--- title: "Loan 120,000: number of installments and rate" subject: "Engineering Economics" chapter: 2 key: "semester_6/engineering_economics:loan-120000-installments-rate" source: https://www.bctengineer.com/q/semester_6/engineering_economics/loan-120000-installments-rate site: BCT Engineering --- # Loan 120,000: number of installments and rate Engineering Economics · Chapter 2: Interest and Time Value of Money - Asked in: 2065 Chaitra, 4+4 marks - Video: [Amortization Loan Formula](https://www.youtube.com/watch?v=lkNJvsy0qU8) — The Organic Chemistry Tutor - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/engineering_economics/loan-120000-installments-rate - Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-2 ## Question Mr. X receives a loan of Rs 120,000 from a bank at an interest rate of 12% per year. i) He wishes to repay the loan in monthly installment with Rs. 3000 per month. How many installments are necessary to complete his payment? ii) What annual interest rate is he paying if the bank asks him to pay Rs 5000 per month for 30 times? ## Short answer i) At 1% monthly, (P/A,1%,n)=40 → n≈51.34, so about 52 installments (51 of Rs 3,000 plus a smaller last). ii) For Rs 5,000/month over 30 months, (P/A,i,30)=24 gives i≈1.505% per month, nominal annual rate ≈ 18.1%. ## Answer Loan P = Rs 1,20,000 received at t = 0; payments are at the end of each month. #### (i) Number of installments at 12% per year Assumption: 12% nominal, compounded monthly, so i = 1% per month. P = A (P/A, i, n) so (P/A, 1%, n) = 1,20,000 / 3,000 = 40 [1 − (1.01)^-n] / 0.01 = 40, so (1.01)^-n = 1 − 0.40 = 0.60 n = −ln(0.60) / ln(1.01) = 0.51083 / 0.00995 = 51.34 months Check: 51 payments have PW = 3,000 × 39.7981 = Rs 119,394, 52 payments have PW = Rs 121,183. So 51 full installments of Rs 3,000 are not enough; a 52nd, smaller payment of about Rs 1,016 clears the loan. **Answer (i): about 52 monthly installments (51 of Rs 3,000 plus a small last one).** #### (ii) Rate for Rs 5,000 per month for 30 months (P/A, i, 30) = 1,20,000 / 5,000 = 24 | i per month | (P/A, i, 30) | vs 24 | |---|---|---| | 1.0% | 25.8077 | too high | | 2.0% | 22.3965 | too low | | 1.5% | 24.0158 | slightly high | | 1.505% | 24.0000 | = 24 | - Nominal annual rate = 12 × 1.505% = **18.06% per year** - Effective annual rate = (1 + 0.01505)^12 − 1 = 19.63% **Answer (ii): about 18.1% per year nominal (19.6% effective).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For 12% per year nominal compounded monthly, what monthly rate is used in part (i)? A. 1% B. 12% C. 0.12% D. 12.68% Answer: A. Nominal 12% divided by 12 months gives i = 1% per month. (Key point: Monthly rate) 2. Which relation is solved to find the number of installments in (i)? A. (F/A, 1%, n) = 120,000 / 3,000 B. (P/F, 1%, n) = 40 C. (P/A, 1%, n) = 120,000 / 3,000 = 40 D. (A/P, 1%, n) = 40 Answer: C. The loan is the present worth of equal monthly payments, so P = A(P/A, i, n). (Key point: Equation used) 3. n computes to 51.34 months. What is the correct conclusion? A. Exactly 51 payments of Rs 3,000 clear the loan B. About 52 payments: 51 of Rs 3,000 plus a smaller last one C. 50 payments are enough D. The loan can never be repaid Answer: B. 51 payments have PW of Rs 119,394, less than the loan, so a small 52nd payment is needed. (Key point: Rounding the result) 4. How is the rate for Rs 5,000 for 30 months found? A. Divide 5,000 by 120,000 directly B. Use (F/A, i, 30) = 24 and read once C. Take i = 12% since the loan rate is unchanged D. Compute (P/A, i, 30) = 24 and find i by trial and interpolation Answer: D. (P/A, i, 30) = 24 lies between 1% (25.8077) and 2% (22.3965); i is about 1.505% per month, i.e. 18.06% nominal. (Key point: Finding the rate in (ii)) ## 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 - [Capital recovery cost; interest on motorbike loan](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Acapital-recovery-cost-motorbike-loan) — Engineering Economics, chapter 2 - [Simple vs compound interest; number of deposits](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Asimple-vs-compound-deposits-20-lakh) — Engineering Economics, chapter 2 - [Monthly deposits, quarterly compounding, balance at year 8](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aequal-monthly-deposits-balance-8-years) — Engineering Economics, chapter 2 - [Constant vs actual dollar; inflation-free rate on loan](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aconstant-actual-dollar-bike-loan) — Engineering Economics, chapter 8, asked 2× - [Deposit now to draw growing monthly amount](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Adeposit-now-growing-withdrawal) — Engineering Economics, chapter 2 ## Related in other subjects - [Capitalised value from Rs 20,000 monthly rent](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Avalue-building-20000-rent-2078) — Construction Management, chapter 13 - [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 - [Best financing option; EBIT Rs 350,000](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Abest-option-two-options-ebit-350000) — 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