S5 Geography Paper: Weather And Climate, Rivers And Coasts, Urbanisation, Industrial And Farming Location, Sustainability And Data Analysis

S5 Geography Paper: Weather and Climate, Rivers and Coasts, Urbanisation, Industrial and Farming Location, Sustainability and Data Analysis

2026-11-13

S5 Geography – climate and landforms

S5 Geography paper: 6 questions, 24 parts, 100 marks, 60 minutes on weather, rivers, coasts, urbanisation, industry and sustainability, with worked answers.

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Paper details　S5 Geography practice paper, 6 questions and 24 parts on weather, rivers, coasts, urbanisation and sustainability, with worked answers and marking points.

Section A: Weather and Climate (20 marks)

Use the rainfall data below. Show your working and give suitable units.

#	Question	Answer and marking notes
1	Monthly rainfall at a station from January to December is 25, 32, 48, 125, 290, 380, 350, 320, 230, 65, 38 and 22 mm. Find the annual total and the mean monthly rainfall to 1 decimal place.	Answer: annual total = 25 + 32 + 48 + 125 + 290 + 380 + 350 + 320 + 230 + 65 + 38 + 22 = 1925 mm; mean monthly = 1925 / 12 = 160.4 mm. (Marking: total 3, mean 3)
2	Find the rainfall total of May to September as a percentage of the annual total, to 1 decimal place, and state one effect of this distribution on water resource management.	Answer: May to September totals 290 + 380 + 350 + 320 + 230 = 1570 mm, which is 1570 / 1925 x 100% = 81.6% of the year; reservoirs must store water in the wet season for the dry season and floods must be prevented. (Marking: total 2, percentage 2, effect 2)
3	Identify the wettest and driest months and find the difference between them in mm.	Answer: wettest is June with 380 mm and driest is December with 22 mm, a difference of 380 – 22 = 358 mm. (Marking: two months 2, difference 2)
4	Explain two reasons why rainfall is concentrated in summer in Hong Kong.	Answer: the prevailing southwest monsoon brings warm moist maritime air, and tropical cyclones plus strong convection with orographic lifting produce heavy rain. (Marking: 2 per reason)

Section B: River and Coastal Landforms (16 marks)

Answer the landform questions below and show all working with units.

#	Question	Answer and marking notes
5	A river has its source at 640 m above sea level and its mouth at 0 m, with a channel length of 42 km. Find the average gradient in m per km to 1 decimal place.	Answer: average gradient = 640 / 42 = 15.2 m per km. (Marking: working 2, answer 2)
6	The river has a mean discharge of 12.6 cubic metres per second and a basin area of 168 square km. Find the runoff per unit area in cubic metres per second per square km to 3 decimal places.	Answer: 12.6 / 168 = 0.075 cubic metres per second per square km. (Marking: working 2, answer 2)
7	A coastline 8 km long retreats 1.5 m each year on average. Find the land area lost in ten years in square km to 2 decimal places.	Answer: retreat in ten years is 15 m, so area = 8000 x 15 = 120000 square metres = 0.12 square km. (Marking: retreat 2, conversion and answer 2)
8	A beach 1000 m long and 30 m wide receives 4800 cubic metres of sand each year. Find the average rise of the beach surface per year in metres to 2 decimal places.	Answer: beach area = 1000 x 30 = 30000 square metres, so rise = 4800 / 30000 = 0.16 m. (Marking: area 2, rise 2)

Section C: Urbanisation and Land Use (16 marks)

Use the population and land data of a city below and show your working.

#	Question	Answer and marking notes
9	The population of a city rose from 1.2 million in 1970 to 4.8 million in 2020. Find the percentage increase.	Answer: increase = (4.8 – 1.2) / 1.2 x 100% = 300%. (Marking: working 2, answer 2)
10	In 2020 the city covered 620 square km. Find its population density in persons per square km to the nearest whole number.	Answer: 4800000 / 620 = 7742 persons per square km. (Marking: working 2, answer 2)
11	Built-up area grew from 150 square km in 1970 to 480 square km in 2020. Find the built-up area density in both years and the percentage rise in density to 1 decimal place.	Answer: 1970 density is 1200000 / 150 = 8000 and 2020 density is 4800000 / 480 = 10000 persons per square km, a rise of 25.0%. (Marking: two densities 2, rise 2)
12	Farmland fell from 300 to 180 square km over the same period. Find the percentage decrease and state two effects of urban expansion on farming.	Answer: decrease = (300 – 180) / 300 x 100% = 40%; farmland is replaced by housing and roads, and remaining plots become fragmented and remote so costs rise. (Marking: decrease 2, effects 2)

Section D: Industrial and Farming Location (16 marks)

Answer the location questions below and show all freight and profit calculations.

#	Question	Answer and marking notes
13	A factory turns 3 tonnes of raw material into 1 tonne of product. The market is 100 km from the raw material site and freight costs 2 dollars per tonne per km. Find the total freight cost of locating at the raw material site, at the market and at the midpoint, and state the cheapest site.	Answer: raw material site = 100 x 1 x 2 = 200 dollars; market = 100 x 3 x 2 = 600 dollars; midpoint = 50 x 3 x 2 + 50 x 1 x 2 = 400 dollars. The raw material site is cheapest at 200 dollars. (Marking: three costs 3, site 1)
14	Which type of location tendency does this industry show? Explain your answer.	Answer: it is a raw-material-oriented or weight-losing industry; the raw material is heavier than the product, so locating at the source avoids transporting waste or water and keeps freight lowest. (Marking: type 2, explanation 2)
15	A farm of 40 hectares yields 5 tonnes of rice per hectare. The price is 2400 dollars per tonne and production costs 8500 dollars per hectare. Find total revenue and profit.	Answer: output is 40 x 5 = 200 tonnes, revenue is 200 x 2400 = 480000 dollars, cost is 40 x 8500 = 340000 dollars, so profit is 140000 dollars. (Marking: revenue 2, profit 2)
16	From the question above, find the profit margin to 1 decimal place and suggest one way to raise it.	Answer: profit margin = 140000 / 480000 x 100% = 29.2%; for example use a higher-yielding variety or cut cost per hectare to raise the return per hectare. (Marking: margin 2, method 2)

Section E: Sustainability and Resources (16 marks)

Answer the energy and carbon questions below and show your working.

#	Question	Answer and marking notes
17	A family of four uses 400 units of electricity a month, 45% of it for air conditioning, at 1.2 dollars per unit. Find the monthly air conditioning cost.	Answer: air conditioning uses 400 x 45% = 180 units, so the cost is 180 x 1.2 = 216 dollars. (Marking: units 2, cost 2)
18	Raising the air conditioner setting by one degree cuts its electricity use by 8%. Find the monthly saving in units and in money to 2 decimal places.	Answer: saving is 180 x 8% = 14.4 units, worth 14.4 x 1.2 = 17.28 dollars. (Marking: units 2, money 2)
19	Each unit of electricity emits 0.7 kg of carbon dioxide. Find the monthly and annual carbon emissions of the family, giving the annual figure in tonnes to 2 decimal places.	Answer: monthly 400 x 0.7 = 280 kg; annual 280 x 12 = 3360 kg = 3.36 tonnes. (Marking: monthly 2, annual 2)
20	From the question above, find how many kg of carbon dioxide are saved each year after raising the setting by one degree, to 2 decimal places, and suggest one other way to save energy at home.	Answer: monthly saving is 14.4 x 0.7 = 10.08 kg, so annual saving is 10.08 x 12 = 120.96 kg; other ways include LED lighting or Grade 1 energy label appliances. (Marking: saving 2, method 2)

Section F: Integrated Data Analysis (16 marks)

Use the population, employment rate and income data of three districts below and show your working.

#	Question	Answer and marking notes
21	Three districts have populations of 250000, 180000 and 90000 with employment rates of 62%, 71% and 55%. Find the total number of employed people.	Answer: 250000 x 62% = 155000, 180000 x 71% = 127800 and 90000 x 55% = 49500, giving a total of 332300 people. (Marking: 1 per district, 1 for the total)
22	The three districts together have 520000 people. Find the overall employment rate to 1 decimal place.	Answer: 332300 / 520000 x 100% = 63.9%. (Marking: working 2, answer 2)
23	Average monthly income in the three districts is 18000, 21500 and 14000 dollars. Find the combined average monthly income to the nearest dollar.	Answer: (250000 x 18000 + 180000 x 21500 + 90000 x 14000) / 520000 = 9630000000 / 520000 = 18519 dollars. (Marking: weighted sum 2, answer 2)
24	Compare the third district with the second and give two possible explanations for the differences.	Answer: the third district has both a lower employment rate and lower income, perhaps because its jobs are mainly farming or low-skilled with few choices, and because weaker education and transport make it hard to commute to other districts. (Marking: 2 each)

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