A man borrows ₹ 15,000 at 12% per annum, compounded annually. If he repays ₹ 4,400 at end of each year, find the amount outstanding against him at the beginning of third year.

For first year :

P = ₹ 15,000

T = 1 year

R = 12%

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

$= \dfrac{15000 \times 12 \times 1}{100}$ = ₹ 1,800.

Amount = P + I = ₹ 15,000 + ₹ 1,800 = ₹ 16,800.

Amount payed at end of first year = ₹ 4,400.

Amount left at beginning of second year = ₹ 16,800 - ₹ 4,400 = ₹ 12,400.

For second year :

P = ₹ 12,400

R = 12%

T = 1 year

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

$= \dfrac{12400 \times 12 \times 1}{100}$ = ₹ 1,488.

Amount = P + I = ₹ 12,400 + ₹ 1,488 = ₹ 13,888.

Amount payed at end of second year = ₹ 4,400.

Amount left at beginning of third year = ₹ 13,888 - ₹ 4,400 = ₹ 9,488.

Hence, amount left at beginning of third year = ₹ 9,488.