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Question
The cost of a machine depreciated by ₹ 900 during the second year and by ₹ 810 during the third year. Find:
- the rate of depreciation.
- the original cost of the machine.
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Solution
Given:
- Depreciation on the machine during the second year = ₹ 900
- Depreciation on the machine during the third year = ₹ 810
Let:
- Original cost of the machine = (P)
- Rate of depreciation = (r%)
Step 1: Depreciation is on the reduced value year by year.
After the first year, value of machine = `P xx (1 - r/100)`
After the second year, value = `P xx (1 - r/100)^2`
After the third year, value = `P xx (1 - r/100)^3`
Step 2: Calculate depreciation amounts in terms of (P) and (r).
Depreciation during the second year = Value at end of first year − Value at end of second year
`P(1 - r/100) - P(1 - r/100)^2 = 900`
Depreciation during the third year = Value at end of second year − Value at end of third year
`P(1 - r/100)^2 - P(1 - r/100)^3 = 810`
Step 3: Divide the two depreciation equations to find (r).
`(P(1 - r/100) - P(1 - r/100)^2)/(P(1 - r/100)^2 - P(1 - r/100)^3) = 900/810`
Simplify numerator and denominator:
`((1 - r/100)[1 - (1 - r/100)])/((1 - r/100)^2[1 - (1 - r/100)]) = 900/810`
The brackets `[1 - (1 - r/100)] = r/100`, so:
`((1 - r/100) xx r/100)/((1 - r/100)^2 xx r/100) = 900/810`
Cancel `r/100`:
`(1 - r/100)/((1 - r/100)^2) = 900/810`
Simplify:
`1/(1 - r/100) = 900/810`
`1/(1 - r/100) = 10/9`
Therefore,
`1 - r/100 = 9/10`
⇒ `r/100 = 1 - 9/10`
⇒ `r/100 = 1/10`
⇒ r = 10
Step 4: Find the original cost (P).
From Step 2’s depreciation formula for the second year:
`P(1 - 10/100) - P(1 - 10/100)^2 = 900`
P × 0.9 – P × 0.92 = 900
P(0.9 – 0.81) = 900
P × 0.9 = 900
`P = 900/0.09`
P = 10,000
