--- title: "Capital recovery cost; interest on motorbike loan" subject: "Engineering Economics" chapter: 2 key: "semester_6/engineering_economics:capital-recovery-cost-motorbike-loan" source: https://www.bctengineer.com/q/semester_6/engineering_economics/capital-recovery-cost-motorbike-loan site: BCT Engineering --- # Capital recovery cost; interest on motorbike loan Engineering Economics · Chapter 2: Interest and Time Value of Money - Asked in: 2066 Magh, 2+6 marks - Video: [Uniform Series Capital Recovery Factor - Engineering Economics Lightboard](https://www.youtube.com/watch?v=wOu-yrxf4LM) — Engineering Economics Guy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/engineering_economics/capital-recovery-cost-motorbike-loan - Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-2 ## Question Define 'Capital Recovery Cost'. Mr. Fox purchased a motorbike which cost Rs. 2,00,000. He pays 30% as down payment. Remaining amount will be paid on installment basis and he wishes to pay Rs. 10,000 per month for 20 months. What annual interest rate is he paying? ## Short answer Capital recovery cost is the uniform annual amount to recover an asset's initial cost minus salvage over its life. Loan 1,40,000, monthly 10,000 for 20 months gives (P/A,i,20)=14; solving yields i ≈ 3.67% per month, nominal annual rate ≈ 44%. ## Answer #### Capital recovery cost Capital recovery (CR) cost is the equivalent uniform annual cost of owning an asset: the annual amount needed to recover the initial investment I over its life N at interest rate i, after allowing for the salvage value S. CR = I (A/P, i, N) − S (A/F, i, N) = (I − S)(A/P, i, N) + S·i It is the sum of annual depreciation and the interest on the unrecovered capital. It is a component of annual worth (AW) and of cost comparisons between alternatives. #### Numerical - Down payment = 30% × 2,00,000 = Rs 60,000 - Loan P = 2,00,000 − 60,000 = Rs 1,40,000 - A = Rs 10,000 per month, n = 20 months ```text +1,40,000 0 1 2 ... 20 (months) ↓ ↓ ↓ 10,000 at the end of each month ``` Equation: 1,40,000 = 10,000 (P/A, i, 20), so (P/A, i, 20) = 14. | i per month | (P/A, i, 20) | vs 14 | |---|---|---| | 3.0% | 14.8775 | too high | | 4.0% | 13.5903 | too low | | 3.667% | 14.0000 | = 14 | Interpolation: i = 3 + 1 × (14.8775 − 14)/(14.8775 − 13.5903) ≈ 3.67% per month. - Nominal annual rate = 12 × 3.667% = **44.0% per year** - Effective annual rate = (1 + 0.03667)^12 − 1 = 54.1% **Answer: Mr. Fox pays about 44% per year nominal (≈ 54% effective). The rate is very high because Rs 2,00,000 is paid back for a Rs 1,40,000 loan within only 20 months.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is capital recovery cost? A. Cost of land B. Operating cost C. Salvage value D. Equivalent uniform annual cost of recovering the investment over life after salvage Answer: D. CR = I(A/P,i,N) - S(A/F,i,N). (Key point: Capital recovery) 2. Loan amount? A. Rs 1,40,000 B. Rs 60,000 C. Rs 2,00,000 D. Rs 2,00,000 + interest Answer: A. 70% of 2,00,000. (Key point: Loan) 3. Equation for rate? A. (P/A, i, 20) = 20 B. (P/A, i, 20) = 14 C. (F/A, i, 20) = 14 D. (P/A, i, 12) = 14 Answer: B. 140,000/10,000. (Key point: Equation) 4. Nominal annual rate? A. 3.67% B. 54% C. About 44% (12 x 3.667%) D. 12% Answer: C. Effective is 54.1%. (Key point: Rate) ## Related in Engineering Economics - [Car loan interest rate and lump sum at year 3](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Acar-loan-installments-lump-sum-year3) — Engineering Economics, chapter 2 - [Loan 120,000: number of installments and rate](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aloan-120000-installments-rate) — Engineering Economics, chapter 2 ## Related in other subjects - [EPS and book value; capital Rs 5,00,000](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Aeps-bv-capital-500000-2000-shares) — Project Engineering, chapter 6 - [EPS; total capital Rs 15,00,000](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Aeps-capital-1500000-pref-loan-a74) — Project Engineering, chapter 6 - [EPS and book value; ABC project Rs 6,00,000](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Aeps-bv-abc-project-600000-b79) — Project Engineering, chapter 6 - [EVA and S-curve; Rs 30,00,000, 24 months](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Aeva-s-curve-24-months-c76) — Project Engineering, chapter 4