HKDSE Geography Paper 1 Practice Paper | 4 questions, 50 marks, 1 hour 15 minutes

2026-10-31

HKDSE Geography Paper 1

HKDSE Geography Paper 1 practice paper: 4 questions, 50 marks, 75 minutes, covering map reading and gradient, river discharge and lag time, population density and dependency ratio, and tropical monsoon rainfall with irrigation demand. Full worked answers and mark schemes are included for S4 to S6 revision.

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Paper details　Level: HKDSE | Subject: Geography | Time: 1 hour 15 minutes | Marks: 50 | Questions: 4

Section A: Map reading and rivers (26 marks)

Answer ALL four questions. Total: 50 marks; time allowed: 1 hour 15 minutes. Show full working for calculation questions and give correct units; method marks are awarded, and an answer without working scores at most half.

#	Question	Answer and marking notes
1	On a 1:20,000 map (schematic) with a contour interval of 20 m, point A is 168 m above sea level and point B is 92 m above sea level. The straight-line map distance from A to B is 7.5 cm; a farmland plot measures 6.0 cm² on the map. (a) Calculate the actual horizontal distance from A to B, in kilometres. (3 marks) (b) Calculate the average gradient between A and B in the form 1:n, where n is an integer. (3 marks) (c) Calculate the actual area of the farmland, in hectares. (3 marks) (d) A footpath near A and B follows a contour. Explain why building a footpath along a contour is better than going straight from A to B. (3 marks)	(a) Actual distance = map distance × scale denominator = 7.5 cm × 20,000 = 150,000 cm = 1.5 km. (3: formula, conversion, answer with unit) (b) Vertical difference = 168 − 92 = 76 m; horizontal distance = 1.5 km = 1500 m. Gradient = 76 ÷ 1500 = 0.050667, i.e. 1:20. (3) (c) Areas scale as the square of the scale denominator: 6.0 cm² × 20,000² = 240,000 m² = 24.0 hectares (= 0.24 km²). (3) (d) A contour path has almost no gradient, so people and vehicles use less energy and soil erosion is reduced. The straight A–B line has a gradient of 1:20, which still makes continuous climbing tiring and is more easily washed out in heavy rain. The contour route is longer but safer and easier. (3)
2	A river cross-section is measured before and after a storm. Before: width 12 m, mean depth 1.5 m, mean velocity 0.8 m s⁻¹. After: width 15 m, mean depth 2.4 m, mean velocity 1.6 m s⁻¹. Rainfall in that 24 hours is 96 mm over a catchment of 24 km². The rainfall peak is at 14:00 and the discharge peak at 20:30. (a) Calculate the discharge before the storm, in m³ s⁻¹. (3 marks) (b) Calculate the total volume of water passing the section in the day before the storm, in m³. (2 marks) (c) Calculate the discharge after the storm and the ratio of the after-storm to the before-storm discharge. (4 marks) (d) Find the lag time of the discharge peak behind the rainfall peak; explain how urbanisation changes the lag time. (3 marks) (e) Using the before-storm discharge, estimate the runoff coefficient (runoff volume ÷ rainfall volume) to 2 decimal places. (2 marks)	(a) Cross-sectional area = 12 × 1.5 = 18 m²; discharge = 18 × 0.8 = 14.4 m³ s⁻¹. (3: area, formula, answer with unit) (b) One day = 86,400 s, so volume = 14.4 × 86,400 = 1,244,160 m³ (about 1.24 × 10⁶ m³). (2) (c) After the storm, area = 15 × 2.4 = 36 m²; discharge = 36 × 1.6 = 57.6 m³ s⁻¹. Ratio = 57.6 ÷ 14.4 = 4 times. (4) (d) Lag time = 20:30 − 14:00 = 6.5 hours. Urbanisation covers the ground with concrete and asphalt, so infiltration falls and surface runoff is drained quickly by channels; straightened, narrowed channels also speed flood travel, so the lag time shortens and the peak discharge rises. (3) (e) Rainfall volume = 96 mm ÷ 1,000 × 24 km² × 10⁶ = 2,304,000 m³; runoff coefficient = 1,244,160 ÷ 2,304,000 = 0.54. (2)

Section B: Population and climate (24 marks)

#	Question	Answer and marking notes
3	A new town has a population of 480,000 in an area of 24 km². Its annual birth rate is 9.5 per 1,000 and death rate 6.2 per 1,000, with a net migration gain of 2,400. Age structure: 0–14 = 72,000, 15–64 = 360,000, 65 and over = 48,000. (a) Calculate the population density in persons per km². (2 marks) (b) Calculate the natural increase in population for the year. (3 marks) (c) Calculate the total annual growth rate to 2 decimal places. (3 marks) (d) Calculate the dependency ratio to 1 decimal place. (2 marks) (e) State one effect of an ageing population on the town’s public services. (2 marks)	(a) Density = 480,000 ÷ 24 = 20,000 persons per km². (2) (b) Natural increase rate = 9.5 − 6.2 = 3.3 per 1,000; natural increase = 480,000 × 3.3 ÷ 1,000 = 1,584 people. (3) (c) Total increase = 1,584 + 2,400 = 3,984; growth rate = 3,984 ÷ 480,000 × 100% = 0.83%. (3) (d) Dependency ratio = (72,000 + 48,000) ÷ 360,000 × 100% = 33.3%, i.e. about 33 dependants per 100 people of working age. (2) (e) People aged 65+ already form 10.0% of the population, so demand rises for hospitals, community care, residential homes and rehabilitation transport, raising health and welfare spending while the working-age share and tax base shrink. (2)
4	Monthly rainfall (mm) at a tropical monsoon place: Jan 12, Feb 18, Mar 35, Apr 90, May 210, Jun 380, Jul 420, Aug 390, Sep 260, Oct 110, Nov 40, Dec 20. A 1.2 hectare field there grows paddy rice; in the dry season (November to February) the crop needs 450 mm of water. (a) Find the annual total and the mean monthly rainfall, to 1 decimal place. (2 marks) (b) Find the percentage of the annual rainfall falling in June to August, to 1 decimal place. (3 marks) (c) Find the dry-season water deficit (mm) and the irrigation water needed for the field (m³). (3 marks) (d) Explain the seasonal distribution of rainfall there. (2 marks) (e) Compare the risk of growing paddy rice and a drought-tolerant crop in the dry season. (2 marks)	(a) Annual total = 1,985 mm; monthly mean = 1,985 ÷ 12 = 165.4 mm. (2) (b) June–August = 380 + 420 + 390 = 1190 mm; share = 1190 ÷ 1,985 × 100% = 59.9%. (3) (c) Dry-season rainfall = 40 + 20 + 12 + 18 = 90 mm; deficit = 450 − 90 = 360 mm. Irrigation = 360 mm ÷ 1,000 × 12,000 m² = 4,320 m³. (3) (d) In summer the moist south-west monsoon blows from sea to land and is forced to rise over hills, and tropical depressions and typhoons add rain, so June to August is wettest; in winter the dry north-east monsoon brings little moisture, so November to February is driest. (2) (e) Paddy rice needs much water, but the dry season brings only 90 mm, a deficit of 360 mm, so without irrigation yields fall or the crop fails, and 4,320 m³ of irrigation must be paid for; a drought-tolerant crop can survive on the limited rain, so the risk is lower. Rice does well in the wet season, when there is a surplus of 950 mm, so the two should be rotated by season. (2)

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