S5 Geography Practice Paper: Industry, Energy and Sustainable Development

An electronics firm plans an assembly plant with monthly output of 30,000 units. Costs for three sites: urban site A (rent 180,000 dollars per month, transport 6 dollars per unit, wages 22 dollars per unit); new town industrial estate B (rent 96,000 dollars, transport 9 dollars, wages 18 dollars); border site C (rent 60,000 dollars, transport 14 dollars, wages 16 dollars).

(a) Calculate the total unit cost at each site. (3 marks)
(b) Choose a site on cost grounds and state its main locational advantage. (2 marks)
(c) If output rises to 45,000 units per month, recalculate the unit cost at B and C and explain how the gap changes. (2 marks)

(a) A: 180,000 / 30,000 + 6 + 22 = 6 + 28 = 34.0 dollars; B: 96,000 / 30,000 + 9 + 18 = 3.2 + 27 = 30.2 dollars; C: 60,000 / 30,000 + 14 + 16 = 2 + 30 = 32.0 dollars. (3 marks)
(b) Site B has the lowest unit cost (30.2 dollars): estate rents are low, wages are below urban levels, the estate is close to new town labour, and transport cost stays moderate. (2 marks)
(c) At 45,000 units, B: 96,000 / 45,000 + 27 = 29.13 dollars; C: 60,000 / 45,000 + 30 = 31.33 dollars. Higher output spreads fixed cost over more units, B also has variable costs 3 dollars below C, so the gap widens from 1.8 to 2.20 dollars. (2 marks)

An export-oriented industrial area receives new orders worth 500 million dollars. Firms spend 60 per cent of each round of income on local purchases and the rest on imported materials.

(a) Use the multiplier formula to find the total output generated. (3 marks)
(b) If each 100 million dollars of output needs 800 jobs, estimate the number of new jobs. (2 marks)
(c) If the local purchasing share falls to 40 per cent, recalculate the multiplier, total output and jobs, and explain the difference. (2 marks)

(a) Multiplier = 1 / (1 – 0.6) = 2.5; total output = 500 million x 2.5 = 12.5 hundred million dollars. (3 marks)
(b) Jobs = 12.5 x 800 = 10,000. (2 marks)
(c) Multiplier = 1 / (1 – 0.4) = 1.67; output = 500 million x 1.67 = 8.33 hundred million; jobs = 6,667. A smaller local purchasing share means income leaks out of the area faster each round, so the multiplier and the employment effect both fall. (2 marks)

Employment data for a city: in 1980 manufacturing was 900,000, services 1,200,000 and total employment 2,400,000; in 2000 manufacturing was 220,000 and total employment 3,300,000; in 2023 manufacturing was 90,000, services 3,300,000 and total employment 3,800,000.

(a) Calculate manufacturing as a percentage of total employment in the three years (one decimal place). (3 marks)
(b) Calculate the service share in 1980 and 2023. (2 marks)
(c) Explain the fall in the manufacturing share using two geographical concepts, with one local example each. (2 marks)

(a) 1980: 37.5%; 2000: 6.7%; 2023: 2.4%. (3 marks)
(b) 1980: 50.0%; 2023: 86.8%. (2 marks)
(c) Deindustrialisation: production moved to lower-cost locations, so manufacturing jobs fell. Structural upgrading: the economy shifted towards finance, logistics and producer services. Local examples: factories relocated to the Pearl River Delta while financial and professional services concentrated in Central and Quarry Bay. (2 marks, 1 each)

An industrial estate has 12 factories. A treatment plant built by each factory costs 2.4 million dollars; a shared central plant costs 18 million dollars in total.

(a) Calculate the total cost of the individual option, the shared option and the difference. (3 marks)
(b) Calculate the cost per factory under the shared option and the saving per factory. (2 marks)
(c) Apart from cost, state two environmental benefits of the shared plant. (2 marks)

(a) Individual total = 12 x 240 = 2880 ten-thousand dollars; shared total = 1800 ten-thousand; difference = 1080 ten-thousand. (3 marks)
(b) Cost per factory = 1,800 / 12 = 150 ten-thousand; saving = 240 – 150 = 90 ten-thousand. (2 marks)
(c) Central treatment removes pollutants before discharge, protecting rivers and coastal water; sludge and chemicals are handled at one monitored site, reducing the risk of accidental leakage. (2 marks)

A city generates 40,000 GWh a year: coal 10,000 GWh, natural gas 16,000 GWh, nuclear 8,000 GWh and renewables 6,000 GWh, with emission factors of 0.82, 0.45, 0.012 and 0.02 kg of carbon dioxide per kWh.

(a) Calculate each fuel share of generation. (3 marks)
(b) Calculate total annual carbon dioxide emissions in million tonnes and the average emission per kWh (three decimals). (3 marks)
(c) If 4,000 GWh of coal is replaced by renewables, calculate the emissions saved and the new total. (2 marks)

(a) Coal 25%; gas 40%; nuclear 20%; renewables 15%. (3 marks)
(b) Coal 8.2, gas 7.2, nuclear 0.096, renewables 0.12 million tonnes. Total = 15.616 million tonnes; average = 0.390 kg per kWh. (3 marks)
(c) Saved = 4,000 x 1,000,000 x 0.80 / 1,000,000,000 = 3.2 million tonnes; new total = 12.416 million tonnes. (2 marks)

An offshore wind farm has 200 MW of installed capacity and generates 480 GWh a year. A solar farm has 100 MW of capacity and a capacity factor of 15 per cent.

(a) Calculate the theoretical maximum annual output of the wind farm and its capacity factor (one decimal place). (3 marks)
(b) If a household uses 4,500 kWh a year, estimate the number of households the wind farm can supply. (3 marks)
(c) Calculate the solar output and compare the two capacity factors, giving one reason for the difference. (2 marks)

(a) Maximum = 200 x 8,760 = 1,752,000 MWh = 1752 GWh; capacity factor = 480 / 1752 = 27.4%. (3 marks)
(b) 480 GWh = 480,000,000 kWh; households = 480,000,000 / 4,500 = 106,667. (3 marks)
(c) Solar = 100 x 8,760 x 0.15 = 131.4 GWh, a capacity factor of 15.0% against 27.4% for wind. Solar output is confined to daylight hours and falls to near zero at night and in cloudy weather, while wind can generate around the clock. (2 marks)

A city uses 1,200 PJ of energy a year (1 PJ equals 1,000,000 GJ), has a population of 7.5 million and households account for 22 per cent of consumption.

(a) Calculate energy use per person in GJ. (3 marks)
(b) Calculate household energy use per person. (2 marks)
(c) If household use per person falls by 15 per cent, calculate the new figure and the reduction. (3 marks)

(a) Total = 1,200 x 1,000,000 = 1.2 billion GJ; per person = 1.2 billion / 7.5 million = 160 GJ. (3 marks)
(b) Household use per person = 160 x 0.22 = 35.2 GJ. (2 marks)
(c) New figure = 35.2 x 0.85 = 29.92 GJ; reduction = 5.28 GJ. (3 marks)

A household uses 400 kWh a month. The basic tariff is 0.98 dollars per kWh, the fuel adjustment is 0.42 dollars per kWh, and a subsidy of 0.30 dollars per kWh applies to the first 200 kWh each month.

(a) Calculate the bill before subsidy, the subsidy and the amount payable. (3 marks)
(b) If use rises to 520 kWh, calculate the amount payable and the increase. (3 marks)
(c) Calculate the marginal cost per unit above 200 kWh and explain how the subsidy affects incentives to save energy. (2 marks)

(a) Unit price = 0.98 + 0.42 = 1.40 dollars; bill = 400 x 1.40 = 560 dollars; subsidy = 200 x 0.30 = 60 dollars; payable = 500 dollars. (3 marks)
(b) 520 x 1.40 = 728; payable = 668 dollars; increase = 168 dollars. (3 marks)
(c) Marginal cost = 168 / 120 = 1.40 dollars. Because the subsidy covers only the first 200 kWh, heavier users still face the full 1.40 dollar price at the margin, while low users have a weaker incentive to cut consumption. (2 marks)

A green manufacturing park costs 2.4 billion dollars, is expected to earn 520 million dollars a year, carries annual environmental restoration costs of 70 million dollars and creates 1,800 jobs.

(a) Calculate annual net income and the payback period (one decimal place). (4 marks)
(b) Calculate jobs created per 100 million dollars invested and the return on investment. (3 marks)
(c) If restoration costs rise to 160 million dollars a year, recalculate the payback period and comment on sustainability. (3 marks)

(a) Net income = 5.2 – 0.7 = 4.5 hundred million dollars; payback = 24 / 4.5 = 5.3 years. (4 marks)
(b) Jobs per 100 million = 1,800 / 24 = 75; return = 4.5 / 24 x 100 = 18.75%. (3 marks)
(c) Net income = 3.6; payback = 24 / 3.6 = 6.7 years. Internalising the environmental cost lengthens payback from 5.3 to 6.7 years, which is still acceptable; a further rise would weaken both financial and environmental sustainability, so charges should follow the polluter pays principle. (3 marks)

A factory makes 50,000 units a year and uses 6,000 MWh of electricity, 400,000 cubic metres of natural gas and 120,000 tonne-kilometres of freight. Emission factors: electricity 0.55 kg per kWh, gas 2.0 kg per cubic metre, freight 0.09 kg per tonne-kilometre.

(a) Calculate emissions from each of the three activities in tonnes. (4 marks)
(b) Calculate total annual emissions and the carbon footprint per unit in kg (one decimal place). (3 marks)
(c) Suggest two measures to cut the footprint per unit and state the emission source each one targets. (3 marks)

(a) Electricity: 6,000,000 x 0.55 / 1,000 = 3,300 tonnes; gas: 400,000 x 2.0 / 1,000 = 800 tonnes; freight: 120,000 x 0.09 / 1,000 = 10.8 tonnes. (4 marks)
(b) Total = 4,110.8 tonnes; per unit = 4,110.8 x 1,000 / 50,000 = 82.2 kg. (3 marks)
(c) Improve electrical efficiency or buy renewable electricity, targeting the largest source; switch to lower-emission fuel or electric trucks and shorten freight distances, targeting gas and freight; cutting packaging and waste also raises output per unit of input. (3 marks)

An industrial district produces 90,000 tonnes of waste a year: 36,000 tonnes recycled, 45,000 tonnes landfilled and 9,000 tonnes incinerated. Landfill costs 800 dollars per tonne.

(a) Calculate the percentage handled by each method. (4 marks)
(b) If the recycling rate rises to 55 per cent with incineration unchanged, calculate the new recycled and landfilled tonnage and the landfill percentage. (3 marks)
(c) Calculate the money saved from the lower landfill tonnage, in ten thousand dollars. (3 marks)

(a) Recycled 40%; landfilled 50%; incinerated 10%. (4 marks)
(b) Recycled = 90,000 x 0.55 = 49,500 tonnes; landfilled = 90,000 – 49,500 – 9,000 = 31,500 tonnes = 35%. (3 marks)
(c) Landfill falls by 13,500 tonnes; saving = 13,500 x 800 / 10,000 = 1,080 ten-thousand dollars. (3 marks)

A city needs 5,000 GWh a year and must choose between: coal at 0.55 dollars per kWh with 0.82 kg of emissions per kWh; gas at 0.72 dollars with 0.45 kg; wind at 0.98 dollars with 0.02 kg. The carbon price is 250 dollars per tonne.

(a) Calculate annual generation cost in 100 million dollars and emissions in million tonnes for the three options. (4 marks)
(b) Including the carbon price, calculate the total cost of each option and choose the cheapest. (3 marks)
(c) If the carbon price rises to 500 dollars per tonne, reassess the cheapest option and explain the policy implication. (3 marks)

(a) Generation cost: coal 27.5, gas 36.0, wind 49.0 hundred million dollars. Emissions: coal 4.10, gas 2.25, wind 0.10 million tonnes. (4 marks)
(b) Carbon cost: coal 10.25, gas 5.625, wind 0.25 hundred million. Total: coal 37.75, gas 41.625, wind 49.25 hundred million dollars; coal is cheapest. (3 marks)
(c) At 500 dollars per tonne: coal 48.0, gas 47.25, wind 49.5 hundred million, so gas becomes cheapest. The carbon price determines which fuel has the cost advantage: at a low price high-carbon fuel still wins, so policy must set a price that reflects the external cost of emissions while also reducing the capital cost of renewables. (3 marks)