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HKDSE Geography Practice Paper: Industrial Location, Urban Land Use and Sustainability (12 questions, 100 marks, 75 minutes)
HKDSE Geography: Industrial Location and Urban Land Use
Original HKDSE Geography practice paper on industrial location, urban land use and sustainability: 12 questions, 100 marks, 75 minutes, with full worked answers and marking points.
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Paper details Level: S6 | Subject: Geography | Time: 75 minutes | Marks: 100 | Questions: 12
Section A (32 marks)
Cost comparison and land-use data questions: show the formula and the unit.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A manufacturer compares the annual total costs of three sites (HK$): Site A: rent 120,000, transport 80,000, wages 300,000; Site B: rent 90,000, transport 140,000, wages 280,000; Site C: rent 150,000, transport 60,000, wages 330,000. (a) Find the annual total cost of each site. (b) State which site has the lowest total cost. (c) Explain why wages have the greatest influence on the location decision. |
(a) A = 120,000 + 80,000 + 300,000 = 500,000; B = 90,000 + 140,000 + 280,000 = 510,000; C = 150,000 + 60,000 + 330,000 = 540,000. (b) Site A has the lowest total cost at 500,000. (c) Wages are 55% to 61% of total cost at the three sites, the largest component, so wage levels dominate the location decision. |
| 2 | An urban district covers 1.8 km2: residential 0.72 km2, commercial 0.18 km2, industrial 0.36 km2 and other uses 0.54 km2. (a) Find each land use as a percentage of the district. (b) Find the ratio of residential to commercial land in its simplest integer form. (c) Decide whether the district is mainly a central business district or a residential area, and give a reason. |
(a) Residential = 0.72 / 1.8 x 100% = 40%; commercial = 10%; industrial = 20%; other = 30%. (b) 0.72 : 0.18 = 4 : 1. (c) Residential land takes 40%, the largest share, while commercial takes only 10%, so the district is mainly residential. |
| 3 | A commuter lives 18 km from work and the average vehicle speed is 24 km/h. (a) Find the one-way commuting time in minutes. (b) If the average speed rises to 30 km/h, find the new one-way time. (c) Find the time saved on each one-way trip. |
(a) Time = 18 / 24 x 60 = 45 minutes. (b) 18 / 30 x 60 = 36 minutes. (c) 45 – 36 = 9 minutes saved. |
| 4 | Population density in a city falls with distance from the centre: 45,000 persons per km2 at 2 km and 18,000 at 6 km. (a) Find the distance between the two points. (b) Find the average fall in density per km of distance from the centre. (c) Explain two reasons for the fall in density with distance. |
(a) 6 – 2 = 4 km. (b) (45,000 – 18,000) / 4 = 6,750 persons per km2. (c) High rent in the centre is affordable only to high-value commerce, pushing housing outwards; the outer areas have more land and lower rent, so housing density falls. |
Section B (36 marks)
Environment and resource data questions: show the working and give a judgement.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | The air quality index (AQI) at a station over seven days was 62, 78, 95, 110, 88, 70 and 55. (a) Find the mean AQI over the seven days (1 decimal place). (b) Find the number of days with AQI above 100 and its percentage of the seven days (1 decimal place). (c) State one land-use factor that raises AQI in urban areas. |
(a) Sum = 558, so 558 / 7 = 79.7. (b) Only one day (110) is above 100, that is 1 day, or 14.3%. (c) Industrial land and heavy traffic release particulates and nitrogen oxides, raising urban AQI. |
| 6 | A landfill has 3.5 million m3 of capacity left and receives 0.25 million m3 of waste a year. (a) Find its remaining life at the present rate. (b) A waste reduction plan cuts the annual intake by 20%. Find the new annual intake. (c) Find how many extra years the reduction adds. |
(a) 3.5 / 0.25 = 14 years. (b) 0.25 x (1 – 0.20) = 0.20 million m3 a year. (c) 3.5 / 0.20 = 17.5 years, adding 3.5 years. |
| 7 | A place uses 48,000 GWh of electricity a year, of which 3,840 GWh comes from renewable sources. (a) Find the renewable share as a percentage. (b) If the 2035 target is 15%, find the renewable electricity needed then, with total demand unchanged. (c) Find the additional renewable electricity required. |
(a) 3,840 / 48,000 x 100% = 8%. (b) 48,000 x 15% = 7,200 GWh. (c) 7,200 – 3,840 = 3,360 GWh. |
| 8 | On one day the urban maximum temperature was 32.4 C and the rural maximum 28.6 C. Each one percentage point of extra urban greenery lowers the maximum by about 0.1 C. (a) Find the urban-rural temperature difference. (b) If urban greenery rises from 12% to 20%, find the expected cooling. (c) Find the improved urban maximum temperature (1 decimal place). |
(a) 32.4 – 28.6 = 3.8 C. (b) (20 – 12) x 0.1 = 0.8 C. (c) 32.4 – 0.8 = 31.6 C. |
Section C (32 marks)
Carrying capacity and sustainability questions: set out the calculation and explain.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A city has 2,400,000 daily trips: rail 1,080,000, bus 720,000, minibus 360,000 and private car 240,000. (a) Find each mode as a percentage of daily trips. (b) Find how many times the rail trips exceed private car trips. (c) Explain the effect of a higher rail share on roadside air quality. |
(a) Rail = 1,080,000 / 2,400,000 x 100% = 45%; bus = 30%; minibus = 15%; private car = 10%. (b) 1,080,000 / 240,000 = 4.5 times. (c) A higher rail share removes private cars and bus trips from the roads, so roadside emissions and particulates fall and air quality improves. |
| 10 | A factory makes 8,500 items a day and uses 12 L of water per item. (a) Find the daily water use in L. (b) Convert the daily water use to m3. (c) A recycling system reuses 35% of the water. Find the daily saving in L. |
(a) 8,500 x 12 = 102,000 L. (b) 102,000 / 1,000 = 102 m3. (c) 102,000 x 35% = 35,700 L saved each day. |
| 11 | A site of 0.6 km2 needs 15 m2 of space per visitor. (a) Find the daily carrying capacity in visitor trips. (b) If 52,000 visitors arrive on a peak day, find the percentage over capacity. (c) Suggest two measures to manage visitor flow. |
(a) 0.6 km2 = 600,000 m2; 600,000 / 15 = 40,000 visitor trips. (b) (52,000 – 40,000) / 40,000 x 100% = 30% over capacity. (c) Introduce online booking quotas and time-banded ticket prices, and add shuttle services to spread arrivals. |
| 12 | Per capita carbon dioxide emissions and populations of three places: A: 4.2 t per person, 0.5 million people; B: 6.0 t, 1.2 million; C: 9.6 t, 0.3 million. (a) Find the combined annual emissions in Mt. (b) Find the combined population in millions and the overall per capita emissions (2 decimal places). (c) State which place should cut emissions first and give a reason. |
(a) A = 4.2 x 0.5 = 2.1 Mt; B = 6.0 x 1.2 = 7.2 Mt; C = 9.6 x 0.3 = 2.88 Mt; combined = 12.18 Mt. (b) Population = 0.5 + 1.2 + 0.3 = 2.0 million; per capita = 12.18 / 2.0 = 6.09 t. (c) Place C first: its per capita emissions of 9.6 t are the highest, so each unit of per capita reduction gives the largest cut. |
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