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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.
Solution
For 1st 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 2nd half year
= Rs.6825 - Rs.2000
= Rs.4825
For 2nd 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 2nd half year = Rs.2000
Balance money for 3rd half year
=Rs.5066.25 - Rs.2000
= Rs.3066.25
For 3rd 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 3rd half year = Rs.2000
Amount outstanding at the end of 3rd payment
= Rs.3219.5625 - Rs.2000
= Rs.1219.5625
= Rs.1220 (nearest rupee).
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