All practice questions

Question 6121Question →

On a topographical map drawn to a scale of $1:20,000$ , a foot trail ascends continuously from Point A at a contour elevation of $150\text{ m}$ to Point B at a contour elevation of $300\text{ m}$ . If the straight-line map distance between Point A and Point B measures $15.0\text{ cm}$ , what is the average gradient of the slope from Point A to Point B?

1 : 10

1 : 20

1 : 200

1 : 500

Show answer & explanation

Answer: $1 : 20$

Answer

The average gradient of the slope from Point A to Point B is

1 : 20

.

To determine topographic gradient, calculate the Vertical Interval (

\text{VI}

) as

300\text{ m} - 150\text{ m} = 150\text{ m}

. Convert the map distance (

15.0\text{ cm}

) to ground distance using the map scale (

1:20,000

):

15.0\text{ cm} \times 20,000 = 300,000\text{ cm} = 3,000\text{ m}

. The ratio

\frac{\text{VI}}{\text{HE}} = \frac{150\text{ m}}{3,000\text{ m}} = \frac{1}{20}

, giving an average gradient of

1 : 20

.

Step-by-Step Solution

1

Calculate the Vertical Interval (VI)

\text{VI} = 300\text{ m} - 150\text{ m} = 150\text{ m}

The vertical interval represents the total vertical height gained between the starting contour line and the destination contour line.

2

Convert map distance to ground distance to determine the Horizontal Equivalent (HE)

\text{HE} = 15.0\text{ cm} \times 20,000 = 300,000\text{ cm} = 3,000\text{ m}

Using the map scale ratio of

1:20,000

, every

1\text{ cm}

on the map corresponds to

20,000\text{ cm}

(

200\text{ m}

) on the actual terrain.

3

Compute the slope gradient ratio

\text{Gradient} = \frac{\text{VI}}{\text{HE}} = \frac{150\text{ m}}{3,000\text{ m}} = \frac{1}{20} = 1 : 20

Gradient expresses the ratio of vertical rise to horizontal distance when both measurements are expressed in identical units.

Key Concept

Topographic Slope Gradient Calculation

Question 6122Question →

Kofar Ltd has an issued share capital of $500,000$ ordinary shares of $\text{₦}1.00$ each and a Share Premium balance of $\text{₦}150,000$ . The company first makes a rights issue of $1$ new ordinary share for every $5$ existing shares held at an issue price of $\text{��}1.50$ per share, which is fully subscribed. Subsequently, the company makes a bonus issue of $1$ new ordinary share for every $4$ shares held, utilizing the Share Premium account to fund the issue as far as possible. What is the remaining balance in the Share Premium account after the capitalization for the bonus issue?

\text{₦}50,000

\text{₦}75,000

\text{₦}0

\text{₦}150,000

Show answer & explanation

Answer: $\text{₦}50,000$

Answer

The remaining balance in the Share Premium account is

\text{₦}50,000

.

The rights issue adds

100,000

shares (

500,000 / 5

) at a premium of

\text{₦}0.50

per share, increasing the Share Premium account by

\text{₦}50,000

to

\text{₦}200,000

and bringing total issued shares to

600,000

. The subsequent

1

-for-

4

bonus issue requires

150,000

shares (

600,000 / 4

) at

\text{₦}1.00

nominal value (

\text{₦}150,000

). Capitalizing this

\text{₦}150,000

from the updated

\text{₦}200,000

Share Premium account leaves a balance of

\text{₦}50,000

.

Step-by-Step Solution

1

Calculate rights issue volume and share premium contribution

Rights shares issued =

500,000 / 5 = 100,000

shares. Premium per share =

\text{₦}1.50 - \text{₦}1.00 = \text{₦}0.50

. Total premium added =

100,000 \times \text{₦}0.50 = \text{₦}50,000

.

Only the excess of issue price over nominal value is credited to the Share Premium account.

2

Determine post-rights share count and updated Share Premium balance

Total issued shares =

500,000 + 100,000 = 600,000

shares. New Share Premium balance =

\text{₦}150,000 + \text{₦}50,000 = \text{₦}200,000

.

Rights shares increase both total shares outstanding and the available Share Premium balance before the bonus issue.

3

Calculate bonus issue volume and required capitalization

Bonus shares issued =

600,000 / 4 = 150,000

shares. Nominal value capitalized =

150,000 \times \text{₦}1.00 = \text{₦}150,000

.

Bonus shares are allocated based on total existing shares after the rights issue and are issued at nominal value.

4

Deduct bonus capitalization from updated Share Premium balance

Remaining Share Premium balance =

\text{₦}200,000 - \text{₦}150,000 = \text{₦}50,000

.

Share Premium is used to fund the nominal value of bonus shares issued.

Key Concept

Accounting for Sequential Rights and Bonus Issues

Estimated Time:2m 0s

Question 6123Question →

Fill in the blanks with the correct names or acronyms of the Nigerian regulatory agencies described.

Fill in the blanks below

In Nigeria, the regulatory body responsible for inspecting, testing, and approving the safety and quality of food, drugs, and cosmetics is , whereas the agency mandated to establish, enforce, and certify quality standards for manufactured industrial goods is .

Show answer & explanation

Answer

blank_1: NAFDAC (National Agency for Food and Drug Administration and Control); blank_2: SON (Standards Organisation of Nigeria)

NAFDAC oversees the regulation, control, and registration of food, drugs, medical devices, packaged water, and cosmetics to prevent fake or adulterated consumables. SON establishes, standardizes, and enforces quality compliance for industrial, mechanical, and general manufactured products.

Step-by-Step Solution

1

Identify the Nigerian regulatory agency in charge of food, drug, and cosmetic regulation.

The National Agency for Food and Drug Administration and Control (NAFDAC) regulates consumables, pharmaceuticals, and cosmetics.

NAFDAC's primary statutory objective is to protect public health by ensuring the safety, quality, and efficacy of food, drugs, medical devices, and cosmetics.

2

Identify the Nigerian agency responsible for setting and enforcing standardization for manufactured industrial goods.

The Standards Organisation of Nigeria (SON) sets national quality standards.

SON is mandated to standardize manufactured items, industrial materials, and equipment through the Nigerian Industrial Standards (NIS) framework.

Key Concept

Statutory functions of consumer protection regulatory bodies (NAFDAC and SON) in Nigeria

Estimated Time:1m 0s

Question 6124Question →

A straight segment of a proposed highway measures $10\text{ cm}$ on a topographic map drawn at a scale of $1 : 50,000$ . On a newly redrawn map, the same segment of highway measures $25\text{ cm}$ . What is the Representative Fraction (R.F.) scale of the new map?

1 : 20,000

1 : 125,000

1 : 8,000

1 : 200,000

Show answer & explanation

Answer: $1 : 20,000$

Answer

The Representative Fraction (R.F.) of the new enlarged map is

1 : 20,000

.

The correct scale is

1 : 20,000

. When a map features are enlarged from

10\text{ cm}

to

25\text{ cm}

, the linear dimension increases by a factor of

2.5

. Because enlargement increases map scale (making features appear larger), the Representative Fraction denominator decreases by dividing the original denominator by

2.5

, giving

50,000 / 2.5 = 20,000

.

Step-by-Step Solution

1

Determine the linear enlargement factor

Linear Enlargement Factor =

\frac{\text{New Distance}}{\text{Original Distance}} = \frac{25\text{ cm}}{10\text{ cm}} = 2.5

Map enlargement increases map distance proportionally by a linear multiplier.

2

Calculate the new Representative Fraction (R.F.) scale denominator

\text{New Denominator} = \frac{\text{Original Denominator}}{\text{Enlargement Factor}} = \frac{50,000}{2.5} = 20,000

Enlarging a map produces a larger scale, which corresponds to a smaller scale denominator.

3

State the new R.F. scale

New R.F. scale =

1 : 20,000

The Representative Fraction expresses the scale ratio with a numerator of 1.

Key Concept

Map Enlargement Linear Scale Calculation

Question 6125Question →

The Purchases Ledger Control Account of Adebayo & Sons showed a credit balance of $\text{₦}186,400$ on 31 December 2025 prior to ledger reconciliation. An investigation of the accounting records revealed the following errors:

1. A credit purchase of inventory for $\text{₦}14,200$ was recorded in the Purchases Day Book as $\text{₦}12,400$ .
2. Discounts received amounting to $\text{₦}3,500$ were correctly posted to individual suppliers' accounts in the Purchases Ledger, but omitted completely from the Control Account.
3. A set-off (contra entry) of $\text{₦}6,000$ between the Sales Ledger and Purchases Ledger was credited to the Purchases Ledger Control Account in error.
4. A credit purchase of office machinery costing $\text{₦}25,000$ was incorrectly entered in the Purchases Day Book.

What is the corrected credit balance of the Purchases Ledger Control Account?

\text{₦}147,700

\text{₦}153,700

\text{₦}154,700

\text{₦}172,700

Show answer & explanation

Answer: $\text{₦}147,700$

Answer

The corrected credit balance of the Purchases Ledger Control Account is

\text{₦}147,700

.

Starting with the unadjusted credit balance of

\text{₦}186,400

, we add

\text{₦}1,800

for the understated Purchases Day Book total. Next, we deduct

\text{₦}3,500

for omitted discounts received,

\text{₦}12,000

to rectify the wrongly credited contra entry on the debit side, and

\text{₦}25,000

for the misclassified machinery purchase. This yields an adjusted credit balance of

\text{₦}147,700

.

Step-by-Step Solution

1

Adjust for the Purchases Day Book under-statement

Add

\text{₦}1,800

to the credit balance

The journal total was recorded as

\text{₦}12,400

instead of

\text{₦}14,200

, understating total credit purchases posted to the control account by

\text{₦}14,200 - \text{₦}12,400 = \text{₦}1,800

.

2

Adjust for omitted discounts received

Debit the control account by

\text{₦}3,500

(deduct from credit balance)

Discount received reduces trade payables and belongs on the debit side of the Purchases Ledger Control Account.

3

Correct the misplaced contra entry

Debit the control account by

\text{₦}12,000

(

2 \times \text{₦}6,000

)

A contra entry reduces liabilities and should be debited. Crediting it erroneously increased the balance by

\text{₦}6,000

, so a debit of

\text{₦}12,000

is needed to eliminate the wrong credit and apply the correct debit.

4

Remove capital expenditure misclassified in the Purchases Day Book

Debit the control account by

\text{₦}25,000

(deduct from credit balance)

Purchases of non-current assets (machinery) belong in the General Journal, not the Purchases Day Book. Including it overstated total credit purchases credited to the control account by

\text{₦}25,000

.

5

Calculate the final adjusted credit balance

\text{₦}186,400 + \text{₦}1,800 - \text{₦}3,500 - \text{₦}12,000 - \text{₦}25,000 = \text{₦}147,700

Summing unadjusted balance and net additions/deductions yields the corrected control account balance.

Key Concept

Control Account Error Corrections

Estimated Time:2m 30s

Question 6126Question →

Read the financial scenario below and calculate the required provision amount to complete the statement.

Fill in the blanks below

A manufacturing firm transfers finished goods from the factory to the trading department at a mark-up of

20\%

on cost. If the closing inventory of finished goods valued at transfer price is

\text{₦}12,000

, the provision for unrealized profit required for the closing inventory is

\text{₦}

.

Show answer & explanation

Answer

The provision for unrealized profit on closing inventory is ₦2,000.

When finished goods are transferred at a mark-up of

20\%

on cost, the transfer price represents

120\%

of manufacturing cost. Therefore, the profit element embedded in closing inventory at transfer price is calculated as

\frac{20}{120} \times \text{₦}12,000 = \text{₦}2,000

.

Step-by-Step Solution

1

Convert the mark-up on cost to the margin on transfer price.

A mark-up of

20\%

on cost (

\frac{20}{100}

) equals a margin of

\frac{20}{100 + 20} = \frac{20}{120} = \frac{1}{6}

on transfer price.

Closing inventory is valued at transfer price, so the profit portion must be determined using the profit margin relative to transfer price.

2

Calculate the unrealized profit included in closing inventory.

Unrealized Profit =

\frac{1}{6} \times \text{₦}12,000 = \text{₦}2,000

.

The provision for unrealized profit eliminates the internal profit element included in unsold inventory at the end of the accounting period.

Key Concept

Provision for Unrealized Profit on Closing Inventory

Question 6127Question →

Complete the statement regarding the accounting settlement for a retiring partner.

Fill in the blanks below

Upon the retirement of a partner, if the total settlement amount due to them is not immediately paid out in cash or bank, the balance of their capital account is transferred to a account.

Show answer & explanation

Answer

loan

When a partner leaves the firm and the business cannot settle their capital account immediately in cash, the credit balance in the retiring partner's capital account is transferred to a loan account in their name, representing a liability for the continuing partnership.

Step-by-Step Solution

1

Determine the final balance payable to the retiring partner after adjusting for goodwill, revaluation, and reserves.

The net amount due to the retiring partner is established in their capital account.

All share of revaluation profit/loss, goodwill adjustments, and accumulated reserves must be closed into the retiring partner's capital account first.

2

Record the transfer of the unpaid capital balance upon retirement.

Debit the retiring partner's capital account and credit the retiring partner's loan account.

Because the retiring partner ceases to be an owner, the unpaid balance cannot remain in the equity section and must be reclassified as an internal liability (loan) owed by the remaining partners.

Key Concept

Settlement of Retiring Partner's Account via Loan Account

Question 6128Question →

In an extensive agricultural region characterized by private land tenure, large individual farm holdings, and mechanized farming, farmers build their homesteads directly on their respective farm plots separated by wide distances. Which type of rural settlement pattern is formed by this spatial arrangement?

Dispersed settlement pattern

Nucleated settlement pattern

Linear settlement pattern

Radial settlement pattern

Show answer & explanation

Answer: Dispersed settlement pattern

Answer

Dispersed settlement pattern

The correct answer is the dispersed settlement pattern. Dispersed (or scattered) rural settlements are characterized by isolated homesteads situated far apart across extensive agricultural fields, commonly occurring in regions with private land tenure, large farm holdings, and peaceful social conditions.

Step-by-Step Solution

1

Analyze the spatial distribution of dwellings described in the scenario

Homesteads are constructed on individual farm plots separated by wide distances across an agricultural plain

Settlement classification depends on the spacing, density, and physical arrangement of dwellings relative to one another

2

Identify the primary socioeconomic and geographical determinants

Private land ownership and extensive mechanized agriculture require farmers to live directly on their large farm holdings

Large farm sizes necessitate broad spatial gaps between individual homesteads

3

Determine the matching rural settlement archetype

The arrangement of widely scattered, isolated houses across farm plots is classified as a dispersed settlement pattern

Dispersed settlements are defined by scattered dwellings without a single dense focal cluster

Key Concept

Dispersed Rural Settlement Pattern

Estimated Time:1m 0s

Question 6129Question →

Geological mapping along an active fold mountain belt reveals a progression of rocks exhibiting distinct foliated textures, ranging from fine-grained slate to coarse-grained gneiss. Which of the following statements correctly explains the process responsible for this texture progression and the resulting economic application of gneiss?

High-grade regional metamorphism driven by intense heat and directed pressure causes progressive mineral recrystallization and foliation, making gneiss valuable as a durable construction aggregate.

Thermal contact metamorphism caused exclusively by localized magma intrusion recrystallizes organic sedimentary deposits into coarse banded rock suitable for agricultural lime production.

In-situ chemical weathering and mass wasting alter fine clay sediments into coarse foliated bands used primarily as raw material for cement manufacturing.

Chemical precipitation of dissolved minerals in shallow lacustrine basins forms coarse banded structures utilized as impermeable petroleum cap rocks.

Show answer & explanation

Answer: High-grade regional metamorphism driven by intense heat and directed pressure causes progressive mineral recrystallization and foliation, making gneiss valuable as a durable construction aggregate.

Answer

High-grade regional metamorphism driven by intense heat and directed pressure causes progressive mineral recrystallization and foliation, making gneiss valuable as a durable construction aggregate.

Regional metamorphism over vast mountain-building zones Subjects parent rocks to intense differential stress and high heat, driving progressive recrystallization from fine slate to coarse, banded gneiss. Gneiss has high compressive strength and durability, making it an excellent material for heavy construction foundations and aggregate.

Step-by-Step Solution

1

Analyze the rock texture progression described in the stem.

The sequence from fine-grained slate to coarse-grained gneiss represents increasing grade of regional metamorphism.

Large-scale tectonic pressure and geothermal heat during mountain building alter minerals dynamically, forming foliated textures.

2

Evaluate the metamorphic process and structural characteristics of gneiss.

Gneiss exhibits distinct mineral banding (alternating light and dark minerals) due to high-grade recrystallization under directed stress.

This sets regional metamorphism apart from contact metamorphism or sedimentary deposition.

3

Determine the economic utility of gneiss based on its physical properties.

Due to its high compressive strength, density, and resistance to weathering, gneiss is extensively quarried for building stone, road metal, and coarse aggregates.

Economic applications directly depend on physical durability derived from metamorphic crystallization.

Key Concept

Regional Metamorphism and Economic Importance of Metamorphic Rocks

Estimated Time:1m 30s

Question 6130Question →

The following balances were extracted from the ledger of a sole trader as at 31st December 2025:
- Plant and Machinery (cost): $\text{₦}1,200,000$
- Accumulated Depreciation on Plant and Machinery: $\text{₦}300,000$

Additional Information:
On 1st July 2025, an additional machine costing $\text{₦}400,000$ was purchased on credit and mistakenly debited to the Purchases Account. Depreciation is to be provided at $20\%$ per annum using the reducing balance method.

What is the total depreciation charge for plant and machinery to be debited to the Profit and Loss Account for the year ended 31st December 2025?

\text{₦}220,000

\text{₦}180,000

\text{₦}260,000

\text{₦}280,000

Show answer & explanation

Answer: $\text{₦}220,000$

Answer

The total depreciation charge to be debited to the Profit and Loss Account is

\text{₦}220,000

.

The correct answer accounts for both the existing asset depreciated on its net book value (

\text{₦}1,200,000 - \text{₦}300,000 = \text{₦}900,000 \times 20\% = \text{₦}180,000

) and the newly purchased machine debited to machinery account and depreciated for 6 months (

\text{₦}400,000 \times 20\% \times \frac{6}{12} = \text{₦}40,000

), yielding a total depreciation of

\text{₦}220,000

.

Step-by-Step Solution

1

Adjust plant and machinery cost for misclassified capital expenditure

Unrecorded plant and machinery addition =

\text{₦}400,000

, acquired on 1st July 2025.

Purchases of fixed assets must be capitalized in the fixed asset account rather than charged to the Purchases Account.

2

Calculate Net Book Value (NBV) and depreciation for existing machinery

NBV =

\text{₦}1,200,000 - \text{₦}300,000 = \text{₦}900,000

. Depreciation =

20\% \times \text{₦}900,000 = \text{₦}180,000

.

Under the reducing balance method, annual depreciation is calculated on the Net Book Value at the beginning of the financial period.

3

Calculate pro-rata depreciation for newly acquired machinery

Depreciation =

20\% \times \text{₦}400,000 \times \frac{6}{12} = \text{₦}40,000

.

Assets acquired during the accounting period are depreciated from the date of acquisition to the financial year-end (6 months from July 1 to December 31).

4

Sum up the total depreciation charge for the year

Total depreciation =

\text{₦}180,000 + \text{₦}40,000 = \text{₦}220,000

.

The total depreciation expense for all machinery held during the year is debited to the Profit and Loss Account.

Key Concept

Adjustments for depreciation on reducing balance method with mid-year asset acquisition and capital expenditure correction

Estimated Time:2m 0s

Question 6131Question →

Kano Head Office operates a dependent branch in Kaduna, supplying all goods at cost price. For the year ended 31 December 2025, the following records were extracted:

Details	Amount (₦)
Branch stock (1 January 2025)	15,000
Goods sent to branch	120,000
Goods returned to head office	5,000
Cash sales at branch	80,000
Credit sales at branch	45,000
Branch stock (31 December 2025)	20,000
Branch operating expenses paid by head office	4,000

What is the net profit of the Kaduna branch for the year in Naira (₦)?

Show answer & explanation

Answer: 11000

Answer

The net profit of the Kaduna branch for the year is ₦11,000.

The net profit is calculated by subtracting total cost of goods sold and operating expenses from total sales (cash + credit sales). Total sales equal ₦125,000, cost of goods sold equals ₦110,000 (Opening stock ₦15,000 + Net goods sent ₦115,000 - Closing stock ₦20,000), leaving a gross profit of ₦15,000. Deducting branch expenses of ₦4,000 yields the net profit of ₦11,000.

Step-by-Step Solution

1

Determine total branch sales

₦125,000

Total revenue includes both cash sales (₦80,000) and credit sales (₦45,000).

2

Calculate net goods sent to branch from head office

₦115,000

Subtract returns to head office from goods sent: ₦120,000 - ₦5,000 = ₦115,000.

3

Calculate cost of goods sold (COGS)

₦110,000

COGS = Opening Stock (₦15,000) + Net Goods Sent (₦115,000) - Closing Stock (₦20,000) = ₦110,000.

4

Calculate branch gross profit

₦15,000

Gross Profit = Total Sales (₦125,000) - COGS (₦110,000) = ₦15,000.

5

Deduct operating expenses to find net profit

₦11,000

Net Profit = Gross Profit (₦15,000) - Operating Expenses (₦4,000) = ₦11,000.

Key Concept

Accounting for Dependent Branches at Cost Price

Question 6132Question →

Coastal landforms evolve through continuous wave processes such as hydraulic action and abrasion along headlands. Which of the following features is formed primarily by coastal marine erosion?

A sea stack

An oxbow lake

A scree slope

A marble deposit

Show answer & explanation

Answer: A sea stack

Answer

A sea stack is formed primarily by coastal marine erosion.

A sea stack is formed by continuous hydraulic action and abrasion on a rocky headland. Wave action undermines caves on both sides of a headland to form an arch, which eventually collapses under gravity, leaving behind a freestanding pillar of rock.

Step-by-Step Solution

1

Identify the process responsible for forming each landform option.

Marine erosion operates along coastlines, while rivers create fluvial landforms, gravity creates slope landforms, and heat/pressure form rocks.

Categorizing features by their main geomorphic agent isolates the correct coastal process.

2

Evaluate the coastal erosional sequence.

Waves attack headland weakness

\rightarrow

sea cave

\rightarrow

sea arch

\rightarrow

arch collapse leaves an isolated pillar called a sea stack.

Sea stacks represent a direct stage in coastal headland erosion.

Key Concept

Coastal Erosional Landforms

Estimated Time:45s

Question 6133Question →

A rocky headland exposed to persistent marine wave action undergoes sequential destructive geomorphic processes over time. Arrange the following coastal features in the correct chronological order of their evolution, starting from the earliest stage of headland erosion to the final remnant landform.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of coastal landform evolution is: (1) Hydraulic action and wave abrasion enlarge sea-level joints into a wave-cut notch and cave, (2) Continuous wave erosion pierces completely through the headland, forming a sea arch, (3) Subaerial weathering and gravity cause the unsupported roof of the arch to collapse, (4) An isolated, vertical column of resistant rock remains standing as a sea stack, and (5) Basal wave undercutting weakens the vertical pillar, causing it to collapse into a sea stump.

Coastal headland degradation follows a strict geomorphic sequence driven by wave action and weathering. Initially, marine processes (hydraulic action and abrasion) enlarge lines of weakness into a cave. As erosion cuts through the headland, an open arch is created. Subaerial weathering then weakens the arch roof until it collapses under gravity, leaving an isolated pillar called a stack. Persistent wave undercutting at the base eventually causes the stack to fall, leaving behind a low-lying stump.

Step-by-Step Solution

1

Identify the initial marine erosional process acting on a pristine cliff face.

Formation of a wave-cut notch and sea cave.

High-energy waves exploit geological joints and lines of weakness near the waterline.

2

Determine the structural change as wave penetration deepens through the headland.

Formation of a sea arch.

Erosion completely perforates the headland, connecting opposite sides under a rock bridge.

3

Identify the subaerial process destabilizing the overarching rock mass.

Collapse of the sea arch roof.

Freeze-thaw, salt weathering, and gravity weaken the unsupported arch roof until it falls.

4

Identify the coastal landform isolated from the mainland cliff after roof collapse.

Formation of a sea stack.

The isolated seaward column remains standing as a free pillar.

5

Determine the final remnant feature resulting from continued wave attack at the base of the pillar.

Reduction to a sea stump.

Under-cutting of the stack base causes top-heavy collapse, leaving only a low-lying rock stump.

Key Concept

Sequential Headland Erosion and Coastal Geomorphology

Question 6134Question →

Coal is an economically vital rock widely utilized as a fuel source for power generation and industrial processes. Through which of the following processes is coal formed?

Thermal metamorphism of pre-existing calcareous rocks under high temperature

Accumulation, decay, and compaction of organic plant remains in marshy environments

Rapid cooling and solidification of basaltic magma on the surface of the Earth

In-situ chemical weathering and downslope gravitational movement of surface minerals

Show answer & explanation

Answer: Accumulation, decay, and compaction of organic plant remains in marshy environments

Answer

The accumulation, decay, and compaction of organic plant remains in marshy environments over geological time.

Coal is an organically formed sedimentary rock derived from accumulated plant debris in swampy conditions. Over millions of years, heat and overburden pressure compress the organic layers, turning them into carbon-rich coal deposits.

Step-by-Step Solution

1

Identify the rock classification of coal.

Coal is categorized as an organically formed sedimentary rock.

Sedimentary rocks can be mechanically, chemically, or organically formed depending on the origin of the constituent materials.

2

Analyze the formation mechanism for organically formed sedimentary rocks.

Dense vegetation in swampy or marshy environments dies, accumulates in oxygen-poor conditions to prevent full decay, and undergoes compaction under overlying sediments.

Heat and pressure over millions of years convert peat into lignite, bituminous coal, and anthracite.

3

Select the option matching this organic sedimentary origin.

The correct response describes the decay, accumulation, and compaction of vegetative plant matter.

Igneous and metamorphic processes do not yield coal.

Key Concept

Organically Formed Sedimentary Rocks

Question 6135Question →

The following transaction balances were extracted from the books of Alabi & Sons for the year ended 31st December 2025:

Transaction Details	Amount (₦)
Opening balance of creditors (1st Jan 2025)	45,000
Payments to creditors by cash and cheque	250,000
Discount received	12,000
Returns outwards	8,500
Returns inwards	6,000
Cash purchases	35,000
Closing balance of creditors (31st Dec 2025)	52,000

What is the total credit purchases for the year?

₦277,500

₦253,500

₦283,500

₦312,500

Show answer & explanation

Answer: ₦277,500

Answer

₦277,500

The correct figure of ₦277,500 is calculated by totaling the debit side of the Purchases Ledger Control Account (payments of ₦250,000 + discount received of ₦12,000 + returns outwards of ₦8,500 + closing balance of ₦52,000 = ₦322,500) and subtracting the opening balance of ₦45,000.

Step-by-Step Solution

1

Filter relevant items for the Purchases Ledger Control Account

Relevant items are opening balance (credit ₦45,000), payments (debit ₦250,000), discount received (debit ₦12,000), returns outwards (debit ₦8,500), and closing balance (debit ₦52,000). Returns inwards and cash purchases are excluded.

Returns inwards belong to the sales ledger control account, and cash purchases do not affect individual creditors' ledger accounts.

2

Sum all debit side entries including the closing balance

Total debit entries = ₦250,000 + ₦12,000 + ₦8,500 + ₦52,000 = ₦322,500

Items reducing creditors and the balance carried down are placed on the debit side of the control account.

3

Deduct opening balance from total debit side to find credit purchases

Credit Purchases = ₦322,500 - ₦45,000 = ₦277,500

The Purchases Ledger Control Account total on the debit side must equal the total on the credit side.

Key Concept

Purchases Ledger Control Account Balancing and Item Selection

Estimated Time:1m 30s

Question 6136Question →

A river system flowing across a mature landscape experiences sudden tectonic uplift, causing a significant base-level drop. This change reactivates intense vertical erosion, prompting the river to cut deeply into its former floodplain and leave step-like benches at identical elevations on both sides of the new valley. Which landform is produced by this geomorphic process?

Paired river terraces

Interlocking spurs

Alluvial fans

Escarpment fault steps

Show answer & explanation

Answer: Paired river terraces

Answer

Paired river terraces are formed when river rejuvenation leads to vertical downcutting into a former floodplain, creating symmetrical bench-like landforms on opposite sides of the valley.

Paired river terraces represent former floodplain levels left standing above the current channel after river rejuvenation. Tectonic uplift or a drop in sea level increases the river's gradient and kinetic energy, driving rapid vertical downcutting. As the river carves a deeper inner gorge, remnants of the old valley floor remain as flat bench-like steps at identical elevations on both sides of the valley.

Step-by-Step Solution

1

Analyze the geomorphic trigger described in the stem.

Tectonic uplift drops the base level, initiating river rejuvenation and accelerating vertical downcutting.

Rejuvenation increases the river's energy, allowing it to incise deeply into its existing channel floor.

2

Examine the spatial arrangement of the resulting valley features.

The old floodplain is abandoned above the active channel, forming flat, bench-like steps at matching elevations across the valley.

When downcutting occurs evenly across symmetrical valley sides, paired terraces are formed.

3

Match the observed features to the correct geomorphic landform.

The step-like remnants at equal heights are identified as paired river terraces.

Unpaired terraces occur with lateral meander migration, whereas equal-height benches on both sides confirm paired terraces.

Key Concept

River Rejuvenation and Terrace Formation

Estimated Time:2m 0s

Question 6137Question →

Match each category of share capital with its correct description in financial accounting.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Authorized Share Capital matches the maximum capital allowed by the Memorandum of Association; Issued Share Capital matches the portion offered to the public; Called-Up Share Capital matches the portion requested by directors; Paid-Up Share Capital matches the actual money received from shareholders.

Each share capital class denotes a distinct accounting stage: Authorized is registered ceiling, Issued is offered shares, Called-Up is demanded payment, and Paid-Up is actual cash collected.

Step-by-Step Solution

1

Identify Authorized Share Capital

Matches the description stating the legal maximum capital specified in the Memorandum of Association.

Authorized capital (also known as nominal or registered capital) sets the upper statutory ceiling of share capital a company can issue.

2

Identify Issued Share Capital

Matches the description stating the portion of authorized capital offered to the public.

Issued capital is the actual portion of nominal capital offered to investors.

3

Identify Called-Up Share Capital

Matches the description stating the portion requested by directors for payment.

Called-up capital represents installments demanded from subscribers.

4

Identify Paid-Up Share Capital

Matches the description stating the actual amount received from shareholders.

Paid-up capital is the net cash received after deducting unpaid calls in arrears from called-up capital.

Key Concept

Classification of Share Capital

Estimated Time:1m 30s

Question 6138Question →

A statistical table shows the annual agricultural crop production for a region in Nigeria as follows:

Crop	Production (metric tonnes)
Cocoa	$540$
Oil Palm	$360$
Rubber	$180$
Groundnut	$120$

If this data is to be represented using a pie chart, what is the exact difference between the central angle allocated to Cocoa and the combined central angle allocated to Rubber and Groundnut?

72^\circ

90^\circ

144^\circ

162^\circ

Show answer & explanation

Answer: $72^\circ$

Answer

The difference between the central angle allocated to Cocoa and the combined central angle allocated to Rubber and Groundnut is

72^\circ

.

The total commodity output equals

1200

tonnes. The sector angle for Cocoa is calculated as

\frac{540}{1200} \times 360^\circ = 162^\circ

. The combined tonnage for Rubber and Groundnut is

180 + 120 = 300

tonnes, giving a combined angle of

\frac{300}{1200} \times 360^\circ = 90^\circ

. The difference between these two values is

162^\circ - 90^\circ = 72^\circ

.

Step-by-Step Solution

1

Calculate the total crop production

\text{Total} = 540 + 360 + 180 + 120 = 1200\text{ metric tonnes}

The total dataset value is required to determine the proportional weight of each commodity.

2

Calculate the central sector angle for Cocoa

\text{Angle}_{\text{Cocoa}} = \left(\frac{540}{1200}\right) \times 360^\circ = 0.45 \times 360^\circ = 162^\circ

A pie chart represents total data as a

360^\circ

circle.

3

Calculate the combined central sector angle for Rubber and Groundnut

\text{Combined Production} = 180 + 120 = 300\text{ metric tonnes}

;

\text{Angle}_{\text{Combined}} = \left(\frac{300}{1200}\right) \times 360^\circ = 0.25 \times 360^\circ = 90^\circ

Summing the metric tonnage of the two minor crops allows direct conversion to their aggregated sector angle.

4

Determine the difference between the angles

162^\circ - 90^\circ = 72^\circ

Subtracting the combined angle from the primary crop angle yields the required angular difference.

Key Concept

Calculation of Sector Angles for Pie Charts

Question 6139Question →

Arrange the following stages of production and distribution for a weight-losing manufacturing industry (such as sugar refining) in their correct logical spatial sequence, starting from the raw material origin to final market consumption:

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct logical sequence is: 1) Harvesting heavy sugarcane at plantations, 2) Crushing and refining sugarcane at a factory adjacent to plantations to eliminate waste, 3) Transporting processed reduced-weight sugar in bulk to regional commercial centers, and 4) Distributing packaged sugar to local retail outlets for final sale.

According to Alfred Weber's Least Cost Theory, weight-losing industries prioritize plant placement close to raw material inputs. The spatial sequence strictly follows raw material harvesting, immediate weight-reducing processing near the source, bulk transit of concentrated goods, and final retail delivery at the consumer market.

Step-by-Step Solution

1

Identify the raw material extraction point.

Harvesting sugarcane occurs first at the agricultural plantation source.

Production begins at the location of the raw input.

2

Apply Alfred Weber's weight-losing industrial location rule.

Primary processing takes place immediately near the farm source to shed bulk and weight (bagasse waste).

Locating weight-losing factories near raw material sources minimizes freight costs.

3

Trace the movement of the finished or semi-finished product.

Refined sugar is transported in bulk to major market centers.

Transporting the lighter, concentrated product incurs far lower freight costs than shipping unrefined sugarcane.

4

Complete the chain at the end consumer level.

Packaged sugar is distributed to retail stores in urban centers.

Final retail sales finalize the spatial supply chain at the market.

Key Concept

Spatial location sequence of weight-losing (raw-material-oriented) manufacturing industries according to Weber's Theory.

Estimated Time:45s

Question 6140Question →

A physical geographer constructs a cross-section across a ridge on a topographic map with a scale of $1 : 25{}000$ . On the vertical axis of the profile, $1\text{ cm}$ represents an elevation of $20\text{ metres}$ . What is the vertical exaggeration ( $VE$ ) of this cross-section profile?

Show answer & explanation

Answer: 12.5

Answer

The vertical exaggeration of the cross-section profile is 12.5.

To find vertical exaggeration (

VE

), convert both the vertical and horizontal scales to Representative Fractions (RF). The vertical scale of

1\text{ cm} = 20\text{ m}

converts to

1\text{ cm} : 2{}000\text{ cm}

(

1 : 2{}000

). The horizontal scale is given as

1 : 25{}000

. Dividing the horizontal scale denominator (

25{}000

) by the vertical scale denominator (

2{}000

) yields

12.5

.

Step-by-Step Solution

1

Convert the vertical scale to a unitless Representative Fraction (RF)

Vertical Scale =

1\text{ cm} : 20\text{ m} = 1\text{ cm} : 2{}000\text{ cm} = 1 : 2{}000

Both horizontal and vertical scales must be expressed in the same unit (centimetres) to enable direct comparison as representative fractions.

2

State the horizontal scale as a Representative Fraction (RF)

Horizontal Scale =

1 : 25{}000

The map scale provides the horizontal ratio between map distance and ground distance.

3

Apply the vertical exaggeration formula

VE = \frac{\text{Vertical Scale (RF)}}{\text{Horizontal Scale (RF)}} = \frac{1 / 2{}000}{1 / 25{}000} = \frac{25{}000}{2{}000}

Vertical exaggeration measures how many times larger the vertical scale is relative to the horizontal scale.

4

Perform the final division

VE = 12.5

Dividing 25,000 by 2,000 yields the unitless exaggeration factor of 12.5.

Key Concept

Vertical Exaggeration of Topographic Profiles

Answer

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