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Secondary 2 Geography Practice Paper 5

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TuitionGoWhere Practice Paper - Geography Secondary 2

TuitionGoWhere Practice Paper (AI) — Version 5

Subject: Geography
Level: Secondary 2 (G2/G3)
Paper: Practice Paper — Map, Graph & Data Skills
Duration: 1 hour 15 minutes
Total Marks: 50

Name: ________________________
Class: ________________________
Date: ________________________

Instructions to Candidates

Answer all questions.
Write your answers in the spaces provided.
The number of marks is given in brackets [ ] at the end of each question or part question.
The total number of marks for this paper is 50.
You may use a calculator.
For map-based questions, refer to the map extract provided in the image placeholder.
Write clearly and legibly.

Section A: Map Reading Skills [15 marks]

Answer all questions in this section.

Question 1

Study the topographic map extract of Area X (scale 1:25,000) provided below.

<image_placeholder> id: Q1-fig1 type: map linked_question: Q1 description: Topographic map extract (1:25,000) showing a coastal area with contour lines, a river, a main road, a hospital, a school, a quarry, and a forested area. Grid lines are labelled with eastings (20–29) and northings (40–49). Key features: Hospital at grid square 2445, School at 2643, Quarry at 2247, River flowing from northwest to southeast, Contour interval 20 m, Spot heights at 120 m (2346) and 80 m (2742). labels: Eastings (20–29), Northings (40–49), Contour lines (20 m interval), Spot heights, River, Main road, Hospital (H), School (Sch), Quarry (Q), Forest (tree symbols), Built-up area (grey shading) values: Scale 1:25,000, Contour interval 20 m, Spot height 120 m at 2346, Spot height 80 m at 2742 must_show: Grid lines with labels, all named features, contour lines with values, scale bar, north arrow </image_placeholder>

(a) State the four-figure grid reference of the Hospital.
[1]

(b) State the six-figure grid reference of the School.
[1]

(c) The Quarry is located at six-figure grid reference 224476. Describe the relief of the land around the Quarry using evidence from the map.
[2]

(d) Calculate the straight-line distance in kilometres between the Hospital and the School.
[2]

Question 2

The map extract shows a river flowing from the northwest to the southeast.

(a) Identify the direction of flow of the river. Explain how you can tell this from the map.
[2]

(b) Measure the length of the river within the map extract (from the northern edge to the southern edge) in kilometres.
[2]

Question 3

(a) What is the contour interval of this map?
[1]

(b) Calculate the average gradient of the slope between spot height 120 m (at 2346) and spot height 80 m (at 2742). Express your answer as a ratio in the form 1 : n.
[3]

(c) A student says: "The land is steeper in the northwest than in the southeast." Using map evidence, state whether you agree or disagree and explain your reasoning.
[2]

Section B: Graph and Data Interpretation [20 marks]

Answer all questions in this section.

Question 4

The table below shows the monthly rainfall (mm) and average temperature (°C) for Station A in 2023.

Month	Jan	Feb	Mar	Apr	May	Jun	Jul	Aug	Sep	Oct	Nov	Dec
Rainfall (mm)	210	150	180	220	190	160	155	170	185	230	250	240
Temperature (°C)	26.5	27.0	27.5	28.0	28.5	28.2	28.0	28.1	27.8	27.5	27.0	26.8

<image_placeholder> id: Q4-fig1 type: graph linked_question: Q4 description: Climate graph for Station A (2023) with dual axes: left axis for rainfall (mm, 0–300), right axis for temperature (°C, 26–29). Bars for monthly rainfall, line for monthly temperature. Months on x-axis (Jan–Dec). labels: Months (Jan–Dec), Rainfall (mm), Temperature (°C), Title: "Climate Graph for Station A, 2023" values: Rainfall: Jan 210, Feb 150, Mar 180, Apr 220, May 190, Jun 160, Jul 155, Aug 170, Sep 185, Oct 230, Nov 250, Dec 240; Temperature: Jan 26.5, Feb 27.0, Mar 27.5, Apr 28.0, May 28.5, Jun 28.2, Jul 28.0, Aug 28.1, Sep 27.8, Oct 27.5, Nov 27.0, Dec 26.8 must_show: Dual-axis climate graph, bars for rainfall, line for temperature, labelled axes, title, legend </image_placeholder>

(a) Which month had the highest rainfall?
[1]

(b) Calculate the annual range of temperature for Station A in 2023.
[1]

(c) Describe the relationship between monthly rainfall and temperature shown in the data. Support your answer with two pieces of evidence from the table.
[3]

(d) Suggest one reason why Station A might experience its highest rainfall in the month you identified in (a).
[2]

Question 5

The divided bar graph below shows the percentage distribution of water sources for four cities (P, Q, R, S) in 2023.

<image_placeholder> id: Q5-fig1 type: chart linked_question: Q5 description: Divided bar graph (100% stacked bar chart) showing water sources for four cities: P, Q, R, S. Four segments per bar: Local Catchment (blue), Imported Water (orange), NEWater (green), Desalinated Water (red). City P: 40% Local, 30% Imported, 20% NEWater, 10% Desalinated. City Q: 10% Local, 50% Imported, 25% NEWater, 15% Desalinated. City R: 60% Local, 10% Imported, 20% NEWater, 10% Desalinated. City S: 5% Local, 20% Imported, 35% NEWater, 40% Desalinated. labels: Cities (P, Q, R, S), Water Sources (Local Catchment, Imported Water, NEWater, Desalinated Water), Percentage (%), Title: "Water Sources for Four Cities, 2023" values: City P: Local 40%, Imported 30%, NEWater 20%, Desalinated 10%; City Q: Local 10%, Imported 50%, NEWater 25%, Desalinated 15%; City R: Local 60%, Imported 10%, NEWater 20%, Desalinated 10%; City S: Local 5%, Imported 20%, NEWater 35%, Desalinated 40% must_show: 100% stacked bar chart, four cities, four coloured segments per bar, labelled axes, legend, percentage values on segments or axis </image_placeholder>

(a) Which city is most dependent on imported water?
[1]

(b) Calculate the combined percentage of NEWater and Desalinated Water for City S.
[1]

(c) Compare the water supply strategies of City P and City S. Use data from the graph to support your answer.
[3]

(d) Explain one advantage and one disadvantage of a high dependence on desalinated water.
[3]

Question 6

The line graph below shows the population growth (in millions) of Country X from 1990 to 2020.

<image_placeholder> id: Q6-fig1 type: graph linked_question: Q6 description: Line graph showing population of Country X (millions) from 1990 to 2020. X-axis: Year (1990–2020, 5-year intervals). Y-axis: Population (millions, 0–50). Data points: 1990: 18.2, 1995: 21.5, 2000: 25.8, 2005: 31.2, 2010: 37.5, 2015: 43.8, 2020: 48.6. Steady upward curve. labels: Year, Population (millions), Title: "Population Growth of Country X, 1990–2020" values: 1990: 18.2M, 1995: 21.5M, 2000: 25.8M, 2005: 31.2M, 2010: 37.5M, 2015: 43.8M, 2020: 48.6M must_show: Line graph with labelled axes, data points marked, title, clear scale </image_placeholder>

(a) What was the population of Country X in 2005?
[1]

(b) Calculate the increase in population between 2000 and 2010.
[1]

(c) Describe the trend in population growth from 1990 to 2020.
[2]

(d) Calculate the average annual population growth rate (in millions per year) between 1990 and 2020.
[2]

(e) Suggest two factors that could explain the population trend shown.
[2]

Question 7

The scatter graph below shows the relationship between GDP per capita (US$) and access to improved water supply (%) for 10 countries.

<image_placeholder> id: Q7-fig1 type: graph linked_question: Q7 description: Scatter graph with GDP per capita (US $, log scale: 1,000–100,000) on x-axis and Access to Improved Water Supply (%) on y-axis (0–100%). 10 data points showing positive correlation. Countries labelled A–J. Trend line upward. labels: GDP per capita (US$ ), Access to Improved Water Supply (%), Country labels (A–J), Title: "GDP per Capita vs Access to Improved Water Supply, 2022" values: Approximate: A (1,200, 55%), B (2,500, 65%), C (5,000, 78%), D (8,000, 85%), E (12,000, 92%), F (20,000, 96%), G (35,000, 98%), H (50,000, 99%), I (70,000, 100%), J (90,000, 100%) must_show: Scatter plot with log x-axis, linear y-axis, 10 labelled points, trend line, axis labels, title </image_placeholder>

(a) Describe the general relationship between GDP per capita and access to improved water supply.
[2]

(b) Country D has a GDP per capita of US$8,000. Estimate its access to improved water supply from the graph.
[1]

(c) Identify one anomaly (outlier) in the data, if any, and suggest a possible reason for it.
[2]

Section C: Data Skills and Geographical Investigation [15 marks]

Answer all questions in this section.

Question 8

A group of Secondary 2 students conducted a fieldwork investigation on traffic flow at a busy junction near their school. They counted vehicles passing through the junction in 15-minute intervals from 7:00 am to 9:00 am on a weekday.

The table below shows their raw data.

Time Interval	Cars	Motorcycles	Buses	Lorries	Total
7:00–7:15	120	45	8	12	185
7:15–7:30	155	60	10	15	240
7:30–7:45	180	72	12	18	282
7:45–8:00	210	85	15	20	330
8:00–8:15	195	78	14	18	305
8:15–8:30	165	65	10	14	254
8:30–8:45	140	52	8	10	210
8:45–9:00	115	42	6	8	171

(a) Complete the Total column for the 7:30–7:45 interval. (The table above shows 282; verify and write the correct total if different.)
[1]

(b) Which 15-minute interval had the highest total vehicle count?
[1]

(c) Calculate the percentage of motorcycles out of the total vehicles for the 7:45–8:00 interval. Round to one decimal place.
[2]

(d) The students want to present the change in total traffic flow over time. Recommend the most suitable type of graph for this purpose and explain why it is appropriate.
[2]

(e) The students concluded: "Traffic congestion is worst at 7:45–8:00 am because that interval has the highest vehicle count." Evaluate this conclusion. State one strength and one limitation of using only vehicle count to measure congestion.
[3]

Question 9

The pie chart below shows the land use distribution of Town Y in 2023.

<image_placeholder> id: Q9-fig1 type: chart linked_question: Q9 description: Pie chart showing land use of Town Y, 2023. Segments: Residential 42%, Commercial 18%, Industrial 15%, Transport 10%, Parks & Recreation 8%, Institutional 7%. Total area = 250 km². labels: Land use categories (Residential, Commercial, Industrial, Transport, Parks & Recreation, Institutional), Percentages (%), Title: "Land Use Distribution of Town Y, 2023" values: Residential 42%, Commercial 18%, Industrial 15%, Transport 10%, Parks & Recreation 8%, Institutional 7%; Total area 250 km² must_show: Pie chart with 6 labelled segments, percentages, title, legend or direct labels </image_placeholder>

(a) Calculate the area (in km²) used for Residential purposes.
[1]

(b) The town planners want to increase Parks & Recreation land to 12% of the total area. How many additional km² of land would be needed?
[2]

(c) Suggest one possible conflict that could arise from converting existing land use to parks.
[2]

Question 10

A student is investigating the relationship between distance from the city centre and building height in a city. She collects data from 8 locations along a transect line.

Distance from City Centre (km)	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0
Average Building Height (storeys)	45	38	30	22	15	10	6	4

(a) Plot the data on the scatter graph axes provided below. The first two points have been plotted for you.

<image_placeholder> id: Q10-fig1 type: graph linked_question: Q10 description: Scatter graph axes for Distance from City Centre (km) vs Average Building Height (storeys). X-axis: 0–4.5 km. Y-axis: 0–50 storeys. Points at (0.5, 45) and (1.0, 38) pre-plotted. Student to plot remaining 6 points. labels: Distance from City Centre (km), Average Building Height (storeys), Title: "Building Height vs Distance from City Centre" values: (0.5, 45), (1.0, 38), (1.5, 30), (2.0, 22), (2.5, 15), (3.0, 10), (3.5, 6), (4.0, 4) must_show: Labelled axes with scales, two pre-plotted points, space for six more points, title </image_placeholder>

(b) Draw a line of best fit through the points.
[1]

(c) Describe the relationship shown by the graph.
[2]

(d) Use your graph to estimate the average building height at 2.8 km from the city centre.
[1]

(e) Explain one geographical reason for this relationship.
[2]

End of Paper

Answers

TuitionGoWhere Practice Paper - Geography Secondary 2 (Answer Key)

TuitionGoWhere Practice Paper (AI) — Version 5
Subject: Geography
Level: Secondary 2 (G2/G3)
Paper: Practice Paper — Map, Graph & Data Skills
Total Marks: 50

Section A: Map Reading Skills [15 marks]

Question 1

(a) 2445
[1]
Marking note: Four-figure grid reference: easting (24) first, then northing (45). Common error: reversing order (4524) or giving six figures.

(b) 263432 (or 263433 depending on exact position within grid square)
[1]
Marking note: Six-figure grid reference: easting 26 + 3 tenths = 263; northing 43 + 2/3 tenths = 432/433. Accept 263432 or 263433. Must have 6 digits. Common error: four-figure only (2643) or wrong subdivision.

(c) The land around the Quarry (224476) is hilly with steep slopes.
Evidence: Contour lines are closely spaced around the quarry, and the spot height at 2346 (120 m) nearby indicates high ground. The contour lines form a V-shape pointing uphill near the river, suggesting a valley, but around the quarry they are concentric and tight, indicating a hill/knoll.
[2]
Mark breakdown: 1 mark for "steep slopes / hilly / high relief"; 1 mark for map evidence (closely spaced contours / spot height 120 m).
Common mistake: Saying "flat" because quarry is dug out — the map shows natural relief before excavation.

(d) 1.0 km (accept 0.95–1.05 km)
[2]
Working:

Hospital at 2445 → centre of grid square ≈ 245455 (easting 24.5, northing 45.5)
School at 2643 → centre ≈ 265435 (easting 26.5, northing 43.5)
Difference in eastings: 2.0 km (2 grid squares × 1 km)
Difference in northings: 2.0 km (2 grid squares × 1 km)
Straight-line distance = √(2² + 2²) = √8 ≈ 2.828 km on ground? Wait — scale is 1:25,000, so 1 cm = 0.25 km. Grid squares are 1 km × 1 km (since 4 cm = 1 km at 1:25,000? No: 1:25,000 → 1 cm = 250 m = 0.25 km. Standard topographic map grid squares are 1 km × 1 km, so 4 cm per grid square. Yes.)
Hospital (24,45) to School (26,43): ΔE = 2 km, ΔN = 2 km
Straight-line distance = √(2² + 2²) = √8 ≈ 2.83 km
Correction: Earlier draft said 1.0 km — that was wrong. Correct answer is 2.83 km (accept 2.8–2.9 km).
[2]
Mark breakdown: 1 mark for correct method (Pythagoras / measuring on map), 1 mark for correct answer with unit (km).
Common mistake: Measuring map distance in cm and forgetting to convert using scale.

Question 2

(a) The river flows from northwest to southeast.
Evidence: Contour lines cross the river in a V-shape pointing upstream (towards northwest). The land is higher in the northwest (spot height 120 m at 2346) and lower in the southeast. Rivers flow from higher to lower ground.
[2]
Mark breakdown: 1 mark for correct direction (NW to SE); 1 mark for valid map evidence (V-shaped contours pointing upstream / spot heights decreasing / contour values decreasing downstream).

(b) Approximately 4.5 km (accept 4.0–5.0 km)
[2]
Method: Use a piece of paper/string to trace the river's meandering path along the map, then measure against the linear scale (1:25,000). The river enters at north edge (~2349) and exits at south edge (~2541), winding ~18 cm on map → 18 × 0.25 = 4.5 km.
Mark breakdown: 1 mark for correct method (string/paper along curves); 1 mark for reasonable answer in km.

Question 3

(a) 20 metres
[1]
Direct from map legend.

(b) 1 : 500
[3]
Working:

Vertical difference (rise) = 120 m – 80 m = 40 m
Horizontal distance (run): Spot height 120 m at 2346, spot height 80 m at 2742.
ΔEasting = 4 grid squares = 4 km = 4,000 m
ΔNorthing = 4 grid squares = 4 km = 4,000 m
Straight-line horizontal distance = √(4000² + 4000²) = √32,000,000 ≈ 5,657 m
Gradient = Rise / Run = 40 / 5,657 ≈ 1 / 141.4 → 1 : 141 (approx)
Wait — gradient is usually measured along the slope line, not straight-line horizontal? In geography, gradient = vertical rise / horizontal distance (along the slope). But if we don't have slope distance, we use straight-line horizontal distance between points as approximation.
Standard school method: Gradient = Vertical Interval / Horizontal Distance (straight-line between points).
So: VI = 40 m, HD = 5,657 m → Gradient = 40 : 5,657 = 1 : 141.4 ≈ 1 : 141
But some syllabi use "horizontal distance along the slope" — but we don't have that. So straight-line horizontal distance is acceptable.
Alternative: If they measure map distance between points: 5.66 cm × 0.25 km/cm = 1.415 km = 1,415 m? No — grid squares are 1 km, so 4 E × 4 N → diagonal = 5.66 km = 5,660 m. Yes.
Answer: 1 : 141 (accept 1 : 140 to 1 : 145)
[3]
Mark breakdown: 1 mark for correct VI (40 m); 1 mark for correct HD calculation (5,657 m or ~5.66 km); 1 mark for correct ratio 1 : n.

(c) Agree. The contour lines are more closely spaced in the northwest (e.g., around 2346, multiple contours within one grid square) compared to the southeast (wider spacing, fewer contours). Closely spaced contours indicate steeper slopes. Spot height 120 m in NW vs lower ground in SE confirms higher relief in NW.
[2]
Mark breakdown: 1 mark for Agree/Disagree with correct choice (Agree); 1 mark for evidence (contour spacing / spot heights).

Section B: Graph and Data Interpretation [20 marks]

Question 4

(a) November (250 mm)
[1]

(b) 2.0 °C (28.5 °C – 26.5 °C)
[1]
Marking note: Annual range = highest monthly temp – lowest monthly temp. May (28.5) – Jan (26.5) = 2.0 °C.

(c) Inverse relationship: Generally, months with higher temperature have lower rainfall, and vice versa.
Evidence 1: May has the highest temperature (28.5 °C) but relatively low rainfall (190 mm).
Evidence 2: November has the highest rainfall (250 mm) but lower temperature (27.0 °C).
[3]
Mark breakdown: 1 mark for identifying inverse/negative relationship; 2 marks for two correct data pairs with months and values.
Common mistake: Saying "positive correlation" or using only one data point.

(d) November is during the Northeast Monsoon season (Dec–Mar), when moisture-laden winds from the South China Sea blow over Singapore, bringing prolonged widespread rain.
[2]
Mark breakdown: 1 mark for identifying monsoon / seasonal wind; 1 mark for mechanism (moisture + uplift → rain).
Alternative acceptable: Convergence of winds / ITCZ position / tropical cyclones nearby.

Question 5

(a) City Q (50% Imported Water)
[1]

(b) 75% (35% NEWater + 40% Desalinated)
[1]

(c) City P relies more on local sources (40% Local Catchment, only 30% Imported), while City S depends heavily on technology-intensive sources (35% NEWater, 40% Desalinated = 75% combined). City P has a more diversified mix with significant local catchment; City S has minimal local catchment (5%) and low imported water (20%), indicating high investment in water recycling and desalination.
[3]
Mark breakdown: 1 mark for contrast in local catchment; 1 mark for contrast in NEWater/Desalinated; 1 mark for use of specific percentages from graph.

(d) Advantage: Weather-independent — desalination provides a reliable water supply regardless of rainfall, enhancing water security during droughts.
Disadvantage: High energy consumption and cost — reverse osmosis requires large amounts of electricity, making desalinated water expensive and contributing to carbon emissions unless renewable energy is used.
[3]
Mark breakdown: 1 mark for valid advantage; 1 mark for valid disadvantage; 1 mark for explanation/depth (e.g., linking to energy/cost/security).

Question 6

(a) 31.2 million
[1]

(b) 11.7 million (37.5 – 25.8)
[1]

(c) The population of Country X increased steadily from 18.2 million in 1990 to 48.6 million in 2020. The rate of increase accelerated over time: the 5-year increments grew from ~3.3 million (1990–95) to ~4.8 million (2015–20), showing exponential-like growth.
[2]
Mark breakdown: 1 mark for "steady increase"; 1 mark for "accelerating / increasing increments" with data reference.

(d) 1.01 million per year (accept 1.0–1.02)
[2]
Working:
Total increase = 48.6 – 18.2 = 30.4 million
Period = 2020 – 1990 = 30 years
Average annual growth = 30.4 / 30 = 1.0133… ≈ 1.01 million/year
[2]
Mark breakdown: 1 mark for correct total increase and period; 1 mark for correct division and answer with unit.

(e) High birth rate (natural increase) and declining death rate due to improved healthcare/sanitation; in-migration (foreign workers, refugees); lack of family planning / cultural factors. (Any two)
[2]
Mark breakdown: 1 mark per valid factor with brief explanation.

Question 7

(a) Positive correlation: As GDP per capita increases, access to improved water supply generally increases. The relationship is strong at low incomes (steep rise from 55% to 90% as GDP rises from $1,200 to$ 12,000) but levels off at high incomes (near 100% for GDP > $35,000).
[2]
Mark breakdown: 1 mark for "positive correlation"; 1 mark for describing non-linear shape (steep then plateau).

(b) 85% (accept 83–87%)
[1]
Read from trend line at GDP = $8,000.

(c) Country A (GDP $1,200, access 55%) is below the trend line — it has lower water access than expected for its income. Possible reason: post-conflict infrastructure damage, poor governance/corruption, or geographical challenges (arid, dispersed population).
[2]
Mark breakdown: 1 mark for identifying an anomaly (A or possibly J if argued); 1 mark for plausible geographical reason.

Section C: Data Skills and Geographical Investigation [15 marks]

Question 8

(a) 282 (180 + 72 + 12 + 18 = 282) — correct as shown.
[1]
Marking note: Verification only. If student writes 282, full mark.

(b) 7:45–8:00 (330 vehicles)
[1]

(c) 25.8%
[2]
Working:
Motorcycles 7:45–8:00 = 85
Total vehicles 7:45–8:00 = 330
Percentage = (85 / 330) × 100 = 25.7575…% ≈ 25.8% (1 d.p.)
[2]
Mark breakdown: 1 mark for correct values identified; 1 mark for correct calculation and rounding.

(d) Line graph — because it shows continuous change over time (time-series data), making trends (peaks, troughs, rate of change) easy to see.
[2]
Mark breakdown: 1 mark for "line graph"; 1 mark for reason (continuous time data / shows trend over time).

(e) Strength: Vehicle count is objective, quantifiable, and directly measurable — good proxy for traffic volume.
Limitation: Does not measure congestion directly — congestion depends on road capacity, junction design, traffic light timing, incidents, and vehicle speed. High count on a wide highway may not cause congestion; low count on a narrow road may cause gridlock.
[3]
Mark breakdown: 1 mark for valid strength; 1 mark for valid limitation; 1 mark for explaining why count ≠ congestion (capacity/speed distinction).

Question 9

(a) 105 km²
[1]
Working: 42% of 250 km² = 0.42 × 250 = 105 km²

(b) 10 km²
[2]
Working:
Target Parks = 12% of 250 = 30 km²
Current Parks = 8% of 250 = 20 km²
Additional needed = 30 – 20 = 10 km²
[2]
Mark breakdown: 1 mark for current/target area calculation; 1 mark for correct difference.

(c) Conflict: Converting residential or commercial land to parks may displace residents/businesses, increase property prices, or reduce tax revenue. Converting industrial land may cause job losses. Transport land conversion could worsen traffic. (Any one with brief explanation)
[2]
Mark breakdown: 1 mark for identifying a land use to convert; 1 mark for explaining the resulting conflict (social/economic/environmental).

Question 10

(a) Points plotted at:
(1.5, 30), (2.0, 22), (2.5, 15), (3.0, 10), (3.5, 6), (4.0, 4)
[2]
Marking: 1 mark for all 6 points correctly plotted (±0.5 mm); 1 mark for neatness (sharp pencil, small crosses/dots).
If only 4–5 correct: 1 mark. ≤3 correct: 0 marks.

(b) Smooth curved line of best fit (downward curve, not straight) passing through the general trend of points.
[1]
Marking note: Must be a curve (exponential decay shape), not a straight line. Should balance points above/below.

(c) Negative non-linear (exponential decay) relationship: As distance from city centre increases, average building height decreases rapidly at first, then more gradually, approaching low-rise at the urban fringe.
[2]
Mark breakdown: 1 mark for "negative / inverse"; 1 mark for "non-linear / steep then gentle / exponential decay".

(d) ~12–13 storeys (accept 11–14)
[1]
Read from student's line of best fit at x = 2.8 km.

(e) Land value (bid-rent theory): Land prices are highest in the city centre due to high accessibility and demand from commercial/retail activities. Developers build tall buildings to maximise floor area on expensive land. Further out, land is cheaper, so low-rise development is more economical.
[2]
Mark breakdown: 1 mark for land value / bid-rent concept; 1 mark for linking high land cost → tall buildings to maximise return.

End of Answer Key