A man borrows rs 10000 at 10% compound interest compounded yearly.At the end of each year, he pays back 20% of the amount for that year.How much money is left unpaid just after the second year?

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{For} \mathrm{first} \mathrm{yeat} \\ \mathrm{Amount}=\mathrm{A}={\left(1+\frac{\mathrm{Rate}}{100}\right)}^1\times \mathrm{P} \\ \mathrm{A}=\left(1+\frac{10}{100}\right)\times 10000=\left(\frac{100+10}{100}\right)\times 10000 \\ \Rightarrow \mathrm{A}=₹11000 \\ \mathrm{He} \mathrm{pays} 20\% \mathrm{of} \mathrm{it} \mathrm{back} \\ \Leftarrow \mathrm{Principal} \mathrm{at} \mathrm{the} \mathrm{start} \mathrm{of} 2\mathrm{nd} \mathrm{year}=\mathrm{A}-20\% \mathrm{of} \mathrm{A}=\mathrm{A}-\frac{20}{100}\mathrm{A}=\mathrm{A}-\frac{\mathrm{A}}{5}=\frac{4\mathrm{A}}{5} \\ =\frac{4}{5}\times 11000=₹8800 \\ \mathrm{For} \mathrm{second} \mathrm{year} \\ \mathrm{Amount}=\mathrm{A}\text{'}={\left(1+\frac{\mathrm{Rate}}{100}\right)}^1\times \mathrm{P} \\ =\left(1+\frac{10}{100}\right)\times 8800=\left(\frac{100+10}{100}\right)\times 8800=₹9680 \\ \mathrm{He} \mathrm{again} \mathrm{pays} 20\% \mathrm{of} \mathrm{it} \\ \Rightarrow \mathrm{Amount} \mathrm{left} \mathrm{just} \mathrm{after} 2\mathrm{nd} \mathrm{year}=9680-20\% \mathrm{of} 9680 \\ =9680-\frac{20}{100}\times 9680 \\ =9680-\frac{1}{5}\times 9680 \\ =9680-1936 \\ =\mathbf{₹}\mathbf{7744}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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