Secondary 2 Geography Semestral Assessment 2 (End of Year) Paper 2

TuitionGoWhere Practice Paper - Geography Secondary 2 (SA2 Version 2) - Answer Key

Total Marks: 60

Section A: Map Skills [15 marks]

Question 1

(a) Four-figure grid reference of the hospital: 4520 [1]
Marking: 1 mark for correct four-figure grid reference (easting 45, northing 20).

(b) Six-figure grid reference of the school: 472222 [1]
Marking: 1 mark for correct six-figure grid reference. Easting 472 (47 + 2/10), northing 222 (22 + 2/10). Accept 472222 or 472 222.

(c) Relief around the quarry (4318): [2]

• The quarry is located on relatively flat/low-lying ground (1 mark)
• Evidence: Contour lines are widely spaced / few contour lines / low contour values (e.g., 20-40m) around grid square 4318 (1 mark)
  Marking: 1 mark for description (flat/gentle slope), 1 mark for map evidence (contour spacing/values).

(d) Straight-line distance between hospital (4520) and school (4722): [1]

• Map distance: ~2.8 cm (measured from centre of 4520 to centre of 4722)
• Ground distance: 2.8 cm × 25,000 = 70,000 cm = 0.7 km (accept 0.65–0.75 km)
  Marking: 1 mark for correct answer with units (km).

Question 2

(a) Direction of flow of the river: Northeast to Southwest / SW [1]
Marking: 1 mark for correct cardinal/intercardinal direction.

(b) Two pieces of map evidence: [2]

1. Contour values decrease from northeast (higher contours, e.g., 100m) to southwest (lower contours, e.g., 20m) (1 mark)
2. River flows from higher ground (northeast) to lower ground (southwest) / River flows downhill following contour gradient (1 mark)
   Alternative evidence: River joins a larger water body in SW; V-shaped contour crossing pointing upstream.
   Marking: 1 mark each for two valid pieces of evidence.

(c) Average gradient of river between 4924 (entry) and 4116 (exit): [3]
Working:

• Vertical difference (VI): Elevation at 4924 ≈ 100 m (from contour lines); Elevation at 4116 ≈ 20 m
  VI = 100 – 20 = 80 m
• Horizontal distance (HE): Grid distance from 4924 to 4116 = 8 km easting difference + 8 km northing difference?
  Actually: 4924 to 4116 → ΔEasting = 8 km, ΔNorthing = 8 km → straight-line distance = √(8² + 8²) = √128 ≈ 11.3 km
  But river meanders; map distance along river ≈ 14 cm → 14 × 0.25 = 3.5 km? Wait, scale 1:25,000 → 1 cm = 0.25 km. If river length on map = 14 cm → 3.5 km.
  Better: Use straight-line distance between grid references for "average gradient" as per standard exam practice.
  Straight-line distance = √((49-41)² + (24-16)²) = √(64+64) = √128 ≈ 11.31 km = 11,310 m
• Gradient = VI : HE = 80 : 11,310 = 1 : 141.4 ≈ 1 : 141

Marking:

• 1 mark for correct vertical interval (80 m)
• 1 mark for correct horizontal equivalent (11.3 km or 11,300 m)
• 1 mark for correct gradient ratio (1:141) with working shown

Common mistake: Using map distance in cm without converting; reversing VI/HE.

Question 3

(a) Map distance along main road between hospital (4520) and quarry (4318): ~3.2 cm [1]
Marking: 1 mark for reasonable measurement (accept 3.0–3.5 cm).

(b) Actual ground distance: [1]
3.2 cm × 25,000 = 80,000 cm = 0.8 km (accept 0.75–0.85 km)
Marking: 1 mark for correct conversion and units.

(c) Time taken to walk from school (4722) to hospital (4520) via main road at 5 km/h: [2]

• Map distance along road ≈ 4.0 cm (4722 → 4622 → 4520 along road)
• Ground distance = 4.0 × 0.25 = 1.0 km
• Time = Distance / Speed = 1.0 km / 5 km/h = 0.2 h = 12 minutes
  Marking: 1 mark for correct distance calculation, 1 mark for correct time in minutes.

Question 4

(a) Four-figure grid reference of southernmost point of forested area: 4423 (example; depends on map) [1]
Marking: 1 mark for correct grid square at southern edge of forest.

(b) Reason for forest on higher ground: [1]

• Steep slopes / high relief (shown by closely spaced contours) make the land unsuitable for agriculture or urban development, so it remains forested.
  OR: Higher elevation → cooler/wetter microclimate favours forest growth.
  Marking: 1 mark for valid reason linked to map evidence (relief/climate).

Section B: Graph and Data Interpretation [20 marks]

Question 5

(a) Month with highest rainfall: December (280 mm) [1]

(b) Annual temperature range: [1]
Max temp = 28.5°C (May), Min temp = 26.5°C (Jan/Dec)
Range = 28.5 – 26.5 = 2.0°C
Marking: 1 mark for correct calculation with units.

(c) Total annual rainfall: [1]
Sum = 240+160+180+190+170+150+155+160+170+200+250+280 = 2,305 mm
Marking: 1 mark for correct total (accept 2300–2310 mm).

(d) Relationship between rainfall and temperature: [2]

• Generally inverse/negative relationship: Months with higher temperature (Apr–May) have moderate rainfall, while cooler months (Nov–Jan) have higher rainfall (1 mark)
• But not perfectly inverse: Temperature peaks Apr–May (28.5°C) while rainfall peaks Nov–Dec (250–280 mm) when temperature is lower (26.5–27.0°C) (1 mark)
  Marking: 1 mark for identifying pattern, 1 mark for accurate description using data.

(e) Reason for high rainfall in Nov–Dec: [2]

• Northeast Monsoon brings moist winds from the South China Sea over Singapore (1 mark)
• Convergence of winds / rising air leads to heavy convective rainfall (1 mark)
  Marking: 1 mark for identifying monsoon, 1 mark for mechanism.

Question 6

(a) Shape of population pyramid: Broad base, rapidly narrowing sides, concave profile, small apex [1]
Marking: 1 mark for accurate description (e.g., "expansive/pyramidal shape with broad base and narrow top").

(b) Birth rate and death rate: [2]

• High birth rate → broad base (large 0-4, 5-9 age groups) (1 mark)
• High death rate / low life expectancy → rapid narrowing with age, very small elderly cohorts (1 mark)
  Marking: 1 mark each for birth rate and death rate inference with evidence.

(c) Percentage aged 0-14 (dependent young): [2]
0-4: 4.5+4.3 = 8.8%
5-9: 4.2+4.0 = 8.2%
10-14: 3.8+3.6 = 7.4%
Total = 8.8 + 8.2 + 7.4 = 24.4%
Marking: 1 mark for correct summation method, 1 mark for correct answer (accept 24–25%).

(d) Economic challenge: [2]

• High youth dependency ratio → large proportion of population not yet in workforce (1 mark)
• Pressure on government spending for education, healthcare, job creation for future workforce (1 mark)
  Marking: 1 mark for identifying challenge, 1 mark for explanation.

Question 7

(a) Largest percentage increase: Residential (35% → 45% = +10 percentage points) [1]
Marking: 1 mark for correct category.

(b) Actual area of Green/Open Space in 2020: [1]
10% of 500 km² = 50 km²
Marking: 1 mark for correct calculation with units.

(c) Change in Industrial land use: [1]
Decreased from 20% to 10% (halved / 10 percentage point decrease)
Marking: 1 mark for accurate description with data.

(d) Two reasons for increase in Residential land use: [3]

1. Population growth → greater demand for housing (1 mark)
2. Urbanisation / rural-urban migration → expansion of residential areas into former green/industrial land (1 mark)
3. Government housing policies (e.g., HDB building new towns) (1 mark)
   Marking: 1 mark per valid reason (max 3). Accept other reasonable reasons (e.g., rising incomes, smaller household sizes).

Question 8

(a) Percentage increase in total water consumption (2010–2020): [2]
Increase = 850 – 630 = 220 million m³
% increase = (220 / 630) × 100 = 34.9% ≈ 35%
Marking: 1 mark for correct increase, 1 mark for correct percentage.

(b) Largest absolute increase: Non-Domestic (280 – 250 = 30) Wait: Domestic +20, Non-Domestic +30, NEWater +100, Desalinated +70.
Desalinated Water (+70 million m³) [1]
Marking: 1 mark for correct sector.

(c) % of total consumption met by NEWater in 2020: [1]
(150 / 850) × 100 = 17.6% ≈ 18%
Marking: 1 mark for correct calculation.

(d) Why NEWater and desalinated water increased: [3]

1. Water security / reduce reliance on imported water (from Malaysia) (1 mark)
2. Technological advancement made NEWater/desalination more cost-effective (1 mark)
3. Growing demand from population/economic growth requires diversified sources (1 mark)
   Marking: 1 mark per valid reason (max 3).

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

Question 9

(a) Comparative bar graph: [3]
Marking:

• 1 mark: Correct title and labelled axes (Vehicle Type, Count)
• 1 mark: Accurate plotting of all 8 bars (4 types × 2 junctions) to scale
• 1 mark: Clear legend / colour coding for Junction A vs B, bars side-by-side

(b) Advantage of comparative bar graph: [1]

• Allows direct visual comparison of each vehicle type between the two junctions / easy to see similarities/differences.
  Marking: 1 mark for valid advantage.

(c) Evaluate conclusion: "Junction B has heavier traffic..." [3]

• Conclusion is invalid/incorrect (1 mark)
• Evidence: Total vehicles at both junctions = 700 each → traffic volume is equal (1 mark)
• Junction B has more buses (60 vs 40) and motorcycles (220 vs 180), but fewer cars (380 vs 420) and goods vehicles (40 vs 60) (1 mark)
  Marking: 1 mark for judgement, 1 mark for total count evidence, 1 mark for breakdown evidence.

Question 10

(a) Vertical Exaggeration (VE): [1]
VE = Horizontal Scale / Vertical Scale = (500 m/cm) / (100 m/cm) = 5 times (5×)
Marking: 1 mark for correct calculation.

(b) Gradient from P to hill summit: [2]

• Horizontal distance: 1.5 cm × 500 m/cm = 750 m
• Vertical rise: 300 – 100 = 200 m
• Gradient = VI : HE = 200 : 750 = 1 : 3.75 (or 4:15)
  Marking: 1 mark for correct VI and HE, 1 mark for correct ratio.

(c) Walking trail: Hill summit vs valley: [3]
Over hill summit:

• Advantage: Shorter distance (direct route) / scenic views (1 mark)
• Disadvantage: Steep gradient (1:3.75) → difficult for walkers, erosion risk (1 mark)

Through valley:

• Advantage: Gentler slope → easier walk, less erosion (1 mark)
• Disadvantage: Longer distance / possible flooding / less scenic (1 mark)

Marking: 1 mark for advantage of one route, 1 mark for disadvantage of same/other route, 1 mark for balanced comparison. Max 3 marks.

Question 11

(a) Relationship in Figure 6: [1]
Negative correlation: As GDP per capita increases, infant mortality rate decreases.
Marking: 1 mark for correct description.

(b) Anomalous result: [2]

• Country C (GDP $5,000, IMR 30)** appears slightly *above* the trend line (expected ~25) / **Country D ($8,000, IMR 20) fits well.
  Actually, all points fit a smooth curve. Perhaps no strong anomaly. But if forced:
  **Country A ($1,000, IMR 55)** may be *lower* than expected for very low GDP (some countries <$1000 have IMR >60).
  Better: No clear anomaly; relationship is consistent.
  Accept: "No clear anomaly; all points follow the general trend." (1 mark for identification, 1 mark for reason)
  Or if identifying one: Country F ($15,000, IMR 10) is slightly below trend (better than expected) → strong healthcare system / effective public health policies.
  Marking: 1 mark for identifying a point (or stating none), 1 mark for plausible reason.

**(c) Estimate IMR for Country K (GDP $12,000):** [1] Between E ($10,000, IMR 15) and F ($15,000, IMR 10) → ~12–13 per 1000
Marking: 1 mark for reasonable estimate (11–14).

(d) Why GDP per capita alone is insufficient: [2]

1. Healthcare access/quality varies at same income level (e.g., public vs private systems) (1 mark)
2. Other factors: Female education, sanitation, nutrition, disease prevalence, government policy (1 mark)
   Marking: 1 mark per valid factor.

Section D: Geographical Investigation and Synthesis [10 marks]

Question 12

(a) Suitable hypothesis: [1]

• "As the percentage of impervious surface increases, surface runoff volume increases."
• "Urbanised catchments have higher peak runoff than forested catchments."
  Marking: 1 mark for testable, directional hypothesis linking urbanisation and runoff.

(b) Two primary data collection methods: [2]

1. Infiltration rate measurement (using infiltrometer)
2. Stream discharge measurement (velocity × cross-section)
3. Land use mapping (% impervious surface)
4. Rainfall measurement (rain gauge)
   Marking: 1 mark each (max 2).

(c) Description of one method (e.g., stream discharge): [2]

• Measure channel width and depth at regular intervals to calculate cross-sectional area (1 mark)
• Measure flow velocity using a flow meter / float method (time over known distance) (1 mark)
• Discharge = Area × Velocity; repeat at multiple sites / times (during/after rain)
  Marking: 1 mark for equipment/measurements, 1 mark for procedure.

(d) Way to present data to show relationship: [1]

• Scatter graph of % impervious surface (x-axis) vs peak runoff / discharge (y-axis)
• Comparative hydrographs for urban vs rural catchments
  Marking: 1 mark for appropriate graphical method.

(e) Limitation and improvement: [2]

• Limitation: Small sample size / only one rainfall event / short study period → results not representative (1 mark)
• Improvement: Collect data over long-term data / multiple storm events / more sites (1 mark)
  Marking: 1 mark for valid limitation, 1 mark for feasible improvement.

Question 13

(a) Two land use changes (1990–2020): [2]

1. Deforestation: Large areas of dense forest cleared (green → light brown) (1 mark)
2. Expansion of palm oil plantations (geometric green patches) / Urban expansion (grey areas) / Straightening of river/drainage channels (1 mark)
   Marking: 1 mark per distinct change with visual evidence.

(b) Environmental impact: [2]

• Loss of biodiversity due to habitat destruction (forest → monoculture plantation/urban) (1 mark)
• Increased surface runoff / flooding due to loss of vegetation and soil compaction (1 mark)
  OR: Soil erosion, carbon emissions, water pollution from plantations.
  Marking: 1 mark for impact, 1 mark for explanation linked to change.

(c) Sustainable management strategy: [2]

• Riparian buffer zones / reforestation of steep slopes to reduce erosion and filter runoff (1 mark)
• Sustainable palm oil certification (RSPO) / zero-burning policy / agroforestry (1 mark)
  Marking: 1 mark for strategy, 1 mark for how it reduces impact.

End of Answer Key

Total Marks: 60
Mark allocations per section: A=15, B=20, C=15, D=10