Q: Economic life comparison — large vs. small chemical process plants The economic life of a large chemical process plant, compared with that of a small plant, is typically:

Introduction / Context: Economic life is the optimal operating period that minimizes the equivalent annual cost or maximizes economic return before replacement. It is influenced by obsolescence, maintenance growth, and performance decline rather than purely by physical durability. Given Data / Assumptions: Compare large vs. small chemical plants under typical market and technology conditions. Economic life is defined by cost optimum, not the last day equipment can physically run. All else equal (maintenance practices, product market), scale primarily affects capital intensity. Concept / Approach: Larger plants enjoy economies of scale in capital and operating costs but face similar technological obsolescence and market shifts as smaller units. Because economic life is dominated by obsolescence and maintenance escalation, scaling up size does not proportionally extend the optimum replacement time. Textbook rules of thumb therefore state that large plants have an economic life only slightly longer than small plants, not dramatically longer. Step-by-Step Solution: Identify dominant factors: obsolescence, maintenance cost growth, regulatory changes.Recognize that these are largely independent of absolute size.Infer that any extension in economic life with size is modest, not “much more.”Select the option reflecting a small increase. Verification / Alternative check: Practice examples show economic life often clustering in comparable ranges (e.g., 6–12 years) across sizes for similar technologies. Why Other Options Are Wrong: Much more: overstates the effect of scale on obsolescence. Slightly less / Almost equal: do not reflect the small but positive tendency for larger units to remain competitive slightly longer. Common Pitfalls: Equating physical life or design life with economic life. Assuming economies of scale automatically extend useful economic life greatly. Final Answer: Only slightly more

Q: Depreciation methods and interest on investment Which of the listed depreciation methods explicitly accounts for the time value of money (interest on investment) in its annual depreciation calculation?

Introduction / Context: Depreciation schedules allocate cost for accounting; financial evaluation methods incorporate the time value of money. It is essential to distinguish a book method of depreciation from economic analyses that include interest. Given Data / Assumptions: Straight-line and declining-balance are book depreciation methods. Interest on investment refers to time value of money, not interest expense as a separate line item. We are asked which depreciation method itself embeds interest in the calculation of annual depreciation charges. Concept / Approach: Straight-line yields a constant annual charge equal to (P − S)/n without discounting. Declining-balance applies a rate to book value; again, it does not discount future charges to present worth. Methods that explicitly include interest are evaluation tools (e.g., sinking fund planning, present-worth and capital recovery factors), not the two book methods listed. Therefore, neither straight-line nor declining-balance includes interest within the depreciation formula. Step-by-Step Solution: Write SL formula: (P − S)/n → no i (interest) term.Write DB concept: Dep(r) = rate * book_value(r−1) → no discounting.Conclude that neither method embeds interest. Verification / Alternative check: Plant economics texts separate book depreciation schedules from discounted cash flow analyses where interest appears in NPV/IRR or capital recovery factor. Why Other Options Are Wrong: Straight-line or declining-balance alone do not feature an interest rate variable. Choosing both would be incorrect for the same reason. Common Pitfalls: Confusing the use of interest in cash-flow evaluations with its absence in standard book depreciation methods. Final Answer: Neither (a) nor (b)

Q: Working capital in process industries Which of the following is a component of working capital investment for a chemical plant?

Introduction / Context: Working capital is the money tied up in day-to-day operations. Correctly classifying costs as capital (fixed) versus working capital is essential for cash-flow planning and profitability analyses. Given Data / Assumptions: Working capital includes items needed to run the plant: inventories, cash, and receivables. Fixed capital covers long-lived assets such as equipment and utilities plants. Depreciation is a non-cash expense. Concept / Approach: Maintenance and repair inventory (spare parts, consumables) is a current asset required to sustain operations and is part of working capital. Utilities plants and process equipment are part of fixed capital investment. Depreciation is an accounting expense reflecting the allocation of fixed capital cost over time and is not itself an investment in working capital. Step-by-Step Solution: List working capital components: raw material, WIP, finished goods, spares, cash, receivables.Identify which option is a current asset: maintenance and repair inventory (spares).Exclude utilities plants and process equipment (fixed capital) and depreciation (non-cash expense). Verification / Alternative check: Standard cost breakdowns (fixed vs. working capital) place spares within inventories under current assets. Why Other Options Are Wrong: Utilities plants: fixed capital asset. Process equipment: fixed capital. Depreciation: accounting charge, not an asset or cash investment. Common Pitfalls: Counting depreciation as an operational cash outlay (it is a non-cash expense affecting taxes). Misclassifying capital projects (utilities, equipment) as working capital. Final Answer: Maintenance and repair inventory (spares)

Q: For compound interest at rate i per period on principal P over n periods, which expression gives the total compounded interest accrued (i.e., amount earned above principal)?

Introduction / Context: Capital budgeting and plant economics frequently require computing the growth of money under compound interest. Distinguishing between the future amount (principal plus interest) and the interest earned alone is essential for cash-flow analysis and depreciation planning. Given Data / Assumptions: Principal = P at time zero. Interest rate per period = i. Number of compounding periods = n. Concept / Approach: The compound amount formula is F = P * (1 + i)^n. The compounded interest (earned amount above principal) equals F - P. Therefore, interest accrued = P * ((1 + i)^n - 1). The other expressions either give the total future value or are incorrect linear/exponential forms for compound growth. Step-by-Step Solution: Start with F = P * (1 + i)^n.Interest accrued, I_c = F - P.I_c = P * (1 + i)^n - P = P * ((1 + i)^n - 1). Verification / Alternative check: For i = 0, the expression gives zero interest: P * ((1 + 0)^n - 1) = 0, which is consistent. For one period (n = 1), I_c = P * ((1 + i) - 1) = P * i, matching simple interest for a single period. Why Other Options Are Wrong: P * (1 + i)^n: This is the future amount, not the interest alone.P * (1 - i)^n: Represents decay, not compound growth at positive i.P * (1 + i * n): Linear growth corresponds to simple interest, not compounding. Common Pitfalls: Confusing future amount with interest portion. Using simple interest formula for multi-period compounding. Forgetting to convert nominal to effective rates when compounding frequency changes. Final Answer: P * ((1 + i)^n - 1)

Question 1

Taxation concepts for cost estimation: Which type of tax is typically assessed on gross production or sales rather than on net income?

Accepted Answer

Introduction / Context: Cost estimators must distinguish taxes that depend on income versus those tied to production or sales. This affects cash flows, pricing, and profitability analyses. Given Data / Assumptions: Gross earnings here refer to gross sales/production basis, not profits. General definitions applied across jurisdictions. Concept / Approach: Excise taxes are levies on the production, sale, or consumption of specific goods (e.g., fuels, alcohol). They are assessed on units or value sold/produced, not on net income after expenses. Step-by-Step Solution: Identify the base: gross production/sales rather than profit.Map to tax type: excise taxes fit this description.Confirm other taxes target different bases (property value, net income, capital appreciation). Verification / Alternative check: Tax guidelines list excise as transaction/commodity-based; income tax is profit-based after deductions. Why Other Options Are Wrong: Property: based on assessed property value. Income: based on net income. Capital gains: based on gain from asset sales. Common Pitfalls: Equating sales tax and excise; both are gross-basis but excise targets specific goods and may be unit-based. Final Answer: Excise tax

Question 2

Utility cost share in chemical manufacturing: For ordinary chemical process plants, utilities (compressed air, steam, water, power, fuel gases, etc.) typically account for approximately what percentage of total product cost?

Accepted Answer

Introduction / Context: Utility costs are an important component of manufacturing cost but, for many ordinary chemical plants, they are a modest share compared with raw materials and labor/overheads. Given Data / Assumptions: Typical continuous chemical processes (non-energy-intensive specialty products). Normal energy prices and efficient operations. Concept / Approach: Rule-of-thumb cost breakdowns commonly place utilities in the low single-digit percent range of total product cost, unless the process is very energy intensive (e.g., electrolytic or high-temperature furnaces). Step-by-Step Solution: Compare ranges: 1–5% aligns with ordinary plants.Larger ranges (10–45%) are typical only for energy-intensive processes or unusual utility price regimes. Verification / Alternative check: Published cost tables in process design texts show utilities often at a few percent for typical chemical production lines. Why Other Options Are Wrong: 10–45% substantially overstates typical utility shares for ordinary plants. Common Pitfalls: Confusing total manufacturing cost with energy-intensive industries; always consider process specifics. Final Answer: 1% to 5%

Question 3

Profitability measure — simple rate of return: A project costs ₹10,00,000 (total investment) and earns ₹1,50,000 profit each year for 10 years. What is the simple rate of return on investment?

Accepted Answer

Introduction / Context: The simple rate of return (accounting rate of return) is a quick screening metric defined as annual profit divided by total investment, expressed as a percentage. It ignores time value but is frequently used for preliminary comparisons. Given Data / Assumptions: Total investment (cost) = ₹10,00,000. Annual profit = ₹1,50,000 (assumed uniform). Project life = 10 years (not required for the simple rate). Concept / Approach: Simple rate of return = (Annual profit / Total investment) * 100%. Do not discount cash flows for this measure. Step-by-Step Solution: Compute ratio: 1,50,000 / 10,00,000 = 0.15Convert to percent: 0.15 * 100% = 15% Verification / Alternative check: If payback is considered, payback time ≈ 10,00,000 / 1,50,000 ≈ 6.67 years, which is consistent with a 15% per-year profit ratio (but payback is a different metric). Why Other Options Are Wrong: 10% and 1.5% use incorrect denominators or decimal placement; 150% confuses ratio with percent. Common Pitfalls: Using cash inflow instead of profit, or mixing after-tax and before-tax figures without consistency. Final Answer: 15%

Question 4

Working capital proportion: For most chemical plants (excluding strongly seasonal products), the ratio of working capital to total capital investment typically lies in which range?

Accepted Answer

Introduction / Context: Working capital finances inventories, receivables, and operating cash. Its ratio to total capital investment is a standard benchmark in project cost estimation. Given Data / Assumptions: Non-seasonal chemical manufacturing. Normal credit and inventory practices. Concept / Approach: Rules of thumb place working capital commonly around one-tenth to one-fifth of total capital investment for typical chemical plants, reflecting inventory cycles and receivables policies. Step-by-Step Solution: Identify realistic range: 10–20% for most non-seasonal plants.Eliminate extremes: below 2% is too small; 50–60% is atypically high. Verification / Alternative check: Design handbooks show typical working capital estimates of 15% ± for steady, continuous operations. Why Other Options Are Wrong: 0.1–2% underestimates; 50–60% grossly overstates normal needs. Common Pitfalls: Ignoring special cases (seasonal demand, long receivables) which can push the ratio higher. Final Answer: 10% to 20%

Question 5

Plant cost breakdown: Construction expenses are roughly what percentage of the total direct cost of a chemical plant (rule of thumb)?

Accepted Answer

Introduction / Context: Estimating a plant's total erected cost involves apportioning direct costs (equipment, installation, piping, instrumentation, buildings) and indirects. Construction expenses are one component within direct costs. Given Data / Assumptions: Typical chemical process facility. Order-of-magnitude or study estimate stage. Average construction complexity. Concept / Approach: Rule-of-thumb fractions are used when detailed quotes are unavailable. Construction expenses (temporary facilities, field supervision, small tools, etc.) are often on the order of ten percent of total direct cost in many estimating guides. Step-by-Step Solution: Consult standard breakdowns: construction expenses commonly ~10% of total direct cost.Exclude extremes: 2% is too low; 30–50% would imply unusual site conditions or scope. Verification / Alternative check: Cross-check with past projects or estimating handbooks showing construction overheads in the single-digit to low-teens percent range. Why Other Options Are Wrong: 2% understates typical field costs; 30% and 50% are excessive for ordinary projects. Common Pitfalls: Confusing construction expenses (a subset of direct cost) with overall indirect costs or contractor's fee. Final Answer: 10%

Question 6

Cost components and typical percentages: Identify the wrong statement about annual percentages used in chemical plant cost estimates.

Accepted Answer

Introduction / Context: Heuristic percentages are widely used for quick cost appraisals. Recognizing which rules are reasonable and which are unrealistic prevents large estimation errors in feasibility studies. Given Data / Assumptions: Percents refer to typical annualized benchmarks. Values are order-of-magnitude guides, not exact for every plant. Concept / Approach: Typical guides suggest single-digit insurance, low single-digit building depreciation, and equipment depreciation near 10% per year (assuming ~10-year straight-line life). However, R&D as 15% of NSR is far too high for most chemical plants; it is usually much lower (often < 5%, commonly 1–3% for many commodity operations). Step-by-Step Solution: Evaluate each statement against common heuristics.(a) 10% equipment depreciation: reasonable order-of-magnitude.(b) Buildings at ~3%: plausible.(c) Insurance ≈1% of FCI: plausible.(d) R&D = 15% of NSR: unrealistically high for chemical plants. Verification / Alternative check: Industry summaries typically place R&D for process industries in low single digits of sales, much less than 15%. Why Other Options Are Wrong: The other percentages are reasonable heuristics for early estimates; exact values depend on asset lives and insurance markets. Common Pitfalls: Assuming high-tech or pharma R&D percentages apply to commodity chemicals; they do not. Final Answer: Research and development costs are about 15% of net sales realisation (NSR).

Question 7

Plant economics — when are annual depreciation charges non-constant? Annual depreciation cost is not constant under which depreciation calculation method commonly used in chemical engineering economics?

Accepted Answer

Introduction / Context: Depreciation spreads the depreciable value of equipment over its useful life for accounting, taxes, and project evaluation. Different methods alter cash flows and profitability projections. A frequent exam point is identifying which methods yield constant year-to-year charges and which do not. Given Data / Assumptions: We compare common methods: straight-line, sinking fund, present worth (time-value analysis approach), and declining balance. Annual depreciation cost here means the booked depreciation expense each year, not accumulated depreciation. No special salvage timing effects are assumed beyond the usual definitions. Concept / Approach: In the straight-line method, the annual charge equals (P − S)/n, which is constant. In the declining-balance method, a fixed percentage is applied to the book value, so the first year is largest and charges reduce each year. The sinking-fund framework explicitly incorporates interest earnings on the fund and yields increasing annual deposits rather than a constant depreciation expense. Present-worth analysis discounts cash flows for comparison but is not itself an annual depreciation schedule. Of the listed choices, the hallmark non-constant pattern identified in cost accounting questions is the declining balance method. Step-by-Step Solution: State what is constant in straight-line: Dep = (P − S)/n each year (constant).Recall declining balance: Dep(r) = rate * book_value(r−1) → decreases annually.Recognize sinking fund: annual deposit typically escalates with interest assumptions; not a constant charge.Conclude the widely accepted non-constant schedule among standard accounting methods is the declining balance method. Verification / Alternative check: Example tables in plant economics texts show declining-balance charges descending year by year, while straight-line remains flat. Why Other Options Are Wrong: Straight-line method: produces a constant annual charge. Sinking fund method: focuses on building a fund; not the typical depreciation expense schedule used in this context. Present worth method: an evaluation technique, not an annual depreciation formula. Common Pitfalls: Confusing time-value evaluation methods (present worth) with depreciation schedules. Assuming all methods yield constant yearly expenses. Final Answer: Declining balance (diminishing balance) method

Question 8

Economic life comparison — large vs. small chemical process plants The economic life of a large chemical process plant, compared with that of a small plant, is typically:

Accepted Answer

Introduction / Context: Economic life is the optimal operating period that minimizes the equivalent annual cost or maximizes economic return before replacement. It is influenced by obsolescence, maintenance growth, and performance decline rather than purely by physical durability. Given Data / Assumptions: Compare large vs. small chemical plants under typical market and technology conditions. Economic life is defined by cost optimum, not the last day equipment can physically run. All else equal (maintenance practices, product market), scale primarily affects capital intensity. Concept / Approach: Larger plants enjoy economies of scale in capital and operating costs but face similar technological obsolescence and market shifts as smaller units. Because economic life is dominated by obsolescence and maintenance escalation, scaling up size does not proportionally extend the optimum replacement time. Textbook rules of thumb therefore state that large plants have an economic life only slightly longer than small plants, not dramatically longer. Step-by-Step Solution: Identify dominant factors: obsolescence, maintenance cost growth, regulatory changes.Recognize that these are largely independent of absolute size.Infer that any extension in economic life with size is modest, not “much more.”Select the option reflecting a small increase. Verification / Alternative check: Practice examples show economic life often clustering in comparable ranges (e.g., 6–12 years) across sizes for similar technologies. Why Other Options Are Wrong: Much more: overstates the effect of scale on obsolescence. Slightly less / Almost equal: do not reflect the small but positive tendency for larger units to remain competitive slightly longer. Common Pitfalls: Equating physical life or design life with economic life. Assuming economies of scale automatically extend useful economic life greatly. Final Answer: Only slightly more

Question 9

Depreciation methods and interest on investment Which of the listed depreciation methods explicitly accounts for the time value of money (interest on investment) in its annual depreciation calculation?

Accepted Answer

Introduction / Context: Depreciation schedules allocate cost for accounting; financial evaluation methods incorporate the time value of money. It is essential to distinguish a book method of depreciation from economic analyses that include interest. Given Data / Assumptions: Straight-line and declining-balance are book depreciation methods. Interest on investment refers to time value of money, not interest expense as a separate line item. We are asked which depreciation method itself embeds interest in the calculation of annual depreciation charges. Concept / Approach: Straight-line yields a constant annual charge equal to (P − S)/n without discounting. Declining-balance applies a rate to book value; again, it does not discount future charges to present worth. Methods that explicitly include interest are evaluation tools (e.g., sinking fund planning, present-worth and capital recovery factors), not the two book methods listed. Therefore, neither straight-line nor declining-balance includes interest within the depreciation formula. Step-by-Step Solution: Write SL formula: (P − S)/n → no i (interest) term.Write DB concept: Dep(r) = rate * book_value(r−1) → no discounting.Conclude that neither method embeds interest. Verification / Alternative check: Plant economics texts separate book depreciation schedules from discounted cash flow analyses where interest appears in NPV/IRR or capital recovery factor. Why Other Options Are Wrong: Straight-line or declining-balance alone do not feature an interest rate variable. Choosing both would be incorrect for the same reason. Common Pitfalls: Confusing the use of interest in cash-flow evaluations with its absence in standard book depreciation methods. Final Answer: Neither (a) nor (b)

Question 10

Working capital in process industries Which of the following is a component of working capital investment for a chemical plant?

Accepted Answer

Introduction / Context: Working capital is the money tied up in day-to-day operations. Correctly classifying costs as capital (fixed) versus working capital is essential for cash-flow planning and profitability analyses. Given Data / Assumptions: Working capital includes items needed to run the plant: inventories, cash, and receivables. Fixed capital covers long-lived assets such as equipment and utilities plants. Depreciation is a non-cash expense. Concept / Approach: Maintenance and repair inventory (spare parts, consumables) is a current asset required to sustain operations and is part of working capital. Utilities plants and process equipment are part of fixed capital investment. Depreciation is an accounting expense reflecting the allocation of fixed capital cost over time and is not itself an investment in working capital. Step-by-Step Solution: List working capital components: raw material, WIP, finished goods, spares, cash, receivables.Identify which option is a current asset: maintenance and repair inventory (spares).Exclude utilities plants and process equipment (fixed capital) and depreciation (non-cash expense). Verification / Alternative check: Standard cost breakdowns (fixed vs. working capital) place spares within inventories under current assets. Why Other Options Are Wrong: Utilities plants: fixed capital asset. Process equipment: fixed capital. Depreciation: accounting charge, not an asset or cash investment. Common Pitfalls: Counting depreciation as an operational cash outlay (it is a non-cash expense affecting taxes). Misclassifying capital projects (utilities, equipment) as working capital. Final Answer: Maintenance and repair inventory (spares)

Question 11

For compound interest at rate i per period on principal P over n periods, which expression gives the total compounded interest accrued (i.e., amount earned above principal)?

Accepted Answer

Introduction / Context: Capital budgeting and plant economics frequently require computing the growth of money under compound interest. Distinguishing between the future amount (principal plus interest) and the interest earned alone is essential for cash-flow analysis and depreciation planning. Given Data / Assumptions: Principal = P at time zero. Interest rate per period = i. Number of compounding periods = n. Concept / Approach: The compound amount formula is F = P * (1 + i)^n. The compounded interest (earned amount above principal) equals F - P. Therefore, interest accrued = P * ((1 + i)^n - 1). The other expressions either give the total future value or are incorrect linear/exponential forms for compound growth. Step-by-Step Solution: Start with F = P * (1 + i)^n.Interest accrued, I_c = F - P.I_c = P * (1 + i)^n - P = P * ((1 + i)^n - 1). Verification / Alternative check: For i = 0, the expression gives zero interest: P * ((1 + 0)^n - 1) = 0, which is consistent. For one period (n = 1), I_c = P * ((1 + i) - 1) = P * i, matching simple interest for a single period. Why Other Options Are Wrong: P * (1 + i)^n: This is the future amount, not the interest alone.P * (1 - i)^n: Represents decay, not compound growth at positive i.P * (1 + i * n): Linear growth corresponds to simple interest, not compounding. Common Pitfalls: Confusing future amount with interest portion. Using simple interest formula for multi-period compounding. Forgetting to convert nominal to effective rates when compounding frequency changes. Final Answer: P * ((1 + i)^n - 1)

Question 12

Among the listed components of fixed capital investment for a chemical plant (equipment installation, electrical installation, piping, and equipment insulation), which typically requires the smallest percentage share?

Accepted Answer

Introduction / Context: Estimating fixed capital investment (FCI) is central to chemical engineering economics. FCI is composed of many cost items: purchased equipment, equipment installation, piping, electrical, instrumentation, buildings, and insulation. Understanding relative magnitudes helps during early cost estimation and value engineering. Given Data / Assumptions: Typical large process plant with standard utilities. No unusual insulation-intensive service (e.g., deep cryogenics across the whole site). Percentages are approximate and industry-typical. Concept / Approach: Empirical cost fractions indicate that equipment installation and piping are major shares of the direct costs. Electrical installations can also be significant depending on motorization and power distribution. In contrast, equipment insulation—although necessary for heat loss reduction and personnel protection—usually represents a relatively small fraction of FCI. Step-by-Step Solution: List candidate components and recall typical ranges.Piping and equipment installation often rank high due to labor, supports, fittings, and complex routing.Electrical installation can vary, but is generally larger than insulation for most plants.Equipment insulation cost is typically the smallest of the four listed items. Verification / Alternative check: Guidelines and historical cost breakdowns in conceptual estimates (order-of-magnitude, factored methods) corroborate that insulation is a minor contributor compared with piping and installation costs. Why Other Options Are Wrong: Electrical installation: Encompasses cabling, switchgear, MCCs—usually higher than insulation.Equipment installation: Includes foundations, setting, alignment—substantial cost item.Piping: Often one of the largest due to materials and labor. Common Pitfalls: Assuming insulation is large because it is applied widely; cost per area is relatively low compared with piping and installation labor. Not considering site-specific exceptions such as cryogenic or very hot plants where insulation scope is extraordinary. Final Answer: Equipment insulation cost.

Question 13

The economically optimum pipe diameter for a given fluid service is chosen primarily on what basis?

Accepted Answer

Introduction / Context: In pipeline design, selecting pipe size impacts both capital and operating costs. A small diameter reduces material cost but increases frictional losses and hence pumping cost; a larger diameter does the opposite. The optimum diameter is an economic trade-off, not a purely hydraulic or purely mechanical decision. Given Data / Assumptions: Steady, incompressible flow in a long pipeline. Pump efficiency and energy price are known or estimated. Pipe cost increases with diameter and material thickness. Concept / Approach: The annualized total cost = annualized capital cost of the pipeline + annual pumping (energy) cost. Frictional head loss is inversely related to approximately D^5 (via Darcy–Weisbach and continuity), so pumping power falls rapidly with diameter, while capital cost increases roughly with D and length. Minimizing the sum yields the economic diameter (e.g., using the well-known optimization known as the economic pipe diameter method). Step-by-Step Solution: Define C_total(D) = C_capital(D) + C_operating(D).C_operating decreases as diameter increases due to lower friction losses.C_capital increases with diameter because of material and installation costs.Choose the diameter D that minimizes C_total(D). Verification / Alternative check: Perform sensitivity analysis by evaluating total cost for a few candidate diameters; the minimum identifies the optimum. This aligns with classical plant design practices. Why Other Options Are Wrong: Viscosity or density alone: These parameters influence head loss but do not by themselves define economic optimum without considering cost trade-offs.None of these: Incorrect because total cost is the accepted criterion. Common Pitfalls: Overemphasis on hydraulic efficiency without capital cost consideration. Ignoring energy pricing and pump efficiency in operating cost calculations. Neglecting installation factors (buried vs aboveground) that affect capital cost. Final Answer: total cost considerations (pumping cost plus fixed cost of the pipe).

Question 14

A reactor costs ₹80,000 initially and has a salvage value of ₹10,000 after a 10-year life. At an annual interest rate of 10%, what is the book value after 5 years using the sinking fund depreciation method?

Accepted Answer

Introduction / Context: The sinking fund method sets aside a uniform annual amount that, when compounded at a given interest rate, will equal the asset’s depreciable amount by the end of its life. This question asks for the intermediate book value after 5 years—useful for financial reporting and replacement planning. Given Data / Assumptions: Original cost, P = 80000 (₹). Salvage value, S = 10000 (₹). Service life, n = 10 years. Annual interest rate, i = 0.10. Concept / Approach: Annual sinking fund deposit A is chosen so that the future worth after n years equals (P - S). Formula: A = (P - S) * i / ((1 + i)^n - 1). Accumulated fund after t years = A * [((1 + i)^t - 1) / i]. The book value at year t is BV(t) = P - accumulated_fund(t). Step-by-Step Solution: Compute depreciable amount: P - S = 80000 - 10000 = 70000.Annual deposit: A = 70000 * 0.10 / ((1.10)^10 - 1) ≈ 4392.18.Accumulated fund at t = 5: F5 = 4392.18 * [((1.10)^5 - 1)/0.10] ≈ 4392.18 * 6.1051 ≈ 26814.68.Book value at year 5: BV(5) = P - F5 ≈ 80000 - 26814.68 ≈ 53185.32.Rounded to nearest option: 53196. Verification / Alternative check: A check at t = 10 yields BV(10) = P - A * [((1.10)^10 - 1)/0.10] = P - (P - S) = S = 10000, confirming the method’s consistency. Why Other Options Are Wrong: 40096, 43196, 60196: These do not match the sinking fund accumulation for the given i and n. They reflect either arithmetic errors or misuse of straight-line or declining-balance methods. Common Pitfalls: Using straight-line depreciation formula instead of sinking fund. Forgetting to compound the annual deposit at rate i. Using (P - S)/n directly, which is not the sinking fund approach. Final Answer: 53196

Question 15

In the straight-line method of depreciation used in plant economics, the value of a property decreases with time in which manner?

Accepted Answer

Introduction / Context: Depreciation methods allocate an asset’s cost over its service life for accounting and economic evaluation. The straight-line method is one of the most common due to its simplicity and uniform annual charges. Recognizing how asset value changes under this method is key for book-keeping and profitability analysis. Given Data / Assumptions: Asset has original cost P and salvage value S. Service life is n years. Annual charge is constant throughout the life. Concept / Approach: In the straight-line method, annual depreciation D is constant: D = (P - S) / n. The book value after t years is BV(t) = P - D * t, a linear function of time. Thus, the decrease in value is linear, not exponential or logarithmic. Step-by-Step Solution: Compute annual depreciation: D = (P - S) / n (constant each year).Book value update: BV(t) = P - D * t.Linear dependence on t indicates a straight-line decline in value from P to S over n years. Verification / Alternative check: Plotting BV(t) versus t yields a straight line from (0, P) to (n, S), confirming linear behavior. This contrasts with declining-balance methods that show faster early depreciation (nonlinear curves). Why Other Options Are Wrong: Non-linearily, exponentially, logarithmically: These describe other depreciation or decay behaviors, not the constant-amount-per-year decrease of straight-line. Common Pitfalls: Confusing book value trends of straight-line with declining-balance methods. Ignoring salvage value when computing annual charges. Mixing economic depreciation (market value changes) with accounting depreciation schedules. Final Answer: linearly

Question 16

Compound interest with sub-annual compounding: A present sum of ₹100 is invested for one year at a half-yearly interest rate of 10% (i.e., 10% every 6 months). What will be the future amount at the end of one year?

Accepted Answer

Introduction / Context: This problem checks understanding of compound interest with multiple compounding periods in a year. When the nominal rate is quoted per half-year (every 6 months), we must compound twice within one year. Many learners mistakenly apply simple interest or compound only once per year, leading to incorrect results. Given Data / Assumptions: Principal P = ₹100 (present sum at time 0). Interest rate per half-year i = 10% = 0.10. Number of half-years in one year n = 2. No withdrawals or additional deposits. Concept / Approach: For compound interest with sub-annual compounding, use the standard formula: F = P * (1 + i)^n, where i is the interest rate per compounding period and n is the number of such periods. Here each period is half a year, so n = 2 within one year. Step-by-Step Solution: Identify i and n: i = 0.10 per half-year, n = 2 half-years.Apply the compound interest formula: F = 100 * (1 + 0.10)^2.Compute the factor: (1.10)^2 = 1.21.Multiply by principal: F = 100 * 1.21 = ₹121. Verification / Alternative check: You can compute period by period: after first 6 months, amount = 100 * 1.10 = ₹110. After second 6 months, reapply 10% on ₹110: 110 * 1.10 = ₹121. This matches the direct formula result. Why Other Options Are Wrong: ₹110: This applies only one half-year of interest instead of two.₹97 and ₹91: These imply negative or incorrect rate applications.₹105: A common guess confusing 10% per year with some averaging; still incorrect. Common Pitfalls: Forgetting the number of compounding periods, using simple interest (F = P * (1 + i_total)) rather than compounding, or mixing nominal annual and per-period rates. Always align i with the period count n. Final Answer: ₹121

Question 17

Simple interest definition check: For n interest periods, with per-period rate i and principal p, the total simple interest accrued equals which expression?

Accepted Answer

Introduction / Context: This question distinguishes simple interest from compound interest. With simple interest, the interest earned each period is calculated only on the original principal, not on accumulated interest. Recognizing the correct formula is essential for quick finance calculations and for setting up more complex cash-flow analyses. Given Data / Assumptions: Principal p is constant over time. Per-period simple interest rate i (dimensionless, e.g., 0.08). Number of periods n. No compounding or reinvestment of interest. Concept / Approach: By definition, total simple interest I over n periods is I = p * i * n. The amount (future value) under simple interest would be A = p + I = p * (1 + i * n). In contrast, compound interest uses A = p * (1 + i)^n, which is not a simple-interest relation. Step-by-Step Solution: Identify what is asked: total interest I (not future amount).Recall definition: I = p * i * n.Select the corresponding option. Verification / Alternative check: Consider p = 100, i = 0.05 per period, n = 3. Simple interest I = 100 * 0.05 * 3 = 15. The future amount would then be 115, matching A = p * (1 + i * n). Why Other Options Are Wrong: p (1 + i * n): That is the future amount, not the interest alone.p (1 + i)^n: Compound interest amount (includes interest on interest).p (1 - i * n): Not meaningful for interest accrual.p / (1 + i * n): Resembles present value, not total interest. Common Pitfalls: Confusing “interest accrued” with “future amount,” or carelessly substituting the compound interest formula where simple interest is specified. Final Answer: p * i * n

Question 18

Quarterly compounding: An investment of ₹1000 earns 10% nominal annual interest compounded quarterly. What is its value at the end of 5 years?

Accepted Answer

Introduction / Context: This exercise reinforces how to handle nominal annual rates with sub-annual compounding. When interest is compounded quarterly, the year is divided into four equal periods, each applying one-quarter of the nominal rate. Over multiple years, the total number of compounding periods grows accordingly. Given Data / Assumptions: Principal P = ₹1000. Nominal annual interest rate r = 10% = 0.10. Quarterly compounding → per-quarter rate i_q = r/4 = 0.025. Time T = 5 years → number of quarters n = 5 * 4 = 20. Concept / Approach: Use the compound interest formula with sub-annual compounding: F = P * (1 + r/m)^(mT). Here m = 4 (quarters per year). The effective annual rate is (1 + 0.10/4)^4 − 1, but for the full 5-year horizon we directly compute the 20-period growth factor. Step-by-Step Solution: Compute per-quarter rate: i_q = 0.10/4 = 0.025.Compute number of periods: n = 20.Apply formula: F = 1000 * (1 + 0.025)^(20).Calculate numeric value: (1.025)^(20) ≈ 1.63862 → F ≈ ₹1,638.62. Verification / Alternative check: Check growth annually using the equivalent effective annual rate i_eff = (1.025)^4 − 1 ≈ 0.1038129. Then F = 1000 * (1 + i_eff)^5 ≈ 1000 * (1.1038129)^5 ≈ ₹1,638.62, confirming the calculation. Why Other Options Are Wrong: 1000 * (1 + 0.10)^(20): Treats 20 years of annual compounding; the period count is wrong.₹1,610: Rough simple-interest style estimate; ignores compounding and underestimates.1000 * (1 + 0.025)^(5): Uses the quarterly rate but only 5 periods instead of 20.₹1,500: A guess that ignores both rate and compounding schedule. Common Pitfalls: Mixing nominal and effective rates, using the quarterly rate with an annual period count, or forgetting to multiply years by the number of compounding periods per year. Final Answer: ₹1,638.62 (i.e., 1000 * (1 + 0.10/4)^(20))

Question 19

Cost classification in estimating: Factory manufacturing cost equals the sum of which components for a chemical plant?

Accepted Answer

Introduction / Context: Understanding standard cost breakdowns is essential in chemical engineering economics. “Factory manufacturing cost” (often called manufacturing cost) aggregates the costs required to manufacture the product before selling and administrative expenses are added to form the total product cost. Given Data / Assumptions: We use conventional textbook definitions (e.g., Peters & Timmerhaus, plant design and economics). Terminology: direct production cost (raw materials, operating labor, utilities, etc.), fixed charges (depreciation, taxes, insurance), plant overhead (supervision, maintenance, payroll overhead). Administrative and selling expenses are excluded from factory manufacturing cost. Concept / Approach: Manufacturing (factory) cost = Direct production cost + Fixed charges + Plant overhead cost. This subtotal precedes the addition of general expenses (administrative, selling, distribution, R&D) to produce the total product cost. Recognizing what is inside or outside each subtotal prevents double-counting or omissions. Step-by-Step Solution: Identify direct charges tied to making the product: direct production cost.Add time-based ownership costs: fixed charges (depreciation, taxes, insurance).Include factory-level support: plant overhead (maintenance, supervision, plant services).Exclude administrative and selling: they belong to general expenses, not factory manufacturing cost. Verification / Alternative check: Cross-check a typical cost tree: Total product cost = Manufacturing cost + General expenses, where Manufacturing cost includes the three components selected above. This partitioning is standard in feasibility studies and detailed estimates. Why Other Options Are Wrong: Direct production cost and plant overhead only: Omits fixed charges, thereby understating factory cost.Plant overhead and administrative expenses: Adds non-manufacturing administrative costs improperly.None of these / Direct + general: Do not match the accepted structure. Common Pitfalls: Confusing “manufacturing cost” with “total product cost,” or misplacing fixed charges into general expenses. Always keep component definitions consistent across estimates. Final Answer: Direct production cost, fixed charges, and plant overhead cost

Question 20

Cost coverage check: The total product cost of a chemical plant does not include which of the following?

Accepted Answer

Introduction / Context: Total product cost represents the ongoing cost to manufacture and sell the product. One-time pre-project investigations—like initial market surveys—belong elsewhere (e.g., development or pre-investment phases) and are not typically included in the recurring total product cost used for annual profitability assessments. Given Data / Assumptions: Total product cost usually includes manufacturing cost plus general expenses (administrative, selling, distribution). Recurring costs of operation and ownership are included (labor, utilities, overheads, depreciation, taxes, insurance). Non-recurring pre-project or exploratory costs are excluded. Concept / Approach: Market survey is a preliminary/occasional expenditure that informs whether to proceed with a project; it is not an annual operating cost tied to the product’s manufacture and sale. In contrast, labor, supervision, supplies, overheads, utilities, and fixed charges occur continually and therefore enter the total product cost. Step-by-Step Solution: List ongoing costs: labor, supervision, supplies, overheads, utilities, depreciation, taxes, insurance.Identify non-recurring study: market survey.Exclude market survey from total product cost; include the rest. Verification / Alternative check: Check a typical cost breakdown in plant design texts: market surveys are part of preliminary evaluation or selling research budgets and not treated as part of annual total product cost once the plant is operating. Why Other Options Are Wrong: Operating labour…: Fundamental manufacturing costs; must be included.Overheads and utilities: Core components of ongoing manufacturing cost.Depreciation, property tax, insurance: Fixed charges, counted within total product cost.Plant maintenance and repair: Part of manufacturing or overhead cost. Common Pitfalls: Confusing up-front project evaluation costs with recurring operating costs; remember total product cost is typically annualized for operating plants. Final Answer: Market survey

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