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HKDSE Geography Paper 1 Data and Graph Analysis Practice Paper (10 Questions, 100 Marks)
HKDSE Geography Paper 1 Data Analysis
HKDSE Geography Paper 1 data and graph analysis practice paper: 10 original questions worth 100 marks in 60 minutes, covering river length, gradient and discharge.
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Paper details Level: S6 (DSE) | Subject: Geography | Time: 60 minutes | Marks: 100 | Questions: 10
Section A Physical Geography Data Analysis (50 marks)
Questions 1-5
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A river is 8 cm long on a map with a scale of 1:25,000. It falls from 120 m to 80 m before reaching a plain. Its width is 6 m, mean depth 0.8 m and mean velocity 1.2 m/s. (a) Calculate the real length of this river section in km. (3 marks) (b) Calculate the gradient of this river section as a percentage. (3 marks) (c) Calculate the discharge (Q = width x depth x velocity). (4 marks) |
Answer: (a) 8 cm x 25,000 = 200,000 cm = 2 km. (b) Drop 40 m / 2,000 m = 0.02 = 2% (1:50). (c) Q = 6 x 0.8 x 1.2 = 5.76 m3/s. Marking notes: show the scale conversion step by step; gradient divides the drop by the horizontal distance, not by the map distance; the unit of discharge must be m3/s. |
| 2 | A new town has a population of 84,000 in an area of 6 km2, with 12,600 people aged below 15 and 9,240 aged 65 or above. In the same year there were 1,050 births and 630 deaths. (a) Calculate the population density in persons per km2. (3 marks) (b) Calculate the dependency ratio to 1 decimal place. (4 marks) (c) Calculate the natural increase rate per 1,000 people to 1 decimal place. (3 marks) |
Answer: (a) 84,000 / 6 = 14,000 persons per km2. (b) (12,600 + 9,240) / (84,000 – 12,600 – 9,240) x 100 = 35.1%. (c) (1,050 – 630) / 84,000 x 1,000 = 5 per 1,000. Marking notes: the denominator of the dependency ratio is the working age population, not the total population; natural increase counts births and deaths only. |
| 3 | A catchment of 15 km2 receives 2,400 mm of rainfall a year. About 35% of the rain becomes runoff, and 80% of that runoff falls in the four wet months. (a) Calculate the total volume of rainfall on the catchment in a year (m3). (3 marks) (b) Calculate the annual runoff volume (m3). (3 marks) (c) Calculate the mean monthly runoff in the wet season (m3). (4 marks) |
Answer: (a) 2.4 m x 15,000,000 m2 = 36,000,000 m3 (3.6 x 10^7). (b) 36,000,000 x 35% = 12,600,000 m3 (1.26 x 10^7). (c) 12,600,000 x 80% / 4 = 2,520,000 m3 (2.52 x 10^6). Marking notes: convert mm to m first (2,400 mm = 2.4 m) and km2 to m2 (15 km2 = 15,000,000 m2); rainfall volume is not the same as runoff volume. |
| 4 | On a 1:20,000 map a slope runs from the 120 m contour to the 220 m contour. The horizontal map distance between the two contours is 4 cm. The vertical scale is 1 cm to 50 m. (a) Calculate the real horizontal distance between the contours (m). (3 marks) (b) Calculate the gradient of the slope as a percentage. (4 marks) (c) Calculate the vertical exaggeration (VE) of the map. (3 marks) |
Answer: (a) 4 cm x 20,000 = 80,000 cm = 800 m. (b) Drop 100 m / 800 m x 100 = 12.5%. (c) Horizontal scale 1:20,000, vertical scale 1:5,000, so VE = 20,000 / 5,000 = 4. Marking notes: gradient uses the horizontal distance as denominator; VE divides the horizontal scale number by the vertical scale number, and a value above 1 means the relief is exaggerated. |
| 5 | In an earthquake the P waves travel at 6.0 km/s and the S waves at 3.5 km/s. At one station the P waves arrive 25 seconds before the S waves. (a) Calculate the distance from the station to the focus (km). (5 marks) (b) Another station is 420 km from the focus; calculate its P-S time lag. (5 marks) |
Answer: (a) Let the distance be d: d / 3.5 – d / 6.0 = 25, so d x (6.0 – 3.5) / 21 = 25 and d = 210 km. (b) 420 / 3.5 – 420 / 6.0 = 120 – 70 = 50 seconds. Marking notes: a bigger lag means a greater distance; both terms divide distance by a wave speed and must use the same unit; state whether the answer is in seconds or minutes. |
Section B Human Geography and Resource Management (50 marks)
Questions 6-10
| # | Question | Answer and marking notes |
|---|---|---|
| 6 | Land in the central business district costs $120,000 per m2 while nearby residential land costs $45,000 per m2. A 30-storey residential block has 8 flats per floor, 3.2 people per flat, a site area of 2,000 m2 and a gross floor area of 24,000 m2. (a) Calculate how many times more expensive the commercial land is to 1 decimal place. (3 marks) (b) Calculate the plot ratio of the residential site. (3 marks) (c) Calculate the population density of the site in persons per hectare. (4 marks) |
Answer: (a) 120,000 / 45,000 = 2.7 times. (b) 24,000 / 2,000 = 12. (c) Population 30 x 8 x 3.2 = 768; site 2,000 m2 = 0.2 ha, so density = 3,840 persons per hectare. Marking notes: plot ratio divides gross floor area by site area; 1 hectare = 10,000 m2 and a conversion error changes the answer by an order of magnitude. |
| 7 | A farm of 12 hectares produces 6.5 tonnes of rice per hectare, sold at $4,200 per tonne. Total production cost for the year is $210,000. (a) Calculate the annual rice output in tonnes. (3 marks) (b) Calculate the annual profit in dollars. (4 marks) (c) Calculate the profit per hectare. (3 marks) |
Answer: (a) 12 x 6.5 = 78 tonnes. (b) Revenue 78 x 4,200 = $327,600; profit = 327,600 – 210,000 = $117,600. (c) 117,600 / 12 = $9,800 per hectare. Marking notes: profit is revenue minus cost, so revenue must be calculated first; profit per hectare divides by area, not by output. |
| 8 | A factory employs 250 workers and produces 18,000 units a month at $85 each. The monthly wage bill is $2,250,000. (a) Calculate the monthly value of output. (3 marks) (b) Calculate labour productivity in units per worker. (3 marks) (c) Calculate the unit labour cost in dollars per unit. (4 marks) |
Answer: (a) 18,000 x 85 = $1,530,000. (b) 18,000 / 250 = 72 units per worker. (c) 2,250,000 / 18,000 = $125 per unit. Marking notes: productivity and unit labour cost are different indicators, the first is better when high and the second when low; never swap numerator and denominator. |
| 9 | A bus route is 12 km long with an average peak speed of 24 km/h; stops and congestion reduce the average speed to 18 km/h. Each bus carries 120 passengers and buses leave every 10 minutes in the peak. Each bus emits 1.2 kg of carbon dioxide per km and the route runs 60 trips per day. (a) Calculate the journey time at the original speed in minutes. (2 marks) (b) Calculate the journey time at 18 km/h and state the delay in minutes. (3 marks) (c) Calculate the peak-hour passenger throughput. (2 marks) (d) Calculate the daily carbon dioxide emissions of the route in kg. (3 marks) |
Answer: (a) 12 / 24 x 60 = 30 minutes. (b) 12 / 18 x 60 = 40 minutes, a delay of 10 minutes. (c) 6 departures x 120 passengers = 720 passengers. (d) 60 x 12 x 1.2 = 864 kg (0.86 tonnes). Marking notes: multiply by 60 to express hours in minutes; departures times passengers gives passenger trips; emissions are trips times km per trip times kg per km. |
| 10 | A beach is 800 m long with a mean usable width of 30 m, and each visitor needs 2 m2. On a peak holiday 18,000 visitors arrive, each producing 0.4 kg of waste, while the daily clean-up capacity is 5,000 kg. (a) Calculate the carrying capacity of the beach in persons. (3 marks) (b) Calculate the percentage by which the peak day exceeds the carrying capacity. (3 marks) (c) Calculate the waste left uncleared that day in kg. (4 marks) |
Answer: (a) 800 x 30 / 2 = 12,000 persons. (b) (18,000 – 12,000) / 12,000 x 100 = 50%. (c) 18,000 x 0.4 = 7,200 kg; 7,200 – 5,000 = 2,200 kg. Marking notes: carrying capacity divides usable area by space per visitor; the overload percentage uses the capacity as denominator; waste needs both generation and treatment capacity. |
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