3.9 Depreciation of Machinery/Asset Value
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Depreciation is mathematically identical to population decline — the value of an asset falls by a fixed percentage of its current (not original) value each year.
V_n = V(1-(r)/(100))^n
Worked Example 1: A machine worth ₹2,00,000 depreciates at 10% per annum. Find its value after 3 years. V₃ = 2,00,000 × (0.9)³ = 2,00,000 × 0.729 = ₹1,45,800.
Worked Example 2: A machine bought for ₹1,50,000 depreciates by 20% in the first year and 15% in the second year. Find its value at the end of two years. V₂ = 1,50,000 × 0.8 × 0.85 = 1,50,000 × 0.68 = ₹1,02,000.
Worked Example 3: A machine depreciates by 10% every year. If its present value is ₹72,900 after 2 years of depreciation, find its original value. Original × (0.9)² = 72,900 → Original × 0.81 = 72,900 → Original = 72,900/0.81 = ₹90,000. (This is the “reverse” application — working backward from a depreciated value to the original, a technique you will reuse heavily in Compound Interest.)
Worked Example 4: The value of a machine depreciates by 20% in the first year and by a further 10% in the second year. If it was purchased for ₹80,000, find the total depreciation (in rupees) over the two years. Value after 2 years = 80,000 × 0.8 × 0.9 = 80,000 × 0.72 = ₹57,600. Total depreciation = 80,000 − 57,600 = ₹22,400. Note that this is not simply 30% of 80,000 (which would be ₹24,000) — the second year’s 10% depreciation acts on the already-reduced value, not the original ₹80,000, which is exactly why the successive-change formula (net change = −28%, i.e., −20 − 10 + (−20×−10)/100 = −30+2 = −28%) gives ₹22,400 (28% of 80,000), matching the direct calculation.
Worked Example 5 (three different annual rates, reverse): A machine depreciates by 10% in the first year, 20% in the second year, and 25% in the third year. If its value after these three years of depreciation is ₹64,800, find its original price. Combined factor = 0.90 × 0.80 × 0.75 = 0.72 × 0.75 = 0.54. Original × 0.54 = 64,800 → Original = 64,800/0.54 = ₹1,20,000.
Worked Example 6: A car was purchased for ₹8,00,000. It depreciates at 12.5% per annum. Find its value after 2 years and the total depreciation over this period. Since 12.5% = 1/8, the retention factor each year is 7/8. Value after 2 years = 8,00,000 × (7/8)² = 8,00,000 × 49/64. 8,00,000 ÷ 64 = 12,500; 12,500 × 49 = ₹6,12,500. Total depreciation = 8,00,000 − 6,12,500 = ₹1,87,500.