Compound Interest-10th Grade-4

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Mr. Harrison bought a car for $400000. If the price of the car after
3 years depreciates to $291600, find the rate of depreciation.

Solution: Let the rate of depreciation be R% per annum.

Time (n) = 3 years

We know, A = P (1 – R /100)n

So, 291600 = 400000 (1 - R /100)3

or, ( 1 - R /100)3 = 291600 / 400000 = 729 / 1000

or, ( 1 - R /100)3 = (9/100)3

or, 1 – R /100 = 9/10

or, R /100 = 1 - 9/10

or, R /100 = 1/10

So, R = 10

So, the required rate of depreciation = 10% per annum.

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