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Form 5 Geography Practice Paper: Map Reading, Weather and Climate, Rivers, Population and Sustainability (12 Questions, 100 Marks, 75 Minutes)
Form 5 Geography – 12 questions – 100 marks – 75 minutes
Form 5 Geography paper: 12 questions, 100 marks, 75 minutes on map scale, lapse rate, river discharge, population density, land use, material index and water use.
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Paper details Form 5 Geography practice paper with 12 structured questions on map scale, lapse rate, river discharge, runoff, population density, land use, material index and water consumption, with full worked answers.
Section A: Map Reading, Weather and Climate (Q1-Q4, 34 marks)
Answer all questions; show the formula, substitution and unit for every calculation.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | 1. On a 1:20 000 topographical map the straight-line distance between points A and B is 6.5 cm, and a rectangular field measures 4.0 cm by 3.0 cm. (a) Find the actual distance between A and B in km. (2 marks) (b) Find the actual area of the field in km2. (3 marks) (c) State what 1 cm represents on this map, and explain why city streets are hard to show one by one at this map scale. (4 marks) | Answer: (a) 6.5 x 20 000 = 130 000 cm = 1.3 km. (b) 4.0 x 20 000 = 80 000 cm = 0.8 km and 3.0 x 20 000 = 60 000 cm = 0.6 km, so the area is 0.8 x 0.6 = 0.48 km2. (c) 1 cm represents 20 000 cm, that is 0.2 km (200 m). A street 10-20 m wide is less than 0.1 cm on the map, so it cannot be drawn to scale and is shown by a single line or symbol. |
| 2 | 2. At the foot of a hill (100 m above sea level) the air temperature is 28.0 C; the summit is 1 900 m above sea level. The mean lapse rate is 6.5 C per 1 000 m of ascent. (a) Find the height difference between the foot and the summit. (2 marks) (b) Find the air temperature at the summit. (3 marks) (c) Recalculate the summit temperature if the lapse rate is 6.0 C per 1 000 m, and explain how falling temperature with height affects natural vegetation above 1 500 m. (4 marks) | Answer: (a) Height difference = 1 900 – 100 = 1 800 m. (b) Drop = 1 800 / 1 000 x 6.5 = 11.7 C, so the summit is 28.0 – 11.7 = 16.3 C. (c) At 6.0 C per 1 000 m the drop is 10.8 C and the summit is 17.2 C; above 1 500 m the short growing season, strong wind and low temperature stop tree growth, so forest gives way to scrub, grassland or bare rock. |
| 3 | 3. In a river reach the channel is 12 m wide, 1.5 m deep on average, with a mean velocity of 0.8 m/s. After a storm the same reach is 14 m wide and 2.0 m deep on average, with a mean velocity of 1.5 m/s. (a) Find the cross-sectional area of the channel in normal flow. (2 marks) (b) Find the discharge in m3/s before and after the storm. (3 marks) (c) Find how many times the storm discharge is greater, and explain the effect of such a surge downstream. (3 marks) | Answer: (a) Cross-sectional area = 12 x 1.5 = 18 m2. (b) Normal discharge = 18 x 0.8 = 14.4 m3/s; after the storm the area is 14 x 2.0 = 28 m2 and the discharge is 28 x 1.5 = 42 m3/s. (c) 42 / 14.4 = 2.92 times. Downstream the level and velocity rise sharply, bank erosion and overbank flooding become likely, and a heavy load of sediment is deposited on the floodplain. |
| 4 | 4. A catchment of 25 km2 receives 2 400 mm of rain a year and has a runoff coefficient of 0.35. (a) Find the total volume of rainfall received in a year in m3. (3 marks) (b) Find the annual runoff volume in m3. (3 marks) (c) Explain how urbanisation changes the runoff coefficient of the same catchment. (2 marks) | Answer: (a) 2 400 mm = 2.4 m, so the volume is 2.4 x 25 x 106 = 6.0 x 107 m3. (b) Runoff = 6.0 x 107 x 0.35 = 2.1 x 107 m3. (c) Concrete and asphalt seal the surface and drains carry rain away quickly, so infiltration and evaporation fall and the runoff coefficient rises. |
Section B: Population, City and Industry (Q5-Q8, 33 marks)
Data: 1 km = 100 000 cm and 1 km2 = 100 hectares; lapse rates are given per 1 000 m of ascent.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | 5. A district has 480 000 people in 30 km2, of whom 72 000 are aged 0-14 and 48 000 are aged 65 or above. (a) Find the population density in persons per km2. (2 marks) (b) Find the number of people aged 15-64. (2 marks) (c) Find the dependency ratio, and explain the pressure a high ratio puts on public services. (4 marks) | Answer: (a) 480 000 / 30 = 16 000 persons per km2. (b) Aged 15-64 = 480 000 – 72 000 – 48 000 = 360 000. (c) Dependency ratio = (72 000 + 48 000) / 360 000 x 100% = 33.33%. With fewer workers per dependant the district needs more schools, clinics, elderly care and hospital beds while the tax base narrows. |
| 6 | 6. A new town covers 5.0 km2 with this land use: residential 55%, roads 18%, open space 12%, commercial 8% and others 7%. It houses 120 000 people. (a) Find the area of residential land and of open space in km2. (3 marks) (b) Find the population density of the whole town and within residential land only, in persons per km2. (3 marks) (c) Explain why about a fifth of the land is still kept for roads and open space. (2 marks) | Answer: (a) Residential = 5.0 x 0.55 = 2.75 km2 and open space = 5.0 x 0.12 = 0.6 km2 (roads 0.9 km2, commercial 0.4 km2, others 0.35 km2, total 5.0 km2). (b) Whole town = 120 000 / 5.0 = 24 000 persons per km2; within residential land = 120 000 / 2.75 = about 43 600 persons per km2. (c) Roads carry commuting traffic and emergency vehicles and hold water, power and telecom pipes, while open space gives ventilation corridors, eases the heat island effect and serves daily recreation. |
| 7 | 7. A steel plant uses 4.8 t of iron ore and 0.9 t of coal to make 1.5 t of steel. (a) Find the total weight of raw materials used. (3 marks) (b) Calculate the material index. (3 marks) (c) Using your answer in (b), state the locational pull of the plant and explain why. (3 marks) | Answer: (a) Total raw material = 4.8 + 0.9 = 5.7 t. (b) Material index = 5.7 / 1.5 = 3.8. (c) The index is above 1 and the ore loses most of its weight in smelting, so the plant is raw-material oriented and locates near the ore field to cut transport cost; coal is used in a much smaller amount, so it pulls less. |
| 8 | 8. A region emits 45 Mt CO2e a year and has 7.5 million people. The government plans to cut the carbon footprint per person by 30% in ten years. (a) Find the present emission per person in t. (3 marks) (b) Find the emission per person and the total emission after the cut. (3 marks) (c) Suggest two measures that can cut emissions from the transport sector. (2 marks) | Answer: (a) 45 Mt / 7.5 million = 45 x 106 t / 7.5 x 106 people = 6 t per person. (b) After the cut, 6 x 0.70 = 4.2 t per person and 45 x 0.70 = 31.5 Mt in total. (c) For example: expand mass transit and raise service frequency, and replace petrol cars with electric vehicles backed by a charging network. |
Section C: Landforms, Farming and Sustainability (Q9-Q12, 33 marks)
Write explanations in full sentences and support each point with calculated values or geographical concepts.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | 9. Two points on a hillside are 50 m apart vertically and 250 m apart horizontally. (a) Express the slope of this section as a percentage. (3 marks) (b) Find the angle of the slope in degrees, to one decimal place. (2 marks) (c) Explain how slope affects the rate of soil erosion there. (3 marks) | Answer: (a) Slope = 50 / 250 x 100% = 20%. (b) Angle = tan-1(50 / 250) = 11.31 degrees. (c) On a steeper slope rain runs off faster and infiltrates less, so splash and sheetwash detach and remove more topsoil; the surface is quickly cut into gullies. |
| 10 | 10. Farm A covers 12 hectares and produces 84 t of rice a year; Farm B covers 8 hectares and produces 48 t a year. (a) Find the yield per hectare of each farm in t/ha. (2 marks) (b) Find by what percentage the yield of Farm A is higher. (3 marks) (c) Suggest two natural factors that explain the difference in yield. (3 marks) | Answer: (a) Farm A = 84 / 12 = 7 t/ha; Farm B = 48 / 8 = 6 t/ha. (b) (7 – 6) / 6 x 100% = 16.67%. (c) For example: soil depth and fertility, rainfall and length of growing season, slope and sunshine (any two). |
| 11 | 11. GPS stations on opposite sides of a fault have moved 0.36 m closer over 10 years. (a) Find the average rate of approach in mm per year. (3 marks) (b) At this rate, find the time needed for a further 500 km of approach, in years, in scientific notation. (3 marks) (c) Describe the earthquakes and landforms typical of this kind of plate boundary. (2 marks) | Answer: (a) 0.36 m = 360 mm, so the rate is 360 / 10 = 36 mm per year. (b) 500 km = 5.0 x 108 mm, so the time is 5.0 x 108 / 36 = 1.39 x 107 years. (c) Two plates converge at a destructive boundary, giving shallow to intermediate earthquakes, fold mountains, a deep-sea trench and a volcanic arc. |
| 12 | 12. A city of 7.5 million people uses 220 L of water per person per day. (a) Find the total daily water use in m3. (3 marks) (b) Find the total annual water consumption in m3. (3 marks) (c) If 20% of the supply comes from desalination, find the average daily output the plant must produce in m3 per day. (3 marks) | Answer: (a) Daily use = 220 x 7.5 x 106 = 1.65 x 109 L = 1.65 x 106 m3. (b) Annual water consumption = 1.65 x 106 x 365 = 6.0225 x 108 m3, about 602 million m3. (c) Desalinated water = 6.0225 x 108 x 0.20 = 1.2045 x 108 m3 a year, so the plant must produce 1.2045 x 108 / 365 = 3.3 x 105 m3 per day, about 330 000 m3 per day. |
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