Editable original practice. Review all changes before classroom use. Use the browser’s print command to save a PDF.
Original practice assessment. The teacher can edit this pack; edits may change its mark total or invalidate its guidance.
Original fictional unfamiliar-resource pack. Work across the resources; each describes the same teaching scenario. This is not an official examination paper or a real project forecast.
Wind: £18 million
Estimated annual output 70.08 GWh
Lifecycle intensity 12 g CO₂e/kWh
Variable generation and landscape effects
Operating costs and grid connection need separate assessment
Solar roofs: £12 million
Estimated annual output 18.4 GWh
Lifecycle intensity 35 g CO₂e/kWh
Seasonal generation and roof agreements
Operating costs and grid connection need separate assessment
Small hydro: £19 million
Estimated annual output 37.67 GWh
Lifecycle intensity 24 g CO₂e/kWh
River ecology and low-flow constraints
Operating costs and grid connection need separate assessment
Grid operator: We need output when demand occurs, not only a yearly total.
Resident: Local environmental effects and affordability need consideration.
Climate adviser: Compare lifecycle boundaries consistently.
| Project | Capacity (MW) | Capacity factor (%) | Annual output (GWh) | Capital cost (£ million) | New land footprint (ha) |
|---|---|---|---|---|---|
| Wind | 20 | 40 | 70.08 | 18 | 24 |
| Solar roofs | 15 | 14 | 18.4 | 12 | 0 |
| Small hydro | 10 | 43 | 37.67 | 19 | 8 |
Annual output = capacity × capacity factor/100 × 8,760 hours ÷ 1,000. Capacity is a power rate; output is an energy quantity.
| Measure | Value | Unit |
|---|---|---|
| Annual local demand | 100 | GWh |
| Wind: lifecycle emissions intensity | 12 | g CO₂e/kWh |
| Solar roofs: lifecycle emissions intensity | 35 | g CO₂e/kWh |
| Small hydro: lifecycle emissions intensity | 24 | g CO₂e/kWh |
Fictional lifecycle intensity estimates use the same boundary. They are not operational emissions only; output timing, storage and grid losses are omitted.
| Option | Capital cost (£ million) | Benefit | Limitation |
|---|---|---|---|
| Wind | 18 | Estimated annual output 70.08 GWh; Lifecycle intensity 12 g CO₂e/kWh | Variable generation and landscape effects; Operating costs and grid connection need separate assessment |
| Solar roofs | 12 | Estimated annual output 18.4 GWh; Lifecycle intensity 35 g CO₂e/kWh | Seasonal generation and roof agreements; Operating costs and grid connection need separate assessment |
| Small hydro | 19 | Estimated annual output 37.67 GWh; Lifecycle intensity 24 g CO₂e/kWh | River ecology and low-flow constraints; Operating costs and grid connection need separate assessment |
Available capital budget: £20 million. Options are alternatives; their benefits cannot simply be added. Operating costs are separate.
Resource limits: Capacity factors are fictional assumptions.; Annual balances do not represent hourly security or whole-system cost.
1 mark1. Calculate the wind annual output in GWh to two decimal places.
1 mark2. What percentage of annual local demand could wind output equal, ignoring timing and losses?
4 marks3. Explain why annual output equal to a large share of demand does not guarantee a secure supply at all times.
1 mark4. Which comparison uses matching quantities?
6 marks5. Recommend an option for energy demand, output and competing priorities. Use at least two resources, compare an alternative and address an uncertainty or stakeholder concern.
The shading represents a percentage of households, not a total. Class boundaries are lower-inclusive and upper-exclusive.
Scroll the diagram sideways to see the full resource.
| District | Reported damage (%) | Households |
|---|---|---|
| A | 2 | 1000 |
| B | 14 | 500 |
| C | 24 | 500 |
| D | 35 | 100 |
A–D use these non-overlapping classes: 0 to less than 10%; 10 to less than 20%; 20 to less than 30%; 30 to less than 40%. A has 1,000 households and D has 100 households.
1 mark1. Which class contains district C?
1 mark2. How many households in A reported damage?
1 mark3. Does the darkest district necessarily have the most affected households?
The profile joins six measured points with straight segments. Horizontal units are kilometres and vertical units are metres. The drawn slope angle is exaggerated by these different scales.
Scroll the diagram sideways to see the full resource.
| Distance (km) | Height (m) |
|---|---|
| 0 | 40 |
| 1 | 240 |
| 2 | 440 |
| 3 | 640 |
| 4 | 440 |
| 5 | 40 |
0 km along the transect: 40 m above datum; 1 km along the transect: 240 m above datum; 2 km along the transect: 440 m above datum; 3 km along the transect: 640 m above datum; 4 km along the transect: 440 m above datum; 5 km along the transect: 40 m above datum
1 mark1. What is the difference in height between the highest and lowest measured points, in metres?
1 mark2. For the section from 2 km to 3 km, express the gradient as 1:n. Enter n.
1 mark3. Can the drawn line angle be used directly as the real slope angle?
Fictional training exercise. A float is timed along the same 5 m reach four times. No student enters the water.
| Trial | Time (s) |
|---|---|
| 1 | 10 |
| 2 | 11 |
| 3 | 10 |
| 4 | 29 |
1 mark1. Which action best investigates the 29-second result?
1 mark2. What was velocity in trial 1, in m/s?
4 marks3. Why might surface floats overestimate mean velocity? Develop two linked points and support them with the resource or a methodological reason.
Six locations were observed once. Several unmeasured factors could influence the number of trees.
Scroll the diagram sideways to see the full resource.
| Distance (km) | Trees within 50 m |
|---|---|
| 1 | 26 |
| 2 | 23 |
| 3 | 21 |
| 4 | 18 |
| 5 | 14 |
| 6 | 13 |
1 km: 26 trees; 2 km: 23 trees; 3 km: 21 trees; 4 km: 18 trees; 5 km: 14 trees; 6 km: 13 trees
1 mark1. Which description best fits the pattern?
1 mark2. Which conclusion is justified?
Fictional training exercise. A damaged counter failed at one of four sites.
| Site | Count in 10 min |
|---|---|
| A | 20 |
| B | 32 |
| C | Missing |
| D | 44 |
1 mark1. What should the missing reading be recorded as?
1 mark2. Calculate the mean of the three observed counts.
4 marks3. How does missing site C affect conclusions? Develop two linked points and support them with the resource or a methodological reason.
The profile joins six measured points with straight segments. Horizontal units are kilometres and vertical units are metres. The drawn slope angle is exaggerated by these different scales.
Scroll the diagram sideways to see the full resource.
| Distance (km) | Height (m) |
|---|---|
| 0 | 20 |
| 1 | 320 |
| 2 | 620 |
| 3 | 920 |
| 4 | 620 |
| 5 | 20 |
0 km along the transect: 20 m above datum; 1 km along the transect: 320 m above datum; 2 km along the transect: 620 m above datum; 3 km along the transect: 920 m above datum; 4 km along the transect: 620 m above datum; 5 km along the transect: 20 m above datum
1 mark1. What is the difference in height between the highest and lowest measured points, in metres?
1 mark2. For the section from 2 km to 3 km, express the gradient as 1:n. Enter n.
1 mark3. Can the drawn line angle be used directly as the real slope angle?
This training rule requires at least 2 hectares, outside protected land and the indicated flood zone. Among eligible sites, choose the lowest stated walk time. It is an explicit exercise rule, not a real planning approval.
Scroll the diagram sideways to see the full resource.
| Site | Protected land | Flood zone | Area (hectares) | Walk to bus stop (min) |
|---|---|---|---|---|
| A | No | No | 2 | 15 |
| B | No | No | 3 | 19 |
| C | Yes | No | 2 | 7 |
| D | No | No | 5 | 11 |
North is up. Blue band: excluded flood zone. Green block: protected land. Sites: A: protected no, flood zone no, area 2 hectares, walk 15 minutes; B: protected no, flood zone no, area 3 hectares, walk 19 minutes; C: protected yes, flood zone no, area 2 hectares, walk 7 minutes; D: protected no, flood zone no, area 5 hectares, walk 11 minutes. Minimum area 2 hectares.
1 mark1. How many sites meet every eligibility condition?
1 mark2. Which site does the stated rule select?
1 mark3. What does the exercise establish?
Rank each variable from smallest to largest. Give tied values their average rank. Spearman’s coefficient is the Pearson correlation of these ranks; the no-ties shortcut is inappropriate here. This task calculates association, not significance.
| Site | Distance (km) | Pedestrians per 10 min |
|---|---|---|
| Site 1 | 1 | 11 |
| Site 2 | 2 | 13 |
| Site 3 | 2 | 13 |
| Site 4 | 4 | 14 |
| Site 5 | 5 | 18 |
| Site 6 | 6 | 16 |
Paired fictional site observations
1 mark1. What rank do the two distance values of 2 km receive?
1 mark2. Calculate Spearman’s coefficient to two decimal places.
1 mark3. What does this calculation establish?
Six locations were observed once. Several unmeasured factors could influence the number of trees.
Scroll the diagram sideways to see the full resource.
| Distance (km) | Trees within 50 m |
|---|---|
| 1 | 10 |
| 2 | 13 |
| 3 | 17 |
| 4 | 20 |
| 5 | 22 |
| 6 | 27 |
1 km: 10 trees; 2 km: 13 trees; 3 km: 17 trees; 4 km: 20 trees; 5 km: 22 trees; 6 km: 27 trees
1 mark1. Which description best fits the pattern?
1 mark2. Which conclusion is justified?
Six locations were observed once. Several unmeasured factors could influence the number of trees.
Scroll the diagram sideways to see the full resource.
| Distance (km) | Trees within 50 m |
|---|---|
| 1 | 22 |
| 2 | 10 |
| 3 | 4 |
| 4 | 4 |
| 5 | 10 |
| 6 | 22 |
1 km: 22 trees; 2 km: 10 trees; 3 km: 4 trees; 4 km: 4 trees; 5 km: 10 trees; 6 km: 22 trees
1 mark1. Which description best fits the pattern?
1 mark2. Which conclusion is justified?
These are broad age categories, not equal five-year bands. The six bars together sum to 100%.
Scroll the diagram sideways to see the full resource.
| Age group | Male (% of total) | Female (% of total) |
|---|---|---|
| 0–14 | 13 | 20 |
| 15–64 | 32 | 26 |
| 65+ | 5 | 4 |
0–14: male 13%, female 20%; 15–64: male 32%, female 26%; 65+: male 5%, female 4%
1 mark1. What percentage of the whole population is aged 0–14?
1 mark2. Calculate dependants per 100 people aged 15–64, using 0–14 and 65+ as dependants.
Fictional training exercise. Two observers rate litter on a five-point scale. Higher scores mean less litter.
| Site | Observer A | Observer B |
|---|---|---|
| Square | 4 | 2 |
| Station | 3 | 3 |
| Park | 5 | 4 |
1 mark1. What is the largest absolute difference between observers?
1 mark2. Which step best improves consistency?
Original fictional model and resources: GeographyHelp.
Use this where a fully checked route-specific profile is not available. No automatic mark or predicted grade is provided.
Original application practice, not a complete examination paper. No assessed investigation, assignment, folio or restricted resource is reproduced.
The linked template fixes its coherent core resources. Resource type and skill selections control the explicitly labelled supplementary geographical skills.
1. Calculate the wind annual output in GWh to two decimal places. (1 mark)
70.08
20 × 40/100 × 8,760 ÷ 1,000.
2. What percentage of annual local demand could wind output equal, ignoring timing and losses? (1 mark)
70.08
70.08 ÷ 100 × 100.
3. Explain why annual output equal to a large share of demand does not guarantee a secure supply at all times. (4 marks)
Demand and output vary through time. Storage, backup, network capacity and losses matter; an annual total cannot show simultaneous availability.
4. Which comparison uses matching quantities? (1 mark)
Annual output in GWh versus annual demand in GWh
Power and energy have different units and meanings.
5. Recommend an option for energy demand, output and competing priorities. Use at least two resources, compare an alternative and address an uncertainty or stakeholder concern. (6 marks)
Different options are defensible. State your priority, use accurate evidence from the energy model, explain the geographical relationships and compare benefits and limitations. Energy security depends on demand, timing, infrastructure, costs and environmental consequences. No option is universally best.
1. Which class contains district C? (1 mark)
20 to less than 30%
24 is at least 20 and below 30. For this 1-mark task, credit the keyed option.
2. How many households in A reported damage? (1 mark)
20
2% of 1,000 = 20. For this 1-mark task, credit the accepted numerical answer in the requested units.
3. Does the darkest district necessarily have the most affected households? (1 mark)
No; the household denominators differ.
A high rate in a small population can represent fewer households than a lower rate in a large population. For this 1-mark task, credit the keyed option.
1. What is the difference in height between the highest and lowest measured points, in metres? (1 mark)
600
640 − 40 = 600 m. For this 1-mark task, credit the accepted numerical answer in the requested units.
2. For the section from 2 km to 3 km, express the gradient as 1:n. Enter n. (1 mark)
5
The horizontal run is 1,000 m and the rise is 200 m. Divide horizontal run by rise. For this 1-mark task, credit the accepted numerical answer in the requested units.
3. Can the drawn line angle be used directly as the real slope angle? (1 mark)
No; the horizontal and vertical scales differ.
Use real vertical and horizontal distances. Vertical exaggeration changes the displayed angle. For this 1-mark task, credit the keyed option.
1. Which action best investigates the 29-second result? (1 mark)
Repeat the trial and check for an obstruction
Retain the raw result, record any obstruction and repeat. A different value alone does not prove an error. For this 1-mark task, credit the keyed option.
2. What was velocity in trial 1, in m/s? (1 mark)
0.5
5 m ÷ 10 s = 0.5 m/s. For this 1-mark task, credit the accepted numerical answer in the requested units.
3. Why might surface floats overestimate mean velocity? Develop two linked points and support them with the resource or a methodological reason. (4 marks)
Friction often slows water near the bed and banks. Surface flow may be faster than the average through the section; wind and float paths also affect the estimate.
Friction often slows water near the bed and banks. Surface flow may be faster than the average through the section; wind and float paths also affect the estimate. Credit up to 2 marks per relevant developed point (1 for the point, 1 for a correct reason, evidence link or implication), up to 4 marks. Original credit guidance: credit relevant explained reasoning; accept accurate alternatives.
1. Which description best fits the pattern? (1 mark)
Negative association
Tree counts generally fall as distance increases. For this 1-mark task, credit the keyed option.
2. Which conclusion is justified? (1 mark)
The observations show an association, but do not establish its cause.
An observational association alone cannot establish causation or universal applicability. For this 1-mark task, credit the keyed option.
1. What should the missing reading be recorded as? (1 mark)
A clearly labelled missing value
Zero is a measured count; missing means no valid observation. Inventing a reading changes the evidence. For this 1-mark task, credit the keyed option.
2. Calculate the mean of the three observed counts. (1 mark)
32
(20 + 32 + 44) ÷ 3 = 32; the missing observation is excluded and the sample size is reported. For this 1-mark task, credit the accepted numerical answer in the requested units.
3. How does missing site C affect conclusions? Develop two linked points and support them with the resource or a methodological reason. (4 marks)
It reduces spatial coverage and may bias the result if C differs systematically. A repeat visit may help, but its different timing must be recorded.
It reduces spatial coverage and may bias the result if C differs systematically. A repeat visit may help, but its different timing must be recorded. Credit up to 2 marks per relevant developed point (1 for the point, 1 for a correct reason, evidence link or implication), up to 4 marks. Original credit guidance: credit relevant explained reasoning; accept accurate alternatives.
1. What is the difference in height between the highest and lowest measured points, in metres? (1 mark)
900
920 − 20 = 900 m. For this 1-mark task, credit the accepted numerical answer in the requested units.
2. For the section from 2 km to 3 km, express the gradient as 1:n. Enter n. (1 mark)
3.3333333333333335
The horizontal run is 1,000 m and the rise is 300 m. Divide horizontal run by rise. For this 1-mark task, credit the accepted numerical answer in the requested units.
3. Can the drawn line angle be used directly as the real slope angle? (1 mark)
No; the horizontal and vertical scales differ.
Use real vertical and horizontal distances. Vertical exaggeration changes the displayed angle. For this 1-mark task, credit the keyed option.
1. How many sites meet every eligibility condition? (1 mark)
3
Apply all three eligibility conditions to each site before comparing journey times. For this 1-mark task, credit the accepted numerical answer in the requested units.
2. Which site does the stated rule select? (1 mark)
D
D is eligible and has the lowest walk time among eligible sites (11 minutes). For this 1-mark task, credit the keyed option.
3. What does the exercise establish? (1 mark)
Which site meets the supplied rule; other planning evidence is still needed.
Overlay screening cannot replace permissions, surveys, consultation or a fuller appraisal. For this 1-mark task, credit the keyed option.
1. What rank do the two distance values of 2 km receive? (1 mark)
2.5
They occupy positions 2 and 3; the average is 2.5. For this 1-mark task, credit the accepted numerical answer in the requested units.
2. Calculate Spearman’s coefficient to two decimal places. (1 mark)
0.94
Using average ranks gives 0.94. Its sign describes the direction of the monotonic association. For this 1-mark task, credit the accepted numerical answer in the requested units.
3. What does this calculation establish? (1 mark)
A rank association in these observations.
A coefficient is not a significance decision. Independence, sample design and an appropriate test would need consideration. For this 1-mark task, credit the keyed option.
1. Which description best fits the pattern? (1 mark)
Positive association
Tree counts generally rise as distance increases. For this 1-mark task, credit the keyed option.
2. Which conclusion is justified? (1 mark)
The observations show an association, but do not establish its cause.
An observational association alone cannot establish causation or universal applicability. For this 1-mark task, credit the keyed option.
1. Which description best fits the pattern? (1 mark)
Curved association
Tree counts fall and then rise. A curved relationship can exist even when a straight-line correlation is zero. For this 1-mark task, credit the keyed option.
2. Which conclusion is justified? (1 mark)
The observations show an association, but do not establish its cause.
An observational association alone cannot establish causation or universal applicability. For this 1-mark task, credit the keyed option.
1. What percentage of the whole population is aged 0–14? (1 mark)
33
Add the male and female percentages for this age group. For this 1-mark task, credit the accepted numerical answer in the requested units.
2. Calculate dependants per 100 people aged 15–64, using 0–14 and 65+ as dependants. (1 mark)
72.41
(33 + 9) ÷ 58 × 100. This is an age-based ratio, not a count of employment status. For this 1-mark task, credit the accepted numerical answer in the requested units.
1. What is the largest absolute difference between observers? (1 mark)
2
At the square, |4 − 2| = 2. For this 1-mark task, credit the accepted numerical answer in the requested units.
2. Which step best improves consistency? (1 mark)
Agree descriptive criteria and practise scoring together
Shared criteria and calibration address interpretation differences while retaining the original observations. For this 1-mark task, credit the keyed option.