Identify the impact of a change to a problem — 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…

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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

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 3

Fred earns his living as a taxi-driver, offering lifts between his home town of Honeytrees and the airport, a distance of 30 km. He charges…

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

Question 4

Fred earns his living as a taxi-driver, offering lifts between his home town of Honeytrees and the airport, a distance of 30 km. He charges…

Accepted Answer

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

Question 5

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

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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 6

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…

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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 7

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…

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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 8

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 9

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 10

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 11

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 12

Spelanskor is a game for two players, played over a number of rounds. The first player to score a total of 60 points or more wins the game.…

Accepted Answer

4(a) Award 1 mark each (maximum 2 marks) for any of the following five 2 possibilities (accept words or digits): SEVEN, EIGHT, NINE, FOUR, TEN SEVEN, NINE, TEN, FOUR, EIGHT EIGHT, NINE, SEVEN, FOUR, TEN NINE, EIGHT, SEVEN, FOUR, TEN TEN, NINE, SEVEN, FOUR, EIGHT 4(b) row 1, to the left (of the fixed E) NE (row 1) 2 row 1, to the right (of the fixed E) EN (row 1) row 2, to the left (of the fixed I) NI (row 2) row 2, to the right (of the fixed I) IN (row 2) Given row 4, to the right (of the fixed O) ON (row 4) Ignore repetition of the location given (i.e. row 2, to the right of the fixed I) Award 1 mark for two or three correct. 4(c) Greg scored 14 points; Ingrid scored 16 points 2 Order : EIGHT – SIX – TEN – FOUR – TWO 1 mark for appreciation that SIX must be on the second row and FOUR on the fourth row. 4(d) H, W and X (it is possible to place all the other available tiles on at least one 2 row) 2 marks for all three letters in any order. Award 1 mark for two correct letters. 4(e)(i) S (the second S – placed by Greg) 1 4(e)(ii) There was not another I in the bag (with which to spell EIGHT). 1 4(e)(iii) Greg scored 11 points [1] 2 Ingrid scored 13 points [1] (Greg completed SIX with the first S and FIVE with the second F; Ingrid completed FOUR with the R and NINE with the N.) Allow 1 mark for 11 points and 13 points, either credited the wrong way round or not specified who scored which. SC: 1 mark for Greg scores 19 and Ingrid scores 5 (if S is disregarded rather than discarded). 4(f) Greg should play his H next to the T (on either side, making EIGHT or 3 THREE) [1] Forcing Ingrid to play either her X (to the right of the I), in which case Greg can use S (to make SIX) [1] or her W (Ito the left of the O), in which case Greg can use T (to make TWO) [1]

Question 13

The British Diplomatic staff in Bolandia are paid a tax-free cost of living allowance to compensate for the extra costs of living abroad. T…

Accepted Answer

2(a) 26+10+17+20+33+15 = $121 1 2(b) Sum of medians is 29 + 15 + 24 + 25 + 34 + 16 = $143 1 2(c) Shampoo at HyperFood [2]. 3 The increases for the three supermarkets would be $18, $17 and $1 (respectively). Cotes does not have any product priced at $18, and Dasamart does not have any product priced at $1 [1]. If 0 scored, award 1 mark for $18, $17 and $1 OR for consideration of totals with a price doubled compared with $152. SC: Bromley Potatoes [1] 2(d)(i) In Cotes, replacing Macassar Oil with Brill-milk reduces the basket price by 1 the same ($3) in both countries, so the allowance will remain the same. (Identification of figures compared required.) 2(d)(ii) The quotient will change from 134/130 (1.0308) to 131/127 (1.0315), so the 2 allowance will go up. Sight of both correct ‘ratios’, with no or incorrect conclusion. [1] SC: 1 mark for correct deduction from ratios involving another supermarket or average. 2(e) Yes. If the UK prices are lower, the allowance will be higher. 1 2(f) Since their salaries in pounds remain the same but buy more dollars, their 1 spending power in Bolandia will increase.

Question 14

The British Diplomatic staff in Bolandia are paid a tax-free cost of living allowance to compensate for the extra costs of living abroad. T…

Accepted Answer

2(a) 26+10+17+20+33+15 = $121 1 2(b) Sum of medians is 29 + 15 + 24 + 25 + 34 + 16 = $143 1 2(c) Shampoo at HyperFood [2]. 3 The increases for the three supermarkets would be $18, $17 and $1 (respectively). Cotes does not have any product priced at $18, and Dasamart does not have any product priced at $1 [1]. If 0 scored, award 1 mark for $18, $17 and $1 OR for consideration of totals with a price doubled compared with $152. SC: Bromley Potatoes [1] 2(d)(i) In Cotes, replacing Macassar Oil with Brill-milk reduces the basket price by 1 the same ($3) in both countries, so the allowance will remain the same. (Identification of figures compared required.) 2(d)(ii) The quotient will change from 134/130 (1.0308) to 131/127 (1.0315), so the 2 allowance will go up. Sight of both correct ‘ratios’, with no or incorrect conclusion. [1] SC: 1 mark for correct deduction from ratios involving another supermarket or average. 2(e) Yes. If the UK prices are lower, the allowance will be higher. 1 2(f) Since their salaries in pounds remain the same but buy more dollars, their 1 spending power in Bolandia will increase.

Question 15

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 16

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 17

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 18

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 19

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 20

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 21

The all-inclusive holiday resort at Cofete provides buffet meals every day, but guests may choose to go for some evening meals in the three…

Accepted Answer

3(a) Saturday [1], B [1 (dependent)], 3 because to visit all three he has to go on Saturday, Monday, and Wednesday, and of those days B is only open on Saturday [1]. 3(b) Arrival on Saturday [1] means that one has to visit B on Friday [1]. 2 3(c) With Monday, Wednesday and Friday all the same, those arriving on 2 Saturday [1] cannot use all three restaurants [1] soi. SC: 1 mark for Friday (A), Sunday (A) and Monday (B) arrivals now also have restrictions on when they can dine. 3(d) 14 is not a multiple of 3. 1 3(e) Monday and Wednesday [1] 2 because either would make the staff break contiguous OR would avoid any restrictions on options / allow choice for the Thursday and Saturday arrivals. 1 mark for either reason

Question 22

In Bolandian currency, $1 is worth 100 cents (¢). Only the following coins are used: 1¢, 2¢, 5¢, 10¢, 20¢ and 50¢ On Monday, Peter has $1.1…

Accepted Answer

Question Answer Marks 1(a) He must have one 50¢ and three 20¢ coins. 1 1(b) The maximum is one 50¢, four 20¢, one 5¢ and four 2¢ coins, [1] 2 making $1.43. [1] SC: 1 mark for £1.41 or £1.39, from miscounting 2¢ coins OR 1 mark for $1.30, considering only 20¢ and 50¢. 1(c) He needs 9 coins: 2 two 50¢, two 20¢, one 10¢, one 5¢, two 2¢ and one 1¢ coins. 1 mark for one extra or omitted coin, or for 9 without a list. 1(d) He should have the same set as Wednesday, but add three 50¢ coins: 2 five 50¢, two 20¢, one 10¢, one 5¢, two 2¢, and one 1¢ coins [1] meaning that he can make any amount up to $2.50 + $0.60 = $3.10. [1] FT both marks from (c) 1(e) The value to weight ratio is very high for 20¢ coins, so a sensible strategy is 2 to maximise the number of these. So he could have e.g. seven 20¢, one 10¢, one 5¢, two 2¢ and one 1¢ coins [1] with a total weight of 65 g. [1] 1(f) Yes, because five 20¢ could be swapped for four 25¢, saving 5 g. 1 FT: if their (e) includes a 50¢ coin: two 25¢ weigh less than one 50¢

Question 23

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 24

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 25

In Bolandian currency, $1 is worth 100 cents (¢). Only the following coins are used: 1¢, 2¢, 5¢, 10¢, 20¢ and 50¢ On Monday, Peter has $1.1…

Accepted Answer

Question Answer Marks 1(a) He must have one 50¢ and three 20¢ coins. 1 1(b) The maximum is one 50¢, four 20¢, one 5¢ and four 2¢ coins, [1] 2 making $1.43. [1] SC: 1 mark for £1.41 or £1.39, from miscounting 2¢ coins OR 1 mark for $1.30, considering only 20¢ and 50¢. 1(c) He needs 9 coins: 2 two 50¢, two 20¢, one 10¢, one 5¢, two 2¢ and one 1¢ coins. 1 mark for one extra or omitted coin, or for 9 without a list. 1(d) He should have the same set as Wednesday, but add three 50¢ coins: 2 five 50¢, two 20¢, one 10¢, one 5¢, two 2¢, and one 1¢ coins [1] meaning that he can make any amount up to $2.50 + $0.60 = $3.10. [1] FT both marks from (c) 1(e) The value to weight ratio is very high for 20¢ coins, so a sensible strategy is 2 to maximise the number of these. So he could have e.g. seven 20¢, one 10¢, one 5¢, two 2¢ and one 1¢ coins [1] with a total weight of 65 g. [1] 1(f) Yes, because five 20¢ could be swapped for four 25¢, saving 5 g. 1 FT: if their (e) includes a 50¢ coin: two 25¢ weigh less than one 50¢

Question 26

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 27

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 28

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 29

The Goodlen Dancing Society holds a dancing competition every Saturday. Ten couples take part each Saturday, and each couple dances the Wal…

Accepted Answer

2(a) 2 Couple A B C D E F G H I J Total score 23 20 16 24 17 24 25 19 22 24 for Waltz Points for 6 4 1 10 2 10 12 3 5 10 Waltz 1 mark for top row, 1 mark for entire bottom row ft from top row 2(b) 1 Couple A B C D E F G H I J Points for 6 4 1 10 2 10 12 3 5 10 Waltz Points for 6 3 2 10 4 7 6 12 1 8 Jive Total 12 7 3 20 6 17 18 15 6 18 D with 20 points ft from (a) 2(c) 2 Couple A B C D E F G H I J Points for 6 4 1 10 2 10 12 3 5 10 Waltz Points for 6 3 2 10 4 7 6 12 1 8 Jive Points for 3 1 2 5 4 7 12 8 6 10 Audience Grand 15 8 5 25 10 24 30 23 12 28 Total G with 30 points 1 mark for audience points correctly assigned SC: 1 mark for G with 39 points (using percentage not points) 2(d) Couple B now have 4 + 3 + 8 = 15 points [1] 2 Couple H have 16 points and are still in 5th place / D, F, G, H and J all have at least 16 points [1] 2(e) Only possibility is that the points for the Jive were: 3 2, 3, 12, 4, 6, 10, 8, 7, 1, 5 giving totals 4, 8, 16, 16, 16, 16, 16, 8, 8, 8 So A with 4 points 1 mark for indication that the couples with 16 points must be C, D, E, F, and G 1 mark for deducing that the four tied (in 6th place) each have 8 points 2(f) One of couples with 16 will have at least 5 points, a total of 21. [1] The couples on 8 points can get a maximum of 12 points, a total of 20. [1] 2(g) Greatest is 55 [1] 3 Least is 15 [2] 1 mark for attempt to find a set of scores that add to 100 with at least 4 adjacent pairs.

Question 30

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 31

The Goodlen Dancing Society holds a dancing competition every Saturday. Ten couples take part each Saturday, and each couple dances the Wal…

Accepted Answer

2(a) 2 Couple A B C D E F G H I J Total score 23 20 16 24 17 24 25 19 22 24 for Waltz Points for 6 4 1 10 2 10 12 3 5 10 Waltz 1 mark for top row, 1 mark for entire bottom row ft from top row 2(b) 1 Couple A B C D E F G H I J Points for 6 4 1 10 2 10 12 3 5 10 Waltz Points for 6 3 2 10 4 7 6 12 1 8 Jive Total 12 7 3 20 6 17 18 15 6 18 D with 20 points ft from (a) 2(c) 2 Couple A B C D E F G H I J Points for 6 4 1 10 2 10 12 3 5 10 Waltz Points for 6 3 2 10 4 7 6 12 1 8 Jive Points for 3 1 2 5 4 7 12 8 6 10 Audience Grand 15 8 5 25 10 24 30 23 12 28 Total G with 30 points 1 mark for audience points correctly assigned SC: 1 mark for G with 39 points (using percentage not points) 2(d) Couple B now have 4 + 3 + 8 = 15 points [1] 2 Couple H have 16 points and are still in 5th place / D, F, G, H and J all have at least 16 points [1] 2(e) Only possibility is that the points for the Jive were: 3 2, 3, 12, 4, 6, 10, 8, 7, 1, 5 giving totals 4, 8, 16, 16, 16, 16, 16, 8, 8, 8 So A with 4 points 1 mark for indication that the couples with 16 points must be C, D, E, F, and G 1 mark for deducing that the four tied (in 6th place) each have 8 points 2(f) One of couples with 16 will have at least 5 points, a total of 21. [1] The couples on 8 points can get a maximum of 12 points, a total of 20. [1] 2(g) Greatest is 55 [1] 3 Least is 15 [2] 1 mark for attempt to find a set of scores that add to 100 with at least 4 adjacent pairs.

Question 32

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 33

The Goodlen Dancing Society holds a dancing competition every Saturday. Ten couples take part each Saturday, and each couple dances the Wal…

Accepted Answer

2(a) 2 Couple A B C D E F G H I J Total score 23 20 16 24 17 24 25 19 22 24 for Waltz Points for 6 4 1 10 2 10 12 3 5 10 Waltz 1 mark for top row, 1 mark for entire bottom row ft from top row 2(b) 1 Couple A B C D E F G H I J Points for 6 4 1 10 2 10 12 3 5 10 Waltz Points for 6 3 2 10 4 7 6 12 1 8 Jive Total 12 7 3 20 6 17 18 15 6 18 D with 20 points ft from (a) 2(c) 2 Couple A B C D E F G H I J Points for 6 4 1 10 2 10 12 3 5 10 Waltz Points for 6 3 2 10 4 7 6 12 1 8 Jive Points for 3 1 2 5 4 7 12 8 6 10 Audience Grand 15 8 5 25 10 24 30 23 12 28 Total G with 30 points 1 mark for audience points correctly assigned SC: 1 mark for G with 39 points (using percentage not points) 2(d) Couple B now have 4 + 3 + 8 = 15 points [1] 2 Couple H have 16 points and are still in 5th place / D, F, G, H and J all have at least 16 points [1] 2(e) Only possibility is that the points for the Jive were: 3 2, 3, 12, 4, 6, 10, 8, 7, 1, 5 giving totals 4, 8, 16, 16, 16, 16, 16, 8, 8, 8 So A with 4 points 1 mark for indication that the couples with 16 points must be C, D, E, F, and G 1 mark for deducing that the four tied (in 6th place) each have 8 points 2(f) One of couples with 16 will have at least 5 points, a total of 21. [1] The couples on 8 points can get a maximum of 12 points, a total of 20. [1] 2(g) Greatest is 55 [1] 3 Least is 15 [2] 1 mark for attempt to find a set of scores that add to 100 with at least 4 adjacent pairs.

Question 34

There is a long-distance cycle route between Princeville and Queda, of total length 1560 km. Eric plans to cycle along the whole route, lea…

Accepted Answer

Question Answer Marks 1(a) 50 × 5 + 30 = 280 km each week. 1 4 complete weeks plus 3 days in August: total = 280 × 4 + 150 = 1270 km AG 1(b) Cycles the remaining distance of 290 km in September. 2 This is one complete week plus one day for the remaining 10 km. Thursday [1] 8 September [1] 1(c) Eric cycles on 27 days in August. Extra 11 km a day means 297 km extra 2 1270 + 297 = 1567 > 1560, so completes in August [Alternatively 297 > 290 so completes in August] 1 mark for 27 or 297 or 1567 soi 1(d)(i) He cycles 400 km in 9 days. (40 + 80 + 0 + 60) × 2 + 40 = 400, so 40 km 1 1(d)(ii) In next 10 days, he will cycle (80 + 0 + 60 + 40) × 2 + 80 + 0 = 440 km 1 ft from (d)(i): 80: 360 + 0 + 60 = 420 0: 360 + 60 + 40 = 460 60: 360 + 40 + 80 = 480 1(e) 1560 = (60 + 40 + 80 + 0) × 8 + 120 [1] 3 120 is achieved by 3 more days (60 + 40 + 80) 35 days in total needed [1] Start must be 35 days before 31 August 27 July OR Cycles maximum of 30 days in August: distance = (40 + 80 + 0 + 60) × 7 + 40 + 80 = 1380 km Therefore, cycles 1560 – 1380 = 180 km in July 31 July is 60 km, which leads to 28 July for 180 km (40 + 80 + 0 + 60) But, starts with a 60 km day, so 27 July 1 mark for 1380 km in August or 180 km in July SC: 1 mark for July 28th

Question 35

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 36

There is a train service between Arba and Boab. Details of the different types of train ticket available are shown in the table below. Type…

Accepted Answer

Question Answer Marks 1(a) $22.50, $24.00, $24.50, $25.50, $27.00 2 1 mark for any three correct 1(b) 4 same-day returns and 1 Weekend special [1] 2 4  $7.50 + $5.00 = $35.00 [1] 1(c)(i) $64 1 SC: Allow $80 1(c)(ii) One week of travel costs $35 2 In the current scheme, her minimum claim = $32 [1] (The minimum satisfies the 10 or more occasions requirement) So costs could be lower by at most $35 – $32 = $3 AG SC: 2 marks for their (b) compared with half of their (c)(i) OR 2 marks for $35 or $45 compared with any of $32, $40 or half of their (c)(i) OR 1 mark for $45 or $40 or $32 or half of their 1(c)(i) seen 1(d)(i) Use the $50 weekly ticket for 7 return journeys in 7 days soi [1] 2 3 weekly + 3 day return: $150 + $27 = $177 1(d)(ii) 12th 1

Question 37

There is a train service between Arba and Boab. Details of the different types of train ticket available are shown in the table below. Type…

Accepted Answer

Question Answer Marks 1(a) $22.50, $24.00, $24.50, $25.50, $27.00 2 1 mark for any three correct 1(b) 4 same-day returns and 1 Weekend special [1] 2 4  $7.50 + $5.00 = $35.00 [1] 1(c)(i) $64 1 SC: Allow $80 1(c)(ii) One week of travel costs $35 2 In the current scheme, her minimum claim = $32 [1] (The minimum satisfies the 10 or more occasions requirement) So costs could be lower by at most $35 – $32 = $3 AG SC: 2 marks for their (b) compared with half of their (c)(i) OR 2 marks for $35 or $45 compared with any of $32, $40 or half of their (c)(i) OR 1 mark for $45 or $40 or $32 or half of their 1(c)(i) seen 1(d)(i) Use the $50 weekly ticket for 7 return journeys in 7 days soi [1] 2 3 weekly + 3 day return: $150 + $27 = $177 1(d)(ii) 12th 1

Question 38

OrienT-8 is a single-player game, played on a 10 × 10 grid displayed on the touch screen of an electronic device. The game consists of five…

Accepted Answer

Question Answer Marks 1(a) 21 2 1 mark for sight of 12 (black squares not already revealed at the start of the round) or 9 (crosses + ticks) SC 1 mark for answer of 57 1(b) 117 2 1 mark for sight of 90 (score at the end of round two) OR 27 (score so far in this round) OR 93 seen (forgets bonus points) 1(c) There are no crosses in squares next to either of these two T-shapes 1 / there are crosses in squares next to each of the other (three) T-shapes. 1(d) 66 and 68 [1] 3 8, 9, 10 and 19 [1] 43, 52, 53 and 54 [1] 1(e) 117 points so far and will score a further 26 in round three = 143 points [1] 2 so 131 needed . ft their 117 + 26 for 143 There are a total of 70 + 80 = 150 points available for the last two rounds so he can afford 19 crosses maximum. Alternatively Tom’s maximum possible game score is 293 ft their 117 + 176 A maximum of 26 crosses allows a score of 274 1 mark for either He has already revealed 7 so can afford 19 more. SC 1 mark for answer of 20

Question 39

Julie sells sweets in her shop. There are five different types of sweet available, each of which is a different colour. All sweets weigh a …

Accepted Answer

2(a) Price per 10 g is $0.70. [1] 2 Total weight is 482 + 507 + 442 = 1431 g $1.00 + 14  $0.70 = $10.80 [1] AG 2(b) Prices for bags would be: 2 Red: $1.00 + 4  $0.30 = $2.20 Yellow: $1.00 + 5  $0.50 = $3.50 Green: $1.00 + 4  $0.70 = $3.80 1 mark for any one calculated correctly Total cost is $2.20 + $3.50 + $3.80 = $9.50 SC 1 mark for answer $6.50 2(c) The cheapest with 2 bags is: 3 R+Y: 989 g, so $1.00 + 9  $0.50 = $5.50 Total cost $3.80 + $5.50 = $9.30 With 3 bags: Add between 8 g and 57 g (inclusive) of yellow to the green bag reduces cost of yellow bag by $0.50 without increasing price of green bag Cheapest possible = $9.00 1 mark for correct calculation for any 2-bag case OR 2 marks for identifying $9.30 as cheapest with 2 bags 2(d)(i) 2 bags should be used, leaving $12.00 to spend on the sweets. 2 The amount paid for the bag containing purple sweets will be at least twice the amount paid for the bag containing only yellow sweets, so $4.00 will be paid for yellow sweets and $8.00 for purple sweets. [1] The maximum weight possible is 899 g for each colour. 1798 g 2(d)(ii) The amount paid for the bag containing purple sweets will be at least four 3 times the amount paid for the bag containing only yellow sweets. $2.40 and $9.60 is not possible, so values would be $2.00 and $10.00. [1] soi Weight in $2.00 bag would be up to 499 g Weight in $10.00 bag would be up to 1099 g Total weight would be 1598 g [1], but must be a multiple of 3 g, so 1596 g (one bag containing 499 g of yellow and one bag containing 33 g of yellow and 1064 g of purple) SC 1 mark for 1497 g (Maximum if only one colour in each bag) 2(e) Every complete 100 g will now cost $0.15 more. [1] 2 This will exceed $1.00 once 700 g has been bought. 2(f) $1.21 1

Question 40

The Bolandian Environment Agency is planning to plant trees on plots of land formerly used for industry. All the plots are rectangular (or …

Accepted Answer

4(a) 25 – 2  2.5 = 20, so 20/2 = 10 gaps, so number of trees = 11 AG 1 4(b) 121 pine trees: cost = $1800 + 4  $95 + $20 = $2200 [1] 3 Number of beech = 4  25  2 = 200 [1] cost $1400 Total cost = $2200 + $1400 = $3600 4(c)(i) 40 – 5 = 35, so 18 trees; 35 – 5 = 30, so 16 trees, 3 total number of pine trees = 18  16 = 288 [1] Cost of pine trees = 2  $1800 + 3  $460 + 2  $95 + 3  $20 = $5230 OR cost of beech trees = 3  $700 = $2100 1 mark for either Total cost = $5230 + $2100 = $7330 4(c)(ii) Cost of beech trees is unchanged. 2 Number of pine trees = 121 + 288 = 409, cost = $7375 [1] Saving = $2200 + $5230 – $7375 = $55 4(d) 5000 = 2  1800 + 3  460 + 20 2 so number of trees = 200 + 75 + 1 = 276 [1] Wilfred will have up to 92 trees; the most he can plant is 90 (9  10) Condone 81 (Thinks ‘same in each row’ means ‘must be square’) 4(e) Common boundary of 25 m: so 100 fewer beech trees needed. [1] 4 Saving on beech trees = $700 Condone $400 for 50 fewer. Number of pine trees per row along the 65 m is 31 instead of (11 + 18 =) 29, so 2 extra for each of 11 rows, so 22 [1] So number of pine trees required is 409 + 22 = 431 Extra cost = $7775- $7375 = $400 [1] Saving is $700 – $400 = $300 Alternative: Beech perimeter price [1] Pine area price [1] (both ft) Total [1] Difference [1] (ft if saving)

Question 41

Julie sells sweets in her shop. There are five different types of sweet available, each of which is a different colour. All sweets weigh a …

Accepted Answer

2(a) Price per 10 g is $0.70. [1] 2 Total weight is 482 + 507 + 442 = 1431 g $1.00 + 14  $0.70 = $10.80 [1] AG 2(b) Prices for bags would be: 2 Red: $1.00 + 4  $0.30 = $2.20 Yellow: $1.00 + 5  $0.50 = $3.50 Green: $1.00 + 4  $0.70 = $3.80 1 mark for any one calculated correctly Total cost is $2.20 + $3.50 + $3.80 = $9.50 SC 1 mark for answer $6.50 2(c) The cheapest with 2 bags is: 3 R+Y: 989 g, so $1.00 + 9  $0.50 = $5.50 Total cost $3.80 + $5.50 = $9.30 With 3 bags: Add between 8 g and 57 g (inclusive) of yellow to the green bag reduces cost of yellow bag by $0.50 without increasing price of green bag Cheapest possible = $9.00 1 mark for correct calculation for any 2-bag case OR 2 marks for identifying $9.30 as cheapest with 2 bags 2(d)(i) 2 bags should be used, leaving $12.00 to spend on the sweets. 2 The amount paid for the bag containing purple sweets will be at least twice the amount paid for the bag containing only yellow sweets, so $4.00 will be paid for yellow sweets and $8.00 for purple sweets. [1] The maximum weight possible is 899 g for each colour. 1798 g 2(d)(ii) The amount paid for the bag containing purple sweets will be at least four 3 times the amount paid for the bag containing only yellow sweets. $2.40 and $9.60 is not possible, so values would be $2.00 and $10.00. [1] soi Weight in $2.00 bag would be up to 499 g Weight in $10.00 bag would be up to 1099 g Total weight would be 1598 g [1], but must be a multiple of 3 g, so 1596 g (one bag containing 499 g of yellow and one bag containing 33 g of yellow and 1064 g of purple) SC 1 mark for 1497 g (Maximum if only one colour in each bag) 2(e) Every complete 100 g will now cost $0.15 more. [1] 2 This will exceed $1.00 once 700 g has been bought. 2(f) $1.21 1

Question 42

The Bolandian Environment Agency is planning to plant trees on plots of land formerly used for industry. All the plots are rectangular (or …

Accepted Answer

4(a) 25 – 2  2.5 = 20, so 20/2 = 10 gaps, so number of trees = 11 AG 1 4(b) 121 pine trees: cost = $1800 + 4  $95 + $20 = $2200 [1] 3 Number of beech = 4  25  2 = 200 [1] cost $1400 Total cost = $2200 + $1400 = $3600 4(c)(i) 40 – 5 = 35, so 18 trees; 35 – 5 = 30, so 16 trees, 3 total number of pine trees = 18  16 = 288 [1] Cost of pine trees = 2  $1800 + 3  $460 + 2  $95 + 3  $20 = $5230 OR cost of beech trees = 3  $700 = $2100 1 mark for either Total cost = $5230 + $2100 = $7330 4(c)(ii) Cost of beech trees is unchanged. 2 Number of pine trees = 121 + 288 = 409, cost = $7375 [1] Saving = $2200 + $5230 – $7375 = $55 4(d) 5000 = 2  1800 + 3  460 + 20 2 so number of trees = 200 + 75 + 1 = 276 [1] Wilfred will have up to 92 trees; the most he can plant is 90 (9  10) Condone 81 (Thinks ‘same in each row’ means ‘must be square’) 4(e) Common boundary of 25 m: so 100 fewer beech trees needed. [1] 4 Saving on beech trees = $700 Condone $400 for 50 fewer. Number of pine trees per row along the 65 m is 31 instead of (11 + 18 =) 29, so 2 extra for each of 11 rows, so 22 [1] So number of pine trees required is 409 + 22 = 431 Extra cost = $7775- $7375 = $400 [1] Saving is $700 – $400 = $300 Alternative: Beech perimeter price [1] Pine area price [1] (both ft) Total [1] Difference [1] (ft if saving)

Question 43

Jez runs his own company, carrying out repairs and routine services on laptops and tablets. His business is very popular, and he always has…

Accepted Answer

4(a) 1 laptop repair @ 6 hours = $350 [1] 2 + 2 services @ $60 = $120 $470 4(b) 3 laptop repairs @ 1 hour each = $300 [1] 2 + 3 tablet repairs @ 1 hour each = $270 + 2 services @ $60 = $120 $690 SC: 1 mark for answer of $800 4(c) Yes with $250 and $360 seen 3 5 8-hour days: least profit per day is ‘$470’ – $100 – 8  $40 = $50, so weekly profit = $250 4 10-hour days: least income comes from 1 laptop repair taking 6 hours + 4 services = $590 Profit per day = $590 – $100 – 10  $40 = $90, so weekly profit $360 OR since salary unchanged: Yes with $1850 and $1960 seen $470 – $100 = $370 and $590 – $100 = $490 per day so 5  $370 = $1850 and 4  $490 = $1960 Award 1 mark for $250 OR $590 OR $360 OR $1850 OR $490 OR $1960 Award 2 marks for $250 AND ($590 OR $360) OR $1850 AND ($490 OR $1960) Alternative methods may look at weekly income v. cost: Yes with $2350 and $110 seen $2350 → $2360, income +$10; cost –$100; so weekly profit increase by $110 Award 1 mark for $2350 or $2360 or $110 Award 2 marks for $2350 and ($2360 or $110) SC: 2 marks for stating Jez is correct based on $250 compared with $400 OR stating Jez is correct based on $1850 compared with $2000 4(d) Least income = $590 (Jez) + 10  $48 (Becky) = $1070 2 Outgoings are $150 + 10  $40 + 10  $25 = $800 1 mark for either Profit = $1070 – $800 = $270 Alternatively: Jez profit = $590 – 10  $40 = $190 Becky profit = 10  $48 – 10  $25 = $230 1 mark for either Other outgoings are $150 Profit = $190 + $230 – $150 = $270 SC: 2 marks for $280 if $600 for Jez seen in 4(c) 4(e) Follow through incorrect value of $600 in 4(c) OR their $270 in 4(d) except for 3 final answer Becky can do one service per hour, at a profit of $23 [1] Profit at 10 hours is $270, so can reduce profit by up to $70 [1] 70 / 23, so 3 hours reduction So 7 hours [1] Alternatively: Least income = $590 + $48x Outgoings = $150 + 10  $40 + $25x Profit per day = $(590 + 48x – 550 – 25x) = $(40 + 23x) [1] 40 + 23x ⩾ 200 [1] requires x to be at least 6.95, so Becky needs to work 7 hours [1] Alternatively: Becky can do one repair per hour, at a profit of $23 [1] Calculation of profit for Becky working a number of hours in the range [5,9] [1] 7 hours [1] 4(f) Least income from Jez is still $590 3 Least income from Becky: 10  routine service = $600 (before discount) so $480 once 20% discount applied [1] Minimum total income is therefore $590 + $480 = $1070 For $200 profit, outgoings must be at most $870 [1] Jez pays himself $400 Other costs are $150 Maximum amount that Becky can be paid is $320 for 10 hours $32 per hour Alternatively: Least income from Becky: 10  routine service = $600 (before discount) so $480 once 20% discount applied [1] Which is the same as before, so the profit is still $270 if Becky earns $25 per hour Maximum possible increase to Becky’s rate of pay is $70/10 [1] New rate of pay is $32 per hour

Question 44

A rural hospital is planning the staff and beds it will need during the forthcoming months. Any patient arriving at the Accident and Emerge…

Accepted Answer

Question Answer Marks 1(a) On average 6 arrivals. 1 assessment per 20 minutes = 3 per Assessor per 1 hour. 6 ÷ 3 = 2 Assessors needed. 1(b) 2 marks for any example with the correct totals AND values within the limits 2 AND must be at least cumulative 6,12,18….. 1 mark for any example with two of these features 1 mark for an example in which any one type of appointment has the correct total and values within the limits 1 2 3 4 5 6 7 8 total U 8 0 0 0 0 0 0 0 8 S 0 0 1 1 1 1 2 2 8 H 2 2 5 5 5 5 5 3 32 1(c) Supervision: 1 an hour for 24 hours, but they start leaving (one in one out) 2 after 6: 6 beds needed [1] Urgents: 1 an hour for 24 hours = 24 beds needed If (at any point in time) the Urgent case classified 24 hours ago was not assessed immediately upon arrival, then 1 more bed is needed Total = 31 SC: 2 marks for final answer 30 1(d) min = 8  2 homes = 16 arrivals [1] 2 max = (8  5 homes) + (2  3 supervision) = 46 arrivals [1] SC: If 0 scored, award 1 mark for max = 44. (from 2  2 supervision) 1(e) If all 15 (max of each) came at the same time then they would take 5 3 Assessor–hours to be seen If this happened every hour for 8 hours, 40 Assessor–hours needed [1] If there were only 3 Assessors, this would take 13 h 20 m to complete [1] Someone arriving at the beginning of the 8th hour could have to wait (13 h 20 m – 7 h – 20 m =) 6 hours [1] oe OR Only 9 patients can be assessed within an hour, so if 15 arrive there will be an overflow of 6 At the start of the 8th hour there could be 7  6 = 42 patients [1] from earlier hours still waiting, plus 15 arrivals for the 8th hour The total number of patients waiting to be seen at the start of the 8th hour could be 57 57 assessments takes 19 hours [1], so with 3 assessors 6 hours 20 minutes are required The assessment takes 20 minutes, so the longest wait would be 6 hours 20 minutes – 20 minutes = 6 hours [1]

Question 45

Fansee is a ball sport in which two teams attempt to hit a target suspended above the ground at their opponents’ end of the pitch. A fansee…

Accepted Answer

2(a) (15  4) + (2  8) + (12  1) = 88 minutes oe 2 1 mark for 12 intervals of 1 minute each between flytes soi SC: 1 mark for a final answer of 89 minutes 2(b) For the scores to be level with both teams having scored the same number of 1 taps, they would need to have scored the same number of holds, which is not possible as there is an odd number of flytes 2(c) The Pentads beat the Treys (33 – 21) 1 2(d) The Hexyls: 1 The Treys did not win any matches OR the Hexyls won one match, but lost to the Aces and the Quartos 2(e) The total number of points scored was 30 (18 + 12) [1] 2 (which means that all the scores were holds,) so every flyte lasted 4 minutes / the maximum possible time [1] OR It was the only match in which there were no taps scored [1] so it lasted the maximum amount of time [1] 1 mark for associating fewer points with more time 2(f) The Aces’ 31 could be scored by 5 taps and 3 holds, 3 taps and 8 holds, or 1 2 tap and 13 holds The Deuces’ 23 could be scored by 3 taps and 4 holds or 1 tap and 9 holds 1 mark for identifying all possibilities for one of the teams The only pair which constitutes 15 flytes is Aces: 5 taps and 3 holds Deuces: 3 taps and 4 holds OR 1 mark for noting 31 + 23 = 54 = 30 + 3t total, 8 taps, so 7 holds. SC: 1 mark for full answer but with names swapped 2(g) 63 points scored by the Quartos must be from 63/(5 + 2) = 9 taps (+ 9 holds) 2 [1] They must have scored 6 taps (and 2 holds) against the Aces, so they scored 3 taps [1] (and 7 holds) against the Hexyls OR 34 against the Aces could be scored by 6 taps and 2 holds, 4 taps and 7 holds, or 2 taps and 12 holds 29 against the Hexyls could be scored by 5 taps and 2 holds, 3 taps and 7 holds, or 1 tap and 12 holds 1 mark for either Only one pair of these is consistent with the opponents’ scores, so they must have scored 6 taps against the Aces and 3 taps [1] against the Hexyls 2(h) 5 flytes won with taps would give them 30 points; if the Pentads won the 2 remaining five with holds they would have 27 1 mark for establishing that either 4 or 3 would not be enough: 4 taps would give the Deuces 25 points, but the other six flytes would give the Treys at least 29 points.3 taps would give the Quartos 20 points, but the other seven flytes would give the Pentads at least 31 points 2(i) Any one of the solutions shown in the table: 2 A v T A v T A v T A v T D v Q D v Q P v H P v H P v H P v H D v Q D v Q T v Q T v Q A v P A v P A v P D v H D v H T v Q D v H A v P T v Q D v H D v H D v H D v H D v H T v Q T v Q A v P A v P A v P A v P T v Q T v Q D v Q D v Q P v H P v H A v T P v H A v T D v Q P v H A v T D v Q A v T 1 mark for a schedule which includes each team twice, but has at most one instance of any of the following: Either Pentads or Quartos in the first match A team name appearing on two consecutive lines A team name not appearing in the last three lines 1 mark for correct answer with final game missing

Question 46

A rural hospital is planning the staff and beds it will need during the forthcoming months. Any patient arriving at the Accident and Emerge…

Accepted Answer

Question Answer Marks 1(a) On average 6 arrivals. 1 assessment per 20 minutes = 3 per Assessor per 1 hour. 6 ÷ 3 = 2 Assessors needed. 1(b) 2 marks for any example with the correct totals AND values within the limits 2 AND must be at least cumulative 6,12,18….. 1 mark for any example with two of these features 1 mark for an example in which any one type of appointment has the correct total and values within the limits 1 2 3 4 5 6 7 8 total U 8 0 0 0 0 0 0 0 8 S 0 0 1 1 1 1 2 2 8 H 2 2 5 5 5 5 5 3 32 1(c) Supervision: 1 an hour for 24 hours, but they start leaving (one in one out) 2 after 6: 6 beds needed [1] Urgents: 1 an hour for 24 hours = 24 beds needed If (at any point in time) the Urgent case classified 24 hours ago was not assessed immediately upon arrival, then 1 more bed is needed Total = 31 SC: 2 marks for final answer 30 1(d) min = 8  2 homes = 16 arrivals [1] 2 max = (8  5 homes) + (2  3 supervision) = 46 arrivals [1] SC: If 0 scored, award 1 mark for max = 44. (from 2  2 supervision) 1(e) If all 15 (max of each) came at the same time then they would take 5 3 Assessor–hours to be seen If this happened every hour for 8 hours, 40 Assessor–hours needed [1] If there were only 3 Assessors, this would take 13 h 20 m to complete [1] Someone arriving at the beginning of the 8th hour could have to wait (13 h 20 m – 7 h – 20 m =) 6 hours [1] oe OR Only 9 patients can be assessed within an hour, so if 15 arrive there will be an overflow of 6 At the start of the 8th hour there could be 7  6 = 42 patients [1] from earlier hours still waiting, plus 15 arrivals for the 8th hour The total number of patients waiting to be seen at the start of the 8th hour could be 57 57 assessments takes 19 hours [1], so with 3 assessors 6 hours 20 minutes are required The assessment takes 20 minutes, so the longest wait would be 6 hours 20 minutes – 20 minutes = 6 hours [1]

Question 47

Fansee is a ball sport in which two teams attempt to hit a target suspended above the ground at their opponents’ end of the pitch. A fansee…

Accepted Answer

2(a) (15  4) + (2  8) + (12  1) = 88 minutes oe 2 1 mark for 12 intervals of 1 minute each between flytes soi SC: 1 mark for a final answer of 89 minutes 2(b) For the scores to be level with both teams having scored the same number of 1 taps, they would need to have scored the same number of holds, which is not possible as there is an odd number of flytes 2(c) The Pentads beat the Treys (33 – 21) 1 2(d) The Hexyls: 1 The Treys did not win any matches OR the Hexyls won one match, but lost to the Aces and the Quartos 2(e) The total number of points scored was 30 (18 + 12) [1] 2 (which means that all the scores were holds,) so every flyte lasted 4 minutes / the maximum possible time [1] OR It was the only match in which there were no taps scored [1] so it lasted the maximum amount of time [1] 1 mark for associating fewer points with more time 2(f) The Aces’ 31 could be scored by 5 taps and 3 holds, 3 taps and 8 holds, or 1 2 tap and 13 holds The Deuces’ 23 could be scored by 3 taps and 4 holds or 1 tap and 9 holds 1 mark for identifying all possibilities for one of the teams The only pair which constitutes 15 flytes is Aces: 5 taps and 3 holds Deuces: 3 taps and 4 holds OR 1 mark for noting 31 + 23 = 54 = 30 + 3t total, 8 taps, so 7 holds. SC: 1 mark for full answer but with names swapped 2(g) 63 points scored by the Quartos must be from 63/(5 + 2) = 9 taps (+ 9 holds) 2 [1] They must have scored 6 taps (and 2 holds) against the Aces, so they scored 3 taps [1] (and 7 holds) against the Hexyls OR 34 against the Aces could be scored by 6 taps and 2 holds, 4 taps and 7 holds, or 2 taps and 12 holds 29 against the Hexyls could be scored by 5 taps and 2 holds, 3 taps and 7 holds, or 1 tap and 12 holds 1 mark for either Only one pair of these is consistent with the opponents’ scores, so they must have scored 6 taps against the Aces and 3 taps [1] against the Hexyls 2(h) 5 flytes won with taps would give them 30 points; if the Pentads won the 2 remaining five with holds they would have 27 1 mark for establishing that either 4 or 3 would not be enough: 4 taps would give the Deuces 25 points, but the other six flytes would give the Treys at least 29 points.3 taps would give the Quartos 20 points, but the other seven flytes would give the Pentads at least 31 points 2(i) Any one of the solutions shown in the table: 2 A v T A v T A v T A v T D v Q D v Q P v H P v H P v H P v H D v Q D v Q T v Q T v Q A v P A v P A v P D v H D v H T v Q D v H A v P T v Q D v H D v H D v H D v H D v H T v Q T v Q A v P A v P A v P A v P T v Q T v Q D v Q D v Q P v H P v H A v T P v H A v T D v Q P v H A v T D v Q A v T 1 mark for a schedule which includes each team twice, but has at most one instance of any of the following: Either Pentads or Quartos in the first match A team name appearing on two consecutive lines A team name not appearing in the last three lines 1 mark for correct answer with final game missing

Question 48

A rural hospital is planning the staff and beds it will need during the forthcoming months. Any patient arriving at the Accident and Emerge…

Accepted Answer

Question Answer Marks 1(a) On average 6 arrivals. 1 assessment per 20 minutes = 3 per Assessor per 1 hour. 6 ÷ 3 = 2 Assessors needed. 1(b) 2 marks for any example with the correct totals AND values within the limits 2 AND must be at least cumulative 6,12,18….. 1 mark for any example with two of these features 1 mark for an example in which any one type of appointment has the correct total and values within the limits 1 2 3 4 5 6 7 8 total U 8 0 0 0 0 0 0 0 8 S 0 0 1 1 1 1 2 2 8 H 2 2 5 5 5 5 5 3 32 1(c) Supervision: 1 an hour for 24 hours, but they start leaving (one in one out) 2 after 6: 6 beds needed [1] Urgents: 1 an hour for 24 hours = 24 beds needed If (at any point in time) the Urgent case classified 24 hours ago was not assessed immediately upon arrival, then 1 more bed is needed Total = 31 SC: 2 marks for final answer 30 1(d) min = 8  2 homes = 16 arrivals [1] 2 max = (8  5 homes) + (2  3 supervision) = 46 arrivals [1] SC: If 0 scored, award 1 mark for max = 44. (from 2  2 supervision) 1(e) If all 15 (max of each) came at the same time then they would take 5 3 Assessor–hours to be seen If this happened every hour for 8 hours, 40 Assessor–hours needed [1] If there were only 3 Assessors, this would take 13 h 20 m to complete [1] Someone arriving at the beginning of the 8th hour could have to wait (13 h 20 m – 7 h – 20 m =) 6 hours [1] oe OR Only 9 patients can be assessed within an hour, so if 15 arrive there will be an overflow of 6 At the start of the 8th hour there could be 7  6 = 42 patients [1] from earlier hours still waiting, plus 15 arrivals for the 8th hour The total number of patients waiting to be seen at the start of the 8th hour could be 57 57 assessments takes 19 hours [1], so with 3 assessors 6 hours 20 minutes are required The assessment takes 20 minutes, so the longest wait would be 6 hours 20 minutes – 20 minutes = 6 hours [1]

Question 49

Fansee is a ball sport in which two teams attempt to hit a target suspended above the ground at their opponents’ end of the pitch. A fansee…

Accepted Answer

2(a) (15  4) + (2  8) + (12  1) = 88 minutes oe 2 1 mark for 12 intervals of 1 minute each between flytes soi SC: 1 mark for a final answer of 89 minutes 2(b) For the scores to be level with both teams having scored the same number of 1 taps, they would need to have scored the same number of holds, which is not possible as there is an odd number of flytes 2(c) The Pentads beat the Treys (33 – 21) 1 2(d) The Hexyls: 1 The Treys did not win any matches OR the Hexyls won one match, but lost to the Aces and the Quartos 2(e) The total number of points scored was 30 (18 + 12) [1] 2 (which means that all the scores were holds,) so every flyte lasted 4 minutes / the maximum possible time [1] OR It was the only match in which there were no taps scored [1] so it lasted the maximum amount of time [1] 1 mark for associating fewer points with more time 2(f) The Aces’ 31 could be scored by 5 taps and 3 holds, 3 taps and 8 holds, or 1 2 tap and 13 holds The Deuces’ 23 could be scored by 3 taps and 4 holds or 1 tap and 9 holds 1 mark for identifying all possibilities for one of the teams The only pair which constitutes 15 flytes is Aces: 5 taps and 3 holds Deuces: 3 taps and 4 holds OR 1 mark for noting 31 + 23 = 54 = 30 + 3t total, 8 taps, so 7 holds. SC: 1 mark for full answer but with names swapped 2(g) 63 points scored by the Quartos must be from 63/(5 + 2) = 9 taps (+ 9 holds) 2 [1] They must have scored 6 taps (and 2 holds) against the Aces, so they scored 3 taps [1] (and 7 holds) against the Hexyls OR 34 against the Aces could be scored by 6 taps and 2 holds, 4 taps and 7 holds, or 2 taps and 12 holds 29 against the Hexyls could be scored by 5 taps and 2 holds, 3 taps and 7 holds, or 1 tap and 12 holds 1 mark for either Only one pair of these is consistent with the opponents’ scores, so they must have scored 6 taps against the Aces and 3 taps [1] against the Hexyls 2(h) 5 flytes won with taps would give them 30 points; if the Pentads won the 2 remaining five with holds they would have 27 1 mark for establishing that either 4 or 3 would not be enough: 4 taps would give the Deuces 25 points, but the other six flytes would give the Treys at least 29 points.3 taps would give the Quartos 20 points, but the other seven flytes would give the Pentads at least 31 points 2(i) Any one of the solutions shown in the table: 2 A v T A v T A v T A v T D v Q D v Q P v H P v H P v H P v H D v Q D v Q T v Q T v Q A v P A v P A v P D v H D v H T v Q D v H A v P T v Q D v H D v H D v H D v H D v H T v Q T v Q A v P A v P A v P A v P T v Q T v Q D v Q D v Q P v H P v H A v T P v H A v T D v Q P v H A v T D v Q A v T 1 mark for a schedule which includes each team twice, but has at most one instance of any of the following: Either Pentads or Quartos in the first match A team name appearing on two consecutive lines A team name not appearing in the last three lines 1 mark for correct answer with final game missing

Question 50

Your Choice is a TV general knowledge quiz. Three contestants take part in each show. The three contestants are asked the same 40 questions…

Accepted Answer

Question Answer Marks 1(a) Trina received $40 for 7 more questions than the other two, so she received 2 $40 for 18 questions (and the other two 11 each) [1] 18  $40 + 22  $30 = $1380 Alternative solution: 1 mark for sight of trial and improvement, beginning with two equal amounts and one slightly larger leading to $1310, $1310 and $1380 1(b) The contestants’ totals add up to ($1290 + $1270 + $1240 =) $3800 [1] 2 They would add up to $4000 [1] if all 40 were answered correctly by at least one contestant Discrepancy 2  $100 1(c)(i) Tracey’s $1240 for 34 correct answers was made up only of a combination of 3 $50 and $30 1 mark for any pair of multiples of $50 and $30 with a sum of 34, OR a total of $1240, evaluated correctly 1 mark for a second trial with an improved result 11  $50 + 23  $30 = $1240 (so) the number of questions answered correctly by all three is 23 Alternative solution: x + y = 34 oe [1] 50x + 30y = 1240 [1] 1(c)(ii) 40 – 2 (nobody) – 1 (Una only) – 23 ft = 14 ft 1 1(d) 5  $990 = $4950 [1] 2 6  $1040 = $6240 So (the minimum number of questions she must answer is) 6 [1] Alternative solution: 1 mark for a correct algebraic expression, e.g. n(1190 – (9 – n) 50) So (the minimum number of questions she must answer is) 6 [1]

Question 51

Grace is a children’s entertainer who can be booked for parties. When she receives a request to perform at a party, Grace collects the foll…

Accepted Answer

2(a) (88 – 10) / 3 = 26 [1] 2 90 % of 26 = 23.4 < 25 [1] Alternative solution: Mollie: (88 – 10) / 3 = 26 [1] 90 % of 26 = 23.4 Frank: 1.1  (99 – 12) / 3 = 31.9 Mollie’s rating of 23.4 is less than Frank’s 31.9, so not chosen [1] 2(b) John’s party has a rating of 31 2 Frank’s party has a rating of 31.9 1 mark for either rating calculated correctly Since 31 < 31.9, John’s party will not be chosen [1] 2(c) 135 – 9 = 126 and 126 ÷ 4 = 31.5 1 31.5 < 31.9 < 31.5 + 3.15 1 or 2 2(d) 31.9  2 = 63.8 [1] 2 The fee must be more than 63.8 + 6 = 69.8 $70 2(e) (69 – 5) ÷ 2 = 32 2 (109 – 5) ÷ 4 = 26 1 mark for both scores So 2 additional hours increases the score by 6 3 for each hour 2(f) 25 + 3  5 = 40 1 2(g) Under the new system, Frank’s party is the best of the others with a score of 2 46.9 [1] The lowest score Wendy’s party could be given is: (135 – 9) ÷ 4 = 31.5 31.5 – 3.15 = 28.35 28.35 + 4  5 = 48.35 [1] 2(h) Polly’s party would have a score of (180 – 5) / 5 = 35, 3 reduced by 10 % to 31.5 [1] plus the fixed amount of 25 = 56.5 To achieve a score of 56.5 would require a fee of (56.5 – 20) [1]  4 + 7 = $153 1 mark for calculating the score for Quentin’s for two choices of fee with improvement towards their value of Polly’s score SC: 2 marks for final answer $133

Question 52

Your Choice is a TV general knowledge quiz. Three contestants take part in each show. The three contestants are asked the same 40 questions…

Accepted Answer

Question Answer Marks 1(a) Trina received $40 for 7 more questions than the other two, so she received 2 $40 for 18 questions (and the other two 11 each) [1] 18  $40 + 22  $30 = $1380 Alternative solution: 1 mark for sight of trial and improvement, beginning with two equal amounts and one slightly larger leading to $1310, $1310 and $1380 1(b) The contestants’ totals add up to ($1290 + $1270 + $1240 =) $3800 [1] 2 They would add up to $4000 [1] if all 40 were answered correctly by at least one contestant Discrepancy 2  $100 1(c)(i) Tracey’s $1240 for 34 correct answers was made up only of a combination of 3 $50 and $30 1 mark for any pair of multiples of $50 and $30 with a sum of 34, OR a total of $1240, evaluated correctly 1 mark for a second trial with an improved result 11  $50 + 23  $30 = $1240 (so) the number of questions answered correctly by all three is 23 Alternative solution: x + y = 34 oe [1] 50x + 30y = 1240 [1] 1(c)(ii) 40 – 2 (nobody) – 1 (Una only) – 23 ft = 14 ft 1 1(d) 5  $990 = $4950 [1] 2 6  $1040 = $6240 So (the minimum number of questions she must answer is) 6 [1] Alternative solution: 1 mark for a correct algebraic expression, e.g. n(1190 – (9 – n) 50) So (the minimum number of questions she must answer is) 6 [1]

Question 53

Grace is a children’s entertainer who can be booked for parties. When she receives a request to perform at a party, Grace collects the foll…

Accepted Answer

2(a) (88 – 10) / 3 = 26 [1] 2 90 % of 26 = 23.4 < 25 [1] Alternative solution: Mollie: (88 – 10) / 3 = 26 [1] 90 % of 26 = 23.4 Frank: 1.1  (99 – 12) / 3 = 31.9 Mollie’s rating of 23.4 is less than Frank’s 31.9, so not chosen [1] 2(b) John’s party has a rating of 31 2 Frank’s party has a rating of 31.9 1 mark for either rating calculated correctly Since 31 < 31.9, John’s party will not be chosen [1] 2(c) 135 – 9 = 126 and 126 ÷ 4 = 31.5 1 31.5 < 31.9 < 31.5 + 3.15 1 or 2 2(d) 31.9  2 = 63.8 [1] 2 The fee must be more than 63.8 + 6 = 69.8 $70 2(e) (69 – 5) ÷ 2 = 32 2 (109 – 5) ÷ 4 = 26 1 mark for both scores So 2 additional hours increases the score by 6 3 for each hour 2(f) 25 + 3  5 = 40 1 2(g) Under the new system, Frank’s party is the best of the others with a score of 2 46.9 [1] The lowest score Wendy’s party could be given is: (135 – 9) ÷ 4 = 31.5 31.5 – 3.15 = 28.35 28.35 + 4  5 = 48.35 [1] 2(h) Polly’s party would have a score of (180 – 5) / 5 = 35, 3 reduced by 10 % to 31.5 [1] plus the fixed amount of 25 = 56.5 To achieve a score of 56.5 would require a fee of (56.5 – 20) [1]  4 + 7 = $153 1 mark for calculating the score for Quentin’s for two choices of fee with improvement towards their value of Polly’s score SC: 2 marks for final answer $133

Question 54

In the Double Triple Quiz there are 4 rounds of questions. In each round each contestant is asked 5 questions. Points are awarded for corre…

Accepted Answer

2(a) Points awarded are 3  (10 + 20 + 30 + 50) = 330 2 Biggest deduction for 2 questions is 0 – 30 – 50 – 70 = (−)150 [1] So least possible total is 330 – 150 = 180 [1] AG 2(b) 2 correct, 2 incorrect and 1 pass 1 2(c) 65, 25 from (4, 0, 1) and (2, 3, 0) 3 40, 0 from (3, 1, 1) and (1, 4, 0) 30, –10 from (3, 0, 2) and (1, 3, 1) 0, –40 from (1, 4, 0) and (1, 0, 4) 2 marks for 3 correct with at most one incorrect OR 2 marks for a list of 4 containing only correct pairs of scores or pairs of scores in which one player answered no questions correctly: 15, –25 from (2,2,1) and (0,5,0), 5, –35 from (2,1,2) and (0,4,1), –5, –45 from (2, 0, 3) and (0, 3, 2), –25, –65 from (0,5,0) and (0,1,4), –35, –75 from (0,4,1) and (0,0,5) OR 1 mark for 2 correct or finding vectors (-2, +3, –1) or (0, +4, –4) 2(d)(i) Alexa (3, 2, 0) 2 Betty (3, 2, 0) Charlie (3, 0, 2) Damon (4, 1, 0) 1 mark for any pair from AC, AD, BC, BD, CD correct 2(d)(ii) If Charlie had given answers (even if incorrect) to the two questions he 1 passed, then he would have at least 30 points more, so a total of at least 300, which is more than 285 2(e) 8 Hard questions, one with Double and one with Triple, 1 so 6  2 + 4 + 6 = 22 AG 2(f) Highest: scores in first 5 questions 1, 2, 1, 2, 1 (7) then Hard Double (4), Easy 2 (1), Hard Triple (6), total 18 [1] Lowest: Valid example for questions 1-5, e.g. 3, 4, 0, 0, 0 or 3, 2, 2, 0, 0 then Hard (2), Easy (1), Hard (2), total 12 [1] SC: 1 mark for 12 and 18 with no/incorrect example given 2(g) • Alexa must have answered her last three questions correctly to score 16 3 with one of them tripled. • Therefore she answered at most 2 questions incorrectly. • The only possibility is that Damon scores 6, 4 and 0 from his final three questions. • This means that he must have answered 3 questions incorrectly. 3 marks for all four steps in the reasoning given. 2 marks for any two given. 1 mark for any one given

Question 55

George and Rachel are playing a game of CounterBid. The equipment consists of a bag of counters, two small trays and two sets of the three …

Accepted Answer

4(a) George has 5 counters in his tray once the counters have been placed. 2 After the first card has been applied, this will increase to 10 counters. [1] The second card will be applied and double this to 20 counters. The final card will not be applied, so he will return the 1 counter to his tray. [1] 4(b) Rachel has 6 counters in her tray once the counters have been placed. 1 After the first card has been applied, this will increase to 12 counters, and once the final card has been applied it will increase to 17 counters. 4(c)(i) George would have two more / 23 counters 1 4(c)(ii) George would have 15 counters [1] 2 Rachel would have 15 counters [1] 4(d)(i) George cannot have placed more counters than Rachel on the first position, 1 as he would have to have more than 24 counters at the end of the round in that case [1] 4(d)(ii) To get 24 counters, he must have gained 5 and lost 2 [1] 2 From the given information, he must have placed at least 1 in each position So 1 and 1 4(e)(i) Score (8 + 5)  2 = 26 [1] 3 Any example with the following features (where Player 1 achieves the maximum possible): Player 1 adds no counters to ‘Your opponent loses 5 counters’ and Player 2 does not apply the ‘Your opponent loses 5 counters’ card [1] Player 1 adds 1 counter to ‘Add 5 counters to your tray’, in an earlier position than 1 counter added to ‘Double the number of counters in your tray’ card [1] 4(e)(ii) Any example in which 5 counters are allocated by the player who ends on 0 3 counters with the following features: For each position where counters are allocated, the opponent has not allocated more counters (so no counters are returned to the tray) [1] The player adds no counters to ‘You gain 5 counters’ [1] The player adds no counters to the Double card [1] OR The opponent will apply the ‘Your opponent loses 5 counters’ card before the player applies the Double card [1] SC: 1 mark for an example in which no counters are allocated to ‘Add 5’ or to ‘Double’

Question 56

In the Double Triple Quiz there are 4 rounds of questions. In each round each contestant is asked 5 questions. Points are awarded for corre…

Accepted Answer

2(a) Points awarded are 3  (10 + 20 + 30 + 50) = 330 2 Biggest deduction for 2 questions is 0 – 30 – 50 – 70 = (−)150 [1] So least possible total is 330 – 150 = 180 [1] AG 2(b) 2 correct, 2 incorrect and 1 pass 1 2(c) 65, 25 from (4, 0, 1) and (2, 3, 0) 3 40, 0 from (3, 1, 1) and (1, 4, 0) 30, –10 from (3, 0, 2) and (1, 3, 1) 0, –40 from (1, 4, 0) and (1, 0, 4) 2 marks for 3 correct with at most one incorrect OR 2 marks for a list of 4 containing only correct pairs of scores or pairs of scores in which one player answered no questions correctly: 15, –25 from (2,2,1) and (0,5,0), 5, –35 from (2,1,2) and (0,4,1), –5, –45 from (2, 0, 3) and (0, 3, 2), –25, –65 from (0,5,0) and (0,1,4), –35, –75 from (0,4,1) and (0,0,5) OR 1 mark for 2 correct or finding vectors (-2, +3, –1) or (0, +4, –4) 2(d)(i) Alexa (3, 2, 0) 2 Betty (3, 2, 0) Charlie (3, 0, 2) Damon (4, 1, 0) 1 mark for any pair from AC, AD, BC, BD, CD correct 2(d)(ii) If Charlie had given answers (even if incorrect) to the two questions he 1 passed, then he would have at least 30 points more, so a total of at least 300, which is more than 285 2(e) 8 Hard questions, one with Double and one with Triple, 1 so 6  2 + 4 + 6 = 22 AG 2(f) Highest: scores in first 5 questions 1, 2, 1, 2, 1 (7) then Hard Double (4), Easy 2 (1), Hard Triple (6), total 18 [1] Lowest: Valid example for questions 1-5, e.g. 3, 4, 0, 0, 0 or 3, 2, 2, 0, 0 then Hard (2), Easy (1), Hard (2), total 12 [1] SC: 1 mark for 12 and 18 with no/incorrect example given 2(g) • Alexa must have answered her last three questions correctly to score 16 3 with one of them tripled. • Therefore she answered at most 2 questions incorrectly. • The only possibility is that Damon scores 6, 4 and 0 from his final three questions. • This means that he must have answered 3 questions incorrectly. 3 marks for all four steps in the reasoning given. 2 marks for any two given. 1 mark for any one given

Question 57

George and Rachel are playing a game of CounterBid. The equipment consists of a bag of counters, two small trays and two sets of the three …

Accepted Answer

4(a) George has 5 counters in his tray once the counters have been placed. 2 After the first card has been applied, this will increase to 10 counters. [1] The second card will be applied and double this to 20 counters. The final card will not be applied, so he will return the 1 counter to his tray. [1] 4(b) Rachel has 6 counters in her tray once the counters have been placed. 1 After the first card has been applied, this will increase to 12 counters, and once the final card has been applied it will increase to 17 counters. 4(c)(i) George would have two more / 23 counters 1 4(c)(ii) George would have 15 counters [1] 2 Rachel would have 15 counters [1] 4(d)(i) George cannot have placed more counters than Rachel on the first position, 1 as he would have to have more than 24 counters at the end of the round in that case [1] 4(d)(ii) To get 24 counters, he must have gained 5 and lost 2 [1] 2 From the given information, he must have placed at least 1 in each position So 1 and 1 4(e)(i) Score (8 + 5)  2 = 26 [1] 3 Any example with the following features (where Player 1 achieves the maximum possible): Player 1 adds no counters to ‘Your opponent loses 5 counters’ and Player 2 does not apply the ‘Your opponent loses 5 counters’ card [1] Player 1 adds 1 counter to ‘Add 5 counters to your tray’, in an earlier position than 1 counter added to ‘Double the number of counters in your tray’ card [1] 4(e)(ii) Any example in which 5 counters are allocated by the player who ends on 0 3 counters with the following features: For each position where counters are allocated, the opponent has not allocated more counters (so no counters are returned to the tray) [1] The player adds no counters to ‘You gain 5 counters’ [1] The player adds no counters to the Double card [1] OR The opponent will apply the ‘Your opponent loses 5 counters’ card before the player applies the Double card [1] SC: 1 mark for an example in which no counters are allocated to ‘Add 5’ or to ‘Double’

Question 58

In the Double Triple Quiz there are 4 rounds of questions. In each round each contestant is asked 5 questions. Points are awarded for corre…

Accepted Answer

2(a) Points awarded are 3  (10 + 20 + 30 + 50) = 330 2 Biggest deduction for 2 questions is 0 – 30 – 50 – 70 = (−)150 [1] So least possible total is 330 – 150 = 180 [1] AG 2(b) 2 correct, 2 incorrect and 1 pass 1 2(c) 65, 25 from (4, 0, 1) and (2, 3, 0) 3 40, 0 from (3, 1, 1) and (1, 4, 0) 30, –10 from (3, 0, 2) and (1, 3, 1) 0, –40 from (1, 4, 0) and (1, 0, 4) 2 marks for 3 correct with at most one incorrect OR 2 marks for a list of 4 containing only correct pairs of scores or pairs of scores in which one player answered no questions correctly: 15, –25 from (2,2,1) and (0,5,0), 5, –35 from (2,1,2) and (0,4,1), –5, –45 from (2, 0, 3) and (0, 3, 2), –25, –65 from (0,5,0) and (0,1,4), –35, –75 from (0,4,1) and (0,0,5) OR 1 mark for 2 correct or finding vectors (-2, +3, –1) or (0, +4, –4) 2(d)(i) Alexa (3, 2, 0) 2 Betty (3, 2, 0) Charlie (3, 0, 2) Damon (4, 1, 0) 1 mark for any pair from AC, AD, BC, BD, CD correct 2(d)(ii) If Charlie had given answers (even if incorrect) to the two questions he 1 passed, then he would have at least 30 points more, so a total of at least 300, which is more than 285 2(e) 8 Hard questions, one with Double and one with Triple, 1 so 6  2 + 4 + 6 = 22 AG 2(f) Highest: scores in first 5 questions 1, 2, 1, 2, 1 (7) then Hard Double (4), Easy 2 (1), Hard Triple (6), total 18 [1] Lowest: Valid example for questions 1-5, e.g. 3, 4, 0, 0, 0 or 3, 2, 2, 0, 0 then Hard (2), Easy (1), Hard (2), total 12 [1] SC: 1 mark for 12 and 18 with no/incorrect example given 2(g) • Alexa must have answered her last three questions correctly to score 16 3 with one of them tripled. • Therefore she answered at most 2 questions incorrectly. • The only possibility is that Damon scores 6, 4 and 0 from his final three questions. • This means that he must have answered 3 questions incorrectly. 3 marks for all four steps in the reasoning given. 2 marks for any two given. 1 mark for any one given

Question 59

George and Rachel are playing a game of CounterBid. The equipment consists of a bag of counters, two small trays and two sets of the three …

Accepted Answer

4(a) George has 5 counters in his tray once the counters have been placed. 2 After the first card has been applied, this will increase to 10 counters. [1] The second card will be applied and double this to 20 counters. The final card will not be applied, so he will return the 1 counter to his tray. [1] 4(b) Rachel has 6 counters in her tray once the counters have been placed. 1 After the first card has been applied, this will increase to 12 counters, and once the final card has been applied it will increase to 17 counters. 4(c)(i) George would have two more / 23 counters 1 4(c)(ii) George would have 15 counters [1] 2 Rachel would have 15 counters [1] 4(d)(i) George cannot have placed more counters than Rachel on the first position, 1 as he would have to have more than 24 counters at the end of the round in that case [1] 4(d)(ii) To get 24 counters, he must have gained 5 and lost 2 [1] 2 From the given information, he must have placed at least 1 in each position So 1 and 1 4(e)(i) Score (8 + 5)  2 = 26 [1] 3 Any example with the following features (where Player 1 achieves the maximum possible): Player 1 adds no counters to ‘Your opponent loses 5 counters’ and Player 2 does not apply the ‘Your opponent loses 5 counters’ card [1] Player 1 adds 1 counter to ‘Add 5 counters to your tray’, in an earlier position than 1 counter added to ‘Double the number of counters in your tray’ card [1] 4(e)(ii) Any example in which 5 counters are allocated by the player who ends on 0 3 counters with the following features: For each position where counters are allocated, the opponent has not allocated more counters (so no counters are returned to the tray) [1] The player adds no counters to ‘You gain 5 counters’ [1] The player adds no counters to the Double card [1] OR The opponent will apply the ‘Your opponent loses 5 counters’ card before the player applies the Double card [1] SC: 1 mark for an example in which no counters are allocated to ‘Add 5’ or to ‘Double’

Question	Answer	Marks	From
1	see sheet	15	9694/31 May/June 2017
2	see sheet	10	9694/31 Oct/Nov 2017
3	see sheet	10	9694/32 Oct/Nov 2017
4	see sheet	10	9694/33 Oct/Nov 2017
5	see sheet	10	9694/31 Oct/Nov 2018
6	see sheet	15	9694/31 Oct/Nov 2018
7	see sheet	15	9694/31 Oct/Nov 2018
8	see sheet	10	9694/33 Oct/Nov 2018
9	see sheet	15	9694/33 Oct/Nov 2018
10	see sheet	15	9694/33 Oct/Nov 2018
11	see sheet	10	9694/31 May/June 2019
12	see sheet	15	9694/31 May/June 2019
13	see sheet	10	9694/32 May/June 2019
14	see sheet	10	9694/33 May/June 2019
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Questions79 pages

Mark scheme59 answers

Another paper, or another topic

Questions as text

Q1 · Square Deal is a game for two players, played over a number of rounds 9694/31 May/June 2017

Q2 · Five days every week, Faridah sells ice cream at the beach from a box on the back of her… 9694/31 Oct/Nov 2017

Q3 · Fred earns his living as a taxi-driver, offering lifts between his home town of… 9694/32 Oct/Nov 2017

Q4 · Fred earns his living as a taxi-driver, offering lifts between his home town of… 9694/33 Oct/Nov 2017

Q5 · Birdnest village school educates children for 5 years 9694/31 Oct/Nov 2018

Q6 · Jaspreet owns a business making suits to order – he only makes the suits once a customer… 9694/31 Oct/Nov 2018

Q7 · David is trying to work out the bonuses that he will pay to his employees for their work… 9694/31 Oct/Nov 2018

Q8 · Birdnest village school educates children for 5 years 9694/33 Oct/Nov 2018

Q9 · Jaspreet owns a business making suits to order – he only makes the suits once a customer… 9694/33 Oct/Nov 2018

Q10 · David is trying to work out the bonuses that he will pay to his employees for their work… 9694/33 Oct/Nov 2018

Q11 · Joshua is a taxi driver who is considering working only on Fridays, doing a 12-hour shift… 9694/31 May/June 2019

Q12 · Spelanskor is a game for two players, played over a number of rounds 9694/31 May/June 2019

Q13 · The British Diplomatic staff in Bolandia are paid a tax-free cost of living allowance to… 9694/32 May/June 2019

Q14 · The British Diplomatic staff in Bolandia are paid a tax-free cost of living allowance to… 9694/33 May/June 2019

Q15 · Electronic passports (e-passports) are only useful if their origin can be checked 9694/31 Oct/Nov 2019

Q16 · Peregrine is spending a week walking along a famous mountain path 9694/31 Oct/Nov 2019

Q17 · Electronic passports (e-passports) are only useful if their origin can be checked 9694/32 Oct/Nov 2019

Q18 · Peregrine is spending a week walking along a famous mountain path 9694/32 Oct/Nov 2019

Q19 · Electronic passports (e-passports) are only useful if their origin can be checked 9694/33 Oct/Nov 2019

Q20 · Peregrine is spending a week walking along a famous mountain path 9694/33 Oct/Nov 2019

Q21 · The all-inclusive holiday resort at Cofete provides buffet meals every day, but guests… 9694/31 May/June 2020

Q22 · In Bolandian currency, $1 is worth 100 cents (¢) 9694/32 May/June 2020

Q23 · Every year at the two-day Chevalier Horse Show teams of four riders from Frogford… 9694/32 May/June 2020

Q24 · Sally is organising a charity concert next month 9694/32 May/June 2020

Q25 · In Bolandian currency, $1 is worth 100 cents (¢) 9694/33 May/June 2020

Q26 · Every year at the two-day Chevalier Horse Show teams of four riders from Frogford… 9694/33 May/June 2020

Q27 · Sally is organising a charity concert next month 9694/33 May/June 2020

Q28 · In a country where too much food is grown it was decided to pay farmers to leave 10% of… 9694/31 Oct/Nov 2020

Q29 · The Goodlen Dancing Society holds a dancing competition every Saturday 9694/31 Oct/Nov 2020

Q30 · In a country where too much food is grown it was decided to pay farmers to leave 10% of… 9694/32 Oct/Nov 2020

Q31 · The Goodlen Dancing Society holds a dancing competition every Saturday 9694/32 Oct/Nov 2020

Q32 · In a country where too much food is grown it was decided to pay farmers to leave 10% of… 9694/33 Oct/Nov 2020

Q33 · The Goodlen Dancing Society holds a dancing competition every Saturday 9694/33 Oct/Nov 2020

Q34 · There is a long-distance cycle route between Princeville and Queda, of total length 1560… 9694/31 May/June 2021

Q35 · John is looking for a hotel near a conference hall 9694/31 May/June 2021

Q36 · There is a train service between Arba and Boab 9694/32 May/June 2022

Q37 · There is a train service between Arba and Boab 9694/33 May/June 2022

Q38 · OrienT-8 is a single-player game, played on a 10 × 10 grid displayed on the touch screen… 9694/31 May/June 2023

Q39 · Julie sells sweets in her shop 9694/32 May/June 2023

Q40 · The Bolandian Environment Agency is planning to plant trees on plots of land formerly… 9694/32 May/June 2023

Q41 · Julie sells sweets in her shop 9694/33 May/June 2023

Q42 · The Bolandian Environment Agency is planning to plant trees on plots of land formerly… 9694/33 May/June 2023

Q43 · Jez runs his own company, carrying out repairs and routine services on laptops and tablets 9694/31 May/June 2024

Q44 · A rural hospital is planning the staff and beds it will need during the forthcoming months 9694/31 Oct/Nov 2024

Q45 · Fansee is a ball sport in which two teams attempt to hit a target suspended above the… 9694/31 Oct/Nov 2024

Q46 · A rural hospital is planning the staff and beds it will need during the forthcoming months 9694/32 Oct/Nov 2024

Q47 · Fansee is a ball sport in which two teams attempt to hit a target suspended above the… 9694/32 Oct/Nov 2024

Q48 · A rural hospital is planning the staff and beds it will need during the forthcoming months 9694/33 Oct/Nov 2024

Q49 · Fansee is a ball sport in which two teams attempt to hit a target suspended above the… 9694/33 Oct/Nov 2024

Q50 · Your Choice is a TV general knowledge quiz 9694/32 May/June 2025

Q51 · Grace is a children’s entertainer who can be booked for parties 9694/32 May/June 2025

Q52 · Your Choice is a TV general knowledge quiz 9694/33 May/June 2025

Q53 · Grace is a children’s entertainer who can be booked for parties 9694/33 May/June 2025

Q54 · In the Double Triple Quiz there are 4 rounds of questions 9694/31 Oct/Nov 2025

Q55 · George and Rachel are playing a game of CounterBid 9694/31 Oct/Nov 2025

Q56 · In the Double Triple Quiz there are 4 rounds of questions 9694/32 Oct/Nov 2025

Q57 · George and Rachel are playing a game of CounterBid 9694/32 Oct/Nov 2025

Q58 · In the Double Triple Quiz there are 4 rounds of questions 9694/33 Oct/Nov 2025

Q59 · George and Rachel are playing a game of CounterBid 9694/33 Oct/Nov 2025