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HKDSE Geography Practice Paper: Rivers, Population and Urban Land Use (12 questions, 100 marks, 75 minutes, with answers)
HKDSE Geography: physical and human geography
Twelve original HKDSE Geography questions in 100 marks and 75 minutes: rivers, runoff, coasts, climate, population, land use, industry, maps and fieldwork. Marking points given.
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Paper details Level: S6 | Subject: Geography | Time: 75 minutes | Marks: 100 | Questions: 12 | Full marking points
Section A: Physical Geography (32 marks)
Answer ALL questions in the answer book.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A river flows from its source at 620 m above sea level to its mouth at 140 m; its total length is 16 km, and it falls 240 m within the first 6 km. (a) Calculate the average gradient of the whole river in m per km. (4 marks) (b) Calculate the gradients of the upper and lower sections, and explain why the valley forms differ. (4 marks) |
(a) Total fall = 620 – 140 = 480 m; average gradient = 480 / 16 = 30 m per km. (b) Upper section = 240 / 6 = 40 m per km; lower section = 240 / 10 = 24 m per km. The steep upper course gives fast flow and strong vertical erosion, producing V-shaped valleys and waterfalls, while the gentle lower course is dominated by lateral erosion and deposition, building a wide flood plain. |
| 2 | A drainage basin covers 25 square kilometres. Annual rainfall is 1800 mm and the runoff coefficient is 0.35. (a) Calculate the total volume of rainfall over the basin in one year. (3 marks) (b) Calculate the annual runoff volume. (3 marks) (c) State where the remaining water goes and give its volume. (2 marks) |
(a) Rainfall volume = 1800 mm / 1000 x 25 square kilometres x 1,000,000 = 45,000,000 cubic metres. (b) Runoff = 45,000,000 x 0.35 = 15,750,000 cubic metres. (c) The remaining 65 per cent, 29,250,000 cubic metres, leaves the basin through evapotranspiration, infiltration and groundwater recharge. |
| 3 | A beach 1200 m long receives 4800 cubic metres of sediment each year from longshore drift. (a) Calculate the average annual sediment gain per metre of coastline. (4 marks) (b) If a groyne is built updrift, explain the erosion that may follow downdrift. (4 marks) |
(a) Annual gain per metre = 4,800 / 1,200 = 4 cubic metres per metre per year. (b) The groyne traps drift, so the supply downdrift falls sharply. Waves continue to remove sand without replacement, so the beach narrows, the berm retreats and overtopping becomes more frequent, creating an erosion hotspot. |
| 4 | A station at 200 m above sea level records 28.0 degrees Celsius. The summit is at 1800 m, and the lapse rate is 6.5 degrees Celsius per kilometre. (a) Calculate the temperature at the summit. (4 marks) (b) If the dew point at the foot of the slope is 20.0 degrees Celsius, calculate the height of the cloud base above the foot of the slope. (4 marks) |
(a) Height difference = (1800 – 200) / 1000 = 1.6 km; cooling = 6.5 x 1.6 = 10.4 degrees; summit temperature = 28.0 – 10.4 = 17.6 degrees Celsius. (b) Cloud base = (28.0 – 20.0) / 6.5 = 1.23 km, about 1,231 m above the foot of the slope. |
Section B: Human Geography (36 marks)
Answer ALL questions and show all working.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | A district has a population of 486,000 in an area of 13.5 square kilometres. In one year there were 5,832 births, 3,402 deaths and a net migration gain of 1,200. (a) Calculate the population density. (3 marks) (b) Calculate the crude birth rate and crude death rate per thousand. (3 marks) (c) Calculate the annual population growth rate in per cent. (3 marks) |
(a) Density = 486,000 / 13.5 = 36,000 persons per square kilometre. (b) Birth rate = 5,832 / 486,000 x 1000 = 12.0 per thousand; death rate = 3,402 / 486,000 x 1000 = 7.0 per thousand. (c) Natural increase = 5,832 – 3,402 = 2,430; total increase = 2,430 + 1,200 = 3,630; growth rate = 3,630 / 486,000 x 100 = 0.75 per cent. |
| 6 | A built-up area of 48.0 square kilometres contains 21.6 of residential land, 4.8 commercial, 7.2 industrial, 6.0 roads and 8.4 open space. (a) Calculate the percentage of each land use. (4 marks) (b) Using the concentric zone model, explain why commercial land concentrates in the city centre. (5 marks) |
(a) Residential 45.0 per cent, commercial 10.0 per cent, industrial 15.0 per cent, roads 12.5 per cent and open space 17.5 per cent, totalling 100 per cent. (b) The centre has the highest accessibility and concentrates footfall and transport. Rent is highest there, so only the highest bidder survives, and commerce needs frequent face-to-face contact and dense information exchange, producing a central business district. |
| 7 | Farm A covers 24 hectares and produces 156 tonnes of rice; Farm B covers 18 hectares and produces 126 tonnes. (a) Calculate the yield per hectare on each farm. (4 marks) (b) Compare the two farms and suggest two reasons for the difference. (5 marks) |
(a) Farm A = 156 / 24 = 6.5 tonnes per hectare; Farm B = 126 / 18 = 7.0 tonnes per hectare. (b) Farm B yields 0.5 tonnes more per hectare, 7.7 per cent higher. Possible reasons: better irrigation and drainage on Farm B, and a higher-yielding variety or greater mechanisation giving more careful field management. |
| 8 | A factory uses 1.6 tonnes of iron ore to make one tonne of steel. Freight costs 8 dollars per tonne per kilometre. The mine is 45 km from the plant and the market is 120 km from the plant. (a) Calculate the freight cost of moving one tonne of ore to the plant. (3 marks) (b) Calculate the total transport cost per tonne of steel. (3 marks) (c) If the plant were built at the mine, calculate the freight cost per tonne of steel to the same market and the saving. (3 marks) |
(a) Ore freight = 45 x 8 = 360 dollars per tonne. (b) Steel freight = 120 x 8 = 960 dollars; total = 1.6 x 360 + 960 = 1,536 dollars per tonne. (c) At the mine the haul to market is 45 + 120 = 165 km, costing 165 x 8 = 1,320 dollars, a saving of 216 dollars per tonne. |
Section C: Skills and Issue-based Enquiry (32 marks)
Data-response and short essay items; support your points with figures.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A map has a scale of 1:20,000. Two points, A and B, are 3.5 cm apart on the map. Their heights are 120 m and 300 m, and the horizontal distance between them is 1.2 km. (a) Calculate the real distance between the two points in metres. (4 marks) (b) Calculate the slope between them as a percentage. (4 marks) |
(a) On the contour map, real distance = 3.5 cm x 20,000 / 100 = 700 m. (b) Height difference = 300 – 120 = 180 m; slope = 180 / 1,200 x 100 = 15.0 per cent. |
| 10 | Monthly rainfall at a station, in millimetres, is: 30, 45, 70, 140, 290, 420, 390, 330, 230, 120, 55, 35. (a) Calculate the annual total and the monthly mean. (4 marks) (b) Calculate the share of the annual total falling from May to September and name the rainfall regime. (4 marks) |
(a) Annual total = 2,155 mm; monthly mean = 2,155 / 12 = 179.6 mm. (b) May to September totals 1,660 mm, which is 1,660 / 2,155 x 100 = 77.0 per cent of the year, a summer maximum regime. |
| 11 | A float travels 20 m in 48 seconds. The correction factor is 0.85. The channel is 6.4 m wide with a mean depth of 0.75 m. (a) Calculate the mean velocity of the river. (4 marks) (b) Calculate the discharge. (4 marks) |
(a) Surface velocity = 20 / 48 = 0.42 m per second; mean velocity = 0.42 x 0.85 = 0.35 m per second. (b) Cross-section = 6.4 x 0.75 = 4.8 square metres; discharge = 0.35 x 4.8 = 1.70 cubic metres per second. |
| 12 | The government must build 4,200 flats within ten years. Option A reclaims 28 hectares of sea; Option B develops 35 hectares of brownfield land. (a) Calculate the plot ratio of each option in flats per hectare. (4 marks) (b) Compare the two options in terms of ecological impact and cost, and give one recommendation. (4 marks) |
(a) Option A = 4,200 / 28 = 150 flats per hectare; Option B = 4,200 / 35 = 120 flats per hectare, so A is 30 per hectare denser. (b) Reclamation destroys seabed habitats and alters currents and water quality, is costly and is irreversible; brownfield development needs land resumption and pollution clean-up but repairs already degraded land. Recommend Option B first as the more sustainable option, with remediation and community facilities, keeping reclamation as a last resort. |
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