Question

Find the compound interest to the nearest rupee on Rs. 10,800 for `2 1/2` years at 10% per annum.

Answer 1

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