IB Geography SL Paper 2 Practice Paper (Structured Questions, 10 Questions, 100 Marks)

2026-11-01

IB Geography SL

IB Geography SL Paper 2 practice paper: population, rivers, urban growth, farming, ice loss, carrying capacity, energy, coasts, floods, HDI. Answers with marking notes.

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Paper details　Level: IB DP　|　Subject: Geography　|　Time: 75 minutes　|　Marks: 100　|　Questions: 10

Section A: Population and urban change (30 marks)

Answer questions 1 to 3. Each question carries 10 marks.

#	Question	Answer and marking notes
1	Q1 A city has a population of 4,800,000 and a land area of 1,200 km². In one year there are 52,000 births, 28,000 deaths and net migration of −6,000. (a) Calculate the population density (people per km²). (b) Calculate the rate of natural increase (%, 1 d.p.). (c) Use the rule of 70 to estimate the doubling time. (d) Explain how net migration changes your answer to (c).	(a) 4,800,000 ÷ 1,200 = 4,000 people per km² (1 mark method, 1 mark answer). (b) Natural increase = 52,000 − 28,000 = 24,000; 24,000 ÷ 4,800,000 × 100 = 0.5% (2 marks). (c) 70 ÷ 0.5 = 140 years (2 marks). (d) Total change = 24,000 − 6,000 = 18,000 (−0.375% per year), so if sustained the doubling time becomes longer than 140 years; positive net migration would shorten it (3 marks: direction, total-change figure, conclusion).
2	Q2 A river channel has a cross-sectional area of 12.0 m² and a mean velocity of 1.5 m/s. (a) Calculate the discharge (m³/s). (b) Calculate annual runoff in km³ (3 d.p.; 1 year = 31,536,000 s). (c) Explain two reasons why discharge rises after heavy rain. (d) Explain how urbanisation changes the flood peak.	(a) 12.0 × 1.5 = 18 m³/s (2 marks). (b) 18 × 31,536,000 = 567,648,000 m³ = 0.568 km³ (2 marks method, 1 mark unit conversion). (c) Any two: rainfall intensity exceeds infiltration capacity; saturated soil raises overland flow; less interception by vegetation; antecedent rainfall raises the water table (2 × 1.5 marks). (d) Impermeable surfaces reduce infiltration and speed up runoff, giving a higher and earlier flood peak (2.5 marks).
3	Q3 A city of 900,000 grows at 3.2% a year for 10 years. (a) Calculate the population after 10 years (900,000 × 1.032¹⁰). (b) Calculate the increase. (c) Identify two pressures on the city and one management strategy for each.	(a) 1.032¹⁰ = 1.37024; 900,000 × 1.37024 = 1,233,217 (3 marks). (b) 1,233,217 − 900,000 = 333,217 (2 marks). (c) Pressures: housing shortage, traffic congestion, inadequate water and sewerage, loss of green space (1 mark each); strategies: high-density public housing, transit-oriented development, district water and sewage works, protected green corridors (1.5 marks each).

Section B: Agriculture, food and ecosystems (30 marks)

Answer questions 4 to 6. Each question carries 10 marks.

#	Question	Answer and marking notes
4	Q4 A farm produces 4,800 tonnes of grain from 600 hectares. (a) Calculate the yield in t/ha. (b) Convert this to kg per m² (1 ha = 10,000 m²). (c) The national average is 2.5 t/ha — how many times the average is this farm? (d) Suggest two sustainable ways to raise output.	(a) 4,800 ÷ 600 = 8 t/ha (2 marks). (b) 8 × 1,000 ÷ 10,000 = 0.8 kg/m² (3 marks). (c) 8 ÷ 2.5 = 3.2 times the average (2 marks). (d) Crop rotation with green manure, drip irrigation, integrated pest management, precision fertiliser application (1.5 marks each with a sustainability reason).
5	Q5 An ice sheet loses 250 Gt of ice a year; 361.8 Gt of ice is roughly equivalent to 1 mm of global sea-level rise. (a) Calculate the annual sea-level rise (mm, 3 d.p.). (b) Calculate the cumulative rise over 30 years (mm). (c) State two further impacts.	(a) 250 ÷ 361.8 = 0.691 mm per year (3 marks). (b) 0.691 × 30 = 20.73 mm (2 marks). (c) Any two: flooding of low-lying coasts, saltwater intrusion into aquifers, stronger storm surges, stress on coral reefs and wetlands, population displacement (2.5 marks each).
6	Q6 A national park of 120 km² receives 480,000 visitors a year; the environmental carrying capacity is 1,500 visitors per km² per year. (a) Calculate the actual visitor density. (b) Calculate the sustainable total number of visitors. (c) Calculate the excess as a percentage (1 d.p.). (d) Suggest one visitor-management strategy.	(a) 480,000 ÷ 120 = 4,000 visitors per km² per year (2 marks). (b) 1,500 × 120 = 180,000.0 visitors (2 marks). (c) (480,000 − 180,000) ÷ 180,000 × 100 = 166.7% (3 marks). (d) Booking quotas, timed entry, shuttle and parking limits, closure of fragile zones, guided tours that disperse visitors (3 marks).

Section C: Energy and coasts (20 marks)

Answer questions 7 and 8. Each question carries 10 marks.

#	Question	Answer and marking notes
7	Q7 A power station has a capacity of 250 MW and a capacity factor of 45%. (a) Calculate its annual output (MWh; 1 year = 8,760 hours). (b) Convert to GWh. (c) If a household uses 4,500 kWh a year, how many households can it supply (whole number)? (d) Give one reason why the capacity factor is below 100%.	(a) 250 × 0.45 × 8,760 = 985,500 MWh (3 marks). (b) 985,500 ÷ 1,000 = 985.5 GWh (2 marks). (c) 985,500 × 1,000 ÷ 4,500 = 219,000 households (3 marks). (d) Maintenance downtime, variation in fuel supply, sunlight or wind speed, output cut at demand troughs (2 marks).
8	Q8 Longshore drift on a coast averages 1.5 m a year from west to east. (a) Calculate the total drift over 60 years. (b) State two factors affecting the strength of drift. (c) Explain two consequences for the beach downstream of a groyne built on the western side.	(a) 1.5 × 60 = 90 m (3 marks). (b) Any two: wave approach angle, wind direction and fetch, beach gradient, sediment supply (2 marks each). (c) The groyne traps sediment moving west to east, so the downstream (eastern) beach loses supply, narrows and erodes while the up-drift side builds up (3 marks; direction and causal chain required).

Section D: Floods and development (20 marks)

Answer questions 9 and 10. Each question carries 10 marks.

#	Question	Answer and marking notes
9	Q9 The probability that a river exceeds 800 m³/s in any year is 0.04. (a) Calculate the return period (years). (b) Calculate the probability of at least one such flood in the next 10 years (1 − 0.96¹⁰, 3 d.p.). (c) Explain why a 25-year return period does not mean a flood every 25 years.	(a) 1 ÷ 0.04 = 25 years (2 marks). (b) 1 − 0.96¹⁰ = 0.335 (4 marks: 2 for the formula, 2 for the answer). (c) A return period is a long-run average probability and events are independent, so two floods may occur within 10 years or none for 40 years; climate change and land-use change also shift the probability itself (4 marks).
10	Q10 A country has a GDP of US$48 billion and a population of 6.0 million; its HDI components are 0.78 (health), 0.71 (education) and 0.69 (income). (a) Calculate GDP per capita (US$). (b) Calculate the HDI as the arithmetic mean of the three (4 d.p.). (c) Explain why GDP per capita alone is an inadequate measure of development.	(a) US$48 billion ÷ 6.0 million = US$8,000 (3 marks). (b) (0.78 + 0.71 + 0.69) ÷ 3 = 0.7267 (3 marks). (c) It ignores income distribution, health and education, environmental costs, non-market production and regional inequality; two countries with the same GDP per capita can have very different HDI values (4 marks; comparison required).

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