Answer :

Value of a machine 3 years ago, P = Rs.60000


Time, n = 3 years


Rate of depreciates, R = 10% per annum


Now,


Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest


P = Present value


R = Annual interest rate


n = Time in years]


Value = P (1 - R/100)n [ Rate decreases]


= 60000 (1 - 10/100)3


= 60000 (1 - 1/10)3


= 60000 (9/10)3


= 60000 × 9/10 × 9/10 × 9/10


= 60 × 9 × 9 × 9


= 43740


Present value of the machine is Rs.43740.

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