HKDSE Geography Paper 1 data-based practice paper

2026-10-31

HKDSE Geography Paper 1

HKDSE Geography Paper 1 data-based practice paper: 6 questions, 100 marks, 165 minutes, covering population, rivers, urban land use, farming, earthquakes and arid-zone energy.

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Paper details　Level: S4-S6 (HKDSE) | Subject: Geography | Time: 165 minutes | Marks: 100 | Questions: 6

Question 1 — Population structure and growth (18 marks)

1. The table shows population data for City A in 2025.

#	Question	Answer and marking notes
1	The population of City A is 4,800,000 and its land area is 320 km2. Calculate the population density (persons per km2).	Density = 4,800,000 / 320 = 15,000 persons per km2. Marking: 2 marks formula, 2 marks value, 1 mark unit.
2	During the year there were 57,600 births and 33,600 deaths. Calculate the birth rate, the death rate (per thousand) and the rate of natural increase (%).	Birth rate = 57,600 / 4,800,000 x 1,000 = 12.0 per thousand; death rate = 33,600 / 4,800,000 x 1,000 = 7.0 per thousand; natural increase = 24,000, rate = 24,000 / 4,800,000 x 100% = 0.5%. Marking: 2 marks each rate, 2 marks natural increase rate.
3	Net migration into the city was 12,000. Calculate the overall population growth rate and decide whether growth is dominated by natural increase or by migration. Explain.	Total growth = 24,000 + 12,000 = 36,000, rate = 36,000 / 4,800,000 x 100% = 0.75%; natural increase contributes 66.7% and net migration 33.3%, so natural increase dominates. Marking: 3 marks calculation, 1 mark judgement, 3 marks comparison.

Question 2 — River discharge and runoff (18 marks)

2. Catchment of River B: area 480 km2, annual rainfall 2,400 mm, annual runoff 6.912 x 10^8 m3.

#	Question	Answer and marking notes
4	Calculate the total annual volume of rainfall over the catchment (m3).	480 km2 = 480 x 10^6 m2; 2,400 mm = 2.4 m; volume = 480 x 10^6 x 2.4 = 1.152 x 10^9 m3. Marking: 2 marks area conversion, 2 marks multiplication.
5	Calculate the runoff coefficient and state where the rest of the rainfall goes.	Runoff coefficient = 6.912 x 10^8 / 1.152 x 10^9 = 0.60 (60%); the other 40% is lost through evaporation, transpiration, infiltration and storage in depressions. Marking: 2 marks coefficient, 2 marks explanation.
6	At the upper station the channel is 40 m wide, 2.5 m deep on average, with a mean velocity of 0.8 m/s. Calculate the discharge (m3/s).	Cross-section = 40 x 2.5 = 100 m2; discharge Q = A x v = 100 x 0.8 = 80 m3/s. Marking: 2 marks cross-section, 2 marks Q = Av, 1 mark unit.
7	At the lower station the channel is 55 m wide, 3.2 m deep and the velocity is 1.1 m/s. Calculate the downstream discharge, the percentage increase, and explain it.	Cross-section = 55 x 3.2 = 176 m2; Q = 176 x 1.1 = 193.6 m3/s; increase = (193.6 – 80) / 80 x 100% = 142.0%. Tributaries join downstream and the catchment area grows, so discharge accumulates. Marking: 2 marks discharge, 2 marks increase, 1 mark explanation.

Question 3 — Map skills and urban land use (17 marks)

3. Study the map of Town C. The scale is 1:20,000.

#	Question	Answer and marking notes
8	Two points on the map are 65 mm apart. Calculate the real distance between them (km).	65 mm x 20,000 = 1,300,000 mm = 1,300 m = 1.3 km. Marking: 2 marks conversion, 2 marks result, 1 mark unit.
9	From point A (120 m above sea level) to point B (180 m), the horizontal distance is 400 m. Calculate the gradient (as a percentage and as 1:n) and judge whether the site suits large-scale housing.	Height difference = 180 – 120 = 60 m; gradient = 60 / 400 x 100% = 15%, i.e. 1:6.7. A slope above 15% is steep: large-scale housing would need costly retaining and drainage works and would raise landslide risk, so it is generally unsuitable. Marking: 1 mark difference, 2 marks percentage, 1 mark ratio, 2 marks judgement.
10	Town C covers 8 km2 and industry occupies 12% of it. Calculate the industrial area (km2 and hectares). Land in the industrial zone costs $18,000/m2 while residential land costs $42,000/m2. Explain the difference.	Industrial area = 8 x 12% = 0.96 km2 = 96 hectares. Residential land lies nearer the centre and transport arteries, with better accessibility and environment, so households bid more and land values are higher; industry needs large flat sites and must keep away from housing to limit nuisance, so it can afford to bid less. Marking: 2 marks area, 1 mark conversion, 3 marks explanation.

Question 4 — Farming and food supply (16 marks)

4. Village D has 12 hectares of farmland yielding 5.4 tonnes of rice per hectare; 180 villagers each need 0.15 tonnes of rice a year.

#	Question	Answer and marking notes
11	Calculate the total annual rice output of Village D (tonnes).	Output = 12 x 5.4 = 64.8 tonnes. Marking: 2 marks formula, 2 marks value, 1 mark unit.
12	Calculate the annual rice requirement of the village (tonnes) and the surplus or shortfall (tonnes).	Requirement = 180 x 0.15 = 27 tonnes; surplus = 64.8 – 27 = 37.8 tonnes. Marking: 3 marks requirement, 3 marks surplus.
13	Calculate the food self-sufficiency ratio (%) and suggest two sustainable ways to raise it.	Ratio = 64.8 / 27 x 100% = 240%, above 100%, so the village produces more than it needs. Measures (any two): higher-yielding varieties, more irrigated area, mechanisation, less post-harvest loss, better storage. Marking: 3 marks ratio, 1 mark per measure.

Question 5 — Plate tectonics, earthquakes and tsunamis (16 marks)

5. An earthquake occurs in Region E. A station records P waves at 6.0 km/s and S waves at 3.5 km/s; the P waves arrive 20 seconds before the S waves.

#	Question	Answer and marking notes
14	Calculate the distance from the station to the focus (km).	Let the distance be d: d / 3.5 – d / 6.0 = 20, so 0.119048d = 20 and d = 168 km. Marking: 3 marks equation, 2 marks solution, 1 mark unit.
15	The focal depth is 30 km. Calculate the epicentral distance (km).	Epicentral distance = sqrt(168^2 – 30^2) = sqrt(28,224 – 900) = sqrt(27,324) = 165.3 km. Marking: 2 marks Pythagoras, 2 marks value, 1 mark unit.
16	The submarine earthquake triggers a tsunami where the sea is 4,000 m deep. Using v = sqrt(gd) (g = 9.81 m/s2), calculate the wave speed (m/s and km/h) and explain why tsunamis travel fast in deep water but cause severe damage near the coast.	v = sqrt(9.81 x 4,000) = 198.1 m/s, about 713 km/h. In deep ocean the wavelength is long and amplitude small, so little energy is lost and the wave moves fast; in shallow coastal water the speed falls and the wavelength shortens, concentrating energy so wave height rises sharply, and dense coastal development makes the damage severe. Marking: 2 marks calculation, 1 mark conversion, 2 marks explanation.

Question 6 — Arid climate and renewable energy (15 marks)

6. Region F is arid: annual rainfall 250 mm, annual evapotranspiration 1,800 mm, average daily solar radiation 5.5 kWh/m2.

#	Question	Answer and marking notes
17	Calculate the annual water balance (mm) and state its effect on local farming.	Water balance = 250 – 1,800 = -1,550 mm, a deficit of 1,550 mm a year; farming must rely on irrigation, otherwise only drought-tolerant crops can be grown and the growing season is limited. Marking: 2 marks value, 2 marks effect.
18	A solar farm of 2 km2 is built in Region F with panels of 18% efficiency. Calculate the daily output (kWh and MWh).	2 km2 = 2,000,000 m2; output per m2 per day = 5.5 x 0.18 = 0.99 kWh; daily output = 2,000,000 x 0.99 = 1,980,000 kWh = 1,980 MWh. Marking: 2 marks area conversion, 2 marks efficiency, 2 marks unit conversion.
19	Explain how solar development in an arid region can reduce land degradation, and give two environmental problems it may also cause.	Reducing degradation: less fuelwood cutting keeps vegetation cover; panel shade lowers evaporation and water collected under panels can support grass, cutting wind erosion. Problems (any two): washing panels uses scarce water; construction breaks surface crust and worsens wind erosion; visual impact and loss of animal habitat. Marking: 2 marks benefits, 1.5 marks per problem.

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