Question

A man borrows Rs.6500 at 10% per annum compound interest payable half-yearly. He repays Rs.2000 at the end of every six months. Calculate the amount outstanding at the end of the third payment. Give your answer to the nearest rupee.

Answer 1

For 1^st half year : P = Rs.6500, R = 10% and T = `(1)/(2)` year

Interest = Rs.`(6500 xx 10 xx 1)/(100 xx 2)`
 =Rs.325
Amount 
 = Rs.6500 + Rs.325
= Rs.6825
Money paid at the end of 1st half year = Rs.2000
Balance money for 2^nd half year
= Rs.6825 - Rs.2000
= Rs.4825
For 2^nd half year : P = Rs.4825; R = 10% and T = `(1)/(2)` year

Interest = Rs.`(4825 xx 10 xx 1)/(100 xx 2)`
= RS.241.25
Amount
= Rs.4825 + Rs.241.25
= Rs.5066.25
Money paid at the end of 2^nd half year = Rs.2000
Balance money for 3^rd half year
 =Rs.5066.25 - Rs.2000
= Rs.3066.25
For 3^rd half year : P = Rs.3066.25; R = 10% and T = `(1)/(2)` year

Interest = Rs.`(3066.25 xx 10 xx 1)/(100 xx 2)`
 =Rs.153.3125
Amount
 =Rs.3066.25 + Rs.153.3125
= Rs.3219.5625
Money paid at the end of 3^rd half year = Rs.2000
Amount outstanding at the end of 3^rd payment
= Rs.3219.5625  - Rs.2000
= Rs.1219.5625
= Rs.1220 (nearest rupee).