HKDSE Geography Practice Paper: River Landforms And Flood Management

HKDSE Geography Practice Paper: River Landforms and Flood Management

2026-11-03

HKDSE Geography: River Landforms and Flood Management

HKDSE Geography practice paper: 12 questions, 100 marks in 75 minutes on river landforms, discharge, flood risk and management, with worked answers and marking points.

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Paper details　Level: S4-S6 | Subject: Geography | Time: 75 minutes | Marks: 100 | Questions: 12

Section A (Questions 1 to 4, 32 marks)

Read each question, quote the formula, substitute the values and give the unit.

#	Question	Answer and marking notes
1	A river flows from a source at 1,250 m above sea level to its mouth, a channel length of 84 km. (a) Find the average gradient of the whole river in m per km. (b) The first 12 km fall 620 m; find the average gradient of this upper section. (c) The last 30 km fall 30 m; find the average gradient of this lower section. (d) Explain how the downstream decrease in gradient affects the work of erosion along the river.	(a) Average gradient = 1,250 / 84 = 14.9 m per km [2]. (b) Upper gradient = 620 / 12 = 51.7 m per km [2]. (c) Lower gradient = 30 / 30 = 1.0 m per km [2]. (d) A steep upper course gives fast flow and high kinetic energy, so vertical downcutting dominates and a V-shaped valley forms; in the gentle lower course velocity and competence fall, so erosion becomes lateral and deposition takes over, building meanders and a floodplain [2]. Marking: gradient is fall divided by horizontal distance; the dominant process changes downstream.
2	A river channel is 18 m wide with a mean depth of 2.5 m and a mean velocity of 1.2 m per second. (a) Find the cross-sectional area. (b) Find the discharge in cubic metres per second. (c) After a storm the width becomes 20 m, the mean depth 3.2 m and the velocity 1.8 m per second; find the new discharge and the percentage increase. (d) Explain the effect of a higher discharge on the bed and the banks.	(a) Cross-sectional area = 18 x 2.5 = 45 square metres [2]. (b) Discharge = 45 x 1.2 = 54 cubic metres per second [2]. (c) New area = 20 x 3.2 = 64 square metres, new discharge = 64 x 1.8 = 115.2 cubic metres per second, an increase of (115.2 – 54) / 54 = 113.3 per cent [2]. (d) A higher discharge means more energy, so shear stress on the bed rises and downcutting and scouring increase; the banks are undercut and collapse, so the channel widens and may shift course [2]. Marking: discharge is area times velocity; more energy strengthens both vertical and lateral erosion.
3	A meandering reach has a straight-line distance of 3.2 km and a channel length, measured along the thalweg, of 5.6 km. (a) Find the sinuosity. (b) If the meander migrates outwards at 1.5 m per year, find the total migration over 30 years. (c) Compare flow velocity and deposition on the outer concave bank and the inner convex bank. (d) Explain how a neck cut-off produces an oxbow lake.	(a) Sinuosity = 5.6 / 3.2 = 1.75 [2]. (b) Total migration = 1.5 x 30 = 45 m [2]. (c) On the outer bank the water is deep and fast, lateral erosion dominates and the bank is steep; on the inner bank the water is shallow and slow, competence falls and sediment is deposited as a point bar [2]. (d) Continued outward migration narrows the neck of land between two bends; in a flood the river cuts through the neck, and the abandoned bend is sealed by sediment at both ends to become an oxbow lake [2]. Marking: a sinuosity above 1.5 is strongly meandering; erosion is on the outer bank and deposition on the inner bank.
4	At a gauging station, water rose above the warning level in 4 of the last 40 years. (a) Find the return period of that level. (b) Find the probability of at least one exceedance in the next 10 years. (c) Explain what a ten-year event means and why it does not mean one event every ten years. (d) Explain how urbanisation changes the frequency of floods.	(a) Return period = 40 / 4 = 10 years [2]. (b) Probability = 1 – (1 – 1/10)^10 = 1 – 0.9^10 = 65.1 per cent [2]. (c) A ten-year event has a 10 per cent chance of being equalled or exceeded in any one year; it is a long-run average, so events are not evenly spaced and two in successive years or none for twenty years are both normal [2]. (d) Urbanisation raises the proportion of impermeable surface and the runoff coefficient and shortens the lag time, so the same rainfall produces a higher peak; the return period of the warning level therefore shortens and floods become more frequent [2]. Marking: exceedance probability is 1 – (1 – 1/T)^n; a return period is an average, not a cycle.

Section B (Questions 5 to 8, 36 marks)

Answer the following, showing full working and the formula used.

#	Question	Answer and marking notes
5	In one catchment the rural rainfall peak is at 14:00 with the flood peak at 20:00; after urbanisation the rainfall peak stays at 14:00 but the flood peak arrives at 16:30. The rural peak discharge is 120 cubic metres per second and the urban peak is 210 cubic metres per second. (a) Find the lag time in each case. (b) Find how many hours earlier the peak now arrives. (c) Find the percentage increase in peak discharge. (d) Explain how impermeable surfaces and channelisation cause these changes. (e) Suggest two soft engineering measures for urban flood management.	(a) Rural lag time = 20:00 – 14:00 = 6 hours; urban lag time = 16:30 – 14:00 = 2.5 hours [2]. (b) The peak arrives 6 – 2.5 = 3.5 hours earlier [2]. (c) The increase is (210 – 120) / 120 = 75 per cent [2]. (d) Impermeable surfaces stop infiltration so most rain becomes surface runoff at once, while channels and storm drains carry water away quickly and shorten the time of concentration, so the peak both arrives earlier and is higher [2]. (e) For example flood risk maps with zoning that limits development on the floodplain, and restoring wetlands, storage ponds and permeable paving to create space for water [1]. Marking: lag time is peak time minus rainfall peak time; urbanisation shortens the lag and raises the peak.
6	A flood defence scheme costs 480 million dollars, is expected to reduce average annual damage by 32 million dollars, costs 4 million dollars a year to maintain and has a life of 60 years. (a) Find the annual net benefit. (b) Find the payback period. (c) Find the total net benefit over 60 years without discounting. (d) Explain why discounting matters when appraising such a scheme. (e) Give one example each of a hard engineering work and a soft engineering measure.	(a) Annual net benefit = 32 – 4 = 28 million dollars [2]. (b) Payback period = 480 / 28 = 17.1 years [2]. (c) Total net benefit = 28 x 60 – 480 = 1,680 – 480 = 1,200 million dollars [2]. (d) Discounting converts future benefits to present value and reflects the opportunity cost of capital; at a high discount rate distant avoided damage is worth little today, so the scheme may compare badly with other uses of the money [1]. (e) A hard work such as a levee or a detention basin, and a soft measure such as a flood warning system, land use zoning or flood insurance [2]. Marking: net benefit is avoided damage minus maintenance; the discount rate decides whether a long lived scheme is worthwhile.
7	A reservoir catchment covers 320 square kilometres, with mean annual rainfall of 2,400 mm and a runoff coefficient of 0.45. (a) Find the annual runoff depth in mm. (b) Find the annual runoff volume in cubic metres. (c) If the reservoir capacity equals 18 per cent of the annual runoff, find the capacity. (d) Explain how urbanisation changes the runoff coefficient and what this means for reservoir design.	(a) Runoff depth = 2,400 x 0.45 = 1,080 mm [2]. (b) The catchment is 320 million square metres, so runoff volume = 320 million x 1.08 = 345.6 million cubic metres [3]. (c) Capacity = 345.6 million x 0.18 = 62.2 million cubic metres [2]. (d) Urbanisation reduces infiltration and raises the runoff coefficient, so the same rainfall yields more runoff and a sharper peak; the reservoir must keep a larger flood storage volume or spillway risk rises [2]. Marking: runoff volume is area times runoff depth; capacity is a volume.
8	A river carries 2.4 million tonnes of sediment a year and an upstream reservoir traps 65 per cent of it. (a) Find the sediment load downstream of the dam. (b) If the delta needs 1.0 million tonnes a year to hold its area, state whether the supply is sufficient and give the shortfall. (c) State two effects of the reduced sediment supply at the coast. (d) Suggest one measure that would reduce the problem.	(a) Downstream load = 2.4 x (1 – 0.65) = 0.84 million tonnes [2]. (b) The shortfall is 1.0 – 0.84 = 0.16 million tonnes a year, so the supply is insufficient and the delta will retreat [2]. (c) Without sediment the delta is eroded by waves and tides and the shoreline retreats; salt water intrudes further upstream, affecting freshwater supply and wetland ecology [3]. (d) For example periodic sediment sluicing, or a bypass channel that releases an artificial flood peak together with its sediment downstream [2]. Marking: the trapping rate reduces the downstream load; a sediment deficit causes delta retreat.

Section C (Questions 9 to 12, 32 marks)

Show fuller working and explanation, with units in every answer.

#	Question	Answer and marking notes
9	A levee is designed for a 50-year flood. (a) Find its annual exceedance probability. (b) Find the probability of at least one exceedance in 25 years. (c) Find the minimum return period needed to keep the 25-year exceedance probability at or below 10 per cent. (d) Explain why relying only on ever higher levees raises long-term risk.	(a) Annual exceedance probability = 1 / 50 = 2 per cent [2]. (b) Probability = 1 – (1 – 0.02)^25 = 1 – 0.98^25 = 39.7 per cent [2]. (c) From 1 – (1 – 1/T)^25 = 0.10, 1/T = 1 – 0.9^(1/25) = 0.00421, so T = 238 years [2]. (d) Higher protection encourages more investment on the floodplain, so a flood above the design standard causes far greater loss than before; sediment raises the bed and increases the height difference across the levee, so the chance of a breach grows, and zoning and flood storage are needed alongside [2]. Marking: annual exceedance probability is 1/T; the levee effect raises potential losses.
10	A floodplain is divided into three zones: high risk with a 5-year return period covering 2.4 square kilometres, medium risk with a 20-year return period covering 5.6 square kilometres, and low risk with a 100-year return period covering 12 square kilometres. (a) Find the total area of the floodplain. (b) State the annual exceedance probability of each zone. (c) There are 180 households in the high risk zone and relocation costs 350,000 dollars per household; find the total cost. (d) Explain how a flood risk map helps land use planning and emergency management.	(a) Total area = 2.4 + 5.6 + 12 = 20 square kilometres [2]. (b) High risk 1/5 = 20 per cent, medium risk 1/20 = 5 per cent, low risk 1/100 = 1 per cent [2]. (c) Total cost = 180 x 350,000 = 63 million dollars [2]. (d) The map shows the extent and depth of flooding for each return period, so planners can restrict development in high risk zones or require raised floor levels and defences, while emergency services can plan evacuation routes and shelters and concentrate resources where risk is greatest [2]. Marking: annual exceedance probability is 1 divided by the return period; the map is the shared basis for planning and response.
11	(a) Describe how a V-shaped valley, an alluvial fan and a meander are formed, and state the dominant process in each. (b) Explain why a point bar forms on the inner bank of a meander. (c) Explain how a waterfall retreats upstream by erosion. (d) State one environmental change indicated by a river terrace.	(a) A V-shaped valley forms where a steep upper course gives fast flow and vertical downcutting dominates while weathered valley sides collapse and retreat. An alluvial fan forms where a stream leaves a steep slope, velocity and competence drop abruptly and coarse gravel is deposited in a fan. A meander forms on a gentle lower course where lateral erosion and deposition alternate as the channel swings sideways [3]. (b) As flow enters a bend, surface water moves towards the outer bank while slower water near the bed carries sediment towards the inner bank; velocity there is too low to move the load, so it settles layer by layer as a point bar [2]. (c) Plunge pool erosion at the foot of the fall undercuts the cliff, the cap rock loses support and collapses, and the fall migrates upstream leaving a gorge [2]. (d) A terrace shows that the river once eroded and deposited at that height and later cut down again, indicating uplift of the land or a fall in sea level [1]. Marking: landform shape follows the dominant process; a terrace records downcutting and uplift.
12	A wetland covers 2.5 square kilometres and can store floodwater to a depth of 0.32 m. (a) Find the volume of floodwater it can store. (b) If a storm produces 1.6 million cubic metres of floodwater, find the percentage of the flood the wetland can absorb. (c) Explain how a wetland lengthens the lag time and lowers the downstream peak. (d) Explain the effect on the downstream peak of draining the wetland for farmland.	(a) The area is 2.5 million square metres, so storage = 2.5 million x 0.32 = 800,000 cubic metres [2]. (b) The wetland can absorb 0.8 / 1.6 = 50 per cent [2]. (c) Vegetation and peat slow surface flow and store runoff, releasing it gradually, so the peak is flattened and delayed and the instantaneous discharge in the channel falls [2]. (d) Draining removes storage and lag, so runoff reaches the channel at once, the peak is higher and earlier, and downstream flooding becomes more extensive and more frequent [2]. Marking: storage is area times depth; a wetland delays and flattens the peak rather than removing the flood.

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