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?

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