Loan Amortization under Compound Interest — Three equal end-of-year payments: A man borrows ₹ 4000 at 7 1/2% p.a. compound interest. At the end of every year he pays ₹ 1500 (covering interest and part principal). How much does he still owe after three such payments?

Introduction / Context:
This is a straightforward amortization with annual compounding: each year the outstanding balance accrues interest, then the borrower pays a fixed amount, reducing the balance. Repeating the process year by year yields the remaining debt after the specified number of payments.

Given Data / Assumptions:

• Initial principal P0 = ₹ 4000
• Annual interest rate i = 7.5% = 0.075
• End-of-year payment each year = ₹ 1500
• Number of payments = 3

Concept / Approach:
Iterate: after interest, subtract the payment. Algebraically: B1 = P0(1 + i) − 1500; B2 = B1(1 + i) − 1500; B3 = B2(1 + i) − 1500. The remaining balance after the 3rd payment is owed to the bank.

Step-by-Step Solution:

Verification / Alternative check:

Why Other Options Are Wrong:

• ₹ 125 and ₹ 469.18 are near-miss computations typically arising from rounding or misordering “interest then payment”.
• ₹ 400 and ₹ 0 do not reflect the calculated amortization path.

Common Pitfalls:

• Subtracting the payment before applying interest; the problem states end-of-year payments (interest applies first).

Final Answer:
Rs. 123.25.