Q. 9

# The value of a ma

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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