2025 CCEA AS-Level Geography 3910 Practice Paper with Answers

A credible in-channel hazard must be identified from the fieldwork context described (river fieldwork on foot, wading to take width/depth/velocity/clast readings): the main physical risks are slipping on wet rocks/uneven bedload, being swept off balance by the current, or cold-water immersion. A named, specific control measure must directly reduce that named hazard, not just be a generic safety statement. Example full-mark answer: 'Hazard: the candidate could slip on wet, uneven bedload clasts while wading across the channel to take mid-channel readings. Control measure: wearing sturdy footwear with a good tread (e.g. wellington boots) and using a ranging pole planted upstream for a third point of balance, combined with working with a partner so that help is immediately available if a fall occurs.' Final answer: Hazard = slipping/losing footing while wading; Control = appropriate footwear plus a balance pole and working in pairs.

[4] total. Hazard identification, specific to river fieldwork and clearly linked to the Fieldwork Statement: [1] named hazard (e.g. slipping, cold water, being swept off feet); [1] brief elaboration of how/when it could occur during this specific investigation. Control measure: [1] named, practical control measure; [1] explanation of how it reduces the specific risk named. Award full marks for any other valid hazard/control pairing consistent with the scenario (e.g. hazard = sudden rise in water level from rain upstream; control = checking weather forecast and having an agreed evacuation signal). Reject generic, unlinked answers (e.g. 'wear a hi-vis vest' with no hazard link) — cap at [1].

Two distinct, developed reasons are required, each tied to the specific aim of testing whether velocity changes systematically with distance downstream. Reason 1 — representativeness: choosing sites at roughly even intervals (150 m, 450 m, 900 m, 1400 m, 2000 m, 2700 m) means the whole course from near-source to lower-course conditions is sampled, rather than risking all sites clustering in one easily-accessible short stretch, which random selection could produce by chance. Reason 2 — validity of the hypothesis test: because the aim is specifically to relate velocity to distance downstream, systematically spacing sites along that one controlled variable removes the risk that a purely random selection leaves distance downstream poorly or unevenly sampled, which would make the Spearman's rank test less reliable. Final answer: reasons = even representation of the whole river course, and a fairer/more reliable test of the specific distance–velocity hypothesis.

[4] total: [2] per developed reason (accept up to 2 reasons). [1] for stating a valid reason (e.g. even coverage of the river course; more reliable/valid test of the specific hypothesis; avoids clustering/bias of purely random site choice); [1] for developing it with explicit reference to this investigation (distance downstream as the key variable, or the six named site distances). Accept other valid reasons, e.g. ease of comparison between evenly-spaced sites, or practical access considerations, if properly explained. Do not credit vague statements such as 'it is more accurate' with no development.

A correctly sequenced description of a recognised primary velocity method is required (float method or flow meter/current meter, both valid). Using the float method as the model answer: (1) a straight, uniform section of channel is measured out (e.g. 5 m) and marked with two ranging poles; (2) a float (e.g. an orange or a cork) is released just upstream of the first pole and a stopwatch started as it passes the first pole, stopped as it passes the second; (3) velocity = distance ÷ time; (4) the reading is repeated at least three times at each site and the results averaged to reduce the effect of anomalous readings; (5) because a float travels on the surface, where friction with the bed and banks is least, the result is multiplied by a correction factor (commonly 0.85) to estimate the mean velocity of the whole cross-section, which is the value that would be recorded in the Table of Data. Final answer: primary method = float-and-stopwatch timing over a measured distance, repeated and averaged, with a surface correction factor applied.

Level of response, [5] marks. Level 1 (1–2 marks): basic naming of a method (e.g. 'used a float') with little or no procedural detail. Level 2 (3–4 marks): a mostly complete, logically sequenced description of the method (measuring a distance, timing, calculating velocity) with limited reference to repeats/averaging or correction factor. Level 3 (5 marks): a full, clearly sequenced description covering measured distance, timing, the velocity formula, repeat readings/averaging to improve reliability, AND the surface correction factor (or equivalent detail if a flow meter is described instead, e.g. taking readings at 0.2 and 0.8 of depth). Accept a flow-meter/current-meter based answer marked to the same standard (correct positioning in the water column, units, repeats).

Distance ranks (1 = smallest): Site 1=1, Site 2=2, Site 3=3, Site 4=4, Site 5=5, Site 6=6 (distance rises monotonically by construction). Velocity ranks (1 = smallest, values 0.28, 0.36, 0.33, 0.47, 0.53, 0.61): Site 1 (0.28)=1, Site 3 (0.33)=2, Site 2 (0.36)=3, Site 4 (0.47)=4, Site 5 (0.53)=5, Site 6 (0.61)=6. So by site: Site1=1, Site2=3, Site3=2, Site4=4, Site5=5, Site6=6. d = distance rank − velocity rank: Site1: 1−1=0; Site2: 2−3=−1; Site3: 3−2=1; Site4: 4−4=0; Site5: 5−5=0; Site6: 6−6=0. d² values: 0, 1, 1, 0, 0, 0. \( \sum d^2 = 2 \). n = 6, so \( n(n^2-1) = 6 \times 35 = 210 \). \( r_s = 1 - \dfrac{6 \times 2}{210} = 1 - \dfrac{12}{210} = 1 - 0.0571 = 0.9429 \), i.e. \( r_s \approx 0.94 \) (2 d.p.). Second-route check: since only sites 2 and 3 swap rank (one transposition, contributing d² of 1+1=2) and all other ranks match exactly, \( \sum d^2 \) must equal 2 — this confirms the table calculation independently. (b) Since the calculated \( r_s = 0.94 \) is greater than the critical value of 0.886 for n = 6 at the 5% significance level, the result is statistically significant. The null hypothesis (H0: there is no relationship between distance downstream and velocity) is rejected, and the alternative hypothesis is accepted: there is a significant, strong positive relationship between distance downstream and velocity in this investigation. Final answer: \( r_s = 0.94 \); significant, reject H0.

(a) [5]: [1] correct ranking of distance downstream; [1] correct ranking of velocity; [1] correct d values (or correctly derived d² column); [1] correct \( \sum d^2 = 2 \); [1] correct substitution and final value \( r_s = 0.94 \) (accept 0.943 or 0.9429, and accept an own-figure-rule value if ranks/d² carried forward consistently from an earlier ranking slip). (b) [2]: [1] correct comparison of the candidate's \( r_s \) value against the critical value 0.886 with a correct significant/not-significant conclusion (own figure rule applies); [1] correct statement in terms of the hypothesis (reject/accept H0, referring to distance downstream and velocity).

The Bradshaw Model predicts that as distance downstream increases: discharge, channel width, channel depth and velocity all increase, while channel bed material (load) particle size and, usually, gradient decrease. The apparent paradox (velocity rising while the energy-giving gradient falls) is explained by channel efficiency. Efficiency is measured by the hydraulic radius (cross-sectional area ÷ wetted perimeter): downstream, the channel becomes wider and deeper (as shown in the Table of Data, width rising from 1.2 m to 5.6 m and depth from 0.18 m to 0.52 m), so the cross-sectional area grows faster than the wetted perimeter, raising the hydraulic radius. A higher hydraulic radius means a smaller proportion of the water mass is in frictional contact with the bed/banks, so less energy is lost to friction. Second, bed material becomes smaller and better sorted downstream (mean clast size falling from 68 mm to 9 mm in the data), reducing bed roughness and further cutting frictional energy loss. With less energy lost to friction, a larger share of the available potential energy (even though gradient itself is lower) is converted into the kinetic energy of flow, so velocity increases. This explains the significant positive Spearman's rank result (rs = 0.94) obtained between distance downstream and velocity. Final answer: velocity rises downstream mainly because increasing hydraulic radius and decreasing bed roughness reduce frictional energy loss, allowing more efficient flow despite a falling gradient — consistent with the Bradshaw Model.

Level of response, [7] marks. Level 1 (1–3): basic, general statement that velocity increases downstream 'because the channel gets bigger/smoother', with little explanation of the mechanism; may simply restate the Bradshaw Model trend without explaining WHY. Level 2 (4–5): sound explanation referring to at least one correct mechanism (e.g. hydraulic radius increasing, OR reduced bed roughness from smaller clast size) with some use of the candidate's own data (widths/depths/clast sizes from the table). Level 3 (6–7): a well-developed explanation that correctly links BOTH increasing hydraulic radius (width/depth increasing faster than wetted perimeter) AND decreasing bed roughness (falling clast size) to reduced frictional energy loss, explicitly resolving why velocity can rise even as gradient falls, with accurate use of specific data values from the Table of Data and correct reference to the Bradshaw Model.

A valid extension must genuinely add new data or a new comparison relevant to the stated aim (testing the Bradshaw Model / distance–velocity relationship), not simply repeat the existing method. Strong example: measure cross-sectional area at each site (by taking depth readings at regular intervals across the channel and using the mid-point/trapezoidal rule) so that discharge = cross-sectional area × velocity can be calculated at each of the six sites; plotting discharge against distance downstream would test another core Bradshaw Model prediction and would help explain whether the velocity increase is linked to increasing discharge, strengthening the conclusions of the original investigation. Other acceptable extensions: repeating the survey after a rainfall event to see whether the velocity–distance relationship still held under higher discharge conditions; adding further sites beyond 2700 m to see if the trend continues into the lower course. Final answer: measure cross-sectional area at each site to calculate discharge and test whether discharge, like velocity, increases downstream.

[3] total: [1] a specific, relevant extension named (not just 'collect more data'); [1] explanation of how it would be carried out; [1] explanation of how it links to/develops the original aim (Bradshaw Model / distance–velocity relationship). Award full marks for any other geographically valid extension meeting all three criteria (e.g. repeating fieldwork in different flow conditions, adding sites further downstream, or measuring bed roughness quantitatively).

Role of the quadrat: it provides a fixed, standard-sized sampling area (commonly 1 m²), within which percentage vegetation cover, bare sand, or the number/type of species present can be recorded systematically; using the same quadrat size at every site along the transect makes the readings directly comparable, so any change in cover between the embryo and fixed dunes can be attributed to real spatial variation rather than to inconsistent sampling area. Why stratified (interval) sampling along a transect is more appropriate than pure random sampling: the aim of the investigation is specifically to show how vegetation cover changes with distance inland (through the succession from embryo to fixed dunes), so sampling needs to be structured along that known environmental gradient. Purely random sampling across the whole dune system could, by chance, cluster all sampling points in one zone (e.g. the fixed dunes) and under-represent or miss the embryo dune zone entirely, making the succession pattern impossible to detect; sampling at fixed, regular intervals along a transect guarantees even coverage of every successional stage, giving a fair and representative test of the change in vegetation with distance from the sea. Final answer: quadrats give a standard, repeatable sampling unit for cover data; interval sampling along a transect ensures every stage of the dune succession is represented, which random sampling cannot guarantee.

[5] total. Quadrat role: [1] correct basic description (standard-area frame for recording cover/species); [1] development (comparability between sites, or specific reference to % cover/species counts). Sampling justification: [1] valid reason stated (ensures all zones/stages represented, or avoids missing a zone by chance); [2] developed explanation clearly linking the reason to the specific aim of showing the successional trend from embryo to fixed dunes, with reference to the risk that random sampling could fail to capture the full gradient. Accept equivalent well-reasoned points for a different but comparable scenario-consistent apparatus (e.g. quadrat frame with sub-divisions for percentage estimation).

(a) n = 6, A = 100 km², d̄ = 3.2 km. First calculate n/A = 6/100 = 0.06. Take the square root: √0.06 = 0.2449 (4 d.p.). Then Rn = 2 × 3.2 × 0.2449 = 6.4 × 0.2449 = 1.5677, which rounds to 1.57 (2 d.p.). Second-route check: compute 2 × d̄ first (2 × 3.2 = 6.4), then multiply by √(n/A) (0.2449), giving the same 6.4 × 0.2449 = 1.5677 ≈ 1.57 — both orders of operation agree, confirming the value. (b) An Rn value of 1.57 lies on the scale between 1.0 (random) and 2.15 (perfectly uniform/regular), and because 1.57 is more than halfway from random towards perfectly regular (57% of the way from 1.0 to 2.15), the pattern is best described as tending towards a regular/dispersed distribution rather than a random one. This supports the stated hypothesis that rural settlements in this area show a regular (dispersed) distribution, though the pattern is not perfectly uniform (which would require Rn = 2.15). Final answer: Rn = 1.57, indicating a dispersed/regular tendency that supports the hypothesis.

(a) [4]: [1] correct calculation of n/A = 0.06; [1] correct square root, √0.06 = 0.24 (or 0.245/0.2449, accept rounding); [1] correct substitution 2 × 3.2 × 0.245 (or equivalent); [1] correct final value Rn = 1.57 (accept 1.56–1.58 depending on rounding stage; own figure rule applies for a mislabelled intermediate step carried through consistently). (b) [2]: [1] correct positioning of the result on the 0–2.15 scale (between random and regular, closer to regular); [1] explicit statement of the implication for the hypothesis (supports the hypothesis of a regular/dispersed pattern, though not perfectly uniform).

A regular/dispersed settlement pattern (Rn = 1.57, tending towards uniform) means the settlements are spread fairly evenly across the study area rather than being clustered together in one part of it. For a planning authority allocating a single new service point (e.g. a GP surgery or primary school), this spatial evenness is directly useful: it suggests that no single settlement is obviously 'central' to a tight cluster of the others, so the optimal location is likely to be at or near the geometric centre of the whole study area, which would minimise the average distance travelled by residents across all six settlements and ensure a more equitable level of access than if the settlements were clustered (in which case locating near the cluster would serve most people, but leave outliers underserved). The result therefore supports a location strategy based on minimising total/average travel distance across a dispersed population, rather than one based on serving a dominant cluster. Final answer: the fairly regular spacing supports siting the new service centrally within the study area, to minimise average travel distance and provide equitable access across all six evenly-spread settlements.

[4] total: [1] correct identification that the pattern is fairly regular/dispersed (not clustered); [1] valid planning implication stated (e.g. central location, equitable access, minimising average travel distance); [2] developed explanation clearly connecting the specific Rn result to the practical decision (why regular spacing changes the optimal location strategy compared with a clustered pattern). Accept alternative valid reasoning, e.g. discussing the need for more than one service point if settlements are far apart, provided it is logically developed from the Rn result.

Stratified sampling requires two linked ideas for full marks: first, that the population or area is divided into pre-identified sub-groups/zones (strata) which are meaningful to the investigation (e.g. dividing a river's length into upper, middle and lower course zones, or a town into distance bands from the CBD); second, that sampling points (or respondents/quadrats) are then chosen from within each stratum, commonly using random or systematic selection within each zone, so that every stratum is represented in the final sample, often in proportion to its size or importance. This differs from simple random sampling, which does not guarantee representation of every sub-group. Final answer: dividing the study area/population into relevant sub-groups (strata) and sampling from within each one so that every sub-group is represented.

[2] total: [1] correct reference to dividing the population/area into sub-groups or zones (strata) relevant to the investigation; [1] correct reference to sampling being carried out within each stratum/sub-group so that all are represented (may be combined into one well-expressed sentence covering both ideas for [2]).

First arrange the seven values in ascending order: 14, 14, 14, 15, 15, 16, 19. Median: with n = 7 (odd), the median is the middle (4th) value in the ordered list, which is 15 °C. Mode: the most frequently occurring value is 14 °C, which appears three times (more than any other value). Range: highest value − lowest value = 19 − 14 = 5 °C. Second-route check: counting occurrences directly from the raw data — 14 appears on days 1, 4 and 7 (three times, confirming the mode); the ordered list confirms the 4th of 7 values is 15 (confirming the median); and 19 − 14 = 5 is a simple subtraction, independently verified. Final answer: median = 15 °C, mode = 14 °C, range = 5 °C.

[3] total: [1] correct median (15 °C), with values shown in ascending order; [1] correct mode (14 °C); [1] correct range (5 °C, with 19 − 14 shown). Award marks independently for each correct statistic even if another is wrong (own figure rule does not apply here as each statistic is calculated independently from the same raw data).

(a) Range = highest − lowest = 76 − 12 = 64. Equal interval method divides this range into 4 equal classes: class width = 64 ÷ 4 = 16. Starting from the minimum value (12), the four class boundaries are: Class 1: 12.0–27.9 (12 + 16 = 28, so the class runs up to just under 28); Class 2: 28.0–43.9; Class 3: 44.0–59.9; Class 4: 60.0–76.0 (covering up to the maximum value). Second-route check: 12 + 16 = 28, 28 + 16 = 44, 44 + 16 = 60, 60 + 16 = 76 — four consecutive additions of the class width from the minimum correctly reach the maximum value (76), confirming the four boundaries are correctly and evenly spaced. (b) Placing each ward's value into a class: Ward A (12) → Class 1; Ward F (20) → Class 1; Ward B (34) → Class 2; Ward E (45) → Class 3; Ward C (58) → Class 3; Ward D (76) → Class 4. The highest class (Class 4, containing Ward D at 76 people/ha) is shaded darkest, and the lowest class (Class 1, containing Ward A at 12 people/ha) is shaded lightest, following the standard choropleth convention that shade intensity darkens as the class value increases, moving through a continuous, non-overlapping, unbroken tonal gradation from light to dark. Final answer: class intervals 12.0–27.9 / 28.0–43.9 / 44.0–59.9 / 60.0–76.0; Ward D darkest, Ward A lightest; shading darkens with increasing value.

(a) [3]: [1] correct range calculation (64) and class width (16); [1] correct first two class boundaries shown; [1] all four correct, non-overlapping, unbroken class boundaries stated (accept minor variation in boundary notation, e.g. 12–27, 28–43, as long as classes are equal width and non-overlapping). (b) [3]: [1] correct identification of Ward D as darkest; [1] correct identification of Ward A as lightest; [1] correct description of the general shading convention (progressively darker shading for higher values, unbroken/non-overlapping gradation).

Two distinct, developed limitations are required. Limitation 1 — false homogeneity within zones: choropleth maps shade each administrative unit (here, each ward) as a single uniform colour based on its average or total value, which implies the value is constant throughout that zone; in reality, population density (or any mapped variable) typically varies considerably within a ward (for example, a ward could contain both a high-density housing estate and low-density open land), so real internal spatial variation is hidden, and the boundaries of the shaded zones are administrative rather than reflecting genuine breaks in the underlying pattern. Limitation 2 — sensitivity to classification choices: the visual pattern produced depends heavily on subjective cartographic decisions such as the number of classes used and the classification method chosen (e.g. equal interval, as used in the previous question, versus quantile or natural breaks); different choices can make the same underlying data look markedly more or less varied/clustered, so two choropleth maps of identical data can look quite different, which is a limitation when a map is meant to objectively represent geographical reality. Other acceptable limitations include: difficulty distinguishing between similar shades on the printed/screen map, or the map showing intensity/density but not the actual number of people involved (a small ward and a large ward in the same class could contain very different total populations). Final answer: (1) it assumes uniform values within each zone, hiding internal variation; (2) the apparent pattern is sensitive to the classification method/number of classes chosen, which is a subjective cartographic decision.

[4] total: [2] per limitation (accept the two strongest of any valid limitations, up to 2). Per limitation: [1] a valid limitation named; [1] a developed explanation of why it is a problem, ideally with reference to this ward/population density context. Valid limitations include: masking of internal variation within zones (false homogeneity); sensitivity of the visual pattern to class-interval method/number of classes chosen; difficulty distinguishing similar shades; shows relative density/intensity but not absolute totals for zones of different size. Do not credit an unexplained single word (e.g. 'inaccurate') without development.