Sudhakar borrows ₹ 22,500 at 10% per annum, compounded annually. If he repays ₹ 11,250 at the end of first year and ₹ 12,550 at the end of the second year, find the amount of loan outstanding against him at the end of the third year.

For first year :

P = ₹ 22,500

T = 1 year

R = 10 %

I = $\dfrac{P \times R \times T}{100}$

$= \dfrac{22500 \times 10 \times 1}{100}$ = ₹ 2,250.

Amount = P + I = ₹ 22,500 + ₹ 2,250 = ₹ 24,750.

Amount payed at end of first year = ₹ 11,250.

Amount left at beginning of second year = ₹ 24,750 - ₹ 11,250 = ₹ 13,500.

For second year :

P = ₹ 13,500

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100}$

$= \dfrac{13500 \times 10 \times 1}{100}$ = ₹ 1,350.

Amount = P + I = ₹ 13,500 + ₹ 1,350 = ₹ 14,850.

Amount payed at end of second year = ₹ 12,550.

Amount left at beginning of third year = ₹ 14,850 - ₹ 12,550 = ₹ 2,300

For third year :

P = ₹ 2,300

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100}$

$= \dfrac{2300 \times 10 \times 1}{100}$ = ₹ 230.

Amount due at the end of third year = P + I = ₹ 2,300 + ₹ 230 = ₹ 2,530.

Hence, the amount outstanding at the end of the third year = ₹ 2,530.