Subtopic: Cost analysis of implements and tractors
Question Statement:
The cost of a tractor is Rs. 340000. Its service life is 10 years and the salvage value is 10% of initial cost. Its depreciation in 2nd year, following constant rate method, in rupees would be
Options:
(A) 30600
(B) 34000
(C) 50070
(D) 55550
Official Answer Key:
D
Step-by-Step Solution & Explanation
Constant (declining) percentage method: book value declines by a fixed fraction r each year such that after n=10 years the salvage value S = 0.10×340000 = 34000 remains. r is found from S = C(1−r)ⁿ ⇒ (1−r) = (S/C)^(1/n) = (0.1)^(0.1) = 0.7943 ⇒ r = 0.2057 (≈20.57%/yr). Book value after year 1: BV₁ = 340000×0.7943 = 270,072. Depreciation in year 2 = r×BV₁ = 0.2057×270,072 ≈ Rs 55,546 ≈ 55,550, matching option (D).
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