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प्रश्न
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.
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उत्तर
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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