DSE Geography Practice Paper: River Landforms And Flood Management (12 Questions, 50 Marks, 75 Minutes)

DSE Geography Practice Paper: River Landforms and Flood Management (12 Questions, 50 Marks, 75 Minutes)

2026-11-02

DSE Geography: Rivers and Flood Management

DSE Geography paper: 12 questions, 50 marks, 75 minutes on river gradients, discharge, lag time and sediment load, with flood management options.

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Paper details　Level: S6 | Subject: Geography | Time: 75 minutes | Marks: 50 | Questions: 12

Section A (Short Questions)

Each question carries 2 marks. Answer ALL questions with units.

#	Question	Answer and marking notes
1	A river flows from point A (480 m above sea level) to point B (360 m), a horizontal distance of 3.0 km. Find the gradient of this reach as 1 : n.	Drop = 480 – 360 = 120 m; horizontal distance = 3,000 m; gradient = 120 / 3,000 = 0.04, i.e. 1 : 25. [1 mark drop, 1 mark gradient]
2	A drainage basin receives 2,100 mm of rainfall a year and covers 24 km2. Find the annual precipitation volume in cubic metres.	Depth of rainfall = 2.1 m; basin area = 24,000,000 m2; volume = 2.1 x 24,000,000 = 50,400,000 m3 (about 50.4 million m3). [1 mark unit conversion, 1 mark calculation]
3	A channel is 40 m wide, 2.5 m deep on average, with a mean velocity of 1.2 m/s. Find the discharge.	Discharge = width x depth x velocity = 40 x 2.5 x 1.2 = 120 m3/s. [1 mark formula, 1 mark answer with unit]
4	Rainfall in three consecutive hours was 45 mm, 38 mm and 62 mm. Find the total and the mean hourly rainfall (1 d.p.).	Total = 45 + 38 + 62 = 145 mm; mean hourly = 145 / 3 = 48.3 mm. [1 mark total, 1 mark mean]
5	A 5 km2 basin receives 100 mm of rain. The runoff coefficient rises from 0.60 before urbanisation to 0.88 after it. Find the increase in surface runoff volume.	Rainfall volume = 5,000,000 x 0.1 = 500,000 m3; runoff rises from 300,000 m3 to 440,000 m3; the increase is 140,000 m3. [1 mark volume, 1 mark comparison]
6	A storm peak occurs at 14:00 and the flood peak at 18:30. (a) Find the lag time. (b) After urbanisation the lag time falls to 2.5 hours; find the percentage reduction (1 d.p.).	(a) Lag time = 18:30 – 14:00 = 4.5 hours. (b) Reduction = (4.5 – 2.5) / 4.5 x 100% = 44.4%. [1 mark each]

Section B (Data Questions)

Each question carries 4 marks. Show working or cite the given data.

#	Question	Answer and marking notes
7	Over 30 km downstream the mean size of bed load falls from 250 mm to 35 mm. (a) Find the percentage decrease. (b) Explain why particles become smaller and rounder downstream.	(a) (250 – 35) / 250 x 100% = 86.0%. (b) The longer the transport distance, the more collisions with other particles and with the bed, so particles become smaller and rounder; falling velocity lets coarse particles settle first, so downstream deposits are finer and better sorted. [2 marks calculation, 2 marks explanation]
8	A river carries 4.5 million tonnes of sediment a year: 12% bed load, 78% suspended load and 10% dissolved load. (a) Find the mass of the suspended load. (b) Explain how an upstream reservoir affects the sediment load downstream.	(a) 4.5 million x 78% = 3.51 million tonnes. (b) The reservoir slows the flow, so coarse sediment settles on the reservoir floor and the released water is sediment-poor; downstream deposition weakens, the channel may incise, and the delta retreats for lack of sediment. [2 marks calculation, 2 marks explanation]
9	Before a dam the 10-year flood peak was 850 m3/s; after the dam the same return period gives 600 m3/s. (a) Find the percentage of peak reduction. (b) State two adverse effects of the dam on the downstream channel.	(a) (850 – 600) / 850 x 100% = 29.4%. (b) Clear water released from the dam scours and deepens the downstream bed, threatening bridge piers and embankment foundations; the delta and coast lose sediment supply and are eroded. [2 marks calculation, 2 marks effects]

Section C (Case Study and Essay Questions)

Follow the instructions and support your points with data or concepts.

#	Question	Answer and marking notes
10	The middle course of the Yellow River crosses the Loess Plateau and carries about 1.6 billion tonnes of sediment a year, of which about 400 million tonnes is deposited on the lower bed. (a) State two natural causes of soil erosion there. (b) State two human causes. (c) Find the percentage deposited on the lower bed and explain the formation of a “perched river”. (d) State one major risk to riverside residents.	(a) Natural: the loess is loose with vertical joints and is easily washed away; summer rainstorms are intense, giving strong slope runoff. (b) Human: historical over-cultivation and overgrazing removed vegetation; road building and mining left loose spoil. (c) 4 / 16 x 100% = 25%. On the flat lower plain the velocity drops sharply, so sediment settles and raises the bed; embankments are then raised for flood defence until the bed stands above the surrounding land. (d) If a levee breaches, floodwater cannot return to the channel and spreads over the plain, sometimes shifting the channel. [(a) 2, (b) 2, (c) 3, (d) 1 marks]
11	A basin compares two flood schemes. Scheme A (hard engineering): 120 km of raised embankments and 3 flood storage areas, costing HK$1.2 billion and protecting 450,000 people. Scheme B (soft engineering): 5,000 hectares of afforestation and land-use planning, costing HK$300 million and protecting 180,000 people. (a) Calculate the cost per person protected for each scheme. (b) Compare the two schemes on flood effectiveness and sustainability (two points each).	(a) Scheme A: 1.2 billion / 450,000 = about 2,667 per person; Scheme B: 300 million / 180,000 = about 1,667 per person. (b) Scheme A protects more people and cuts peaks immediately, but costs more, needs constant maintenance and may encourage further development on the floodplain; Scheme B is cheaper and raises infiltration and storage, so it is more sustainable, but works slowly and protects a smaller area. [1 mark per cost, 2 marks comparison each, 1 mark conclusion]
12	A city now has a 100-year flood peak of 1,200 m3/s. Experts expect short-duration rainfall intensity to rise by 20% under climate change while the runoff coefficient rises from 0.55 to 0.70. (a) Estimate the new peak discharge and its percentage increase. (b) Rank three responses – raising embankments, restoring floodplains and wetlands, restricting development in the basin – and justify the order.	(a) New peak = 1,200 x (0.70 / 0.55) x 1.20 = about 1,833 m3/s, an increase of about 52.7%. (b) Restoring floodplains and wetlands first, then restricting development, then raising embankments. Floodplain storage is cheap, adds detention capacity and improves ecology; land-use control attacks the runoff coefficient at source; raising embankments only lifts the defence level without reducing the peak and locks the city into ever-higher walls and accumulating risk. [4 marks estimate, 5 marks ranking with reasons]

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