Question

A man borrows Rs. $1,200$ at $10$ percent per annum compound interest. If he repays Rs. $250$ at the end of each year, find the amount of loan outstanding at the beginning of the fourth year.

Answer 1

For the first year

$P=Rs1,200$
$N=1year$
$R=10$ %
We have $S.I.=\frac{PNR}{100}=\frac{1,200\times 1\times 10}{100}=Rs120$
And Amount at the end of first year $P+S.I.=Rs1,200+Rs120=Rs1,320$

Now, for the second year

$P=Rs1,320-Rs250=Rs1,070$
$N=1year$
$R=10$ %
We have $S.I.=\frac{PNR}{100}=\frac{1,070\times 1\times 10}{100}=Rs107$
And Amount at the end of second year $P+S.I.=Rs1,070+Rs107=Rs1,177$

Also, for the third year

$P=Rs1,177-Rs250=Rs927$
$N=1year$
$R=10$ %

We have $S.I.=\frac{PNR}{100}=\frac{927\times 1\times 10}{100}=Rs92.70$
And Amount at the end of third year $P+S.I.=Rs927+Rs92.70=Rs1019.70$
At the beginning of the fourth year, amount of loan outstanding $=Rs1019.70-Rs250=Rs769.70$