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प्रश्न
Find the compound interest to the nearest rupee on Rs. 10,800 for `2 1/2` years at 10% per annum.
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उत्तर
Given : P = Rs. 10,800 ; Time = `2 1/2` years and Rate = 10% p.a.
For 2 years
A = P`( 1 + r/100)^n` = 10,800( 1 + 10/100 )^2 = Rs. 13,068
For `1/2` year
∴ A = P`( 1 + r/[ 2 xx 100 ])^( n xx 2) = 13068( 1 + 10/[ 2 xx 100 ])^( 1/2 xx 2)`
= 13068 x `21/20` = Rs. 13721.40 = Rs. 13721 ( nearest rupee)
∴ Rs.13,721 - Rs.10,800 = Rs.2,921
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