Secondary 4 Geography Practice Paper: Plate Tectonics and Earthquakes, Slope Processes and Landslides, Weather and Climate, Population and Migration, Urban Land Use, Agriculture and Food Supply

2026-11-14

Geography – Secondary 4 – Practice Paper

S4 Geography paper, 24 parts: tectonic hazards, slope failure, humidity and lapse rate, dependency ratios, urban density, crop yields and food supply.

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Paper details　S4 Geography practice paper with 24 parts and marking points on plate tectonics, slope processes, weather and climate, population structure, urban land use and food supply.

Section A: Plate Tectonics and Earthquakes (16 marks)

Source A: Table 1 shows two seismic events. Event A had magnitude 6.8 and a focal depth of 12 km; Event B had magnitude 5.2 and a focal depth of 8 km.
Source B: Magnetic anomaly stripes are symmetrical on both sides of a mid-ocean ridge. The same stripe lies 240 km apart across the ridge and its age is 3.2 million years.
Source C: The 2004 Indian Ocean tsunami travelled across ocean of average depth 4,000 m. Its speed can be derived from gravitational acceleration (9.81 m per second squared) and water depth.

#	Question	Answer and marking notes
1	(a) With reference to Source A, calculate how many times more seismic energy Event A released than Event B (1 decimal place).	Answer: The magnitudes differ by 1.6, so the energy ratio is 10 to the power of 2.4, about 251.2 times (2 marks). Marking points: correct logarithmic relationship (1 mark); correct computation (1 mark); answer to 1 decimal place with the word times (2 marks). A common error is treating the magnitude difference as the ratio.
2	(b) Compare how focal depth and magnitude affect the damage at the surface, giving one reason for each.	Answer: Magnitude controls total energy released, roughly 31.6 times more for each unit, and hence the size of the damaged area (2 marks). Focal depth controls attenuation before energy reaches the surface: shallow events (less than 70 km) shake the ground strongly and cause concentrated damage, while deep events lose energy on the way up (2 marks). Marking points: mechanism for each factor (1 mark each) plus a specific comparison from the sources (1 mark each).
3	(c) With reference to Source B, calculate the average plate movement in centimetres per year (1 decimal place) and name the type of plate boundary.	Answer: Half-spreading distance is 120 km, equal to 12,000,000 cm, divided by 3,200,000 years, giving about 3.8 cm per year (3 marks). The boundary is divergent (constructive) because new crust forms symmetrically on both sides of the ridge (1 mark). Marking points: halving for symmetry (1 mark); correct unit conversion (1 mark); answer to 1 decimal place (1 mark).
4	(d) With reference to Source C, calculate the tsunami speed in kilometres per hour (1 decimal place) and explain why wave height rises near the shore.	Answer: Speed equals the square root of gravitational acceleration times depth, about 198.1 m per second, that is 713.1 km per hour (2 marks). Height rises near shore because shallower water slows the wave and shortens its wavelength, so the same energy is packed into a shorter water column; seabed friction also steepens the wave (2 marks). Marking points: formula and conversion (1), figure (1), energy concentration (1), wavelength shortening (1).

Section B: Slope Processes and Landslides (18 marks)

Source A: Table 2 shows a slope with a horizontal distance of 250 m, a vertical drop of 60 m and a topsoil thickness of 1.8 m.
Source B: A rainstorm delivered 60 mm of rain in 45 minutes. Resistance of the soil fell from 96 to 72 kilonewtons per square metre while the driving force stayed at 58 kilonewtons per square metre.
Source C: The district recorded 14 landslides in ten years; 11 of them followed rainfall above 150 mm.

#	Question	Answer and marking notes
5	(a) With reference to Source A, calculate the slope angle in degrees (1 decimal place).	Answer: The angle is the inverse tangent of 60 divided by 250, that is the inverse tangent of 0.24, about 13.5 degrees (3 marks). Marking points: correct ratio (1), use of inverse tangent (1), answer to 1 decimal place with unit (1).
6	(b) With reference to Source B, calculate the factor of safety before and after the rainstorm (2 decimal places) and judge whether the slope became less stable.	Answer: The factor of safety is resistance divided by driving force. Before the storm it was 96 divided by 58, giving 1.66; after the storm 72 divided by 58, giving 1.24 (2 marks). Resistance fell by 25.0 per cent, so the slope is still above 1 but moving towards failure and may fail in a further storm (2 marks). Marking points: one mark per calculation and two marks for interpreting a value above 1 correctly.
7	(c) Explain three mechanisms by which heavy rain triggers landslides.	Answer: First, infiltration raises pore water pressure and lowers effective stress, reducing shear strength (2 marks). Second, added water increases soil weight and therefore the driving force (2 marks). Third, loss of cohesion and surface runoff undercut the slope toe (1 mark). Marking points: award marks for cause and effect rather than naming a phenomenon; credit reference to the 150 mm cumulative rainfall in Source C.
8	(d) Evaluate the effectiveness of drainage channels with vegetation compared with a retaining wall in reducing landslide risk.	Answer: Drainage with vegetation is cheaper and quicker and lowers pore pressure while roots reinforce topsoil, suiting large natural slopes of moderate risk (3 marks), but it adds limited resistance against deep failure and needs maintenance (2 marks). A retaining wall gives direct resistance on steep slopes near buildings but is costly, takes land and needs structural design (2 marks). The conclusion should match the measure to the risk level rather than replace one with the other (1 mark). Marking points: 2 to 3 marks per option with a comparative judgement; a list of benefits alone cannot earn full marks.

Section C: Weather and Climate (16 marks)

Source A: Air temperature was 25 degrees, vapour pressure 21.0 hPa and the saturated vapour pressure at that temperature 31.7 hPa.
Source B: The air temperature at the foot of the hill was 28.0 degrees. The summit is 1,500 m higher and the mean lapse rate is 0.6 degree per 100 m.
Source C: A tropical cyclone had a central pressure of 950 hPa, pressure of 1,010 hPa at 300 km from the centre, and a ten-minute mean wind speed of 22 m per second near the centre.

#	Question	Answer and marking notes
9	(a) With reference to Source A, calculate the relative humidity (1 decimal place) and state whether the air is close to saturation.	Answer: Relative humidity is vapour pressure divided by saturated vapour pressure times 100, that is 21.0 divided by 31.7, giving 66.2 per cent (3 marks). The air is well below saturation and would need cooling of about 6 degrees to reach the dew point (1 mark). Marking points: formula (1), figure (1), per cent sign and 1 decimal place (1), sensible judgement (1).
10	(b) With reference to Source B, calculate the temperature at the summit (1 decimal place) and explain why temperature falls with altitude.	Answer: Over 1,500 m the fall is 9.0 degrees at 0.6 degree per 100 m, so the summit is 28.0 minus 9.0, that is 19.0 degrees (2 marks). Reasons: the troposphere is heated mainly by long-wave radiation from the ground, so less heat is received with height; rising air expands as pressure falls and cools (2 marks). Marking points: calculation (2), reasons (1 each).
11	(c) With reference to Source C, calculate the pressure gradient from centre to periphery in hPa per km (2 decimal places) and convert the wind speed to km per hour (1 decimal place).	Answer: The difference is 60 hPa over 300 km, so the gradient is 0.2 hPa per km (2 marks); 22 m per second times 3.6 gives 79.2 km per hour (2 marks). Marking points: 2 marks each, full marks only with correct units and decimal places.
12	(d) Explain two reasons why a tropical cyclone weakens rapidly after landfall.	Answer: First, the supply of moisture and latent heat from the warm sea is cut off, so convection cannot be sustained (2 marks). Second, greater land friction reduces surface evaporation and orographic lifting disrupts the circulation, so the central pressure rises (2 marks). Marking points: 2 marks per reason, with the causal link from energy supply or structural disruption.

Section D: Population Structure and Migration (16 marks)

Source A: Birth rate was 11.2 per thousand and death rate 6.4 per thousand in a year.
Source B: The population comprised 780,000 aged 0 to 14, 3,900,000 aged 15 to 64 and 320,000 aged 65 and above.
Source C: 42,000 people moved in and 18,500 moved out in the same year; 62% of arrivals were aged 25 to 44.

#	Question	Answer and marking notes
13	(a) With reference to Source A, calculate the rate of natural increase per thousand (1 decimal place) and the doubling time in years (1 decimal place).	Answer: Natural increase is 11.2 minus 6.4, that is 4.8 per thousand or 0.48 per cent (2 marks). By the rule of 70 the doubling time is 70 divided by 0.48, about 145.8 years (2 marks). Marking points: per thousand to per cent conversion (1), figure (1), use of rule of 70 (1), answer (1).
14	(b) With reference to Source B, calculate the total dependency ratio (1 decimal place) and state what a low ratio implies.	Answer: The dependent population is 1,100,000 (780,000 plus 320,000) divided by 3,900,000 working age, times 100, giving 28.2 per cent (3 marks). A low ratio means a lighter burden on workers, a demographic dividend and more potential for saving and growth (1 mark). Marking points: correct numerator and denominator (1 each), figure (1), implication (1).
15	(c) With reference to Source C, calculate net migration for the year and explain its effect on population structure.	Answer: Net migration is 42,000 minus 18,500, giving 23,500 people (2 marks). Since 62 per cent of arrivals are aged 25 to 44, the working age share rises, the dependency ratio falls and ageing may be eased (2 marks). Marking points: calculation (2), two structural effects (1 each).
16	(d) Explain two pressures that population ageing places on public finances.	Answer: First, rising demand for elderly medical care and long-term care raises public health and residential care spending (2 marks). Second, a smaller workforce narrows the tax base while retirement protection and allowances cost more, increasing the risk of fiscal imbalance (2 marks). Marking points: 2 marks per pressure with a specific link to spending or revenue.

Section E: Urban Land Use (18 marks)

Source A: An urban district covers 28.5 square kilometres with a population of 486,000.
Source B: A redevelopment site of 18,200 square metres has 148,000 square metres of gross floor area. The whole district covers 640 hectares, of which 96 hectares are green space.
Source C: Land use in the district is 38% residential, 12% commercial, 9% industrial, 18% roads, 15% open space and 8% other.

#	Question	Answer and marking notes
17	(a) With reference to Source A, calculate population density in persons per square kilometre (whole number) and compare it with the territory average of about 6,800 to classify the district.	Answer: Density is 486,000 divided by 28.5, about 17,053 persons per square kilometre (3 marks). That is roughly 2.5 times the territory average, so it is a high density urban district dominated by high-rise residential and commercial uses (2 marks). Marking points: calculation and rounding (3), comparison (1), classification (1).
18	(b) With reference to Source B, calculate the plot ratio (1 decimal place) and the proportion of green space (1 decimal place).	Answer: Plot ratio is gross floor area divided by site area, 148,000 divided by 18,200, giving 8.1 (3 marks); green space is 96 divided by 640 times 100, giving 15.0 per cent (2 marks). Marking points: plot ratio formula (2) and figure (1); green space (2) with correct unit and decimals.
19	(c) Explain two reasons why industrial land moves from the city centre to the outskirts.	Answer: First, high land prices and rents in the centre make large factory and warehouse space unaffordable, while suburban land is cheaper (2 marks). Second, congestion and tighter environmental controls in the centre contrast with outer areas near highways and container terminals, lowering logistics costs and meeting pollution zoning (2 marks). Marking points: 2 marks per reason covering both push and pull factors.
20	(d) Evaluate the feasibility of raising the open space share from 15 to 20 per cent in old district redevelopment.	Answer: Feasibility is limited by fragmented lots and multiple owners, which make land assembly slow (1 mark); a higher open space share cuts saleable floor area and project viability (1 mark); multi-level or roof gardens can meet the target in three dimensions (1 mark). It becomes more workable with land exchange or compensation and with a per capita open space standard rather than a single ratio (1 mark). Marking points: 1 mark per sound consideration with a clear conclusion.

Section F: Agriculture and Food Supply (16 marks)

Source A: Farm output in a region was 27,600 tonnes from 2,400 hectares of farmland.
Source B: Local production was 34,500 tonnes while total food demand was 138,000 tonnes.
Source C: Agriculture abstracted 45 million cubic metres of irrigation water, of which 32.4 million cubic metres were used effectively. Farmland fell from 4,200 to 3,780 hectares in ten years.

#	Question	Answer and marking notes
21	(a) With reference to Source A, calculate the yield per hectare in tonnes (1 decimal place).	Answer: Yield per hectare is 27,600 divided by 2,400, that is 11.5 tonnes per hectare (4 marks). Marking points: correct output over area (2), figure (1), unit and 1 decimal place (1).
22	(b) With reference to Source B, calculate the food self-sufficiency rate (1 decimal place) and state two risks of a low rate.	Answer: Self-sufficiency is 34,500 divided by 138,000 times 100, giving 25.0 per cent (2 marks). Risks: import prices and exchange rates pass straight through to local food prices; supply chain disruption from shipping, disease or export bans can cause shortages (2 marks). Marking points: calculation (2), risks (1 each).
23	(c) With reference to Source C, calculate irrigation water use efficiency (1 decimal place) and the percentage change in farmland area (1 decimal place).	Answer: Efficiency is 3,240 divided by 4,500, giving 72.0 per cent (2 marks); the change is 3,780 minus 4,200 divided by 4,200, giving -10.0 per cent, a fall of 10 per cent over ten years (2 marks). Marking points: 2 marks each, with the negative sign or the word fall shown.
24	(d) Explain two constraints on raising local food self-sufficiency.	Answer: First, land is scarce and farmland keeps being lost to housing and infrastructure, so the cultivated area cannot easily expand (2 marks). Second, high local production costs and labour shortages make prices uncompetitive against imports, while climate hazards destabilise output (2 marks). Marking points: 2 marks per constraint linked to the farmland loss or cost evidence in Source C.

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