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Mathematics

A man borrows ₹6000 at 5% compound interest. If he repays ₹1200 at the end of each year, find the amount outstanding at the beginning of the third year.

Compound Interest

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Answer

Principal for first year = ₹6000, rate = 5%.

Interest for first year = 6000×5×1100=30000100\dfrac{6000 \times 5 \times 1}{100} = \dfrac{30000}{100} = ₹300.

Amount after first year = ₹6000 + ₹300 = ₹6300.

Money refunded at the end of first year = ₹1200.

Principal for second year = ₹6300 - ₹1200 = ₹5100.

Interest for second year = 5100×5×1100=25500100\dfrac{5100 \times 5 \times 1}{100} = \dfrac{25500}{100} = ₹255.

Amount after second year = ₹5100 + ₹255 = ₹5355.

Money refunded at the end of second year = ₹1200.

Principal for third year = ₹5355 - ₹1200 = ₹4155.

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

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