All practice questions

Question 6081Question →

Prior to ledger reconciliation, the Sales Ledger Control Account of Tari Traders showed a debit balance of $₦482,000$ on 31 December 2025. An audit of the books revealed the following errors:

1. A sales invoice of $₦18,400$ issued to a customer was completely omitted from the sales journal.
2. A cash refund of $₦5,200$ made to a debtor was mistakenly credited to the Sales Ledger Control Account.
3. The total of the sales returns journal was overcast by $₦3,500$ .
4. Discount allowed totaling $₦2,800$ had been posted to the debit side of the Sales Ledger Control Account instead of the credit side.

What is the adjusted balance of the Sales Ledger Control Account?

₦508,700

₦506,300

₦519,900

₦492,100

Show answer & explanation

Answer: ₦508,700

Answer

The adjusted balance of the Sales Ledger Control Account is ₦508,700.

The correct adjusted balance of ₦508,700 is derived by systematically applying double-entry control account principles to each error: adding ₦18,400 for the unrecorded invoice, adding ₦10,400 to reverse and correctly debit the cash refund, adding ₦3,500 to correct overcredited sales returns, and deducting ₦5,600 to remove the erroneous debit and apply the credit for discount allowed.

Step-by-Step Solution

1

Identify the unadjusted starting debit balance

Unadjusted Debit Balance =

₦482,000

This is the initial ledger control balance prior to adjusting for discovered errors.

2

Adjust for omitted sales invoice

Add Debit of

₦18,400

An omitted sales invoice means total sales posted to the control account was understated, requiring a debit entry to increase the control balance.

3

Correct cash refund credited in error

Add Debit of

₦10,400

(

₦5,200 \times 2

)

A cash refund to a customer increases the debtor balance (debit). Crediting it by mistake requires a debit of twice the amount—once to cancel the wrong credit and once to record the correct debit.

4

Correct overcast sales returns journal

Add Debit of

₦3,500

Sales returns reduce debtors (credited to control account). Because the returns journal total was overcast, too much was credited, so debiting

₦3,500

restores the correct balance.

5

Correct discount allowed wrongly debited

Subtract Credit of

₦5,600

(

₦2,800 \times 2

)

Discount allowed reduces debtors and belongs on the credit side of the control account. Posting it to the debit side requires a credit entry of twice the amount (

₦5,600

) to rectify.

6

Compute the final adjusted debit balance

Adjusted Balance =

₦482,000 + ₦18,400 + ₦10,400 + ₦3,500 - ₦5,600 = ₦508,700

Summing all corrective debit and credit entries yields the accurate adjusted Sales Ledger Control Account balance.

Key Concept

Control Account Error Correction and Wrong-Side Adjustment Rules

Question 6082Question →

In an agricultural district characterized by rugged terrain and individual homesteads built directly on private, fragmented farm holdings, which rural settlement pattern typically emerges, and what is its primary socio-economic drawback?

Dispersed settlement, which increases the per-capita cost of providing public utilities and infrastructure

Linear settlement, which concentrates dwellings exclusively along major drainage channels and river banks

Nucleated settlement, which acts as an economic push factor driving farmers away from fertile agricultural soils

Radial settlement, which forces communal farming activities onto steep highland slopes

Show answer & explanation

Answer: Dispersed settlement, which increases the per-capita cost of providing public utilities and infrastructure

Answer

Dispersed settlement, which increases the per-capita cost of providing public utilities and infrastructure

In regions where physical relief is rugged and land tenure favors living on separate family farm holdings, dwellings become widely scattered, forming a dispersed settlement pattern. A major drawback of dispersed rural settlements is the inefficiency and high expense associated with extending public infrastructure like road networks, electricity, and clean water to widely separated households.

Step-by-Step Solution

1

Analyze the spatial distribution of dwellings described in the scenario

The scenario describes isolated homesteads built on individual private farm plots across rugged terrain.

Living directly on widely separated farm holdings creates a scattered spatial arrangement of dwellings.

2

Identify the corresponding rural settlement pattern

The spatial pattern described is a dispersed (or scattered) settlement.

Dispersed settlements occur where farmsteads are separated by considerable distances across agricultural land.

3

Evaluate the socio-economic implications of this pattern

Because dwellings are spread far apart, providing centralized infrastructure such as electricity networks, piped water, and schools requires greater distances per household, significantly increasing per-capita costs.

Spatial isolation directly increases infrastructural development and service delivery expenses.

Key Concept

Dispersed Rural Settlement Characteristics and Disadvantages

Question 6083Question →

A geological survey of a sedimentary basin identifies a fine-grained, impermeable rock formed from the compaction of mud and clay particles, which frequently acts as an effective cap rock for trapping petroleum. Which rock type exhibits these specific origin and structural characteristics?

Shale

Marble

Granite

Quartzite

Show answer & explanation

Answer: Shale

Answer

Shale is the fine-grained sedimentary rock formed from compressed mud and clay that serves as an impermeable cap rock in petroleum reservoirs.

Shale is a mechanically formed, fine-grained argillaceous sedimentary rock produced through the diagenesis and compaction of mud, clay, and silt particles. Because of its microscopic pore spaces and low permeability, shale acts as an impermeable seal or cap rock that prevents crude oil and natural gas from escaping upward out of porous reservoir rocks.

Step-by-Step Solution

1

Analyze the rock characteristics given in the stem: fine-grained texture, origin from mud/clay compaction, and low permeability serving as petroleum cap rock.

Identified that the target rock belongs to the argillaceous group of mechanically formed sedimentary rocks.

Sedimentary rocks derived from clay and silt deposits compact into fissile, fine-grained argillaceous rocks.

2

Match these physical properties and economic importances to the correct rock classification.

Shale fits all criteria (argillaceous sedimentary origin, fine-grained mud/clay composition, impermeable structure trapping hydrocarbons).

Metamorphic rocks (marble, quartzite) and igneous rocks (granite) have different origins and structural properties.

Key Concept

Sedimentary rock classification by origin, texture, and economic significance in hydrocarbon trapping

Estimated Time:1m 0s

Question 6084Question →

Zainab and Babatunde are partners in a haulage enterprise sharing profits and losses in the ratio of $3:2$ . Prior to the admission of Chijioke, their capital balances were $\text{₦}400,000$ and $\text{₦}300,000$ respectively. Upon admitting Chijioke to a $\frac{1}{5}$ share in the profits, the firm's assets were revalued, resulting in an appreciation of $\text{₦}80,000$ on machinery and a depreciation of $\text{₦}20,000$ on inventory. Additionally, goodwill was valued at $\text{₦}150,000$ , and Chijioke brought in his share of goodwill premium in cash to be distributed to the existing partners. What is the updated balance of Zainab's capital account immediately following these admission adjustments?

Show answer & explanation

Answer: 454000

Answer

The updated balance of Zainab's capital account immediately following the admission adjustments is ₦454,000.

Zainab's capital account begins with ₦400,000. The net gain on revaluation (₦80,000 appreciation minus ₦20,000 depreciation = ₦60,000) is split in the old ratio of 3:2, adding ₦36,000 to her account. The goodwill premium paid by Chijioke is 1/5 of ₦150,000, which equals ₦30,000; Zainab receives 3/5 of this premium (₦18,000). Adding these yields an updated capital balance of ₦454,000.

Step-by-Step Solution

1

Calculate the net gain or loss on revaluation of assets

Net Revaluation Gain = ₦80,000 - ₦20,000 = ₦60,000

Appreciation increases asset value while depreciation decreases it. The net figure represents total gain shared by existing partners.

2

Distribute the net revaluation gain to Zainab using the old profit sharing ratio

Zainab's Share of Revaluation Gain = 3/5 × ₦60,000 = ₦36,000

Revaluation gains belong to existing partners in their old profit sharing ratio.

3

Determine Chijioke's share of goodwill and allocate premium to Zainab

Goodwill Premium = 1/5 × ₦150,000 = ₦30,000; Zainab's Share = 3/5 × ₦30,000 = ₦18,000

The incoming partner pays premium for goodwill proportional to their profit share, which is credited to existing partners in their sacrificing ratio.

4

Sum Zainab's initial capital and all admission credits

Updated Capital Balance = ₦400,000 + ₦36,000 + ₦18,000 = ₦454,000

Capital accounts increase with revaluation gains and goodwill premium credits.

Key Concept

Adjustment of Partner's Capital Account on Admission

Question 6085Question →

Match each transportation mode with its primary operational advantage.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Pipeline transport pairs with continuous long-distance fluid transport; Water transport pairs with cheapest heavy freight movement over ocean routes; Air transport pairs with fastest long-distance transport for perishable/high-value cargo; Road transport pairs with highest flexibility and door-to-door delivery service.

Each mode of transportation is correctly matched to its distinct operational capability: pipelines for fluid materials, water for heavy ocean bulk, air for high-speed valuable/perishable transport, and roads for direct door-to-door accessibility.

Step-by-Step Solution

1

Identify the key characteristic of pipeline transport.

Pipelines specialize in moving liquids and gases continuously.

Pipelines operate constantly without vehicle traffic interference.

2

Identify the key characteristic of water transport.

Water transport provides maximum bulk carrying capacity at minimal cost per ton-kilometer.

Ships experience minimal friction relative to mass compared to overland transport modes.

3

Identify the key characteristic of air transport.

Air transport delivers rapid long-distance movement.

Air travel avoids terrestrial obstacles and terrain friction.

4

Identify the key characteristic of road transport.

Road transport provides direct door-to-door connectivity.

Road networks penetrate urban and rural areas to connect directly with individual origins and destinations.

Key Concept

Operational characteristics and trade-offs of major transportation modes

Question 6086Question →

An accountant extracted the following ledger summary for the trading period of Chukwu Enterprises:

- Debtors balance (1 Jan): Debit balance $\text{₦}12,500$ , Credit balance $\text{₦}400$
- Credit sales: $\text{₦}55,000$
- Cash and cheques received from debtors: $\text{₦}48,200$
- Returns inwards: $\text{₦}1,600$
- Discount allowed: $\text{₦}950$
- Dishonoured cheques from customer accounts: $\text{₦}1,200$
- Bad debts written off: $\text{₦}800$
- Provision for doubtful debts: $\text{₦}500$
- Set-off contra entry with Purchases Ledger: $\text{₦}1,100$
- Cash refunds to credit customers: $\text{₦}350$
- Debtors credit balance (31 Dec): $\text{₦}250$

What is the closing debit balance of the Sales Ledger Control Account at 31 December?

₦16,250

₦15,750

₦13,850

₦18,150

Show answer & explanation

Answer: ₦16,250

Answer

The closing debit balance of the Sales Ledger Control Account is ₦16,250.

The correct closing debit balance is ₦16,250. Summing all debit side postings gives ₦12,500 (opening Dr) + ₦55,000 (credit sales) + ₦1,200 (dishonoured cheques) + ₦350 (cash refunds) + ₦250 (closing Cr balance) = ₦69,300. Subtracting the credit side postings of ₦400 (opening Cr balance) + ₦48,200 (cash received) + ₦1,600 (returns inwards) + ₦950 (discount allowed) + ₦800 (bad debts) + ₦1,100 (contra set-off) = ₦53,050 leaves a closing debit balance of ₦16,250. Provision for doubtful debts is correctly ignored.

Step-by-Step Solution

1

Identify items to be posted to the Debit side of the Sales Ledger Control Account

Debit items: Opening Debit Balance (

\text{₦}12,500

), Credit Sales (

\text{₦}55,000

), Dishonoured Cheques (

\text{₦}1,200

), Cash Refunds to Debtors (

\text{₦}350

), and Closing Credit Balance (

\text{₦}250

). Total Debit Side =

\text{₦}69,300

.

Debit side represents increases in total amounts owed by trade debtors plus balances brought down.

2

Identify items to be posted to the Credit side of the Sales Ledger Control Account

Credit items: Opening Credit Balance (

\text{₦}400

), Cash and Cheques Received (

\text{₦}48,200

), Returns Inwards (

\text{₦}1,600

), Discount Allowed (

\text{₦}950

), Bad Debts Written Off (

\text{₦}800

), and Contra Set-off (

\text{₦}1,100

). Total Known Credit Items =

\text{₦}53,050

.

Credit side represents reductions in total amounts owed by trade debtors.

3

Exclude non-control account items

Provision for doubtful debts (

\text{₦}500

) is excluded completely from the Sales Ledger Control Account.

Provision for doubtful debts is an adjustment made in final accounts and ledger, not entered into the sales ledger control account.

4

Calculate the closing debit balance

Closing Debit Balance = Total Debit Side - Total Credit Side =

\text{₦}69,300 - \text{₦}53,050 = \text{₦}16,250

.

The balancing figure on the credit side represents the closing debit balance of trade debtors.

Key Concept

Sales Ledger Control Account preparation and item classification

Estimated Time:1m 30s

Question 6087Question →

Match each topographical contour line pattern with the correct relief feature or slope characteristic it depicts on a map.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: Widely spaced at top / closely spaced at base matches Convex slope; Closely spaced at top / widely spaced at base matches Concave slope; V-shaped contours pointing downhill match Spur; Asymmetrical contour spacing on opposite sides matches Escarpment.

Each contour pattern corresponds to a fundamental geometric principle of map work: contour density reflects gradient steepness (wide = gentle, close = steep), and V-shaped orientation indicates direction of slope projection.

Step-by-Step Solution

1

Analyze contour spacing relative to slope profiles.

Wide spacing indicates a gentle gradient, whereas close spacing indicates a steep gradient.

Topographical slope profiles (convex vs. concave) are determined by how gradient changes from high elevation to low elevation.

2

Differentiate between convex and concave slopes.

Convex slope = gentle top (wide spacing) to steep base (close spacing). Concave slope = steep top (close spacing) to gentle base (wide spacing).

Curvature of terrain determines where gradient transitions occur along elevation contours.

3

Analyze V-shaped contour orientation for relief features.

V-shapes pointing toward lower elevation represent spurs projecting downhill; V-shapes pointing toward higher elevation represent river valleys.

Contour lines bend around highland projections (spurs) differently than valley depressions.

4

Evaluate asymmetrical ridge patterns.

A steep slope on one side paired with a gentle slope on the other represents an escarpment (scarp and dip slopes).

Escarpments arise from tilted rock strata, resulting in contrasting contour densities across the ridge crest.

Key Concept

Interpretation of Relief and Slope Profiles from Topographical Contour Configurations

Question 6088Question →

On a topographical map, Point X is situated at an elevation of $200\text{ m}$ and Point Y is situated at an elevation of $500\text{ m}$ . If the horizontal ground distance between the two points is $6\text{ km}$ , what is the gradient of the slope from Point X to Point Y expressed as a ratio?

1 : 20

1 : 2

1 : 12

20 : 1

Show answer & explanation

Answer: $1 : 20$

Answer

The gradient of the slope between Point X and Point Y is

1 : 20

.

The correct ratio

1 : 20

is obtained by subtracting the lower elevation (

200\text{ m}

) from the higher elevation (

500\text{ m}

) to obtain a vertical interval of

300\text{ m}

, converting the horizontal distance of

6\text{ km}

into

6,000\text{ m}

, and simplifying the fraction

\frac{300}{6000}

to

\frac{1}{20}

.

Step-by-Step Solution

1

Calculate the Vertical Interval (VI)

\text{VI} = 500\text{ m} - 200\text{ m} = 300\text{ m}

Vertical interval is the difference in elevation between the higher and lower points.

2

Convert the Horizontal Equivalent (HE) to the same units as VI

\text{HE} = 6\text{ km} \times 1,000 = 6,000\text{ m}

Both vertical difference and horizontal distance must be expressed in identical units (meters).

3

Compute the slope gradient as a ratio

\text{Gradient} = \frac{\text{VI}}{\text{HE}} = \frac{300\text{ m}}{6,000\text{ m}} = \frac{1}{20} = 1 : 20

Gradient is calculated by dividing Vertical Interval by Horizontal Equivalent and simplifying to a ratio of 1 in N.

Key Concept

Calculating topographic gradient using Vertical Interval (VI) and Horizontal Equivalent (HE)

Estimated Time:45s

Question 6089Question →

In arid and semi-arid rural environments, dwellings are frequently clustered closely together around a localized natural resource rather than widely spread across the landscape. Which geographical factor is primarily responsible for creating this nucleated rural settlement pattern?

Concentration of scarce water supplies at specific wet-point sites

Development of continuous transportation networks along river channels

Availability of extensive, uniform plains for large-scale mechanized farming

Prevalence of severe environmental push factors forcing out-migration to urban centers

Show answer & explanation

Answer: Concentration of scarce water supplies at specific wet-point sites

Answer

Concentration of scarce water supplies at specific wet-point sites

The concentration of scarce water supplies at specific wet-point sites (such as springs, wells, or oases in arid environments) acts as a powerful site factor. Settlers cluster around these limited water points for survival and domestic use, resulting in a compact, nucleated settlement pattern.

Step-by-Step Solution

1

Analyze the settlement pattern described in the stem

The stem describes houses clustered closely together, which defines a nucleated (compact) settlement pattern.

Identifying the settlement pattern type is essential before determining its environmental driver.

2

Evaluate the environmental context of arid and semi-arid regions

In arid regions, water is the primary limiting resource for survival and agriculture.

Physical site factors heavily influence where people build homesteads in harsh climates.

3

Connect the resource location to spatial distribution

Populations gather around isolated water sources (wet-point sites), leading directly to a nucleated spatial arrangement.

Wet-point attraction pulls settlers together into a single cluster around the water source.

Key Concept

Wet-Point Sites and Nucleated Settlement Patterns

Estimated Time:1m 0s

Question 6090Question →

A topographic map with an original scale of $1 : 50,000$ is reduced to half its linear size. On the reduced map, the measured distance between Point X (elevation $750\text{ m}$ ) and Point Y along a slope is $5.0\text{ cm}$ . If Point Y lies $8$ contour intervals below Point X on a map with a contour interval of $25\text{ m}$ , what is the gradient between Point X and Point Y expressed as a ratio?

1 \text{ in } 6.25

1 \text{ in } 25

1 \text{ in } 40

1 \text{ in } 20

Show answer & explanation

Answer: $1 \text{ in } 25$

Answer

The gradient between Point X and Point Y is

1 \text{ in } 25

.

The gradient expressed as

1 \text{ in } 25

is correct because reducing a

1 : 50,000

map to half its linear size yields a scale of

1 : 100,000

. At this scale,

5.0\text{ cm}

corresponds to a Horizontal Equivalent (HE) of

5,000\text{ m}

. With

8

contour intervals at

25\text{ m}

each, the Vertical Interval (VI) is

200\text{ m}

. Dividing VI by HE gives

\frac{200}{5,000} = \frac{1}{25}

.

Step-by-Step Solution

1

Determine the new map scale after linear reduction

New Scale =

1 : 100,000

Reducing a map to half its linear size doubles the scale denominator (

50,000 \times 2 = 100,000

).

2

Calculate the Horizontal Equivalent (HE) ground distance

\text{HE} = 5.0\text{ cm} \times 100,000 = 500,000\text{ cm} = 5,000\text{ m}

Multiply map distance by the new scale factor and convert centimeters to meters.

3

Calculate the Vertical Interval (VI) between Point X and Point Y

\text{VI} = 8 \times 25\text{ m} = 200\text{ m}

Multiply the number of contour intervals by the contour interval value.

4

Compute the gradient ratio

\text{Gradient} = \frac{\text{VI}}{\text{HE}} = \frac{200\text{ m}}{5,000\text{ m}} = \frac{1}{25}

or

1 \text{ in } 25

Divide Vertical Interval by Horizontal Equivalent in identical units to find the simple ratio.

Key Concept

Topographic Gradient and Map Scale Reduction

Estimated Time:3m 0s

Question 6091Question →

A farm settlement measures $6\text{ cm}$ by $8\text{ cm}$ on a topographical map drawn at a scale of $1 : 40,000$ . If the map is enlarged to a new scale of $1 : 10,000$ , what is the area of the farm settlement on the enlarged map?

768\text{ cm}^2

192\text{ cm}^2

12\text{ cm}^2

3\text{ cm}^2

Show answer & explanation

Answer: $768\text{ cm}^2$

Answer

The area of the farm settlement on the enlarged map is

768\text{ cm}^2

.

First, calculate the original area on the map:

6\text{ cm} \times 8\text{ cm} = 48\text{ cm}^2

. Next, determine the linear scale factor

n

by comparing the scale denominators:

\frac{40,000}{10,000} = 4

. Because area scale varies with the square of the linear scale, the area scale multiplier is

n^2 = 4^2 = 16

. Multiplying the original map area by

16

yields

48\text{ cm}^2 \times 16 = 768\text{ cm}^2

.

Step-by-Step Solution

1

Calculate the original area of the farm settlement on the map.

\text{Area} = 6\text{ cm} \times 8\text{ cm} = 48\text{ cm}^2

.

The area on a rectangular grid is found by multiplying width by length.

2

Determine the linear scale factor (

n

) of the enlargement.

n = \frac{\text{Old Scale Denominator}}{\text{New Scale Denominator}} = \frac{40,000}{10,000} = 4

.

Enlarging a map decreases the scale denominator, increasing linear dimensions by a factor of

n

.

3

Calculate the area enlargement factor.

\text{Area Factor} = n^2 = 4^2 = 16

.

Area changes proportionally to the square of the linear scale factor.

4

Multiply the original map area by the area enlargement factor.

\text{Enlarged Area} = 48\text{ cm}^2 \times 16 = 768\text{ cm}^2

.

To find the new surface area on the enlarged map, multiply the initial area by

n^2

.

Key Concept

Map Area Enlargement and Reduction Ratio (

n^2

rule)

Estimated Time:1m 30s

Question 6092Question →

The following ledger details were extracted from the single-entry records of Adebayo Traders for the year ended 31 December 2025:

- Opening balance of debtors (1 Jan 2025): ₦15,000
- Cash received from debtors during the year: ₦68,000
- Discount allowed to debtors: ₦2,000
- Bad debts written off: ₦1,500
- Returns inwards: ₦2,500
- Closing balance of debtors (31 Dec 2025): ₦18,000

What is the total amount of credit sales for the year?

Show answer & explanation

Answer: 77000

Answer

₦77,000

Reconstructing the Sales Ledger (Debtors) Control Account shows a total credit side of ₦92,000 (comprising cash received of ₦68,000, discount allowed of ₦2,000, bad debts of ₦1,500, returns inwards of ₦2,500, and closing debtors balance of ₦18,000). Deducting the opening balance of ₦15,000 gives the missing total credit sales figure of ₦77,000.

Step-by-Step Solution

1

Calculate the total of all credit entries in the Debtors Control Account including the closing balance.

₦68,000 + ₦2,000 + ₦1,500 + ₦2,500 + ₦18,000 = ₦92,000

Cash received, discount allowed, bad debts written off, returns inwards, and closing balance all reduce trade debtors and belong on the credit side of the Debtors Control Account.

2

Subtract the opening debtors balance from the credit side total to derive the missing credit sales figure.

₦92,000 - ₦15,000 = ₦77,000

Credit sales increase trade debtors and represent the debit balancing figure required to equate both sides of the account.

Key Concept

Debtors Control Account for Missing Figures

Question 6093Question →

The following balances were extracted from the accounting records of Folake Enterprises, a sole trader, as at 31 December 2025:

Account Title	Amount (₦)
Premises	450,000
Motor Vehicles	180,000
Trade Debtors	65,000
Trade Creditors	42,000
Bank Overdraft	15,000
Accrued Rent Payable	8,000
Prepaid Insurance	5,000
Closing Inventory	32,000
10-Year Bank Loan	100,000
Owner's Drawings	12,000

What is the working capital of Folake Enterprises as at 31 December 2025?

₦37,000

₦52,000

₦217,000

₦25,000

Show answer & explanation

Answer: ₦37,000

Answer

The working capital of Folake Enterprises as at 31 December 2025 is ₦37,000.

Working capital represents net current assets, derived by subtracting total current liabilities from total current assets. Summing Trade Debtors (₦65,000), Prepaid Insurance (₦5,000), and Closing Inventory (₦32,000) yields ₦102,000 in Current Assets. Summing Trade Creditors (₦42,000), Bank Overdraft (₦15,000), and Accrued Rent Payable (₦8,000) yields ₦65,000 in Current Liabilities. The difference is ₦37,000.

Step-by-Step Solution

1

Identify and sum all Current Assets

Current Assets = Trade Debtors (₦65,000) + Prepaid Insurance (₦5,000) + Closing Inventory (₦32,000) = ₦102,000

Current assets comprise cash and items expected to be converted into cash or consumed within one accounting year.

2

Identify and sum all Current Liabilities

Current Liabilities = Trade Creditors (₦42,000) + Bank Overdraft (₦15,000) + Accrued Rent Payable (₦8,000) = ₦65,000

Current liabilities are short-term obligations payable within one accounting period. Long-term bank loan (10-year) is excluded as it is a non-current liability.

3

Compute Working Capital

Working Capital = Total Current Assets - Total Current Liabilities = ₦102,000 - ₦65,000 = ₦37,000

Working capital measures the short-term liquidity and operational efficiency of the enterprise.

Key Concept

Working Capital Calculation and Balance Sheet Element Classification

Question 6094Question →

Bello and Kabir are partners in a trading firm sharing profits and losses in the ratio of $4:1$ . They admit Danladi into the partnership, giving him a $\frac{1}{6}$ share of future profits. The goodwill of the firm is valued at $\text{₦}120,000$ , and Danladi brings in his required share of goodwill in cash. What is the amount of goodwill premium (in Naira) to be credited to Kabir's capital account?

Show answer & explanation

Answer: 4000

Answer

The amount of goodwill premium credited to Kabir's capital account is ₦4,000.

The total goodwill of the firm is

\text{₦}120,000

. Danladi's

\frac{1}{6}

share of goodwill is

\frac{1}{6} \times \text{₦}120,000 = \text{₦}20,000

. This premium is shared between Bello and Kabir in their sacrificing ratio of

4:1

. Therefore, Kabir receives

\frac{1}{5} \times \text{₦}20,000 = \text{₦}4,000

.

Step-by-Step Solution

1

Determine the new partner's share of goodwill

Danladi's share of goodwill =

\frac{1}{6} \times \text{₦}120,000 = \text{₦}20,000

The incoming partner pays a premium for goodwill proportional to their share of future profits.

2

Apportion the goodwill premium to the existing partners

Kabir's share =

\frac{1}{5} \times \text{₦}20,000 = \text{₦}4,000

Goodwill premium brought in cash by a new partner is credited to existing partners in their sacrificing ratio (which equals their old profit-sharing ratio of

4:1

when no new ratio among old partners is specified).

Key Concept

Valuation and distribution of goodwill premium upon admission of a new partner

Question 6095Question →

Ade Enterprises records its cash and bank transactions in a two-column cash book. On 1 June 2025, the opening cash balance was ₦45,000 and the bank balance was ₦120,000. During the month, the following transactions took place:
- Cash sales: ₦30,000
- Paid a supplier by cheque: ₦50,000
- Withdrew cash from the bank for office use: ₦15,000
- Paid office rent in cash: ₦10,000

What is the closing cash balance at the end of June 2025?

₦50,000

₦65,000

₦80,000

₦30,000

Show answer & explanation

Answer: ₦80,000

Answer

₦80,000

The closing cash balance is ₦80,000. It is calculated by adding cash receipts (cash sales of ₦30,000 and cash withdrawn from bank of ₦15,000) to the opening cash balance (₦45,000), giving total cash available of ₦90,000, and deducting cash payments (office rent of ₦10,000). Payments by cheque affect the bank column only.

Step-by-Step Solution

1

Identify all transactions affecting the Cash column of the Two-Column Cash Book.

Opening cash balance = ₦45,000 (Debit); Cash sales = ₦30,000 (Debit); Cash drawn from bank for office use = ₦15,000 (Debit - Contra entry); Rent paid in cash = ₦10,000 (Credit). The cheque payment of ₦50,000 affects the Bank column only.

Only transactions involving physical cash receipts and cash payments are recorded in the cash column.

2

Calculate total debit entries (total cash receipts) for the month.

Total Cash Receipts = ₦45,000 + ₦30,000 + ₦15,000 = ₦90,000.

Cash sales and cash withdrawn from bank increase available cash in hand.

3

Calculate total credit entries (total cash payments) and compute the closing balance.

Closing Cash Balance = ₦90,000 - ₦10,000 = ₦80,000.

Deducting cash payments from total cash receipts gives the balance carried down at the end of the period.

Key Concept

Two-Column Cash Book Posting and Contra Entries

Estimated Time:1m 30s

Question 6096Question →

An administrative state in West Africa covers a total area of $120,000\text{ km}^2$ . Environmental surveys indicate that $65\%$ of the terrain is non-arable (comprising rocky highlands, water bodies, and urban infrastructure), while the remainder is arable land suitable for farming. The state has a total population of $16,800,000$ inhabitants, of which $75\%$ are actively engaged in rural agricultural production. What is the agricultural population density of this state in persons per square kilometer of arable land?

Show answer & explanation

Answer: 300

Answer

The agricultural population density of the state is

300\text{ persons/km}^2

.

Agricultural population density measures the ratio of the agricultural population to the area of arable land. First, determine the arable land area:

100\% - 65\% = 35\%

of

120,000\text{ km}^2

, which equals

42,000\text{ km}^2

. Next, determine the rural agricultural population:

75\%

of

16,800,000

, which gives

12,600,000

agricultural inhabitants. Finally, dividing the agricultural population by the arable land area (

12,600,000 / 42,000

) yields

300\text{ persons/km}^2

.

Step-by-Step Solution

1

Calculate Arable Land Area

Arable land area is

42,000\text{ km}^2

.

Agricultural density requires using arable land area rather than total surface area.

2

Calculate Agricultural Population

Agricultural population is

12,600,000

inhabitants.

Agricultural density considers only the segment of the population engaged in farming rather than the total population.

3

Calculate Agricultural Density

Agricultural population density is

300\text{ persons/km}^2

.

Dividing the agricultural population by the arable land area yields the density of farmers per unit of cultivated land.

Key Concept

Agricultural Population Density

Estimated Time:3m 0s

Question 6097Question →

Musa, Emeka, and Audu are partners in a firm sharing profits and losses equally. Upon Audu's death, his capital account balance stands at $\text{₦}45,000$ . The revaluation of assets resulted in a profit where Audu's share is $\text{₦}5,000$ , and his share of goodwill is valued at $\text{₦}10,000$ . What is the total amount payable to Audu's executor?

Show answer & explanation

Answer: 60000

Answer

The total amount payable to Audu's executor is

\text{₦}60,000

.

The total amount due to the deceased partner's executor is calculated by adding all credit adjustments (revaluation profit share and goodwill share) to the partner's opening capital balance:

\text{₦}45,000 + \text{₦}5,000 + \text{₦}10,000 = \text{₦}60,000

.

Step-by-Step Solution

1

Add the deceased partner's initial capital balance, share of revaluation profit, and share of goodwill.

\text{₦}45,000 + \text{₦}5,000 + \text{₦}10,000 = \text{₦}60,000

When a partner dies, their capital account is credited with their balance of capital, share of asset revaluation profits, and share of goodwill before transferring the total amount to the executor's account.

Key Concept

Deceased Partner Settlement

Question 6098Question →

Ebonyi Processing Company operates two departments: Milling and Bakery. During the financial year ended 31 December 2025, the Milling Department transferred flour to the Bakery Department at cost plus a mark-up of $33\frac{1}{3}\%$ . At the close of the year, the Bakery Department held closing inventory valued at $\text{N}120,000$ , of which $75\%$ represented flour transferred from the Milling Department. Calculate the amount of provision for unrealized profit on closing inventory to be recognized in the general profit and loss account.

Show answer & explanation

Answer: 22500

Answer

The provision for unrealized profit to be recognized in the general profit and loss account is N22,500.

To calculate the provision for unrealized profit, first isolate the value of transferred stock in closing inventory:

75\% \times \text{N}120,000 = \text{N}90,000

. Next, convert the mark-up of

33\frac{1}{3}\%

(

\frac{1}{3}

) on cost to margin on transfer price:

\frac{1}{3+1} = \frac{1}{4}

(

25\%

). Multiplying the margin by the transferred stock value gives

25\% \times \text{N}90,000 = \text{N}22,500

.

Step-by-Step Solution

1

Determine the value of transferred goods included in closing inventory

Transferred inventory = 75% * N120,000 = N90,000

Only the portion of inventory transferred internally contains unrealized profit.

2

Convert mark-up percentage on cost to margin percentage on transfer price

Margin = Mark-up / (1 + Mark-up) = (1/3) / (1 + 1/3) = 1/4 or 25%

Closing inventory is valued at transfer price (selling price), requiring margin to extract the profit element.

3

Calculate the provision for unrealized profit

Provision = 25% * N90,000 = N22,500

Unrealized profit is the profit margin embedded in unsold transferred inventory remaining at year-end.

Key Concept

Provision for Unrealized Profit on Inter-departmental Transfers

Question 6099Question →

Sola and Musa are partners in a consulting firm sharing profits and losses in the ratio of $3:1$ . They admit Obinna into the partnership with a $\frac{1}{5}$ share of future profits. Obinna brings in $\text{₦}120,000$ in cash as goodwill premium to be shared by the existing partners. How much goodwill premium will be credited to Musa's capital account?

\text{₦}90,000

\text{₦}40,000

\text{₦}30,000

\text{₦}24,000

Show answer & explanation

Answer: $\text{₦}30,000$

Answer

Musa will be credited with

\text{₦}30,000

as his share of the goodwill premium.

Goodwill premium paid by an incoming partner is distributed among existing partners in their sacrificing ratio. Since no special agreement is mentioned, Sola and Musa sacrifice in their original profit sharing ratio of

3:1

. Musa's share is

\frac{1}{4} \times \text{₦}120,000 = \text{₦}30,000

.

Step-by-Step Solution

1

Determine the sacrificing ratio of the existing partners

Since no specific sacrifice terms are given, Sola and Musa sacrifice in their old profit sharing ratio of

3:1

.

When a new partner is admitted without a specified new ratio, existing partners sacrifice in proportion to their old ratio.

2

Calculate Musa's share of the goodwill premium

\text{Musa's share} = \frac{1}{3 + 1} \times \text{₦}120,000 = \frac{1}{4} \times \text{₦}120,000 = \text{₦}30,000

.

Goodwill premium brought in cash by an incoming partner is credited to existing partners' capital accounts in their sacrificing ratio.

Key Concept

Distribution of Goodwill Premium upon Admission of a New Partner

Estimated Time:1m 30s

Question 6100Question →

A topographical map extract drawn at a scale of $1:50,000$ delineates a drainage basin. The total area of the basin measured on the map is $160\text{ cm}^2$ . Morphometric analysis reveals the following total stream segment lengths measured directly from the map across all stream orders:
- 1st-order stream segments: $48\text{ cm}$
- 2nd-order stream segments: $20\text{ cm}$
- 3rd-order stream segments: $12\text{ cm}$
- 4th-order stream segments: $8\text{ cm}$

Calculate the drainage density of the river basin in $\text{km/km}^2$ .

Show answer & explanation

Answer: 1.1

Answer

The drainage density of the river basin is

1.1\text{ km/km}^2

.

Drainage density (

D

) measures the ratio of total stream channel length to the total drainage basin area (

D = \frac{\sum L}{A}

). Converting map measurements using the scale

1:50,000

(

1\text{ cm} = 0.5\text{ km}

), the total ground stream length is

88\text{ cm} \times 0.5\text{ km/cm} = 44\text{ km}

, and the actual ground area is

160\text{ cm}^2 \times (0.5)^2 = 40\text{ km}^2

. Dividing ground length by area produces

1.1\text{ km/km}^2

.

Step-by-Step Solution

1

Determine linear ground scale from map scale

1\text{ cm on map} = 0.5\text{ km on ground}

Scale

1:50,000

means

1\text{ cm} = 50,000\text{ cm} = 500\text{ m} = 0.5\text{ km}

.

2

Calculate ground area of the basin

Basin Area

A = 40\text{ km}^2

Area conversion requires squaring the linear scale factor:

160\text{ cm}^2 \times (0.5\text{ km/cm})^2 = 160 \times 0.25 = 40\text{ km}^2

.

3

Calculate total actual stream length in the basin

Total Stream Length

\sum L = 44\text{ km}

Total map stream length

= 48 + 20 + 12 + 8 = 88\text{ cm}

. Ground length

= 88\text{ cm} \times 0.5\text{ km/cm} = 44\text{ km}

.

4

Compute Drainage Density (

D

)

D = 1.1\text{ km/km}^2

Drainage density is the total channel length divided by total basin area:

D = \frac{\sum L}{A} = \frac{44\text{ km}}{40\text{ km}^2} = 1.1\text{ km/km}^2

.

Key Concept

Drainage Density Calculation from Topographical Maps

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