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A man borrow Rs. 4000 at 15%, compound rate of interest. At the end of each year he pays back Rs. 1500. How much amount should be pay at the end of the third year to clear all his dues ?

a) Rs. 874.75
b) Rs. 824.50
c) Rs. 924.25
d) Rs. 974.25

Answer: a
Explanation:
$$\eqalign{ & {\text{Amount after }}{{\text{1}}^{{\text{st}}}}{\text{ year}} \cr & {\text{ = Rs}}{\text{. }}\left[ {4000\left( {1 + \frac{{15}}{{100}}} \right) – 1500} \right] \cr & = {\text{Rs}}{\text{. }}\left[ {\left( {4000 \times \frac{{23}}{{20}}} \right) – 1500} \right] \cr & = {\text{Rs}}{\text{. }}\left( {4600 – 1500} \right) \cr & = {\text{Rs}}{\text{. }}3100 \cr & {\text{Amount after }}{{\text{2}}^{{\text{nd}}}}{\text{ year}} \cr & {\text{ = Rs}}{\text{. }}\left[ {3100\left( {1 + \frac{{15}}{{100}}} \right) – 1500} \right] \cr & = {\text{Rs}}{\text{. }}\left[ {\left( {3100 \times \frac{{23}}{{20}}} \right) – 1500} \right] \cr & = {\text{Rs}}{\text{. }}\left( {3565 – 1500} \right) \cr & = {\text{Rs}}{\text{. }}2065 \cr & {\text{Amount after }}{{\text{3}}^{{\text{rd}}}}{\text{ year}} \cr & {\text{ = Rs}}{\text{. }}\left[ {2065\left( {1 + \frac{{15}}{{100}}} \right) – 1500} \right] \cr & = {\text{Rs}}{\text{. }}\left[ {\left( {2065 \times \frac{{23}}{{20}}} \right) – 1500} \right] \cr & = {\text{Rs}}{\text{. }}\left( {2374.75 – 1500} \right) \cr & = {\text{Rs}}{\text{. }}874.75 \cr} $$

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