O Level Geography Practice Paper 2

TuitionGoWhere Practice Paper - Geography O-Level (Map, Graph & Data Skills) - Answer Key

Version: 2 of 5
Topic: Map, Graph & Data Skills

Section A: Map Reading and Interpretation

1. Give the four-figure grid reference for the square containing the church.
Answer: 4683
Marking: [1] for correct reference.

2. Calculate the straight-line distance in kilometres between the school (Grid Ref 4582) and the church (Grid Ref 4683).
Answer:

• The squares are adjacent diagonally.
• Distance across one square (1km x 1km) diagonal = $\sqrt{1^2 + 1^2} = \sqrt{2} \approx 1.41$ km.
• Alternatively, if measured on a 1:25,000 map, the distance between centers of adjacent diagonal squares is approx 1.4 km.
• Note: In a real exam, students would use a ruler. Here, logical deduction of grid geometry is accepted.
• Accept answers between 1.4 km and 1.5 km.
  Marking: [1] for correct method/calculation, [1] for correct unit and value. Total [2].

3. Describe the relief (height and slope) of the land in Grid Square 4683.
Answer:

• Height: The land is relatively low-lying, with a spot height of 45m. Contour lines (if visible in full extract) would likely show heights between 40-50m.
• Slope: The presence of a spot height and lack of closely packed contours suggests the land is gently sloping or flat.
  Marking:
• [1] for referencing height (e.g., approx 45m / low lying).
• [1] for referencing slope (e.g., gentle / flat).
• [1] for evidence (e.g., spot height / wide spacing of contours). Total [3].

4. Identify the likely direction of flow of the river shown in the extract and give one piece of evidence from the map to support your answer.
Answer:

• Direction: South-West (or from North-East to South-West).
• Evidence:
  • Contour lines form a "V" shape pointing upstream (North-East).
  • Spot heights decrease in the South-West direction.
  • Tributaries join the main river at an angle pointing upstream.
    Marking:
• [1] for correct direction.
• [1] for valid evidence 1.
• [1] for valid evidence 2 (if required, but question asks for one, so max 2 marks for evidence if detailed).
• Correction based on standard marking: [1] Direction, [1] Evidence. Total [2] usually, but question is [4] marks? Let's re-read prompt. Prompt says [4].
• Revised Marking for [4]:
  • [1] Direction (South-West).
  • [1] Evidence 1 (e.g., Contours bend upstream).
  • [1] Explanation of Evidence 1 (e.g., V-shape points to higher ground).
  • [1] Second piece of evidence or elaboration (e.g., Decreasing spot heights).
  • Total [4].

Section B: Graphical Skills and Data Presentation

5. Suggest the most appropriate type of graph... and give two reasons.
Answer:

• Graph Type: Grouped Bar Chart (or Multiple Bar Chart).
• Reason 1: Allows for easy comparison of discrete categories (Locations A, B, C).
• Reason 2: Clearly shows the difference in magnitude between the three locations.
• Note: A line graph is inappropriate as the locations are not continuous time series or sequential spatial points in a way that implies trend continuity between them.
  Marking:
• [1] for correct graph type.
• [1] for first reason.
• [1] for second reason. Total [3].

6. (a) Describe the trend in pedestrian numbers at Intersection Y from 09:00 to 18:00.
Answer:

• The number of pedestrians increases sharply from 09:00 (45) to a peak at 12:00 (120).
• It then decreases steadily to 18:00 (60), but remains higher than the morning starting value.
  Marking:
• [1] for identifying the peak/increase.
• [1] for identifying the subsequent decrease. Total [2].

6. (b) Calculate the percentage increase in pedestrian numbers at Intersection X between 09:00 and 12:00.
Answer:

• Initial value (09:00) = 12
• Final value (12:00) = 35
• Increase = $35 - 12 = 23$
• Percentage Increase = $(23 / 12) \times 100$
• Calculation: $1.9166... \times 100 = 191.7\%$
  Marking:
• [1] for correct working (showing subtraction and division).
• [1] for correct answer (approx 192% or 191.7%). Total [2].

6. (c) Explain one advantage of using a divided bar chart over a simple bar chart for this specific dataset.
Answer:

• A divided bar chart allows for the comparison of the total volume of pedestrians at each time slot (height of bar) while simultaneously showing the proportion contributed by each intersection (if stacked) or simply comparing totals if side-by-side.
• Correction: The question asks about divided bar chart vs simple bar chart for this dataset (two intersections over time). A divided bar chart is typically used for parts of a whole. If the students want to show the total footfall and the split between X and Y, a divided bar is useful.
• Advantage: It shows both the total number of pedestrians at each time period and the relative contribution of Intersection X and Y to that total in a single visual element.
  Marking:
• [1] for identifying it shows totals.
• [1] for identifying it shows proportions/composition. Total [2].

7. Sketch a pie chart... Show calculation for at least one sector angle.
Answer:

• Calculations:
  • Residential: $50\% \times 360^\circ = 180^\circ$
  • Commercial: $30\% \times 360^\circ = 108^\circ$
  • Industrial: $10\% \times 360^\circ = 36^\circ$
  • Green Space: $10\% \times 360^\circ = 36^\circ$
• Sketch: A circle divided into four sectors. One semi-circle (180°) for Residential. One large sector (108°) for Commercial. Two small equal sectors (36°) for Industrial and Green Space. Labels must be present.
  Marking:
• [1] for correct calculation shown.
• [1] for accurate sketch (proportions visually correct).
• [1] for correct labels. Total [3].

Section C: Fieldwork Data Analysis and Evaluation

8. Calculate the mean Dissolved Oxygen level for Site 3.
Answer:

• $(6.5 + 6.7 + 6.6) / 3 = 19.8 / 3 = 6.6$ mg/L.
  Marking: [1] for correct answer.

9. Describe the relationship between the location of the sites and the Mean Dissolved Oxygen levels.
Answer:

• There is a clear spatial pattern: DO levels are highest upstream (8.3 mg/L at Site 1).
• They drop significantly at the industrial estate (3.0 mg/L at Site 2).
• They partially recover downstream (6.6 mg/L at Site 3) but do not return to upstream levels.
• This suggests the industrial estate is a source of pollution that reduces oxygen levels.
  Marking:
• [1] for describing high upstream.
• [1] for describing drop at Site 2.
• [1] for describing partial recovery downstream. (Data references required for full marks). Total [3].

10. Explain why using both chemical tests and biological indicators provides a more reliable assessment.
Answer:

• Chemical tests provide a "snapshot" of water quality at the specific moment of testing. Conditions can fluctuate rapidly (e.g., a sudden discharge).
• Biological indicators reflect the water quality over a longer period (weeks/months). Organisms like stonefly nymphs cannot survive in polluted water for long, so their presence indicates sustained good quality.
• Using both allows cross-verification: if chemical tests are good but biological indicators are poor, it suggests intermittent pollution or past damage. This reduces the risk of anomalous results affecting the conclusion.
  Marking:
• [1] for explaining chemical = snapshot/instantaneous.
• [1] for explaining biological = long-term/cumulative indicator.
• [1] for explaining cross-verification/reliability.
• [1] for clarity/coherence. Total [4].

11. Evaluate the reliability of the data collection method.
Answer:

• Strengths (Reliability):
  • Repeated Trials: Taking three trials at each site and calculating a mean helps to identify and minimize anomalies/random errors, increasing reliability.
  • Digital Equipment: Using digital pH meters and DO probes is generally more precise and less prone to human reading error than manual test kits.
• Weaknesses (Limitations):
  • Timing (Single Day/Time): Collecting all data between 10:00 and 11:00 on a single day limits reliability. Water quality can vary diurnally (e.g., temperature changes affecting DO) or due to daily industrial cycles. The data may not be representative of typical conditions.
  • Sample Size (Spatial): Only three sites were sampled. This is a small sample size for a river system, potentially missing other pollution sources or variations.
• Conclusion: The data is moderately reliable due to repeated trials and good equipment, but the limited temporal scope (one day) reduces its overall validity for generalizing about the river's health.
  Marking:
• [1] for evaluating repeated trials (positive).
• [1] for evaluating equipment (positive).
• [1] for evaluating timing/single day (negative).
• [1] for evaluating spatial sample size (negative).
• [1] for balanced judgement/conclusion.
• [1] for structure/clarity. Total [6].

12. Explain how heavy rainfall 24 hours prior could act as a confounding variable.
Answer:

• Heavy rainfall increases surface runoff and river discharge.
• This can lead to dilution of pollutants, potentially making the water at Site 2 or 3 appear cleaner (higher DO) than it actually is under normal flow conditions.
• Conversely, runoff can wash additional pollutants (fertilizers, sediments) from the land into the river, potentially lowering water quality at Site 3 independently of the industrial estate.
• Therefore, the changes observed might be due to the weather event rather than the industrial discharge, making it difficult to isolate the industrial estate as the sole cause.
  Marking:
• [1] for identifying dilution effect OR increased runoff.
• [1] for explaining impact on data (e.g., artificially high/low readings).
• [1] for linking to confounding variable concept (alternative cause).
• [1] for concluding impact on validity of hypothesis. Total [4].