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HKDSE Tourism and Hospitality Studies Practice Paper: Introduction to Tourism, Destination Marketing and Hospitality Operations
HKDSE Tourism and Hospitality – 12 questions – 100 marks – 75 minutes
HKDSE Tourism and Hospitality Studies practice paper: 12 questions on visitor growth, hotel occupancy and RevPAR, package pricing, carrying capacity and NPS, with a full marking scheme.
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Paper details HKDSE Tourism and Hospitality Studies paper: visitor growth, occupancy and RevPAR, package pricing, carrying capacity, NPS, commission, food cost and carbon footprint, with a marking scheme.
Section A: Introduction to Tourism and Operating Data (Q1-Q4)
Show the formula and working; answers alone earn no method marks.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A destination recorded 12,500,000 inbound visitor arrivals last year and 14,000,000 this year. Average spending per visitor this year was HK$6,800. (a) Calculate the year-on-year growth rate of arrivals (one decimal place). (2 marks) (b) Calculate total tourism receipts this year and express the answer in hundred millions of HK$. (3 marks) (c) If arrivals grow a further 8% next year, find the forecast number of arrivals. (3 marks) |
(a) Growth = (14,000,000 – 12,500,000) / 12,500,000 x 100% = 12.0%. (1 mark formula, 1 mark answer) (b) Receipts = 14,000,000 x 6,800 = HK$95,200,000,000, that is HK$95.2 billion. (2 marks formula, 1 mark answer) (c) 14,000,000 x 1.08 = 15,120,000 arrivals. (2 marks formula, 1 mark answer) |
| 2 | A hotel has 200 rooms. In a 30-day month it sold 4,650 room nights and earned room revenue of HK$4,185,000. (a) Find the available room nights and the occupancy rate (one decimal place). (3 marks) (b) Find the average daily rate (ADR). (2 marks) (c) Find revenue per available room (RevPAR) and check it against ADR x occupancy. (3 marks) |
(a) Available room nights = 200 x 30 = 6,000; occupancy = 4,650 / 6,000 x 100% = 77.5%. (1.5 marks each) (b) ADR = 4,185,000 / 4,650 = HK$900. (2 marks) (c) RevPAR = 4,185,000 / 6,000 = HK$697.5; check 900 x 77.5% = 697.5, consistent. (2 marks + 1 mark) |
| 3 | A travel agent packages a 4-day Japan trip: return airfare HK$2,400; 3 hotel nights totalling JPY 36,000; local transport JPY 9,000. The exchange rate is HK$1 = JPY 19.5. The agency requires a 15% mark-up on cost. (a) Find the land cost per passenger in HK$ (two decimal places). (3 marks) (b) Find total cost per passenger and the selling price at a 15% mark-up, rounded up to the nearest HK$10. (3 marks) (c) Find the actual gross margin percentage (one decimal place). (2 marks) |
(a) Land cost = 36,000 + 9,000 = JPY 45,000; converted 45,000 / 19.5 = HK$2,307.69. (2 marks + 1 mark) (b) Total cost = 2,400 + 2,307.69 = HK$4,707.69; price = 4,707.69 x 1.15 = HK$5,413.85, rounded up to HK$5,420. (2 marks + 1 mark) (c) Gross profit = 5,420 – 4,707.69 = HK$712.31; margin = 712.31 / 5,420 x 100% = 13.1%. (1 mark + 1 mark) |
| 4 | A scenic site has a daily environmental carrying capacity of 3,000 visitors. In the 45-day peak season it actually received 3,600 visitors per day. (a) Find total peak-season visits and the ceiling implied by carrying capacity. (3 marks) (b) Find the excess visits and the percentage overload. (3 marks) (c) Give one visitor-flow management measure and one limitation of it. (2 marks) |
(a) Total visits = 3,600 x 45 = 162,000; ceiling = 3,000 x 45 = 135,000. (1.5 marks each) (b) Excess = 162,000 – 135,000 = 27,000 visits; overload = 27,000 / 135,000 x 100% = 20%. (2 marks + 1 mark) (c) Example: a timed online booking quota capping daily entry at 3,000; limitation: quotas may be scalped and off-peak revenue falls. (1 mark measure, 1 mark limitation) |
Section B: Destination Marketing and Visitor Behaviour (Q5-Q8)
Support your analysis with the case data and define the concepts you use.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | Three source markets supplied 480,000, 320,000 and 200,000 visitors respectively. The tourism board has a promotion budget of HK$12,000,000, allocated in proportion to market share. (a) Find the share of each market as a percentage. (3 marks) (b) Find the budget allocated to each market and the promotion cost per visitor. (3 marks) (c) If market C targets 25% growth next year at the same cost per visitor, find the extra budget required. (3 marks) |
(a) Total 1,000,000 visitors; shares A 48%, B 32%, C 20%. (3 marks) (b) A HK$5,760,000, B HK$3,840,000, C HK$2,400,000; cost per visitor = 12,000,000 / 1,000,000 = HK$12. (2 marks + 1 mark) (c) Extra visitors = 200,000 x 25% = 50,000; extra budget = 50,000 x 12 = HK$600,000. (2 marks + 1 mark) |
| 6 | A cultural attraction recorded 40,000, 52,000, 61,000, 63,000 and 64,000 visits in years one to five. (a) Find the year-on-year growth rate in year two and in year five (one decimal place). (3 marks) (b) Identify the stage of the tourism area life cycle (TALC) and support your answer with the data. (3 marks) (c) Suggest two marketing strategies for that stage and state one risk. (3 marks) |
(a) Year two growth = 12,000 / 40,000 x 100% = 30.0%; year five = 1,000 / 63,000 x 100% = 1.6%. (2 marks + 1 mark) (b) The site is in stagnation: growth has fallen from 30.0% to 1.6% and visits are near its carrying capacity. (1 mark stage, 2 marks evidence) (c) Strategies: develop off-season and in-depth guided products, reposition the image to reach new segments; risk: reinvestment may accelerate decline, or staged authenticity may alienate core visitors. (2 marks strategies, 1 mark risk) |
| 7 | A resort collected 500 questionnaires: mean expectation 4.4 and mean perception 3.9 on a 5-point scale; 260 promoters, 90 detractors and 150 passives. (a) Find the service quality gap (perception minus expectation). (2 marks) (b) Calculate the net promoter score (NPS). (4 marks) (c) Suggest one improvement for the gap and explain how its effect would be measured. (3 marks) |
(a) Gap = 3.9 – 4.4 = -0.5, perception below expectation. (2 marks) (b) NPS = (260 – 90) / 500 x 100 = 34. (2 marks formula, 1 mark substitution, 1 mark answer) (c) Example: follow up complaints at check-in and reply within 24 hours; measure with repeat-stay rate and quarterly NPS and perception scores. (1 mark measure, 2 marks measurement) |
| 8 | A hotel sells a room direct at HK$1,000 with direct costs (payment handling and promotion) of 6% of rate. The same room is sold through an OTA at HK$1,150 on 18% commission. (a) Find the net revenue per room night for both channels. (4 marks) (b) Compare the two and state one non-monetary cost of the OTA channel. (3 marks) (c) If 2,000 room nights are sold through the OTA each year, find the net revenue difference against direct selling. (2 marks) |
(a) Direct net = 1,000 x (1 – 6%) = HK$940; OTA net = 1,150 x (1 – 18%) = HK$943. (2 marks each) (b) The OTA yields HK$3 more per room night; non-monetary cost: guest data and the direct relationship sit with the OTA, limiting later marketing. (2 marks + 1 mark) (c) 2,000 x 3 = HK$6,000; the sum is small, so the loss of customer equity matters too. (1 mark + 1 mark) |
Section C: Hospitality Operations and Sustainable Tourism (Q9-Q12)
Answer in full sentences and state at least one limitation for each recommendation.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A restaurant charges HK$180 per cover with a 32% food cost ratio. The supplier raises prices by 12% next month while the menu price stays unchanged. (a) Find the current food cost per cover and the gross profit. (3 marks) (b) Find the new food cost ratio and gross profit ratio (two decimal places). (3 marks) (c) Find the new menu price needed to keep a 32% food cost ratio. (2 marks) |
(a) Food cost = 180 x 32% = HK$57.6; gross profit = 180 – 57.6 = HK$122.4. (2 marks + 1 mark) (b) New cost = 57.6 x 1.12 = HK$64.512; ratio = 64.512 / 180 x 100% = 35.84%; gross margin = 64.16%. (2 marks + 1 mark) (c) New price = 64.512 / 32% = HK$201.6. (2 marks) |
| 10 | A conference has 600 delegates and needs one staff member per 25 delegates. Costs: venue HK$120,000, catering HK$380 per head, audio-visual HK$85,000 and promotion HK$45,000. The registration fee is HK$900 per head. (a) Find the number of staff required (round up to a whole number). (2 marks) (b) Find total expenditure, total income and the surplus. (4 marks) (c) Find the minimum registration fee that avoids a loss (two decimal places). (2 marks) |
(a) Staff = 600 / 25 = 24. (2 marks) (b) Catering = 600 x 380 = HK$228,000; total expenditure = 120,000 + 228,000 + 85,000 + 45,000 = HK$478,000; income = 600 x 900 = HK$540,000; surplus = HK$62,000. (3 marks + 1 mark) (c) Minimum fee = 478,000 / 600 = HK$796.67. (2 marks) |
| 11 | A 4-day tour involves a 3,000 km return flight (0.15 kg CO2e per passenger-km) and 200 km by coach (0.05 kg CO2e per passenger-km) for 250 passengers. One tree absorbs 21 kg CO2e a year. (a) Find the flight and coach emissions per passenger. (3 marks) (b) Find the total emissions of the group in tonnes. (3 marks) (c) Find the number of trees needed to offset one year of the group emissions (round up). (2 marks) |
(a) Flight = 3,000 x 0.15 = 450 kg; coach = 200 x 0.05 = 10 kg. (2 marks + 1 mark) (b) Per passenger 460 kg; group = 460 x 250 = 115,000.0 kg = 115 tonnes. (2 marks + 1 mark) (c) 115,000.0 / 21 = 5,476.19, rounded up to 5,477 trees. (2 marks) |
| 12 | A destination tourism recovery index (2019 = 100) reads 8 in 2020, 12 in 2021, 27 in 2022, 55 in 2023 and 78 in 2024. (a) Find the recovery ratio in 2024 and the shortfall against 2019. (2 marks) (b) Find the compound annual growth rate from 2020 to 2024 (one decimal place). (4 marks) (c) Does a rising index mean tourism receipts have recovered equally? Give one reason. (2 marks) |
(a) Recovery = 78 / 100 = 78%, still 22% below 2019. (2 marks) (b) (78 / 8)^(1/4) – 1 = 9.75^0.25 – 1 = 0.767, that is 76.7%. (2 marks formula, 1 mark working, 1 mark answer) (c) Not necessarily: visitor numbers may recover while spending per visitor or length of stay falls, and receipts also depend on exchange rates and fares. (1 mark judgement, 1 mark reason) |
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