at any time between 06:00 and 09:00 3(e) At 10:20 the first walkers will have walked for 4 hours 20 minutes = 13 km [1] 4 He has walked 1 hour = 3 km. So 16 km [1] between hostels, so halfway point is 8 km in. This will take Peregrine 2 hours 40 minutes to accomplish: 12:00 [1] Picnickers leave at 10:20 in order to reach halfway point at 13:00 They will meet at 12:30 3(f)(i) The family must be at the hostel by nightfall: 1 14:00 to 18:00 = 4 hours = 8 km = 2 hours 40 minutes for Peregrine 3(f)(ii) If the family sets off at 6 am, and walks without stopping, they walk ‘2t’ km in 1 the hours after 6 am and he walks 3(t – 3) km. These are equal when t = 9 : at 15:00 3(g) Maximum distance to next hostel = 7 hours = 21 km 2 For the runner this is 21/4 [1] hours at 4 km/h. He must have set off no later than 12:45

Question 37

This map shows the ferry routes on Lake Veronica, together with the time taken to sail from each stop to the next. Carleton 31 mins 15 mins…

Accepted Answer

4(a) Via Detlie, earliest arrival time is 09:41 + 31 mins = 10:12. 2 Via Toth, earliest is 09:43 + 21 mins = 10:04, so earliest is 10:04. 1 mark for either time correct AND an appropriate conclusion from their times if two are given. SC: 1 mark for 10:18 or 10:10 (+6 minutes, uses departure times) SC: 1 mark for 10:06 or 09:58 (–6 minutes, uses sum of travel times from Munro) SC :1 mark for 10:25 or 10:30 (using the departure times from Carleton) 4(b) Earliest arrival times are 16:30 + 21 mins = 16:51 aboard Frances and 16:01 2 (from Ockelman Island) + 12 mins = 16:13 aboard Marie, so 38 mins. 1 mark for either time correct OR 44 minutes (6 minutes on Ockelman Island omitted) 4(c) By 13:00, Marie had stopped five times at the island. According to the 2 record, 23 + 28 + 37 + 45 + 34 = 167 visitors had arrived and (0 +) 7 + 11 + 18 + 26 = 62 had departed, so there were 105 visitors on the island at that time. 1 mark for recognition that the first five rows of figures (and only the first five) are involved in the calculation (indicated by ‘167 arrived’ or ‘62 departed’), or for calculating the number on the island after the 3rd (70), or 4th (97) or 6th (109) ferries have left. 4(d) Cost of Day Roamer tickets = (2 × $15) + (2 × $9) = $48 3 Extraction of ticket prices $5.10 and $3.00 (Munro – Toth) and $6.30 and $3.80 (Ockelman Is.) 2 × ($5.10 + $3.00 + $6.30 + $3.80) = $36.40 Difference $11.60 1 mark each for any of the following (max 2): $48 [whole family day roamer] $18.20/$36.20 [half family/whole family ticket by ticket] $3.60/$7.20 [adult(s) difference] $2.20/$4.40[child(ren) difference] SC: 2 marks for final answer of $5.80 (uses one-way tickets OR only buys tickets for 1 × adult + child) 4(e) 52 × $15 + 31 × $9 = $1059 2 Number of adult tickets = 52 Search method: The criteria for the search are tickets = 83 and income = $1059 1 mark for an initial search that meets either one of the criteria AND for an adjustment that gets closer to the solution Algebraic method (1 mark for parsing algebraically) a + c = 83 AND 15a + 9c = 1059 OR 15x + 9(83 – x) = 1059 4(f) Detlie first [CDMTC] 4 Depart C 10:25, depart D 13:27, arrive M 13:53 Depart M 17:20 3 hours 27 minutes / 207 minutes at Munro Toth first [CTMDC] Depart C 10:30, depart T 12:57, arrive M 13:14 Depart M 16:24 3 hours 10 minutes / 190 minutes at Munro She should therefore visit the towns in the order Detlie, Munro, Toth. 4 marks for 3 hours 27 minutes / 207 minutes and 3 hours 10 minutes / 190 minutes AND statement of the order Detlie, Munro, Toth. 3 marks for 3 hours 27 minutes / 207 minutes OR 3 hours 10 minutes / 190 minutes. 2 marks for three or more correct arrival/departure times in Munro. 1 mark for two correct arrival/departure times in Munro (allowing the 6 minute discrepancy) SC: using the departure times from the table rather than the journey times (i.e. 6 minutes later): 1 mark for arrival and departure times from Munro [13:59 and 17:20, 13:20 and 16:24] OR the time spent on Munro for either route [3 h 21 m or 3 h 04 m] 2 marks for 3 h 21 m AND 3 h 04 m

Question 38

Every year at the two-day Chevalier Horse Show teams of four riders from Frogford, Hockingham and Witherston Horse Clubs take part in a jum…

Accepted Answer

2(a) Andrew 9th; Dilani 3rd; Graham 6th; Sana 2nd 2 1 mark (max) for any of the following: • two or three positions correct • sight of correct number of points for all four riders (1, 12, 6, and 15 respectively) • all four correct positions without the names of the riders 2(b) 12 (Tamsin), 20 (Natalie) and 15 (Brian) makes 1 47 (points) 2(c) Biscuit (ridden by Brian, Tamsin and Sana) 1 2(d) Graham (rode Meteor, Sapphire and Harvey from Witherston) 1 2(e) Andrew would have scored 4 instead of 6; 2 Dilani would have scored 0 instead of 1. 1 mark for either Sana would have scored 12 instead of 0. 12 – 2 – 1 = 9 2(f) 1 mark for each of the following: 4 • None of the (seven) riders who have fewer than 21 points at the end of the fourth round can win. • At least one of the (four) riders with 41 points will score (because only three score no points). • All the riders with 41 points who score will score a different number of points. • Mahela could total 44 points (if he wins the round), but none of those with 41 points could tie with him (because it is not possible to score 3 points). 2(g) 4 marks for any assignment of positions to the riders that does not violate 4 the order for the four known riders and is consistent with the correct total points. If 4 not scored: A three-way tie requires each team to have (5 × 78 ÷ 3 =) 130 points. [1] OR (78 – 30 =) 48 points to split between them, so 16 each [1] Therefore Frogford need (16 + 13 =) 29 points, Hockingham 16 points and Witherston (16 + 17=) 33 points. [1] 1 mark for any valid allocation of points that gives the right total for each team, e.g. 15 + 10 + 4 + 0 = 29; 8 + 6 + 2 + 0 = 16; 20 + 12 + 1 + 0 = 33.

Question 39

Tridaw is a game played between 2 teams of 3 players. Each round of a match is played by one player from each of the teams. A match consist…

Accepted Answer

3(a) There is a total of 3 × 1 + 3 × 3 + 3 × 5 = 27 points available, so the winning 1 team must score at least 14. 3(b) 3 2 1 2 1 3 2 2 1 3 1 mark for a completed grid in which there are no repetitions in any row OR no repetitions in any column. 3(c) Any example that is either two allocations to the same group or two 1 allocations in the same row or column. 3(d) Griffins: 2×1 + 2×5 = 12 points. [1] 2 Hawks: 1×1 + 3×3 + 1×5 = 15 points. [1] SC: 1 mark for 12 and 15 with no indication of teams. 3(e)(i) 16 2 1 mark for evidence of different decision leading to Roger scoring 8 points instead of 6 OR Tom scoring 3 points instead of 1. 3(e)(ii) Specifying group 2 would force all the remaining rounds to be as required / If 2 the captain of the Griffins had specified group 3 for this round then the remaining rounds would not have been determined [1] and so the other captain’s selection might have put the other Griffin wins into group 1 rather than group 2 (giving the Griffins a score of 12). [1]

Question 40

Sally is organising a charity concert next month. She has booked the hall and is deciding on the price that she should set for tickets. She…

Accepted Answer

4(a) The increase of price between the cheapest and most expensive tickets was 1 $20 and the drop in sales was approximately 80. The sales for $20, $30, $35 and $40 are consistent with a drop of sales by 20 for each additional $5 on the price (the $30 are consistent as the capacity had been reached, so it is possible that another 20 customers would have bought tickets). The $25 row is the inconsistent one (allow any clear identification of the row). 4(b) The concert with tickets $35 made the largest donation to charity. [1] 3 The income was a total of $3535. [1] The cost of the venue was $800, so the donation to charity was $2735. 4(c) The options for the amount that Sally earns are: 2 Ticket price Number sold Amount made $20 150 $3000 $25 140 $3500 $30 120 $3600 $35 100 $3500 $40 80 $3200 The total income reduces as the price increases. The most that Sally can receive from sales is therefore $3600. 1 mark for any correct total income calculated. 4(d) Sally’s model predicts that 60 tickets would sell at $45 and only 40 would 3 sell at $50, so the best choice of price for the more expensive tickets would be $45. [1] For the remaining tickets, the options are shown in the table below. Ticket price Number sold Amount made $20 90 $1800 $25 80 $2000 $30 60 $1800 $35 40 $1400 $40 20 $800 1 mark for a correct calculation with the adjusted number sold. It is therefore best to set the cheaper price as $25. [1] 4(e)(i) 30 tickets would not be sold at the higher price, reducing the income by 2 $1350. But 10 additional standard tickets can be sold for an extra $250. The expected income would be reduced by $1100. 1 mark for $600 or $1350. ft their $25 in (d). 4(e)(ii) Working through the different options, starting with the higher price: 4 Higher Number Lower Number Total Income Income price sold price sold income $30 60 $1800 $25 80 $2000 $3800 $35 50 $1750 $25 90 $2250 $4000 $40 40 $1600 $30 80 $2400 $4000 $45 30 $1350 $30 90 $2700 $4050 $50 20 $1000 $35 80 $2800 $3800 $55 10 $550 $35 90 $3150 $3700 Sally should therefore charge $45 for the more expensive tickets and $30 for the cheaper tickets. 1 mark for identifying the correct total income for any combination of ticket prices. 1 mark for finding the best lower price to go with a given higher price. 1 mark for reaching a combination that generates an income of at least $4000.

Question 41

Every year at the two-day Chevalier Horse Show teams of four riders from Frogford, Hockingham and Witherston Horse Clubs take part in a jum…

Accepted Answer

2(a) Andrew 9th; Dilani 3rd; Graham 6th; Sana 2nd 2 1 mark (max) for any of the following: • two or three positions correct • sight of correct number of points for all four riders (1, 12, 6, and 15 respectively) • all four correct positions without the names of the riders 2(b) 12 (Tamsin), 20 (Natalie) and 15 (Brian) makes 1 47 (points) 2(c) Biscuit (ridden by Brian, Tamsin and Sana) 1 2(d) Graham (rode Meteor, Sapphire and Harvey from Witherston) 1 2(e) Andrew would have scored 4 instead of 6; 2 Dilani would have scored 0 instead of 1. 1 mark for either Sana would have scored 12 instead of 0. 12 – 2 – 1 = 9 2(f) 1 mark for each of the following: 4 • None of the (seven) riders who have fewer than 21 points at the end of the fourth round can win. • At least one of the (four) riders with 41 points will score (because only three score no points). • All the riders with 41 points who score will score a different number of points. • Mahela could total 44 points (if he wins the round), but none of those with 41 points could tie with him (because it is not possible to score 3 points). 2(g) 4 marks for any assignment of positions to the riders that does not violate 4 the order for the four known riders and is consistent with the correct total points. If 4 not scored: A three-way tie requires each team to have (5 × 78 ÷ 3 =) 130 points. [1] OR (78 – 30 =) 48 points to split between them, so 16 each [1] Therefore Frogford need (16 + 13 =) 29 points, Hockingham 16 points and Witherston (16 + 17=) 33 points. [1] 1 mark for any valid allocation of points that gives the right total for each team, e.g. 15 + 10 + 4 + 0 = 29; 8 + 6 + 2 + 0 = 16; 20 + 12 + 1 + 0 = 33.

Question 42

Tridaw is a game played between 2 teams of 3 players. Each round of a match is played by one player from each of the teams. A match consist…

Accepted Answer

3(a) There is a total of 3 × 1 + 3 × 3 + 3 × 5 = 27 points available, so the winning 1 team must score at least 14. 3(b) 3 2 1 2 1 3 2 2 1 3 1 mark for a completed grid in which there are no repetitions in any row OR no repetitions in any column. 3(c) Any example that is either two allocations to the same group or two 1 allocations in the same row or column. 3(d) Griffins: 2×1 + 2×5 = 12 points. [1] 2 Hawks: 1×1 + 3×3 + 1×5 = 15 points. [1] SC: 1 mark for 12 and 15 with no indication of teams. 3(e)(i) 16 2 1 mark for evidence of different decision leading to Roger scoring 8 points instead of 6 OR Tom scoring 3 points instead of 1. 3(e)(ii) Specifying group 2 would force all the remaining rounds to be as required / If 2 the captain of the Griffins had specified group 3 for this round then the remaining rounds would not have been determined [1] and so the other captain’s selection might have put the other Griffin wins into group 1 rather than group 2 (giving the Griffins a score of 12). [1]

Question 43

Sally is organising a charity concert next month. She has booked the hall and is deciding on the price that she should set for tickets. She…

Accepted Answer

4(a) The increase of price between the cheapest and most expensive tickets was 1 $20 and the drop in sales was approximately 80. The sales for $20, $30, $35 and $40 are consistent with a drop of sales by 20 for each additional $5 on the price (the $30 are consistent as the capacity had been reached, so it is possible that another 20 customers would have bought tickets). The $25 row is the inconsistent one (allow any clear identification of the row). 4(b) The concert with tickets $35 made the largest donation to charity. [1] 3 The income was a total of $3535. [1] The cost of the venue was $800, so the donation to charity was $2735. 4(c) The options for the amount that Sally earns are: 2 Ticket price Number sold Amount made $20 150 $3000 $25 140 $3500 $30 120 $3600 $35 100 $3500 $40 80 $3200 The total income reduces as the price increases. The most that Sally can receive from sales is therefore $3600. 1 mark for any correct total income calculated. 4(d) Sally’s model predicts that 60 tickets would sell at $45 and only 40 would 3 sell at $50, so the best choice of price for the more expensive tickets would be $45. [1] For the remaining tickets, the options are shown in the table below. Ticket price Number sold Amount made $20 90 $1800 $25 80 $2000 $30 60 $1800 $35 40 $1400 $40 20 $800 1 mark for a correct calculation with the adjusted number sold. It is therefore best to set the cheaper price as $25. [1] 4(e)(i) 30 tickets would not be sold at the higher price, reducing the income by 2 $1350. But 10 additional standard tickets can be sold for an extra $250. The expected income would be reduced by $1100. 1 mark for $600 or $1350. ft their $25 in (d). 4(e)(ii) Working through the different options, starting with the higher price: 4 Higher Number Lower Number Total Income Income price sold price sold income $30 60 $1800 $25 80 $2000 $3800 $35 50 $1750 $25 90 $2250 $4000 $40 40 $1600 $30 80 $2400 $4000 $45 30 $1350 $30 90 $2700 $4050 $50 20 $1000 $35 80 $2800 $3800 $55 10 $550 $35 90 $3150 $3700 Sally should therefore charge $45 for the more expensive tickets and $30 for the cheaper tickets. 1 mark for identifying the correct total income for any combination of ticket prices. 1 mark for finding the best lower price to go with a given higher price. 1 mark for reaching a combination that generates an income of at least $4000.

Question 44

In a country where too much food is grown it was decided to pay farmers to leave 10% of their land unused. This is called a set-aside subsi…

Accepted Answer

Question Answer Marks 1(a) Two hectares from each 20 hectare field, so least are 2 + 3 and 1 + 1 [1] 2 = 7, Original was 155 so 15.5 expected 15.5 – 7 = 8.5. 1 mark for 7 or 15.5 seen SC: 1 mark for treating as one field: 1 + 1 + 1 + 2 = 5 hectares out of action. So, 10.5 more than expected. 1(b) Least four hectares of the combined area produce 1 + 1 + 1 + 2 = 5, 1 so 7 – 5 = 2 OR 150 – 148 = 2 FT their 7 OR their 148 from (a) 1(c) $900 × 15% × 155 = $20 925 2 1 mark for 15% correctly combined with one of other numbers. 1(d) $1999 ignore cents. 2 Accept ⩽ $2000 FT (b) × $1000 1 mark for $2001 or > $2000 1(e) The largest percentage loss is 10% [1] 2 from a field of uniform production. The threshold is when 10% of the new price is 15% of the old, or $1350. 1(f) 100m of 4 × 8 (down) + 7 × 5 (across) = 6700 m 1

Question 45

In a country where too much food is grown it was decided to pay farmers to leave 10% of their land unused. This is called a set-aside subsi…

Accepted Answer

Question Answer Marks 1(a) Two hectares from each 20 hectare field, so least are 2 + 3 and 1 + 1 [1] 2 = 7, Original was 155 so 15.5 expected 15.5 – 7 = 8.5. 1 mark for 7 or 15.5 seen SC: 1 mark for treating as one field: 1 + 1 + 1 + 2 = 5 hectares out of action. So, 10.5 more than expected. 1(b) Least four hectares of the combined area produce 1 + 1 + 1 + 2 = 5, 1 so 7 – 5 = 2 OR 150 – 148 = 2 FT their 7 OR their 148 from (a) 1(c) $900 × 15% × 155 = $20 925 2 1 mark for 15% correctly combined with one of other numbers. 1(d) $1999 ignore cents. 2 Accept ⩽ $2000 FT (b) × $1000 1 mark for $2001 or > $2000 1(e) The largest percentage loss is 10% [1] 2 from a field of uniform production. The threshold is when 10% of the new price is 15% of the old, or $1350. 1(f) 100m of 4 × 8 (down) + 7 × 5 (across) = 6700 m 1

Question 46

In a country where too much food is grown it was decided to pay farmers to leave 10% of their land unused. This is called a set-aside subsi…

Accepted Answer

Question Answer Marks 1(a) Two hectares from each 20 hectare field, so least are 2 + 3 and 1 + 1 [1] 2 = 7, Original was 155 so 15.5 expected 15.5 – 7 = 8.5. 1 mark for 7 or 15.5 seen SC: 1 mark for treating as one field: 1 + 1 + 1 + 2 = 5 hectares out of action. So, 10.5 more than expected. 1(b) Least four hectares of the combined area produce 1 + 1 + 1 + 2 = 5, 1 so 7 – 5 = 2 OR 150 – 148 = 2 FT their 7 OR their 148 from (a) 1(c) $900 × 15% × 155 = $20 925 2 1 mark for 15% correctly combined with one of other numbers. 1(d) $1999 ignore cents. 2 Accept ⩽ $2000 FT (b) × $1000 1 mark for $2001 or > $2000 1(e) The largest percentage loss is 10% [1] 2 from a field of uniform production. The threshold is when 10% of the new price is 15% of the old, or $1350. 1(f) 100m of 4 × 8 (down) + 7 × 5 (across) = 6700 m 1

Question 47

John is looking for a hotel near a conference hall. This is a list of the hotels available online, and a graph of the cost in $ against dis…

Accepted Answer

3(a)(i) Lisbet, Elizabeth, Liza, Libby (in any order) 2 1 mark for any three correct and no more than four given. 3(a)(ii) 0.0 < x ⩽ 2.0 , 70 < y ⩽ 80 – Lisbet 3 2.0 < x ⩽ 3.0 , 60 < y ⩽ 70 – Elizabeth 3.0 < x ⩽ 4.5 , 55 < y ⩽ 60 – Liza 4.5. < x ⩽ 7.3 , 0 < y ⩽ 55 – Libby 1 mark for any suitable x, y 1 mark for second suitable x, y from a different range 1 mark for both the matching hotels 3(b) Liza & Elizabeth 1 3(c)(i) Award up to 2 marks for 3 Algebraic Inequality between any pair of hotels [1] The fixed charge is constant for all cases and can be ignored explicitly identified [1]. The (cheapest) nearest is always included, in this case Lisbet. [1] The steepest rate of change (is from there to Elizabeth:) $10 for 1 km [1] The critical combination is Lisbet and Elizabeth. [1] There are two trips, so $5 per km. SC: 2 marks for $10 (per km) final answer 3(c)(ii) Elizabeth 1

Question 48

eXeL is a game for two players, played over at least two rounds on the electronic device shown below. eXeL SCORES ROUND TOTAL X Y START The…

Accepted Answer

4(a)(i) The maximum round score for a player after five turns is 8 + 8 + 8 + 7 + 7 1 = 38. A sixth turn is therefore needed to reach 40. 4(a)(ii) If the player achieving 40 has played first, the other player may only have 2 had five turns, scoring at least 1 + 1 + 1 + 2 + 2 = 7. 1 mark for an answer of 9, which assumes that both will have the same number of turns OR 1 mark ft from (i): 5→5, 7→9, 8→12 only. 4(b) A grid with the numbers 1–8 three times each, with no number appearing 2 twice in the same row or column. [1] (Excluding the example in the QP) A grid with a different number (1–8) in each of the squares shown below with a number in bold. [1] e.g. 3 1 4 2 8 8 2 7 6 3 5 4 XY 1 6 7 5 3 8 4 2 6 1 5 7 4(c)(i) The numbers missing from the grid are: 4, 5, 5, 6, 6 and 7. These add up 2 to 33, so Max’s round score is 33 – 16 = 17. 1 mark for identification of 4, 5, 5, 6, 6 and 7 scored so far, or that the grid total would have been 108 to begin with and is now 75. 4(c)(ii) 5 & 6 are in the same column as Y and 7 & 8 are in the same row as Y, Bc 1 must have contained a 4. 4(c)(iii) First turn: 7 from square Ed [1] 2 Second turn: 5 from square Ce [1] OR Ed then Ce [1] If [0], award 1 mark for clear evidence that Katy’s three scores must have been 4, 5 and 7 (or Max’s three scores must have been 5, 6 and 6). 4(c)(iv) Max / X has 3 squares to choose from (Aa, Dd, Ec); Katy / Y has 6 2 squares to choose from (Aa, Ae, Ca, Db, Dd, Ec). 3 × 6 = 18, but this includes 3 appearances that cannot occur because they would require both X and Y to occupy the same square [1], so the number of possibilities is 15. 1 mark for 3 × 6 (implied by 18) OR subtracting 3 at end 4(d) Lydia must move to square Cd, to force Trixie to move to square Eb, which 3 will force her to go to Ee on her next turn, from where no further moves are possible. Lydia can only win the game this round if she scores the bonus of 20 / has a round score of (exactly) 40. She can reach 40 by scoring / moving to, in order: 1 / Cd, 4 / Ec, 6 / Cb, 3 / Ea (or 2 / Bd, 1 / Ba). 3 marks for fully correct solution. 2 marks for first three moves correct. 1 mark for L must move first to Cd. OR 1 mark for recognition that Lydia must score 40 exactly.

Question 49

Helena and Daisy are thinking about how they can select 6 cards from a box of 100 cards. They wish to have a fair method which involves bot…

Accepted Answer

3(a)(i) 11 → 5 → 8 [1] 2 8 → 4 → 7 7 → 3 → 6 [1] Accept → 4 and → 3 not stated 3(a)(ii) Start with 100 cards 1 Remove 11 initially Remove three sets of 3 during the process 100 – 11 – 9 = 80 3(b) 35 → 17 → 20 2 20 → 10 → 13 [1] 13 → 6 → 9 9 → 4 → 7 7 → 3 → 6 So 5 divisions 3(c) 14 2 1 mark for accurately testing any of 12, 13, 15 and 16 as a starting number. This may be done by reducing to previously considered case. 3(d) The smallest initial number of cards to require 5 divisions is 22 [1] 3 For 6 divisions it is 38 and for 7 divisions it is 70. The largest possible starting number must therefore require 7 divisions [1] 7 divisions will involve the addition of 21 cards from the box, so the initial number taken must be 100 – 21 = 79 1 division: 7 (3 additional cards needed from pack) 2 divisions: 8–9 (6 additional cards needed from pack) 3 divisions: 10–13 (9 additional cards needed from pack) 4 divisions: 14–21 (12 ...) 5 divisions: 22–37 (15 ...) 6 divisions: 38–69 (18 ...) 7 divisions: 70–133 (21 …) SC: 3 marks for 78 if requiring at least one card to be left

Question 50

Carpenters sometimes use ropes with carefully positioned ribbons on them to measure distances. They do this by measuring a distance from th…

Accepted Answer

3(a) 1, 3 [1] 2 2 [1] (ignore 4, 6, 7). Only award 2 marks if no incorrect lengths given. 3(b) Correct ropes: 2 [1,2,5], [1,3,7], [1,4,6], [1,5,7], [2,4,7], [3,6,7], [2,3,7], [1,5,6] 1 mark for any correct rope. 1 mark for clear identification of all lengths for that rope. 3(c) A 15 metre rope with ribbons at 7 m and 13 m (or 2 m and 8 m) will allow 2 this. 1 mark for any rope length with 3 ribbons that achieves all the required lengths. 3(d) 21 1 3(e) An 11 m rope with ribbons at [2,7,8] (= [3,4,9]) OR [1,4,9] (= [2,7,10]) [2] 3 can be used to measure 1, 2, 3, 4, 5, 6, 7, 8, 9 or 11 metres. [1] 1 mark for any rope and set of 3 ribbons that achieves 9 identified different lengths. OR 2 marks for any rope and set of 3 ribbons that achieves10 identified different lengths.

Question 51

Matchboxes are often made from a cardboard ‘tray’ (the five faces of a cuboid without a top), inserted into a ‘sleeve’ (the four faces of a…

Accepted Answer

4(a) Tray = 35 cm2 [1] 2 Sleeve = 45 cm2 [1] 4(b) 4 × H × H = L × H or 4H = L 1 Any answer for which the length = 4 × height 4(c) 3 marks for any answer where L = 2H 3 OR 1 mark for each of the following (max 2): • an algebraic representation of the area of painted tray • an algebraic representation of the area of painted sleeve 4(d) 8 × 6 × 3 3 1 mark for each of the following (max 2): • a correct algebraic expression of the equation (that simplifies to 8W + 8H = 2HW + 4H2) • substitute a value for H or W and deduce matching other value • one pair of tray and sleeve correctly calculated • another pair of tray and sleeve correctly calculated 4(e) Tray net has dimensions 18 × 12 2 Sleeve net has dimensions 21 × 12 Can fit trays to cover 90 × 96 on one sheet, and sleeves to cover 84 × 96 1 mark for either So sleeves will generate more waste [1]. 4(f)(i) Any matchbox with dimensions 1 L = 2W + H, where 2W + 3H ⩽ 100 OR L = W + 2H, where L + 2W + 5H ⩽ 100 4(f)(ii) 1 mark for finding a dimension of the net pair that is a factor of 100. 3 1 mark for finding the other dimension of the net pair that wastes no more than 400 cm2. 10 × 4 × 3 or 19 × 8 × 3 L = W + 2H layout 3 10 3 3 4 3 4 3 3 4 3 OR L = 2W + H layout 3 8 3 19 3 19 3

Question 52

Carpenters sometimes use ropes with carefully positioned ribbons on them to measure distances. They do this by measuring a distance from th…

Accepted Answer

3(a) 1, 3 [1] 2 2 [1] (ignore 4, 6, 7). Only award 2 marks if no incorrect lengths given. 3(b) Correct ropes: 2 [1,2,5], [1,3,7], [1,4,6], [1,5,7], [2,4,7], [3,6,7], [2,3,7], [1,5,6] 1 mark for any correct rope. 1 mark for clear identification of all lengths for that rope. 3(c) A 15 metre rope with ribbons at 7 m and 13 m (or 2 m and 8 m) will allow 2 this. 1 mark for any rope length with 3 ribbons that achieves all the required lengths. 3(d) 21 1 3(e) An 11 m rope with ribbons at [2,7,8] (= [3,4,9]) OR [1,4,9] (= [2,7,10]) [2] 3 can be used to measure 1, 2, 3, 4, 5, 6, 7, 8, 9 or 11 metres. [1] 1 mark for any rope and set of 3 ribbons that achieves 9 identified different lengths. OR 2 marks for any rope and set of 3 ribbons that achieves10 identified different lengths.

Question 53

Matchboxes are often made from a cardboard ‘tray’ (the five faces of a cuboid without a top), inserted into a ‘sleeve’ (the four faces of a…

Accepted Answer

4(a) Tray = 35 cm2 [1] 2 Sleeve = 45 cm2 [1] 4(b) 4 × H × H = L × H or 4H = L 1 Any answer for which the length = 4 × height 4(c) 3 marks for any answer where L = 2H 3 OR 1 mark for each of the following (max 2): • an algebraic representation of the area of painted tray • an algebraic representation of the area of painted sleeve 4(d) 8 × 6 × 3 3 1 mark for each of the following (max 2): • a correct algebraic expression of the equation (that simplifies to 8W + 8H = 2HW + 4H2) • substitute a value for H or W and deduce matching other value • one pair of tray and sleeve correctly calculated • another pair of tray and sleeve correctly calculated 4(e) Tray net has dimensions 18 × 12 2 Sleeve net has dimensions 21 × 12 Can fit trays to cover 90 × 96 on one sheet, and sleeves to cover 84 × 96 1 mark for either So sleeves will generate more waste [1]. 4(f)(i) Any matchbox with dimensions 1 L = 2W + H, where 2W + 3H ⩽ 100 OR L = W + 2H, where L + 2W + 5H ⩽ 100 4(f)(ii) 1 mark for finding a dimension of the net pair that is a factor of 100. 3 1 mark for finding the other dimension of the net pair that wastes no more than 400 cm2. 10 × 4 × 3 or 19 × 8 × 3 L = W + 2H layout 3 10 3 3 4 3 4 3 3 4 3 OR L = 2W + H layout 3 8 3 19 3 19 3

Question 54

Carpenters sometimes use ropes with carefully positioned ribbons on them to measure distances. They do this by measuring a distance from th…

Accepted Answer

3(a) 1, 3 [1] 2 2 [1] (ignore 4, 6, 7). Only award 2 marks if no incorrect lengths given. 3(b) Correct ropes: 2 [1,2,5], [1,3,7], [1,4,6], [1,5,7], [2,4,7], [3,6,7], [2,3,7], [1,5,6] 1 mark for any correct rope. 1 mark for clear identification of all lengths for that rope. 3(c) A 15 metre rope with ribbons at 7 m and 13 m (or 2 m and 8 m) will allow 2 this. 1 mark for any rope length with 3 ribbons that achieves all the required lengths. 3(d) 21 1 3(e) An 11 m rope with ribbons at [2,7,8] (= [3,4,9]) OR [1,4,9] (= [2,7,10]) [2] 3 can be used to measure 1, 2, 3, 4, 5, 6, 7, 8, 9 or 11 metres. [1] 1 mark for any rope and set of 3 ribbons that achieves 9 identified different lengths. OR 2 marks for any rope and set of 3 ribbons that achieves10 identified different lengths.

Question 55

Matchboxes are often made from a cardboard ‘tray’ (the five faces of a cuboid without a top), inserted into a ‘sleeve’ (the four faces of a…

Accepted Answer

4(a) Tray = 35 cm2 [1] 2 Sleeve = 45 cm2 [1] 4(b) 4 × H × H = L × H or 4H = L 1 Any answer for which the length = 4 × height 4(c) 3 marks for any answer where L = 2H 3 OR 1 mark for each of the following (max 2): • an algebraic representation of the area of painted tray • an algebraic representation of the area of painted sleeve 4(d) 8 × 6 × 3 3 1 mark for each of the following (max 2): • a correct algebraic expression of the equation (that simplifies to 8W + 8H = 2HW + 4H2) • substitute a value for H or W and deduce matching other value • one pair of tray and sleeve correctly calculated • another pair of tray and sleeve correctly calculated 4(e) Tray net has dimensions 18 × 12 2 Sleeve net has dimensions 21 × 12 Can fit trays to cover 90 × 96 on one sheet, and sleeves to cover 84 × 96 1 mark for either So sleeves will generate more waste [1]. 4(f)(i) Any matchbox with dimensions 1 L = 2W + H, where 2W + 3H ⩽ 100 OR L = W + 2H, where L + 2W + 5H ⩽ 100 4(f)(ii) 1 mark for finding a dimension of the net pair that is a factor of 100. 3 1 mark for finding the other dimension of the net pair that wastes no more than 400 cm2. 10 × 4 × 3 or 19 × 8 × 3 L = W + 2H layout 3 10 3 3 4 3 4 3 3 4 3 OR L = 2W + H layout 3 8 3 19 3 19 3

Question 56

George has his own window cleaning business. His charges to customers depend on the type of building and the number of windows. These charg…

Accepted Answer

4(a)(i) (30  $36) + (30  $3  3) + (10  $20) = $1550 AG 1 4(a)(ii) (30  40) + (30  4  3) + (10  25) = 1810 mins 1 4(b)(i) Houses give greatest income per minute [1] 2 In 7 hours (= 420 mins), George can clean 10 houses + 1 apartment, giving an income of $372 4(b)(ii) Apartments give least income per minute, but only 15 [1] 3 15 apartments leave 90 minutes. Arrangement of houses and bungalows between 90 and 120 minutes [1] Least income from 1 house + 2 bungalows = $180 + $76 = $256 4(c) 10 houses + 1 apartment take 7 hours to clean, 240 houses + 24 apartments 3 take 168 hours. Remaining 96 apartments take 32 hours, so total time needed is 168 + 32 = 200 hours [1] 35 hours per week, so 200/35 oe [1] = 5.7... Number of employees needed is 6. [1] 3 marks for final answer 6 if 200 hours oe seen 4(d)(i) 1500/30 = 50 houses per week. 1 Time taken = 50  40 = 2000 mins (33 hours 20 mins) 4(d)(ii) Profit = $1500 – $1000 – $250 = $250 per employee, so for 10 employees, 1 $2500 4(e) In 6 hours, one employee can clean 360/30 (6/0.5) = 12 houses with a daily 3 income of 12  $30  1.10 = $396 Total annual income = $396  5 = $1980 per week for 45 weeks [1] Outgoings = $1000 + $500 = $1500 per employee for 52 weeks [1] Profit = $1980  45 – $1500  52 = $11 100 per year per employee Number of employees for this to be $80 000 in 52 weeks = ($80 000/11 100) > 7 < 8. so least number of employees required is 8. [1] 3 marks for final answer 8 if 11 100 seen

Question 57

George has his own window cleaning business. His charges to customers depend on the type of building and the number of windows. These charg…

Accepted Answer

4(a)(i) (30  $36) + (30  $3  3) + (10  $20) = $1550 AG 1 4(a)(ii) (30  40) + (30  4  3) + (10  25) = 1810 mins 1 4(b)(i) Houses give greatest income per minute [1] 2 In 7 hours (= 420 mins), George can clean 10 houses + 1 apartment, giving an income of $372 4(b)(ii) Apartments give least income per minute, but only 15 [1] 3 15 apartments leave 90 minutes. Arrangement of houses and bungalows between 90 and 120 minutes [1] Least income from 1 house + 2 bungalows = $180 + $76 = $256 4(c) 10 houses + 1 apartment take 7 hours to clean, 240 houses + 24 apartments 3 take 168 hours. Remaining 96 apartments take 32 hours, so total time needed is 168 + 32 = 200 hours [1] 35 hours per week, so 200/35 oe [1] = 5.7... Number of employees needed is 6. [1] 3 marks for final answer 6 if 200 hours oe seen 4(d)(i) 1500/30 = 50 houses per week. 1 Time taken = 50  40 = 2000 mins (33 hours 20 mins) 4(d)(ii) Profit = $1500 – $1000 – $250 = $250 per employee, so for 10 employees, 1 $2500 4(e) In 6 hours, one employee can clean 360/30 (6/0.5) = 12 houses with a daily 3 income of 12  $30  1.10 = $396 Total annual income = $396  5 = $1980 per week for 45 weeks [1] Outgoings = $1000 + $500 = $1500 per employee for 52 weeks [1] Profit = $1980  45 – $1500  52 = $11 100 per year per employee Number of employees for this to be $80 000 in 52 weeks = ($80 000/11 100) > 7 < 8. so least number of employees required is 8. [1] 3 marks for final answer 8 if 11 100 seen

Question 58

Hugh is planning to hold an executive meeting in a boardroom that contains a large circular table surrounded by 25 seats. The table and the…

Accepted Answer

3(a) 9 executives are necessary [1] 2 If there are 2 empty seats between every pair of executives, then with 8 executives this would only account for 8 + 2  8 = 24 seats. [1] 3(b) She is correct, because 18 executives could sit in 6 separate groups of 3 2 (or 5 groups of 3, a pair and a single), but, no matter where they sit, the nineteenth executive will have to create a group of at least 4. Award 1 mark for recognising that 18 executives could sit down without 4 executives sitting consecutively. Alternatively: 6 separate groups of three people plus a gap would use up 6  4 = 24 seats, but there are 25 seats, so there must be another person in a seat somewhere, making a group of 4. Award 1 mark for evidence of 6  4. 3(c) 7 people sitting together with one empty seat can happen 3 times [1] 2 so with 22 executives (only 3 empty seats) a group of (at least) 8 must occur. 3(d) 10 managers could sit consecutively on each table, which would mean that 1 Hugh must invite at least 10  (12 – 2) = 100 managers. 3(e) 4 managers are necessary in order to prevent more than two groups of 3 3 consecutive empty seats. [1] In fact, only 4 managers, suitably deployed, are needed to prevent 3 consecutive empty seats. [1] So the total number of managers needed is 4  10 = 40.

Question 59

Hugh is planning to hold an executive meeting in a boardroom that contains a large circular table surrounded by 25 seats. The table and the…

Accepted Answer

3(a) 9 executives are necessary [1] 2 If there are 2 empty seats between every pair of executives, then with 8 executives this would only account for 8 + 2  8 = 24 seats. [1] 3(b) She is correct, because 18 executives could sit in 6 separate groups of 3 2 (or 5 groups of 3, a pair and a single), but, no matter where they sit, the nineteenth executive will have to create a group of at least 4. Award 1 mark for recognising that 18 executives could sit down without 4 executives sitting consecutively. Alternatively: 6 separate groups of three people plus a gap would use up 6  4 = 24 seats, but there are 25 seats, so there must be another person in a seat somewhere, making a group of 4. Award 1 mark for evidence of 6  4. 3(c) 7 people sitting together with one empty seat can happen 3 times [1] 2 so with 22 executives (only 3 empty seats) a group of (at least) 8 must occur. 3(d) 10 managers could sit consecutively on each table, which would mean that 1 Hugh must invite at least 10  (12 – 2) = 100 managers. 3(e) 4 managers are necessary in order to prevent more than two groups of 3 3 consecutive empty seats. [1] In fact, only 4 managers, suitably deployed, are needed to prevent 3 consecutive empty seats. [1] So the total number of managers needed is 4  10 = 40.

Question 60

Hugh is planning to hold an executive meeting in a boardroom that contains a large circular table surrounded by 25 seats. The table and the…

Accepted Answer

3(a) 9 executives are necessary [1] 2 If there are 2 empty seats between every pair of executives, then with 8 executives this would only account for 8 + 2  8 = 24 seats. [1] 3(b) She is correct, because 18 executives could sit in 6 separate groups of 3 2 (or 5 groups of 3, a pair and a single), but, no matter where they sit, the nineteenth executive will have to create a group of at least 4. Award 1 mark for recognising that 18 executives could sit down without 4 executives sitting consecutively. Alternatively: 6 separate groups of three people plus a gap would use up 6  4 = 24 seats, but there are 25 seats, so there must be another person in a seat somewhere, making a group of 4. Award 1 mark for evidence of 6  4. 3(c) 7 people sitting together with one empty seat can happen 3 times [1] 2 so with 22 executives (only 3 empty seats) a group of (at least) 8 must occur. 3(d) 10 managers could sit consecutively on each table, which would mean that 1 Hugh must invite at least 10  (12 – 2) = 100 managers. 3(e) 4 managers are necessary in order to prevent more than two groups of 3 3 consecutive empty seats. [1] In fact, only 4 managers, suitably deployed, are needed to prevent 3 consecutive empty seats. [1] So the total number of managers needed is 4  10 = 40.

Question 61

Visits to sites in Antarctica by tourists are strictly controlled. The rules state that not more than one ship may visit each site each day…

Accepted Answer

Question Answer Marks 1(a) 93 were from B, so 1603 from others, [1] 2 each max 100, so 17 others and thus 18 [1] SC: 1 mark for final answer 17 (rounding up but not noting 93 separate) 1(b) Been to A&C, W&Y are closed, D and G do not have enough slots left, so 1 F,K,T 1(c) W had at least 2 visits [1] 2 so Y must have had at least 3 Two could have been singletons, so at most 44 1(d) B visits K and F 2 S visits D and F OR B visits K and T S visits D and F OR B visits K and F S visits D and T 1 mark for B visits K OR S visits D OR for correct pairings but not necessarily going to both sites OR not specifying ships. 1(e) Sensible assumption in line with the information given [1] 3 Calculation using data relevant to assumption (possibly rounded) [1] Consistent final answer that is at least 1863 and at most 3000 [1] Allow substantial rounding anywhere For example: Assume: 4 sites per cruise Total expected landings this season = 8644 2161 tourists Assume: Proportion landing from B and S are typical 150 tourists: 2  (187+103) = 580 landings So 3.89 landings per visitor Other ships landings 8344 – (187 + 103) = 8054 150 + 8054/3.89 = 2233 tourists SC: 1 mark for Minimum possible: 1863 + 150 = 2013.

Question 62

A new form of long jump competition is being trialled. In the first round of the event, each athlete has up to three attempts to jump a cer…

Accepted Answer

2(a) Nathan. He has fewer fails. (7, Matt has 8) 1 2(b) For example: 3 5.0 5.2 5.4 5.6 5.8 M ✓ ✓ ✓ ✓  N ✓ ✓ ✓ ✓  O ✓ ✓ ✓  (–) P ✓ ✓ ✓  (–) Q ✓ ✓ ✓  (–) Quentin must have 9 fails [1] Matt has at most 4 fails OR Ollie has at most 7 fails [1] Nathan (5), Pete(8) and both of Matt/Ollie correct [1] Each row must be valid () ()✓ then  2(c) B E A H D 2 1 mark for 3 or 4 names in correct positions OR 4 in correct order 2(d)(i) 5.8 m with all of 8, 9, 10, 11, 12, 13 fails 2 1 mark for 5.8 m with at least one of 8,9,10,11,12,13 fails 2(d)(ii) 5.6 m with all of 3, 4 or 5 fails 2 1 mark for 5.6 m with at least one of 3, 4 or 5 fails SC: 2 marks 5.6 with 3 (ways) if 5.8 and 6 (ways) in 2(d)(i) SC2: 5.8 with 3-5 fails if 2(d)(i) 6.0 with 8–13 fails 2e Den had 9 fails. 3 at 5.4 m, [1] 2 2 at each of 5.2 m, 5.0 m and 4.8 m makes 4.8 m the maximum possible starting distance [1] SC: 1 mark for 4.6 m with first 3 fails at 5.2 m 2f 3 Adi Ben Cal Den Eric Fran Greg Haz Ido Josh Ken 3rd 1st 8th 5th 2nd 9th 11th 4th 7th 10th 6th 6th 5th 8th 11th 7th 9th 4th 2nd 3rd 10th 1st 9 11 4 7 10 3 1 8 5 2 6 6 7 4 1 5 3 8 10 9 2 11 15 18 8 8 15 6 9 18 14 4 17 Haz & Ben have equal maximum points (18) [1] Haz has (6 + 5 =) 11 fails; Ben has (8 + 8 =) 16 fails [1] Ken has 17 points OR in third place behind Haz & Ben. [1]

Question 63

Visits to sites in Antarctica by tourists are strictly controlled. The rules state that not more than one ship may visit each site each day…

Accepted Answer

Question Answer Marks 1(a) 93 were from B, so 1603 from others, [1] 2 each max 100, so 17 others and thus 18 [1] SC: 1 mark for final answer 17 (rounding up but not noting 93 separate) 1(b) Been to A&C, W&Y are closed, D and G do not have enough slots left, so 1 F,K,T 1(c) W had at least 2 visits [1] 2 so Y must have had at least 3 Two could have been singletons, so at most 44 1(d) B visits K and F 2 S visits D and F OR B visits K and T S visits D and F OR B visits K and F S visits D and T 1 mark for B visits K OR S visits D OR for correct pairings but not necessarily going to both sites OR not specifying ships. 1(e) Sensible assumption in line with the information given [1] 3 Calculation using data relevant to assumption (possibly rounded) [1] Consistent final answer that is at least 1863 and at most 3000 [1] Allow substantial rounding anywhere For example: Assume: 4 sites per cruise Total expected landings this season = 8644 2161 tourists Assume: Proportion landing from B and S are typical 150 tourists: 2  (187+103) = 580 landings So 3.89 landings per visitor Other ships landings 8344 – (187 + 103) = 8054 150 + 8054/3.89 = 2233 tourists SC: 1 mark for Minimum possible: 1863 + 150 = 2013.

Question 64

A new form of long jump competition is being trialled. In the first round of the event, each athlete has up to three attempts to jump a cer…

Accepted Answer

2(a) Nathan. He has fewer fails. (7, Matt has 8) 1 2(b) For example: 3 5.0 5.2 5.4 5.6 5.8 M ✓ ✓ ✓ ✓  N ✓ ✓ ✓ ✓  O ✓ ✓ ✓  (–) P ✓ ✓ ✓  (–) Q ✓ ✓ ✓  (–) Quentin must have 9 fails [1] Matt has at most 4 fails OR Ollie has at most 7 fails [1] Nathan (5), Pete(8) and both of Matt/Ollie correct [1] Each row must be valid () ()✓ then  2(c) B E A H D 2 1 mark for 3 or 4 names in correct positions OR 4 in correct order 2(d)(i) 5.8 m with all of 8, 9, 10, 11, 12, 13 fails 2 1 mark for 5.8 m with at least one of 8,9,10,11,12,13 fails 2(d)(ii) 5.6 m with all of 3, 4 or 5 fails 2 1 mark for 5.6 m with at least one of 3, 4 or 5 fails SC: 2 marks 5.6 with 3 (ways) if 5.8 and 6 (ways) in 2(d)(i) SC2: 5.8 with 3-5 fails if 2(d)(i) 6.0 with 8–13 fails 2e Den had 9 fails. 3 at 5.4 m, [1] 2 2 at each of 5.2 m, 5.0 m and 4.8 m makes 4.8 m the maximum possible starting distance [1] SC: 1 mark for 4.6 m with first 3 fails at 5.2 m 2f 3 Adi Ben Cal Den Eric Fran Greg Haz Ido Josh Ken 3rd 1st 8th 5th 2nd 9th 11th 4th 7th 10th 6th 6th 5th 8th 11th 7th 9th 4th 2nd 3rd 10th 1st 9 11 4 7 10 3 1 8 5 2 6 6 7 4 1 5 3 8 10 9 2 11 15 18 8 8 15 6 9 18 14 4 17 Haz & Ben have equal maximum points (18) [1] Haz has (6 + 5 =) 11 fails; Ben has (8 + 8 =) 16 fails [1] Ken has 17 points OR in third place behind Haz & Ben. [1]

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Questions133 pages

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Q61 · Visits to sites in Antarctica by tourists are strictly controlled 9694/31 Oct/Nov 2023

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Q63 · Visits to sites in Antarctica by tourists are strictly controlled 9694/32 Oct/Nov 2023

Q64 · A new form of long jump competition is being trialled 9694/32 Oct/Nov 2023

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