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Form 5 Geography Practice Paper: Map Reading, Weather and Population (12 questions, 100 marks, 75 minutes)
Form 5 Geography – 12 questions – 100 marks – 75 minutes
Form 5 Geography paper: 12 questions, 100 marks in 75 minutes, covering map scale, gradient, discharge, rainfall, lapse rate, monsoon, pressure gradient, population density and urban growth.
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Paper details Form 5 Geography practice paper with 12 structured questions, full answers and mark scheme (100 marks, 75 minutes).
Section A: Map Reading and the Natural Environment
Answer all parts; show all working and units in calculations.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A map of a district has a scale of 1:20000. (a) Two points are 6.5 cm apart on the map. Find the real distance in km. (b) A lake measures 4 cm² on the map. Find its real area in km². (c) If the same district were redrawn at 1:50000, what fraction of its map area would the lake occupy? Explain. |
(a) Real distance = 6.5 cm x 20000 = 130000 cm = 1.3 km. (b) 1 cm on the map represents 0.2 km, so 1 cm² represents 0.04 km²; the lake covers 4 x 0.04 = 0.16 km². (c) Area scales as the square of the linear scale: the new scale is 0.40 of the old one, so the map area becomes 0.16 of the original, about 16%. A larger denominator means a smaller map, so the same feature covers less paper. |
| 2 | On a 1:25000 map, points A and B are 7.2 cm apart. A is 120 m above sea level and B is 480 m. (a) Find the horizontal distance between A and B in km. (b) Find the average gradient from A to B, expressed both as a percentage and as 1:n. (c) State whether the slope is steep or gentle, and explain the effect on hikers and road design. |
(a) Horizontal distance = 7.2 cm x 25000 = 180000 cm = 1.80 km. (b) Vertical rise = 480 – 120 = 360 m over 1800 m horizontal; gradient = 360/1800 = 0.20, i.e. 20%, or 1:5 (1 m rise for every 5 m horizontal). (c) 20% is a steep slope (15% is the usual threshold). Hikers tire quickly and slip on descents, so roads must zigzag or tunnel, with guard rails and drains added. |
| 3 | A river runs 12.0 cm from source to mouth on a 1:50000 map, and its basin covers 40 cm² on the same map. (a) Find the real length of the river in km. (b) Find the real area of the basin in km². (c) Contours in the valley form V shapes pointing upstream. Explain this, and compare the cross-profiles of the upper and lower valley. |
(a) River length = 12.0 cm x 50000 = 600000 cm = 6.0 km. (b) 1 cm represents 0.5 km, so 1 cm² represents 0.25 km²; basin area = 40 x 0.25 = 10.0 km². (c) The river erodes its bed, so valley-floor contours bend towards higher ground upstream; the V points upstream, and sharper, closer V shapes mean a narrower, deeper valley. The upper valley is a narrow V with steep sides; downstream lateral erosion and deposition widen and flatten it into a broad shallow U with a flood plain. |
| 4 | A river reach is 20 m wide with a mean depth of 3 m and a mean velocity of 1.5 m/s. (a) Calculate the discharge in m³/s. (b) A dam upstream cuts the velocity to 0.9 m/s with the cross-section unchanged. Find the new discharge and express it as a percentage of the original. (c) Explain the effect of the slower flow on the river’s transporting power and on the downstream bed. |
(a) Discharge = width x depth x velocity = 20 x 3 x 1.5 = 90 m³/s. (b) New discharge = 20 x 3 x 0.9 = 54 m³/s, which is 60% (0.60) of the original. (c) Transporting power falls sharply with velocity, so coarser sand and gravel settle first, building bars and shoals. The downstream bed silts up, the flow area shrinks, flood risk rises and the channel needs regular dredging. |
Section B: Weather and Climate
Answer all parts; give numerical answers to the stated precision.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | The table shows monthly rainfall (mm) for a coastal city over six months: January 320, February 410, March 380, April 250, May 120, June 40. (a) Find the total rainfall for the six months. (b) Find the mean monthly rainfall, correct to 0.1 mm. (c) Find the ratio of the wettest month to the driest month. (d) Identify the climate type and give two reasons from the data. |
(a) Total rainfall = 320 + 410 + 380 + 250 + 120 + 40 = 1520 mm. (b) Mean monthly rainfall = 1520 / 6 = 253.3 mm. (c) Wettest is February (410 mm) and driest is June (40 mm); ratio = 410 / 40 = 10.25, about 10.3 times. (d) Tropical monsoon climate: the first three months are very wet while June brings only 40 mm, so there is a clear wet season and dry season, the seasonal rainfall pattern typical of monsoon regions. |
| 6 | Air at the foot of a mountain is 28°C and cools at 0.6°C per 100 m while unsaturated. (a) Find the temperature at the 1800 m summit. (b) A parcel rises from the ground to 2500 m and begins to condense at 1000 m, after which it cools at only 0.5°C per 100 m. Find the temperature at 2500 m. (c) Explain why saturated air cools more slowly and what this means for rainfall on a hillside. |
(a) Cooling = 1800 / 100 x 0.6 = 10.8°C, so the summit is 28 – 10.8 = 17.2°C. (b) At 1000 m the air is 28 – 10 x 0.6 = 22.0°C; a further 1500 m at 0.5°C per 100 m gives 7.5°C of cooling, so at 2500 m the temperature is 14.5°C. (c) Condensation releases latent heat, which offsets part of the cooling by expansion, so the saturated lapse rate (0.5°C per 100 m) is lower than the dry rate (about 1°C per 100 m). Forced ascent keeps condensing, giving heavy rain on the windward slope, while descending air warms and dries on the leeward side to form a rain shadow. |
| 7 | Hong Kong is affected by the northeast monsoon in winter. (a) State the winter wind direction and explain the pressure pattern between the Asian interior and the Pacific. (b) Describe two weather characteristics the winter monsoon brings to Hong Kong. (c) How does the summer monsoon differ in direction and cause, and what is the effect on rainfall? |
(a) The winter wind is northeasterly. The Asian interior (Siberia and Mongolia) cools rapidly and forms a strong cold high, while the warmer Pacific and South China Sea keep lower pressure; air flows down the pressure gradient from land to sea, reaching Hong Kong after crossing warmer water. (b) Dry weather with little cloud and plenty of sunshine; low temperatures, with cold surges bringing spells below 10°C in the city, colder still inland in the New Territories, and occasional frost. (c) The summer monsoon is southwesterly: the heated interior forms a low and the cooler ocean a high, reversing the gradient. It arrives from the sea loaded with moisture, and strong convection plus orographic lifting produce the rainy season, whereas the winter monsoon comes from a dry interior and brings far less rain. |
| 8 | Two stations 150 km apart differ in pressure by 6 hPa. (a) Calculate the pressure gradient in hPa/km. (b) State the initial direction of air movement and how the Coriolis force turns the wind in the northern hemisphere. (c) Compare wind speeds where isobars are closely packed with where they are widely spaced, and explain. |
(a) Pressure gradient = 6 / 150 = 0.040 hPa/km. (b) Air starts moving from high pressure towards low pressure; in the northern hemisphere the Coriolis force deflects it to the right, so the wind blows nearly at right angles to the gradient, anticlockwise around a low. (c) Closely spaced isobars mean a large pressure difference over a short distance, so the pressure gradient force and the wind speed are both greater; widely spaced isobars give light winds. |
Section C: Population and Urban Studies
Answer all parts; support essay answers with data or examples.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | District A has 1,250,000 people in 500 km²; district B has 800,000 people in 400 km². (a) Calculate the population density of each district. (b) How many times denser is district A than district B? (c) Apart from population size, give two factors that affect population density and explain each. |
(a) A = 1,250,000 / 500 = 2,500 people per km²; B = 800,000 / 400 = 2,000 people per km². (b) 2,500 / 2,000 = 1.25, so A is about 1.25 times as dense as B. (c) Relief: flat coastal plains allow housing and farming, so density is high, while steep uplands leave little usable land. Economic activity: business districts with many jobs attract migrants and raise density, whereas remote or farming areas offer few jobs and lose people. Climate, transport access and public facilities also matter. |
| 10 | A place starts the year with 2,000,000 people, a birth rate of 12 per thousand and a death rate of 7 per thousand. (a) Find the rate of natural increase. (b) Ignoring migration, find the population after one year. (c) Net migration is an outflow of 3,000. Find the actual year-end population and explain how an ageing population affects future natural increase. |
(a) Rate of natural increase = 12 – 7 = 5 per thousand. (b) Growth = 2,000,000 x 5/1000 = 10,000; population = 2,000,000 + 10,000 = 2,010,000.0. (c) Year-end population = 2,000,000 + 10,000 – 3,000 = 2,007,000.0. An ageing population shrinks the number of women of childbearing age and lowers fertility, while the death rate rises as elderly people become a larger share, so natural increase falls and may turn negative. |
| 11 | A country of 7,200,000 people sees its urban share rise from 45% to 78%. (a) By how many percentage points did the urban share rise? (b) Find the urban population before and after the change, and the increase. (c) State two common urban problems linked to urbanisation, with one practical measure for each. |
(a) The rise is 78% – 45% = 33 percentage points. (b) Before: 7,200,000 x 45% = 3,240,000; after: 7,200,000 x 78% = 5,616,000; an increase of 2,376,000 people. (c) Housing shortage and squatter settlements, where arrivals outpace housing supply: plan public housing on the urban fringe and upgrade old districts. Traffic congestion and air pollution, from long commutes and concentrated vehicles: extend rail and bus lanes and link stations with walking and cycling routes to cut short trips. |
| 12 | The government plans housing on land at the edge of a country park, which is disputed. (a) Give one main position and reason for each of: the government, nearby residents, and an environmental group. (b) State the root cause of this kind of land-use conflict. (c) Propose one sustainable option that balances housing need with conservation, and justify it. |
(a) Government: in favour, because urban land is scarce, housing demand is pressing and the site already has roads and drains nearby, keeping development costs low. Nearby residents: mostly opposed, fearing construction noise and dust, blocked views and reduced ventilation, and traffic beyond road capacity. Environmental group: opposed, because the site is part of a wildlife corridor; clearing trees fragments habitats, disrupts animal movement and weakens the buffer role of the country park. (b) The root cause is that land is finite while different groups rank the same site’s uses differently: housing is a social need, ecology and landscape are public environmental goods, and residents care about living quality, so all three cannot be fully satisfied at once. (c) A compact-development option: build high density only on the strip nearest the road, keep the remaining woodland and a corridor linking to the country park, and set height, ventilation and greening rules. Concentrating units houses more people on less land, limits habitat fragmentation and, with traffic and drainage assessments, contains the impact on residents. |
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