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HKDSE Geography Practice Paper: River Landforms, Flooding and Water Management (12 questions, 100 marks, 75 minutes)
HKDSE Geography: River Landforms, Flooding and Water Management
Original HKDSE Geography practice paper on river landforms, flooding and water management: 12 questions, 100 marks, 75 minutes, with full worked answers and marking points.
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Paper details Level: S6 | Subject: Geography | Time: 75 minutes | Marks: 100 | Questions: 12
Section A (32 marks)
Catchment, gradient and sediment questions: show the formula and the unit.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A catchment covers 120 km2 with a mean annual rainfall of 2,100 mm and a runoff coefficient of 0.45. (a) Calculate the annual runoff volume in cubic metres. (b) Calculate the mean discharge in m3/s (2 d.p.). (c) State where the remaining rainfall goes. |
(a) Runoff = 120,000,000 x 2.1 x 0.45 = 113,400,000 m3. (b) Mean discharge = 113,400,000 / 31,536,000 = 3.60 m3/s. (c) About 55% is lost to evaporation, transpiration, infiltration and soil storage and never becomes surface runoff. |
| 2 | A river falls from 250 m at its source to a gauging station 8 km downstream. (a) Express the average gradient as a ratio. (b) State whether this reach is upper, middle or lower course and justify. (c) Explain how gradient affects river processes. |
(a) Gradient = 250 / 8,000 = 0.03125, that is 1:32. (b) Upper course: steep, fast flow, dominated by vertical erosion, with V-shaped valleys and waterfalls. (c) A steep gradient raises velocity and shear stress, so erosion and transport are stronger and bed material is coarser. |
| 3 | A station records 4,800 tonnes per year of suspended sediment, and bedload is 15% of the total load. (a) Calculate the total load in tonnes per year (whole number). (b) Calculate the bedload. (c) Explain why reservoir design must consider both loads. |
(a) Total load = 4,800 / 0.85 = 5,647 tonnes per year. (b) Bedload = 5,647 – 4,800 = 847 tonnes per year. (c) Bedload builds a delta at the reservoir head while suspended load settles in the basin; both consume storage, so both must be estimated. |
| 4 | A gauging station has an annual exceedance probability of 4% for a flood of a given magnitude. (a) Calculate the return period. (b) Calculate the probability of at least one such flood in 30 years (1 d.p.). (c) Explain whether a 1 in 100 year flood means one flood every 100 years. |
(a) Return period = 1 / 0.04 = 25 years. (b) Probability = 1 – (1 – 0.04)^30 = 70.6%. (c) No. It means a 1% annual exceedance probability; it can happen in any year, and the chance of at least one in 30 years is about 26%. |
Section B (36 marks)
Hydrological and water-quality calculations: show the steps and give a judgement.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | A straight reach has hydraulic radius 1.8 m, water-surface slope 0.0015, Manning n of 0.035 and cross-section 72 m2. (a) Use the Manning equation to find the mean velocity (2 d.p.). (b) Find the discharge (1 d.p.). (c) If dredging lowers n to 0.030, predict the change in velocity and flood peak. |
(a) V = (1/0.035) x 1.8^(2/3) x 0.0015^0.5 = 1.64 m/s. (b) Q = 1.64 x 72 = 117.9 m3/s. (c) Lower roughness raises velocity and discharge, so more water passes at the same stage, but the flood peak arrives earlier and is higher. |
| 6 | A reservoir has a total capacity of 54 million m3 and siltation of 0.36 million m3 per year. (a) Calculate its useful life before it fills with sediment. (b) If upstream afforestation halves siltation, find the new life. (c) Explain the downstream effects of reservoir siltation. |
(a) Life = 54 million / 0.36 million = 150 years. (b) With siltation halved: 54 million / 0.18 million = 300 years. (c) Trapping sediment starves the downstream channel, causing bed incision, delta retreat and coastal erosion. |
| 7 | An irrigation scheme diverts 50 million m3 per year, and crops use 32.5 million m3 effectively. (a) Calculate the irrigation efficiency as a percentage. (b) Calculate the annual losses. (c) Suggest one field measure that raises efficiency and explain why it works. |
(a) Efficiency = 32.5 / 50 x 100% = 65%. (b) Losses = 50 – 32.5 = 17.5 million m3 per year. (c) Drip or sprinkler irrigation cuts seepage and evaporation by delivering water straight to the root zone. |
| 8 | A treatment plant discharges 0.5 m3/s at BOD 250 mg/L into a river flowing at 4.5 m3/s with BOD 3 mg/L. (a) Calculate the mixed BOD (1 d.p.). (b) The standard is 20 mg/L: state whether the river complies. (c) Explain why loads, not just concentrations, must be compared. |
(a) Mixed BOD = (0.5 x 250 + 4.5 x 3) / 5.0 = (125 + 13.5) / 5 = 27.7 mg/L. (b) 27.7 exceeds 20 mg/L, so the river does not comply and further treatment is needed. (c) Concentration must be multiplied by flow to give a load (mass per time) before adding, otherwise the mixed concentration is misestimated. |
Section C (32 marks)
Water demand and catchment management: set out the calculation and explain its geographical meaning.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A city of 480,000 people uses 220 litres per person per day. (a) Calculate the daily domestic demand in m3. (b) With 12% leakage, calculate the water that must be supplied. (c) Suggest one measure to cut per-capita use. |
(a) 480,000 x 220 = 105,600,000 litres = 105,600 m3 per day. (b) Supply = 105,600 / 0.88 = 120,000 m3 per day. (c) Low-flow fittings, tiered tariffs and rainwater harvesting all cut per-capita use. |
| 10 | A 180 mm storm falls on a catchment. The runoff coefficient is 0.30 before urbanisation and 0.62 after. (a) Calculate the runoff depth before and after, in mm. (b) Calculate the ratio after divided by before (2 d.p.). (c) Explain why urbanisation raises the runoff coefficient. |
(a) Before: 180 x 0.30 = 54 mm; after: 180 x 0.62 = 111.6 mm. (b) Ratio = 111.6 / 54 = 2.07. (c) Impermeable surfaces replace vegetation and soil, so infiltration and evapotranspiration fall and rain becomes surface runoff quickly. |
| 11 | A meander reach has a channel length of 14.4 km across a valley floor 9.0 km long. (a) Calculate the sinuosity. (b) Describe the degree of meandering and its effect on velocity. (c) Name the landform most likely to form at the outer bend. |
(a) Sinuosity = 14.4 / 9.0 = 1.6. (b) The reach is meandering; velocity is higher at the outer bank and lower at the inner bank, producing helical flow. (c) A river cliff at the outer bank, with a point bar on the inner bank. |
| 12 | After soil conservation, erosion falls from 1,800 to 720 tonnes per km2 per year in a 45 km2 catchment. (a) Calculate the percentage reduction. (b) Calculate the tonnes of soil saved each year. (c) State two benefits for a downstream reservoir. |
(a) Reduction = (1,800 – 720) / 1,800 x 100% = 60%. (b) Saved = 1,080 x 45 = 48,600 tonnes per year. (c) Less siltation extends reservoir life, and lower turbidity cuts water-treatment costs. |
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