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Mathematics

A man borrows ₹ 6000 at 5 percent C.I. per annum. If he repays ₹ 1200 at the end of each year, find the amount of the loan outstanding at the beginning of the third year.

Compound Interest

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Answer

For first year :

P = ₹ 6000

T = 1 year

R = 5 %

I = P×R×T100=6000×5×1100\dfrac{P \times R \times T}{100} = \dfrac{6000 \times 5 \times 1}{100} = ₹ 300

Amount = ₹ 6000 + ₹ 300 = ₹ 6300

Amount payed at end of first year = ₹ 1200

Amount left at beginning of second year = ₹ 6300 - ₹ 1200 = ₹ 5100.

For second year :

P = ₹ 5100

R = 5%

T = 1 year

I = P×R×T100=5100×5×1100\dfrac{P \times R \times T}{100} = \dfrac{5100 \times 5 \times 1}{100} = ₹ 255

Amount = ₹ 5100 + ₹ 255 = ₹ 5355

Amount payed at end of second year = ₹ 1200

Amount left at beginning of third year = ₹ 5355 - ₹ 1200 = ₹ 4155.

Hence, amount left at beginning of third year = ₹ 4155.

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