Math, asked by mahadheer, 1 month ago

A machine depreciated at the rate of 20% on reducing balance.The original cost of the machine was rs.100000 and it's ultimate scrap value is rs.30000. find the efective life of machine.

Answers

Answered by durgeshpandey83
1

Step-by-step explanation:

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Answered by mohammedrairaz
2

Step-by-step explanation:

Cost of machine (P) = Rs 1,00,000

Scrap value (A) = Rs 30,000

Rate of Depreciation = 20% per annum on reducing value

The effective life of the machine in years is the number of years in which P (Rs 1,00,000) would reduce to A (scrap value Rs 30,000) reducing at the rate of 20% per annum of the value at the start of that year year.

Value of the machine at time t= 0 years = P

The depreciated cost at end of one year = P[1 — 20%] = P[1 — 0.2] = P × 0.8

At the end of second year = P × 0.8²

At the end of 3rd year = P × 0.8³

And so on.

Let after n years the value depreciate to scrap value. We are required to find n.

P(0.8)^n = A

1,00000 (0.8)^n = 30,000

=> (0.8)^n = (30,000)/(1,00,000) = 0.3

Taking log of both sides

n log (0.8) = log (0.3)

=>n × (-0.09691) = (-0.52288)

=> n = (-0.52288)/(-0.09691)= 5.396 year ~5.4 years

The effective life of machine in years is about 5 year and 5 months.

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