All practice questions

Question 6181Question →

An agricultural cooperative organization lists bulk unprocessed palm oil on a dedicated digital trading platform, which is subsequently purchased by an industrial soap manufacturing enterprise for factory production. Which e-commerce model is demonstrated in this transaction?

Business-to-Business (B2B)

Business-to-Consumer (B2C)

Consumer-to-Business (C2B)

Consumer-to-Consumer (C2C)

Show answer & explanation

Answer: Business-to-Business (B2B)

Answer

Business-to-Business (B2B)

The transaction occurs between two commercial entities—an agricultural cooperative selling bulk raw materials online and a manufacturing enterprise purchasing them for factory processing. Electronic trade conducted between business organizations is classified as Business-to-Business (B2B).

Step-by-Step Solution

1

Identify the nature of the selling entity in the transaction.

The agricultural cooperative represents a commercial producer entity selling bulk raw materials.

Determining the seller's institutional status establishes the primary side of the e-commerce transaction model.

2

Identify the nature of the purchasing entity in the transaction.

The industrial manufacturing enterprise is a corporate buyer acquiring inputs for processing.

Determining the buyer's status completes the structural classification of the trade relationship.

3

Classify the commercial relationship model.

Commercial transactions conducted electronically between two business enterprises constitute Business-to-Business (B2B) e-commerce.

Inter-firm trade conducted via online networks falls directly under the B2B framework.

Key Concept

Business-to-Business (B2B) E-Commerce Model

Question 6182Question →

Arrange the following sequential stages of timber extraction and transport in the equatorial rainforests of Central and West Africa in the correct chronological order from first to last:

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct chronological order begins with locating and marking mature commercial hardwood trees, followed by felling the selected trees, then hauling or skidding the heavy logs to riverbanks or roadsides, and finally floating or trucking the logs to coastal sawmills and export ports.

Lumbering in African rainforests follows a clear chronological workflow: locating scattered hardwood trees, felling them, dragging/skidding the logs to collection points, and finally transporting them via river or truck to coastal mills and ports.

Step-by-Step Solution

1

Identify the initial field activity required before harvesting.

Locating and marking mature commercial hardwood trees.

Because commercial timber species in African tropical rainforests grow far apart from one another, prospectors must locate suitable trees first.

2

Determine the extraction step following tree selection.

Felling the selected trees and removing surrounding branches.

Once a tree is identified and marked, lumbermen fell it and prepare the trunk into manageable log sections.

3

Identify the local transport step inside the forest.

Hauling or skidding the heavy logs through forest tracks to riverbanks or roadsides.

Logs must be moved out of the dense forest interior to an accessible collection point.

4

Determine the final transport step to industrial destinations.

Floating or trucking the logs to coastal sawmills and export ports.

Waterways and heavy timber trucks carry the logs downstream or overland to coastal sawmills and shipping ports.

Key Concept

Sequential stages of timber extraction and transport in African equatorial forests

Estimated Time:45s

Question 6183Question →

On July 1, 2026, an enterprise operating a petty cash imprest system had a remaining cash balance of $₦6,500$ in the petty cash box from the previous month. The chief cashier replenished the float to its agreed limit of $₦40,000$ . During July, disbursements were made for office stationery ( $₦12,300$ ), bus fares ( $₦7,500$ ), and courier services ( $₦9,400$ ). What is the amount of cash in Naira ( $₦$ ) remaining in the petty cash box at the end of July 2026 before any reimbursement is made?

Show answer & explanation

Answer: 10800

Answer

The amount of cash remaining in the petty cash box at the end of July 2026 before reimbursement is

₦10,800

.

Under the imprest system, the float is restored to the agreed limit at the start of the accounting period (

₦40,000

). Total expenses paid during July equal

₦12,300 + ₦7,500 + ₦9,400 = ₦29,200

. Therefore, the cash remaining in the petty cash box before any reimbursement is

₦40,000 - ₦29,200 = ₦10,800

.

Step-by-Step Solution

1

Calculate the total petty cash expenses disbursed during the month

Total Disbursements =

₦12,300 + ₦7,500 + ₦9,400 = ₦29,200

Summing all paid petty cash vouchers gives the total cash spent out of the petty cash box during July.

2

Calculate the remaining cash balance before reimbursement

Remaining Cash = Imprest Float - Total Disbursements =

₦40,000 - ₦29,200 = ₦10,800

Under the imprest system, cash remaining in the box equals the total agreed imprest float minus total cash spent.

Key Concept

Imprest System Remaining Cash Balance Calculation

Question 6184Question →

Match each Köppen climate classification code on the left with its correct characteristic rainfall and temperature regime on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Af pairs with Tropical wet climate with high temperatures and rainfall year-round; BWh pairs with Hot desert climate with extreme aridity; Cs pairs with Mediterranean climate featuring dry summers and rainy winters; ET pairs with Tundra climate with cool summers and cold winters.

Each Köppen climate code corresponds to specific temperature and precipitation thresholds: Af represents constantly wet equatorial regions, BWh denotes hyper-arid hot deserts, Cs represents dry-summer Mediterranean zones, and ET represents cold polar tundras.

Step-by-Step Solution

1

Identify the primary temperature/moisture group represented by each capital letter.

A = Tropical humid, B = Dry/Arid, C = Mild temperate, E = Polar.

The first capital letter in the Köppen system defines the broad global thermal and moisture zone.

2

Interpret the lower-case and second capital modifier letters.

f = no dry season, W = desert (Wüste), h = hot, s = dry summer, T = tundra.

Secondary letters refine the climate type by seasonal rainfall pattern and thermal intensity.

3

Match each combined code to its corresponding description.

Af to Equatorial rainforest, BWh to Hot desert, Cs to Mediterranean, ET to Tundra.

Connecting the code meanings to geographic rainfall and temperature regimes establishes the correct pairs.

Key Concept

Köppen Climate Classification Scheme

Question 6185Question →

Match each urban settlement problem listed on the left with its primary defining characteristic or environmental manifestation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Urban Sprawl matches with uncontrolled horizontal expansion into rural hinterlands; Squatter Settlements match with makeshift residential developments on illegally occupied land lacking utilities; Traffic Congestion matches with over-saturation of road transit networks exceeding design capacity; and Urban Heat Island Effect matches with elevated ambient temperatures in central urban areas due to dense built infrastructure.

Each urban settlement challenge is paired accurately with its defining physical, socio-economic, or micro-climatic manifestation.

Step-by-Step Solution

1

Analyze the spatial growth problem of cities

Uncontrolled outward spatial expansion defines Urban Sprawl.

Rapid population growth often drives cities to engulf peripheral agricultural land.

2

Examine the social housing challenge arising from rapid urbanization

Informal housing on unserviced, unauthorized land defines Squatter Settlements.

Housing deficits in growing metropolitan areas force low-income migrants into self-built informal housing.

3

Assess urban transportation constraints

Vehicular gridlock exceeding road carrying capacity defines Traffic Congestion.

High automobile density combined with inadequate public transit networks results in bottleneck delays.

4

Evaluate micro-climatic environmental impacts of built environments

Thermal absorption by concrete and asphalt creating warmer urban microclimates defines the Urban Heat Island Effect.

Replacing natural land cover with artificial structures reduces evapotranspiration and traps solar radiation.

Key Concept

Urban Settlement Problems and Environmental Consequences

Estimated Time:1m 30s

Question 6186Question →

Pairs of non-profit entity financial transactions and accounting procedures are given below. Match each financial item to its exact treatment in the Income and Expenditure Account.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Honorarium paid to a guest lecturer matches with being debited as a revenue expenditure; Unrestricted legacy matches with being credited as general revenue income; Purchase price of a new bus matches with being excluded from the Income and Expenditure Account as a capital expenditure; Net proceeds from a special charity dinner matches with being credited as net income after deducting direct activity costs.

The Income and Expenditure Account operates on accrual accounting rules for revenue items only. Operational costs like honoraria are debited as revenue expenditure; unrestricted legacies are credited as general revenue income; capital purchases like vehicles are excluded and listed on the Balance Sheet; and special event earnings are credited net of direct activity costs.

Step-by-Step Solution

1

Distinguish between capital expenditure and revenue expenditure for non-profit entities

Identify that acquiring a long-term asset (club bus) is capital expenditure excluded from the Income and Expenditure Account, whereas an honorarium is an operating revenue expense debited to the account.

The Income and Expenditure Account strictly records revenue operations belonging to the current financial year.

2

Classify general non-recurring receipts according to accounting standards for NPOs

Categorize unrestricted legacies as revenue income credited directly to the Income and Expenditure Account.

Unless a bequest is explicitly earmarked for an endowment or capital fund by the donor, it forms part of general operational income.

3

Apply net presentation rules for special fund-raising activities

Match net charity dinner proceeds with the net profit entry credited to the Income and Expenditure Account.

Expenses directly tied to running a special social or fund-raising event are set off against the gross income of that specific event before transferring the surplus to the main account.

Key Concept

Revenue versus Capital Item Classification in Non-Profit Income and Expenditure Accounts

Question 6187Question →

In transportation geography, graph theory and topological metrics are used to measure network efficiency, structure, and connectivity. Match each network connectivity metric or concept on the left with its corresponding mathematical formula or structural definition on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Beta Index matches with link-to-vertex ratio calculation; Gamma Index matches with planar graph completeness calculation; Alpha Index matches with circuitness/cycle ratio calculation; Nodal Accessibility matches with the ease of reaching a vertex from all other points.

Each transport graph index uniquely describes a topological dimension of a transportation network. The Beta Index (

\beta = e/v

) measures simple connectivity/density; the Gamma Index (

\gamma

) evaluates network completeness against maximum planar connections; the Alpha Index (

\alpha

) measures circuit redundancy; and Nodal Accessibility describes the ease of reaching a vertex from all other nodes.

Step-by-Step Solution

1

Identify the definition of Beta Index

Beta Index (

\beta

) is computed as

\beta = \frac{e}{v}

, representing basic network linkage per node.

Beta index evaluates simple network structure without considering planar geometry limits.

2

Identify the definition of Gamma Index

Gamma Index (

\gamma

) compares actual edges

e

to maximum planar edges

3(v-2)

.

In a planar network graph, no two edges cross without creating a vertex; maximum link potential is

3(v-2)

.

3

Identify the definition of Alpha Index

Alpha Index (

\alpha

) evaluates circuitness via

\alpha = \frac{e - v + 1}{2v - 5}

.

Redundant routes or loops in a network are measured by independent cycles

u = e - v + 1

relative to maximum possible cycles.

4

Identify Nodal Accessibility

Nodal Accessibility represents the spatial centrality and path convenience of a specific vertex in relation to the whole system.

Accessibility is a property of a single node relative to the entire connectivity matrix of the transport network.

Key Concept

Topological Network Connectivity Indices in Transport Geography

Estimated Time:2m 0s

Question 6188Question →

Environmental degradation manifests differently across Nigeria's ecological zones. While gully erosion severely threatens the southeastern states and desertification impacts the far north, oil spillage poses a severe threat to the Niger Delta ecosystem. Which specialized federal regulatory body in Nigeria is specifically mandated to monitor, detect, and respond to oil spills?

National Oil Spill Detection and Response Agency (NOSDRA)

National Environmental Standards and Regulations Enforcement Agency (NESREA)

Federal Environmental Protection Agency (FEPA)

Forestry Research Institute of Nigeria (FRIN)

Show answer & explanation

Answer: National Oil Spill Detection and Response Agency (NOSDRA)

Answer

The National Oil Spill Detection and Response Agency (NOSDRA) is the specialized regulatory agency in Nigeria mandated to monitor, detect, and coordinate response to oil spills.

The correct option is National Oil Spill Detection and Response Agency (NOSDRA). It was created in 2006 under the Ministry of Environment as a specialized institutional framework to implement the National Oil Spill Contingency Plan for Nigeria, focusing on environmental safety and restoration in oil-producing communities.

Step-by-Step Solution

1

Identify the specific environmental hazard mentioned in the prompt

The target environmental hazard is oil spillage occurring in the Niger Delta oil-producing region of Nigeria.

Different environmental hazards in Nigeria fall under specialized institutional mandates.

2

Evaluate the statutory mandate of Nigerian environmental regulatory agencies

The National Oil Spill Detection and Response Agency (NOSDRA) was established under Decree 15 of 2006 specifically for oil spill management, whereas NESREA handles broader environmental enforcement outside the oil sector.

Distinguishing between general environmental enforcement agencies and specialized sector-specific agencies is essential in resource management.

Key Concept

Institutional Frameworks for Environmental Management in Nigeria

Estimated Time:50s

Question 6189Question →

A ball of mass $0.20\text{ kg}$ moving horizontally towards a vertical wall at a speed of $15\text{ m s}^{-1}$ rebounds in the opposite direction at $10\text{ m s}^{-1}$ . If the impact with the wall lasts for $0.020\text{ s}$ , what is the magnitude of the average force exerted on the ball by the wall?

50\text{ N}

250\text{ N}

1250\text{ N}

0.10\text{ N}

Show answer & explanation

Answer: $250\text{ N}$

Answer

The magnitude of the average force exerted on the ball by the wall is

250\text{ N}

.

The average force is determined by Newton's second law (

F = \frac{\Delta p}{\Delta t}

). Because the ball rebounds in the opposite direction, velocity changes from

+15\text{ m s}^{-1}

to

-10\text{ m s}^{-1}

, yielding a total velocity change magnitude of

25\text{ m s}^{-1}

. Multiplying by mass (

0.20\text{ kg}

) gives an impulse magnitude of

5.0\text{ N s}

. Dividing impulse by the contact time (

0.020\text{ s}

) yields

250\text{ N}

.

Step-by-Step Solution

1

Establish vector direction and assign initial and final velocities

Initial velocity

u = +15\text{ m s}^{-1}

, final velocity

v = -10\text{ m s}^{-1}

Velocity is a vector quantity, so reversing direction requires a opposite sign convention.

2

Calculate the change in momentum (impulse)

\Delta p = m(v - u) = 0.20 \times (-10 - 15) = 0.20 \times (-25) = -5.0\text{ N s}

Impulse is equal to the change in linear momentum.

3

Calculate the magnitude of the average force

F = \frac{|\Delta p|}{\Delta t} = \frac{5.0\text{ N s}}{0.020\text{ s}} = 250\text{ N}

By Newton's second law, average force equals rate of change of momentum (F = \Delta p / \Delta t).

Key Concept

Impulse-Momentum Theorem and Vector Nature of Momentum

Estimated Time:1m 0s

Question 6190Question →

At a weather observation station in Jos, Nigeria, a meteorologist recorded the following daily temperature readings using a Six's maximum and minimum thermometer over a four-day period:

- Day 1: Maximum = $31.0^\circ\text{C}$ , Minimum = $19.0^\circ\text{C}$
- Day 2: Maximum = $33.5^\circ\text{C}$ , Minimum = $17.5^\circ\text{C}$
- Day 3: Maximum = $28.0^\circ\text{C}$ , Minimum = $16.0^\circ\text{C}$
- Day 4: Maximum = $29.5^\circ\text{C}$ , Minimum = $18.5^\circ\text{C}$

Calculate the mean diurnal (daily) temperature range for this four-day period in degrees Celsius ( $^\circ\text{C}$ ).

Show answer & explanation

Answer: 12.75

Answer

The mean diurnal temperature range over the four-day period is

12.75^\circ\text{C}

.

The mean diurnal range measures the average difference between daily peak heat and night cooling over a given period. Subtracting minimum from maximum for each day yields

12.0^\circ\text{C}

,

16.0^\circ\text{C}

,

12.0^\circ\text{C}

, and

11.0^\circ\text{C}

. The average of these four values is

12.75^\circ\text{C}

.

Step-by-Step Solution

1

Find the diurnal range for each individual day

Day 1:

12.0^\circ\text{C}

, Day 2:

16.0^\circ\text{C}

, Day 3:

12.0^\circ\text{C}

, Day 4:

11.0^\circ\text{C}

The diurnal temperature range is defined as the difference between the maximum and minimum temperatures recorded in a single 24-hour period.

2

Calculate the sum of all daily ranges

51.0^\circ\text{C}

To compute an average over multiple days, the individual daily ranges must first be aggregated.

3

Compute the arithmetic mean across the 4 days

12.75^\circ\text{C}

Dividing the aggregate sum by the total number of observation days (4) yields the mean diurnal range.

Key Concept

Diurnal Temperature Range and Mean Calculation

Question 6191Question →

Adeola Traders operates a dual ledger system. On 1st July 2025, the debit balance of the Sales Ledger Control Account was $\text{N}145,000$ . During July, credit sales amounted to $\text{N}82,000$ , cash collected from debtors was $\text{N}94,000$ , discount allowed was $\text{N}4,000$ , and bad debts written off totaled $\text{N}3,000$ . A contra entry of $\text{N}16,500$ was set off between the sales ledger and purchases ledger. What is the closing debit balance of the Sales Ledger Control Account at the end of July 2025?

Show answer & explanation

Answer: 109500

Answer

109,500 Naira

Contra entries represent mutual set-offs between customer and supplier accounts. In the Sales Ledger Control Account, a contra entry is credited because it reduces the overall indebtedness of customers. Therefore, subtracting cash received, discounts allowed, bad debts, and the contra entry from total debits (opening balance plus credit sales) yields a closing debit balance of 109,500 Naira.

Step-by-Step Solution

1

Sum all debit entries in the Sales Ledger Control Account

Total Debits =

\text{N}145,000 + \text{N}82,000 = \text{N}227,000

Opening debit balance and credit sales increase the total receivables balance.

2

Sum all credit entries, including the contra set-off

Total Credits =

\text{N}94,000 + \text{N}4,000 + \text{N}3,000 + \text{N}16,500 = \text{N}117,500

Cash receipts, discounts allowed, bad debts written off, and contra entries reduce the debt owed by debtors.

3

Calculate the closing debit balance

Closing Balance =

\text{N}227,000 - \text{N}117,500 = \text{N}109,500

Subtracting total credits from total debits yields the net remaining debit balance.

Key Concept

Treatment of Contra Entries in Sales Ledger Control Account

Question 6192Question →

On a topographical survey map drawn to a scale of $1:40,000$ , Point X lies at an elevation of $250\text{ m}$ on a hillside, and Point Y lies further uphill along a straight spur at an elevation of $650\text{ m}$ . If the distance between Point X and Point Y measured on the map is $5\text{ cm}$ , the gradient of the slope between the two points can be expressed in the ratio form $1 : x$ . What is the numerical value of $x$ ?

Show answer & explanation

Answer: 5

Answer

The numerical value of

x

is 5.

The difference in elevation (Vertical Interval) between the two points is

400\text{ m}

. Converting the

5\text{ cm}

map measurement to ground distance using the

1:40,000

scale gives

2,000\text{ m}

(Horizontal Equivalent). Dividing the Vertical Interval by the Horizontal Equivalent yields

\frac{400}{2000} = \frac{1}{5}

, meaning the gradient expressed as

1 : x

has

x = 5

.

Step-by-Step Solution

1

Calculate the Vertical Interval (VI)

400 m

The vertical interval represents the elevation difference between Point Y and Point X:

650\text{ m} - 250\text{ m} = 400\text{ m}

.

2

Calculate the Horizontal Equivalent (HE)

2,000 m

At a map scale of

1:40,000

,

1\text{ cm}

on the map corresponds to

40,000\text{ cm} = 400\text{ m}

on the ground. A map measurement of

5\text{ cm}

equals

5 \times 400\text{ m} = 2,000\text{ m}

.

3

Determine the Gradient Ratio

1 : 5 (x = 5)

Gradient is determined by dividing Vertical Interval by Horizontal Equivalent:

\frac{400\text{ m}}{2,000\text{ m}} = \frac{1}{5}

. Writing this ratio as

1 : x

yields

x = 5

.

Key Concept

Slope and Gradient Calculation from Topographical Contour Maps

Question 6193Question →

Two adjacent drainage basins, Basin A and Basin B, experience identical climatic conditions and precipitation levels. A geomorphic survey records the following morphometric data:

Drainage Basin	Total Channel Length ( $L$ )	Total Basin Area ( $A$ )
Basin A	$180\text{ km}$	$60\text{ km}^2$
Basin B	$50\text{ km}$	$50\text{ km}^2$

Based on the drainage density of each basin, which of the following statements accurately compares their hydrological and geological characteristics?

Basin A has a higher drainage density (

3.0\text{ km/km}^2

), indicating predominantly impermeable surface rocks, higher surface runoff, and a faster flood response compared to Basin B.

Basin A has a lower drainage density (

0.33\text{ km/km}^2

) than Basin B (

1.0\text{ km/km}^2

), indicating greater surface permeability and enhanced groundwater infiltration in Basin A.

Basin A's drainage density of

3.0\text{ km/km}^2

is primarily produced by lower-course floodplain depositional features such as oxbow lakes and meander loops that lengthen channels.

Basin B has a higher drainage density than Basin A because V-shaped contour lines across its main channels point downstream towards the basin outlet.

Show answer & explanation

Answer: Basin A has a higher drainage density ( $3.0\text{ km/km}^2$ ), indicating predominantly impermeable surface rocks, higher surface runoff, and a faster flood response compared to Basin B.

Answer

Basin A has a higher drainage density (

3.0\text{ km/km}^2

), indicating predominantly impermeable surface rocks, higher surface runoff, and a faster flood response compared to Basin B.

The option identifying Basin A as having a higher drainage density (

3.0\text{ km/km}^2

) is correct because

D_d

is calculated by dividing total stream length (

180\text{ km}

) by total basin area (

60\text{ km}^2

). High values of drainage density reflect dense channel development, typical of impermeable rock strata where water cannot easily infiltrate, resulting in high surface runoff velocity.

Step-by-Step Solution

1

Calculate the drainage density (

D_d

) for Basin A using the formula

D_d = \frac{L}{A}

.

D_d(\text{Basin A}) = \frac{180\text{ km}}{60\text{ km}^2} = 3.0\text{ km/km}^2

.

Drainage density measures the total channel length per unit basin area.

2

Calculate the drainage density (

D_d

) for Basin B.

D_d(\text{Basin B}) = \frac{50\text{ km}}{50\text{ km}^2} = 1.0\text{ km/km}^2

.

Provides the baseline comparison value for Basin B.

3

Interpret the geomorphic and hydrological implications of high vs. low drainage density.

Basin A (

3.0\text{ km/km}^2 > 1.0\text{ km/km}^2

) has a higher stream network concentration, which signifies impermeable bedrock/clay soils, reduced infiltration, and rapid surface runoff leading to intense flood peaks.

High drainage density correlates with surface impermeability, sparse vegetation, and efficient surface runoff networks.

Key Concept

Drainage Density (

D_d

) and Basin Hydrology

Estimated Time:2m 0s

Question 6194Question →

In humid tropical regions with heavy seasonal rainfall, intense chemical alteration of granite bedrocks produces deep, clay-rich regolith layers in situ. Following a prolonged rainstorm, a large volume of this water-saturated regolith suddenly loses cohesion and moves rapidly downhill under the direct influence of gravity. Which of the following processes accounts for this rapid downhill movement of the saturated regolith?

Hydrolysis

Mudflow

Frost shattering

Metamorphic recrystallization

Show answer & explanation

Answer: Mudflow

Answer

Mudflow

The scenario describes mass wasting where gravity pulls heavy, water-saturated weathered material down a slope. A mudflow specifically refers to the rapid movement of fine-grained, highly saturated regolith down steep slopes after intense rainfall.

Step-by-Step Solution

1

Distinguish between weathering and mass wasting mechanisms described in the scenario.

Chemical weathering (hydrolysis) formed the clay-rich regolith in situ, whereas gravity caused the subsequent downhill movement.

Weathering involves stationary breakdown, while mass wasting involves downslope displacement driven by gravity.

2

Identify the specific mass wasting type matching the movement characteristics.

A rapid movement of heavily saturated, fluid-like clay regolith following heavy rainfall is categorized as a mudflow.

Mudflows occur when fine-grained weathered material becomes saturated with water and rapidly flows down slopes.

Key Concept

Mass wasting processes vs. in-situ weathering

Estimated Time:1m 15s

Question 6195Question →

Arrange the sequential stages involved in the process of frost shattering (freeze-thaw weathering) in chronological order, starting from the initial entry of moisture to the final disintegration of the rock face.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence begins with liquid water percolating into rock joints during warmer temperatures, followed by water freezing and expanding by about

9\%

as temperatures fall below

0^\circ\text{C}

. This frost wedging progressively widens rock fractures over repeated freeze-thaw cycles, eventually causing angular rock fragments to dislodge and form scree slopes at the base of the rock face.

The process of frost shattering follows a clear mechanical progression: liquid water must first occupy pre-existing fractures in the rock face; sub-zero temperatures then cause the trapped water to freeze and expand by roughly

9\%

, creating intense lateral pressure; recurrent freeze-thaw cycles continuously strain and widen these micro-fractures; and ultimately, angular fragments detach from the parent cliff and accumulate downslope as scree deposits.

Step-by-Step Solution

1

Identify the initial moisture entry requirement.

Water must first collect inside pre-existing rock joints and fissures.

Physical freeze-thaw weathering cannot take place without trapped liquid water inside open rock spaces.

2

Determine the physical change triggered by freezing conditions.

Water turns to ice below

0^\circ\text{C}

and expands by

9\%

.

The anomalous expansion of freezing water exerts immense outward pressure on fracture walls.

3

Trace the structural deterioration over time.

Repeated thermal cycling widens and extends internal fractures.

Continuous pressure fluctuations weaken the cohesive strength of the rock along lines of weakness.

4

Establish the end product of the weathering process.

Angular rock fragments break free and accumulate as talus or scree at the mountain base.

Complete mechanical failure occurs when fractures sever the fragment from the main outcrop.

Key Concept

Mechanism and Stages of Frost Shattering (Freeze-Thaw Weathering)

Estimated Time:1m 15s

Question 6196Question →

In evaluating agricultural systems and pastoral transhumance across Nigeria's latitudinal ecological zones, livestock herders in the northern Sudan and Sahel savannas routinely migrate southward toward the Guinea savanna and major river basins during the dry season, but return northward at the onset of the rainy season. Which of the following correctly identifies the primary environmental driver of this seasonal southward migration and the critical ecological constraint that necessitates their wet-season northward retreat?

Dry-season migration is driven by acute pasture and water shortages in the northern savanna, while the wet-season retreat is necessitated by the proliferation of tsetse flies and trypanosomiasis risk in humid vegetation zones.

Dry-season migration is driven by the harvesting labor demands of tree cash crops in the Sudan belt, while the wet-season retreat is necessitated by extensive salt flats flooding the Guinea savanna.

Dry-season migration is driven by the northward shift of the Inter-Tropical Discontinuity (ITD), while the wet-season retreat is necessitated by rapid soil salinization across the Benue Trough.

Dry-season migration is driven by severe crude oil contamination of northern rangelands, while the wet-season retreat is necessitated by extreme frost formation on the Jos Plateau.

Show answer & explanation

Answer: Dry-season migration is driven by acute pasture and water shortages in the northern savanna, while the wet-season retreat is necessitated by the proliferation of tsetse flies and trypanosomiasis risk in humid vegetation zones.

Answer

Dry-season migration is driven by acute pasture and water shortages in the northern savanna, while the wet-season retreat is necessitated by the proliferation of tsetse flies and trypanosomiasis risk in humid vegetation zones.

The correct choice accurately reflects the ecological dynamics of transhumance in Nigeria: dry-season movement southward is driven by the search for water and pasture in the Guinea savanna and river valleys, while the wet-season movement back northward is driven by the need to avoid tsetse fly infestation (which causes trypanosomiasis in cattle) and crop land conflicts in humid southern zones.

Step-by-Step Solution

1

Identify dry-season environmental conditions in the far north of Nigeria.

The tropical continental air mass (Harmattan) brings prolonged drought, drying up streams and desiccating grasslands in the Sudan and Sahel savannas.

Faced with feed and water scarcity, pastoral Fulani herdsmen migrate southward toward wetter ecological belts (Guinea savanna and floodplains).

2

Identify wet-season environmental constraints in the middle and southern belts.

With the retreat of the dry season and arrival of the moisture-laden tropical maritime air mass, vegetation in the middle belt becomes dense and humid.

High humidity and thick foliage foster vector breeding grounds for tsetse flies (Glossina spp.), which transmit fatal nagana (trypanosomiasis) to cattle, forcing herdsmen back to drier northern zones.

3

Evaluate options against verified geographic distributions and ecological facts.

The option pairing northern dry-season pasture/water deficits with southern wet-season tsetse fly infestation correctly captures the dual mechanism of transhumance.

Distractors contain spatial misallocations, such as placing tree cash crops or crude oil fields in northern savanna zones.

Key Concept

Ecological controls of pastoral transhumance and crop zoning in Nigeria

Question 6197Question →

Arrange the following steps involved in completing an electronic funds transfer via mobile banking in the correct sequential order from transaction initiation to final settlement:

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence begins with entering transfer details, followed by PIN authorization, originating bank debit and interbank switch routing, and ends with beneficiary account crediting and confirmation.

The electronic fund transfer workflow logically begins with entering transaction parameters, followed by user authentication via secret PIN, then originating bank debiting with interbank switch routing, and finishes with beneficiary account crediting and confirmation generation.

Step-by-Step Solution

1

Identify transaction initiation stage

Capturing transfer parameters on the banking application.

A transaction must be defined before security credentials can be applied to authorize it.

2

Identify security authorization stage

Entering PIN or biometric validation.

Authentication verifies customer identity and authorizes the banking system to execute the specified payment request.

3

Identify origin bank processing and routing stage

Account debiting and interbank payload transmission.

The originating bank must confirm solvency and reserve/debit funds locally before clearing through the central switch system (e.g., NIBSS).

4

Identify destination settlement stage

Crediting recipient account and issuing confirmation status.

Final settlement occurs when funds are successfully posted to the destination account and receipt notifications are issued.

Key Concept

Electronic Funds Transfer Process and Interbank Settlement Workflow

Question 6198Question →

Which climatic factor is primarily responsible for lower atmospheric temperatures recorded in highland regions compared to nearby lowlands situated at the same latitude?

Altitude

Distance from the sea

Ocean currents

Latitude

Show answer & explanation

Answer: Altitude

Answer

Altitude

Altitude is the main climatic control involved. As altitude increases, atmospheric density and pressure decrease, reducing the atmosphere's capacity to absorb heat re-radiated from the Earth's surface. Consequently, temperatures decline at an average normal lapse rate of about

6.5^\circ\text{C}

per

1,000\text{ m}

.

Step-by-Step Solution

1

Identify the variable factor between the two locations described.

The two places share the same latitude, meaning solar angle is identical, but one location is higher in elevation than the other.

This isolates elevation (altitude) as the primary physical control responsible for the temperature difference.

2

Apply the concept of the Normal Lapse Rate.

Within the troposphere, temperature decreases with height at an average rate of approximately

6.5^\circ\text{C}

per

1,000\text{ meters}

(

3.5^\circ\text{F}

per

1,000\text{ feet}

).

Higher altitude air is less dense and absorbs less terrestrial radiation, resulting in lower temperatures.

Key Concept

Altitude as a Climatic Control and Environmental Lapse Rate

Estimated Time:45s

Question 6199Question →

A manufacturing plant utilizes $400\text{ tonnes}$ of localized raw material and $100\text{ tonnes}$ of ubiquitous raw material (such as water) to manufacture $100\text{ tonnes}$ of a finished product. Based on Alfred Weber's Least Cost Theory of Industrial Location, what is the Material Index ( $MI$ ) of this production process, and where is the optimal site for locating this factory to minimize total transportation costs?

Material Index is

4.0

; the factory should be located at the raw material site.

Material Index is

4.0

; the factory should be located at the market.

Material Index is

5.0

; the factory should be located at the raw material site.

Material Index is

0.25

; the factory should be located at the market.

Show answer & explanation

Answer: Material Index is $4.0$ ; the factory should be located at the raw material site.

Answer

The Material Index is

4.0

, indicating a weight-losing manufacturing process that should be located at the raw material site to minimize total transport cost.

Under Alfred Weber's Least Cost Theory, the Material Index (

MI

) is defined as the ratio of localized raw material weight to finished product weight:

MI = \frac{400}{100} = 4.0

. Ubiquitous inputs (like water) are excluded because they are universally accessible. Because

MI > 1.0

, substantial weight is lost during production, creating a material-oriented industry that minimizes transport expenses by locating adjacent to the raw material extraction site.

Step-by-Step Solution

1

Identify the weights of localized raw materials, ubiquitous materials, and finished products.

Localized raw material weight =

400\text{ tonnes}

; Ubiquitous raw material weight =

100\text{ tonnes}

; Finished product weight =

100\text{ tonnes}

.

Alfred Weber's Material Index considers only localized raw materials because ubiquitous materials are available everywhere at zero transport cost differential.

2

Calculate Weber's Material Index (

MI

) using the formula

MI = \frac{\text{Weight of Localized Raw Material}}{\text{Weight of Finished Product}}

.

MI = \frac{400\text{ tonnes}}{100\text{ tonnes}} = 4.0

.

This formula measures the degree of weight loss or weight gain during the manufacturing process.

3

Apply Weber's location decision rule based on the calculated

MI

.

Since

MI = 4.0 > 1.0

, the industry is weight-losing (gross material).

Transporting

400\text{ tonnes}

of raw material to market costs substantially more than transporting

100\text{ tonnes}

of finished product, making the raw material source the least-cost location.

Key Concept

Weber's Material Index and Industrial Location Principle

Estimated Time:2m 0s

Question 6200Question →

Match each fundamental GIS data model or remote sensing concept on the left with its correct defining characteristic on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Raster Data Model corresponds to the continuous grid of square pixels or cells; Vector Data Model corresponds to discrete points, lines, and polygon boundaries; Passive Remote Sensing corresponds to detecting naturally occurring radiation; Active Remote Sensing corresponds to emitting its own energy signal and measuring the reflected response.

Each concept is matched correctly according to standard GIS and remote sensing definitions: Raster uses pixel grids; Vector uses points/lines/polygons; Passive sensing detects natural sunlight/thermal radiation; Active sensing transmits and records its own artificial energy pulse.

Step-by-Step Solution

1

Identify the core structural difference between spatial data models.

Raster uses continuous pixel grids, whereas vector uses coordinate-based points, lines, and polygons.

Raster data stores attributes cell-by-cell across space, while vector data outlines discrete geographical entities explicitly.

2

Distinguish between remote sensing systems based on their energy source.

Passive sensing measures existing natural radiation (sunlight/heat), while active sensing supplies its own illumination signal (radar pulses).

The distinction hinges entirely on whether the sensor generates energy or passively records available environmental radiation.

Key Concept

GIS Data Structure Fundamentals and Remote Sensing Energy Sources

Answer

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