Question 65

# 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

= $$5100 + (\frac{5100 \times 5 \times 1}{100}) - 1200$$

= $$5100 + 255 - 1200 = 4155$$