A man borrows Rs.$8000$ at $10$ percent compound interest. If he repays Rs.$1500$ at the end of the first year and Rs.$3200$ at the end of the second year, find the amount of loan outstanding at the beginning of the third year.

Hint: Calculate the amount to be paid at the end of two years by applying the formula of compound interest. Deduce the sum of money paid by him. The remaining money will be outstanding at the beginning of third year.

Complete step-by-step answer: Formula of compound interest is given by
$A = P{\left( {1 + \dfrac{r}{{100}}} \right)^t}$ (1)
Where A is amount, P is principal , r is rate of interest and t is the number of years.
We are given that $P = 8000$
$r = 10 p.a.$
Since, we need to calculate the money at the end of two years,
$t = 2$
Now, we have to calculate A.
Substituting the values of P,r,t in equation (1),
We get
$
  A = 8000{\left( {1 + \dfrac{{10}}{{100}}} \right)^2} \\
   = 8000{\left( {1 + 0.1} \right)^2} \\
   = 8000{\left( {1.1} \right)^2} \\
   = 8000\left( {1.21} \right) \\
   = 9680 \\
 $
Total amount= Rs.$9680$
Also, given that he repays Rs.$1500$ at the end of first year and Rs.$3200$ at the end of second year.
So, the amount already paid by him in two years is $3200 + 1500 = 4700$
Therefore, amount of loan outstanding at the beginning of third year will be
$\text{total amount} – \text{amount paid}$
$
   = 9680 - 4700 \\
   = 4980 \\
 $


Note: These types of questions could be easily solved by applying the formula of compound interest. We need to take special care while writing the value of A and P in the formula as students often confuse them.