Question

Ram borrows $Rs\,22500$ at $10$ percent per annum, compounded annually. If he repays $Rs\,11250$ at the end of first year and $Rs\,12550$ at the end of the second year, find the amount of loan outstanding against him at the end of third year.

Answer 1

Hint: In order to find the outstanding loan at the end of the third year, we need to calculate the amount need to pay at the end of first year, then subtracting with the amount paid actually at the end of the year, and the resultant becomes the principal for second year. Similarly following these steps till the amount to be paid at the end of third year obtained.

Formula used:
$A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}$,
where $A$ is the amount to be paid at the end of the year, $r$ is the rate per annum, and $n$ is the time period/year.

Complete step by step answer:
Since, the loan borrowed is $Rs\,22500$, So, the Principal Amount $\left( P \right) = Rs\,22500$, the time given is $n = 3$, and the rate is $10\% $ per annum. We would calculate the amount year by year. For the first year, Principal Amount is $Rs\,22500$, time period is $1$ year:

Amount to be paid at the end of the first year$ = P{\left( {1 + \dfrac{r}{{100}}} \right)^n} = 22500{\left( {1 + \dfrac{{10}}{{100}}} \right)^1} = 22500\left( {\dfrac{{100 + 10}}{{100}}} \right) = 22500 \times \dfrac{{110}}{{100}} = Rs\,24750$
But the amount paid at the end of first year is $Rs11250$.

So, the remaining balance is $Rs24750 - Rs11250 = Rs13500$, which becomes the Principal Amount for second year and time period for second year$\left( n \right) = 1$.
Amount to be paid at the end of the second year$ = 13500{\left( {1 + \dfrac{{10}}{{100}}} \right)^1} = 13500 \times {\left( {\dfrac{{110}}{{100}}} \right)^1} = Rs14850$
But the amount paid at the end of second year is $Rs12550$. So, the remaining balance is $Rs14850 - Rs12550 = Rs2300$, which becomes the Principal Amount for the third year.
Finally, the Amount to be paid at the end of the third year $ = 2300{\left( {1 + \dfrac{{10}}{{100}}} \right)^1} = 2300 \times {\left( {\dfrac{{110}}{{100}}} \right)^1} = Rs\,2530$

Therefore, the amount of loan outstanding at the end of third year is $Rs\,2530$.

Note:It’s always important to go step by step for each year rather than directly stepping in the last year, as the borrower was paying the money some amount of money at the end of every year. Cross check your answer once always, before coming to a conclusion.