Question

A man borrows Rs. $5,800$ at $12$% per annum compound interest. He repays Rs. $1,800$ at the end of every six months. Calculate the amount standing at the end of the third payment. Give your answer to the nearest rupee.

Answer 1

For the first year
$P=Rs5,800$
$N=\frac{1}{2}year$
$R=12$ %
We have $S.I.=\frac{PNR}{100}=\frac{5,800\times \frac{1}{2}\times 12}{100}=Rs348$
And Amount at the end of first 6 months $P+S.I.=Rs5,800+Rs348=Rs6,148$
Now, for the second half year
$P=Rs6,148-Rs1,800=Rs4,348$
$N=\frac{1}{2}year$
$R=12$ %
We have $S.I.=\frac{PNR}{100}=\frac{4,348\times \frac{1}{2}\times 12}{100}=Rs260.88$
And Amount at the end of second half - year $P+S.I.=Rs4,348+Rs260.88=Rs4,608.88$
Also, for the third half year
$P=Rs4,608.88-Rs1,800=Rs2,808.88$
$N=\frac{1}{2}year$
$R=12$ %
We have $S.I.=\frac{PNR}{100}=\frac{2,808.88\times \frac{1}{2}\times 12}{100}=Rs168.5328$
And Amount at the end of third half - year $P+S.I.=Rs2,808.88+Rs168.5328=Rs2977.4128$ or approximately $Rs2977$
And after the third payment amount outstanding $=2977.4128-1,800=Rs1177.4128$ or approximately $Rs1177$