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Question 8
A man borrows 6000 at 5% interest, on reducing balance, at the start of the year. If he repays 1200 at the end of each year, find the amount of loan outstanding, in , at the beginning of the third year.
Solution
Amount man gets after 1 year
= $$6000 + (\frac{6000 \times 5 \times 1}{100}) - 1200$$
= $$6000 + 300 - 1200 = 5100$$
$$\therefore$$ Amount at the beginning of third year, i.e. after 2 years
= $$5100 + (\frac{5100 \times 5 \times 1}{100}) - 1200$$
= $$5100 + 255 - 1200 = 4155$$
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