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O Level Geography Practice Paper 1

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TuitionGoWhere Practice Paper - Geography O-Level

TuitionGoWhere Exam Practice (AI)

Subject: Geography (2279)
Level: O-Level
Paper: Practice Paper 1 (Version 1 of 5)
Topic Focus: Map, Graph & Data Skills
Duration: 1 Hour
Total Marks: 40

Name: ________________________
Class: ________________________
Date: ________________________

Instructions to Candidates

Write your Name, Class, and Date in the spaces provided.
Answer all questions.
The number of marks is given in brackets [ ] at the end of each question or part question.
You may use a calculator for any calculations.
Where maps, graphs, or photographs are provided, refer to them explicitly in your answers.

Section A: Data Collection and Representation (10 Marks)

1. A group of students is conducting fieldwork to investigate the variation in river velocity at three different sites (Site A, Site B, and Site C) along a river course. They measured the velocity (m/s) at five points across the channel at each site and calculated the mean velocity for each site.

Table 1.1: Mean River Velocity at Three Sites

Site	Distance from Source (km)	Mean Velocity (m/s)
A	2.0	0.45
B	5.5	0.82
C	12.0	1.15

(a) Suggest the most appropriate type of graph to display the data in Table 1.1 to show the relationship between distance from source and mean velocity. [1]

(b) Explain one reason why the graph type you selected in (a) is appropriate for this data. [1]

(c) The students also collected data on river bedload size (pebble length in cm) at the same three sites. They want to compare the mean velocity and the mean pebble length on a single graph.

Describe how they could construct this graph to allow for easy comparison of both variables. [2]

(d) Identify one potential source of error when measuring river velocity using a float method (e.g., an orange or dog biscuit) and a stopwatch. [1]

(e) State one way the students could improve the reliability of their velocity data at each site. [1]

(f) The students recorded the following individual velocity readings (m/s) at Site B: 0.75, 0.80, 0.95, 0.82, 0.78

Calculate the mean velocity for these readings. Show your working. [2]

Answer: _______________ m/s

(g) One reading of 0.95 m/s was taken near the center of the channel, while the others were taken near the banks. Explain why this reading might be considered an anomaly or outlier in the context of river channel geometry. [2]

Section B: Statistical Analysis and Interpretation (15 Marks)

2. Students conducted a survey to assess the environmental quality of two different urban neighborhoods: Area X (City Center) and Area Y (Suburban Residential). They used an Environmental Quality Survey (EQS) where participants scored 5 categories on a scale of 0 to 5 (0 = Very Poor, 5 = Excellent).

Table 2.1: Environmental Quality Scores

Category	Area X (City Center) Scores	Area Y (Suburban) Scores
Noise Levels	1, 2, 1, 2, 1	4, 5, 4, 4, 5
Air Quality	2, 1, 2, 1, 2	4, 4, 5, 4, 4
Litter	1, 1, 2, 1, 1	5, 5, 4, 5, 5
Green Space	1, 1, 1, 2, 1	3, 3, 4, 3, 3
Traffic Congestion	1, 1, 1, 1, 2	4, 4, 4, 5, 4

(a) Calculate the median score for "Noise Levels" in Area X. [1]

Answer: _______________

(b) Calculate the mean score for "Green Space" in Area Y. Show your working. [2]

Answer: _______________

(c) Describe the difference in environmental quality between Area X and Area Y based on the data in Table 2.1. Use specific data from the table to support your answer. [3]

(d) The students decided to calculate a weighted score for "Traffic Congestion" to account for the severity of the issue. They assigned the following weights:

Score 1-2 (High Congestion): Weight x 2
Score 3 (Moderate): Weight x 1
Score 4-5 (Low Congestion): Weight x 0.5

Calculate the total weighted score for Area X's Traffic Congestion data: 1, 1, 1, 1, 2. [3]

Answer: _______________

(e) Evaluate the reliability of using an Environmental Quality Survey (EQS) with a scale of 0-5 to compare two different neighborhoods. Consider one strength and one limitation of this method. [4]

Strength:

Limitation:

(f) Suggest one additional piece of quantitative data the students could collect to support their findings on air quality, other than a survey score. [2]

Section C: Map and Graph Interpretation (15 Marks)

3. Study Figure 3.1, which shows the distribution of annual rainfall (mm) across a hypothetical island, and Figure 3.2, which shows the location of three weather stations (P, Q, and R).

(Note: In a real exam, Figure 3.1 would be a map with isohyets. Figure 3.2 would be a line graph showing monthly rainfall for Station P.)

Figure 3.1 Description:

The island has a mountain range running North-South in the center.
West Coast (Windward side): Annual rainfall > 2500 mm.
East Coast (Leeward side): Annual rainfall < 1000 mm.
Station P is on the West Coast.
Station Q is at the mountain peak.
Station R is on the East Coast.

Figure 3.2 Description (Station P Monthly Rainfall):

Jan: 200mm, Feb: 180mm, Mar: 220mm, Apr: 250mm, May: 300mm, Jun: 320mm
Jul: 310mm, Aug: 290mm, Sep: 260mm, Oct: 240mm, Nov: 210mm, Dec: 200mm

(a) Describe the pattern of annual rainfall distribution shown in Figure 3.1. [2]

(b) Using Figure 3.2, calculate the total rainfall for Station P during the second quarter of the year (April, May, June). [2]

Answer: _______________ mm

(c) Calculate the range of monthly rainfall for Station P based on Figure 3.2. [2]

Answer: _______________ mm

(d) Explain why Station P (West Coast) receives significantly more rainfall than Station R (East Coast), referring to the physical features shown in Figure 3.1. [3]

(e) The students want to present the monthly rainfall data for all three stations (P, Q, and R) on one graph to compare seasonal variations.

(i) Name the most suitable graph type for this comparison. [1]

(ii) Explain why a pie chart would be unsuitable for displaying this monthly rainfall data. [2]

(f) A student claims: "The data in Figure 3.2 proves that Station P has a monsoon climate."

Evaluate this statement. Is the data sufficient to make this conclusion? Give two reasons for your answer. [3]

Answers

TuitionGoWhere Practice Paper - Geography O-Level (Answer Key)

Topic: Map, Graph & Data Skills
Version: 1 of 5

Section A: Data Collection and Representation

1. (a) [1 mark]

Line graph.
Accept: Scatter graph (if trend line is mentioned), but Line graph is standard for continuous change over distance.

1. (b) [1 mark]

Line graphs are suitable for showing continuous data / trends / changes over a continuous variable (distance).
Must mention: trend, continuous nature, or relationship between two continuous variables.

1. (c) [2 marks]

Use a dual-axis graph (compound bar chart or line graph with two y-axes).
One y-axis for velocity (e.g., left side) and one y-axis for pebble length (e.g., right side).
Use different colors/styles (e.g., line for velocity, bars for pebble length) and a clear legend.
1 mark for method (dual axis/compound), 1 mark for clarity (legend/labels).

1. (d) [1 mark]

Wind affecting the float (pushing it faster/slower than water).
Human reaction time error with the stopwatch.
Float getting stuck on rocks/vegetation.
Any valid source of error.

1. (e) [1 mark]

Take multiple readings at each site and calculate the mean.
Use a flow meter instead of a float (more precise instrument).
Ensure the same distance is used for timing at all sites.

1. (f) [2 marks]

Working: $(0.75 + 0.80 + 0.95 + 0.82 + 0.78) \div 5$
Sum = $4.10$
Mean = $4.10 \div 5 = 0.82$
Answer: 0.82 m/s
1 mark for correct sum or setup, 1 mark for correct final answer.

1. (g) [2 marks]

River velocity is typically fastest in the center/deepest part of the channel due to less friction with the bed and banks.
Readings near the banks are slower due to higher friction.
Therefore, 0.95 is likely the true maximum velocity, while the lower readings near banks are not anomalies but expected variations; however, if the aim was to measure average cross-section velocity, mixing center and bank readings without weighting creates an inaccurate mean.
Accept: Explanation of friction/depth affecting velocity.

Section B: Statistical Analysis and Interpretation

2. (a) [1 mark]

Data: 1, 2, 1, 2, 1 $\rightarrow$ Ordered: 1, 1, 1, 2, 2
Median: 1

2. (b) [2 marks]

Data: 3, 3, 4, 3, 3
Working: $(3+3+4+3+3) \div 5 = 16 \div 5$
Answer: 3.2

2. (c) [3 marks]

Area Y has significantly higher environmental quality scores than Area X.
Example: Noise levels in Area Y are 4-5 (Excellent/Good), while Area X are 1-2 (Poor).
Example: Litter scores in Area Y are 4-5, while Area X are 1-2.
1 mark for general comparison, 2 marks for specific data references.

2. (d) [3 marks]

Data: 1, 1, 1, 1, 2
Weights: Scores 1-2 get weight x2.
Calculation:
- Four scores of 1: $4 \times (1 \times 2) = 8$ ? No, the weight applies to the score or the count?
- Clarification of Template Method: Usually, weighted score = $\sum (\text{Frequency} \times \text{Weight})$ . Or $\sum (\text{Score} \times \text{Weight Factor})$ .
- Let's assume the prompt implies: Value $\times$ Weight.
- Score 1 (Weight 2): $1 \times 2 = 2$ . There are four 1s. $4 \times 2 = 8$ .
- Score 2 (Weight 2): $2 \times 2 = 4$ . There is one 2. $1 \times 4 = 4$ .
- Total = $8 + 4 = 12$ .
- Alternative Interpretation: If weight is applied to frequency:
  - Freq of 1-2 is 5. Weight 2. Total = 10.
- Standard Geography Fieldwork Method: Usually, you sum the scores. If weighted, you multiply the score by a factor.
- Let's stick to the simplest interpretation of "Weighted Score" in O-Level context:
- Scores: 1, 1, 1, 1, 2. All fall in "1-2" category.
- Weight for this category is x2.
- Sum of raw scores = $1+1+1+1+2 = 6$ .
- Weighted Sum = $6 \times 2 = 12$ .
- Or: If the weight replaces the score:
- Score 1 becomes 2. Score 2 becomes 2.
- $2+2+2+2+2 = 10$ .
- Marking Note: Accept 10 or 12 if working is shown clearly. The key is applying the weight consistently.
- Preferred Answer: 10 (If weight replaces the value for severity) or 12 (If weight multiplies the value). Let's provide 10 as the "Severity Index" approach often used in EQS.
- Revised Working for 10: All 5 readings are in the 1-2 range. Weight is 2. $5 \text{ readings} \times 2 = 10$ .

2. (e) [4 marks]

Strength: Allows for quick comparison of multiple factors; quantifies subjective opinions (makes qualitative data quantitative); easy to administer.
Limitation: Subjective/Biased (different people have different standards for "clean" or "quiet"); small sample size may not be representative; does not explain why the score is low.
2 marks for strength (1 for point, 1 for explanation), 2 marks for limitation (1 for point, 1 for explanation).

2. (f) [2 marks]

Use a particulate matter sensor (PM2.5/PM10 monitor).
Use a gas sensor (e.g., for $CO_2$ , $NO_2$ , $SO_2$ ).
Must be a quantitative instrument, not a survey.

Section C: Map and Graph Interpretation

3. (a) [2 marks]

Rainfall is higher on the West Coast (>2500mm) and lower on the East Coast (<1000mm).
Rainfall decreases from West to East across the island.
1 mark for identifying high/low areas, 1 mark for the gradient/direction.

3. (b) [2 marks]

April (250) + May (300) + June (320).
$250 + 300 + 320 = 870$ .
Answer: 870 mm.

3. (c) [2 marks]

Highest: 320 mm (June).
Lowest: 180 mm (February).
Range: $320 - 180 = 140$ mm.
Answer: 140 mm.

3. (d) [3 marks]

Prevailing winds blow from the West (sea) towards the land.
Moist air is forced to rise over the central mountain range (Orographic uplift).
Air cools, condenses, and rains on the Windward side (West/Station P).
Air descends on the Leeward side (East/Station R), warms up, and holds more moisture (Rain Shadow effect), resulting in less rain.
1 mark for prevailing wind/uplift, 1 mark for condensation/rain on windward, 1 mark for rain shadow/leeward dryness.

3. (e) (i) [1 mark]

Multiple line graph (or Compound line graph).
Accept: Grouped bar chart.

3. (e) (ii) [2 marks]

Pie charts show parts of a whole (percentages/proportions) for a single point in time or category.
They cannot effectively show trends over time (months) or compare multiple variables (3 stations) across 12 time periods simultaneously.
1 mark for "parts of a whole" limitation, 1 mark for "cannot show trends/time series".

3. (f) [3 marks]

No / Insufficient.
Reason 1: Only one year of data (or unspecified duration) is shown; climate requires long-term averages (usually 30 years).
Reason 2: Temperature data is missing; monsoon climates are defined by both rainfall and temperature patterns (or wind direction).
Reason 3: Data from only one station (P) is insufficient to define the climate of the whole region/island.
1 mark for judgment, 2 marks for valid reasons.