Question

The compound interest on a sum of money for 2 years is Rs 832 and the simple interest on the same sum for the same period is Rs 800. The difference between the compound interest and the simple interest for 3 years will be?

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{P}\mathrm{be}\mathrm{the}\mathrm{sum} \\ \mathrm{then}\mathrm{C}.\mathrm{I}=\mathrm{A}-\mathrm{P}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^n-\mathrm{P} \\ \Rightarrow \mathrm{C}.\mathrm{I}=\mathrm{P}\left[{\left(1+\frac{\mathrm{R}}{100}\right)}^n-1\right] \\ \Rightarrow 832=\mathrm{P}\left[{\left(1+\frac{\mathrm{R}}{100}\right)}^2-1\right]\mathbf{}\left\{\mathbf{as}\mathbf{}\mathbf{C}\mathbf{.}\mathbf{I}\mathbf{}\mathbf{=}\mathbf{}\mathbf{Rs}\mathbf{}\mathbf{832}\mathbf{}\mathbf{\&}\mathbf{}\mathbf{time}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathbf{years}\right\} \\ \Rightarrow 832=\mathrm{P}\left[1+\frac{{\mathrm{R}}^2}{10000}+\frac{\mathrm{R}}{50}-1\right] \\ \Rightarrow 832=\mathrm{P}\left[\frac{{\mathrm{R}}^2}{10000}+\frac{\mathrm{R}}{50}\right] \\ \Rightarrow 832=\mathrm{PR}\left[\frac{\mathrm{R}}{10000}+\frac{1}{50}\right]..............\left(1\right) \\ \mathrm{Now},\mathrm{S}.\mathrm{I}=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100} \\ \Rightarrow 800=\frac{\mathrm{P}\times \mathrm{R}\times 2}{100}\mathbf{}\left\{\mathbf{as}\mathbf{}\mathbf{S}\mathbf{.}\mathbf{I}\mathbf{}\mathbf{=}\mathbf{}\mathbf{Rs}\mathbf{}\mathbf{800}\mathbf{}\mathbf{\&}\mathbf{}\mathbf{time}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathbf{years}\right\} \\ \Rightarrow 800=\frac{\mathrm{PR}}{50} \\ \Rightarrow \mathrm{PR}=40000.................\left(2\right) \end{gathered}$$

Putting the value of PR in (1), we get,
$$\begin{gathered} 832=40000\left[\frac{\mathrm{R}}{10000}+\frac{1}{50}\right] \\ \Rightarrow 832=4\mathrm{R}+800 \\ \Rightarrow 4\mathrm{R}=832-800=32 \\ \Rightarrow \mathrm{R}=\frac{32}{4}=8\% \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{R}\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get}, \\ \mathrm{P}\times 8=40000 \\ \Rightarrow \mathrm{P}=\frac{40000}{8}=5000 \\ \mathrm{Hence},\mathrm{P}=\mathrm{Rs}5000\& \\ \mathrm{R}=8\% \\ \mathbf{To}\mathbf{}\mathbf{find}\mathbf{}\mathbf{the}\mathbf{}\mathbf{difference}\mathbf{}\mathbf{between}\mathbf{}\mathbf{C}\mathbf{.}\mathbf{I}\mathbf{}\mathbf{\&}\mathbf{}\mathbf{S}\mathbf{.}\mathbf{I}\mathbf{}\mathbf{for}\mathbf{}\mathbf{three}\mathbf{}\mathbf{years}\mathbf{.} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathrm{C}.\mathrm{I}=\mathrm{A}-\mathrm{P} \\ \Rightarrow \mathrm{C}.\mathrm{I}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^n-\mathrm{P} \\ \Rightarrow \mathrm{C}.\mathrm{I}=\mathrm{P}\left[{\left(1+\frac{\mathrm{R}}{100}\right)}^n-1\right] \\ \Rightarrow \mathrm{C}.\mathrm{I}=5000\left[{\left(1+\frac{8}{100}\right)}^3-1\right]\left\{\mathrm{as}\mathit{}n=3\mathrm{years}\right\} \\ \Rightarrow \mathrm{C}.\mathrm{I}=5000\left[{\left(\frac{108}{100}\right)}^3-1\right] \\ \Rightarrow \mathrm{C}.\mathrm{I}=5000\left[{\left(\frac{27}{25}\right)}^3-1\right] \\ \Rightarrow \mathrm{C}.\mathrm{I}=5000\left[\frac{19683}{15625}-1\right] \\ \Rightarrow \mathrm{C}.\mathrm{I}=5000\left[\frac{19683-15625}{15625}\right] \\ \Rightarrow \mathrm{C}.\mathrm{I}=5000\times \frac{4058}{15625}=\frac{8\times 4058}{25}=\frac{32464}{25}=\mathrm{Rs}1298.56 \\ \mathrm{S}.\mathrm{I}=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}=\frac{5000\times 8\times 3}{100}=\mathrm{Rs}1200 \end{gathered}$$

$\mathrm{Hence},\mathrm{difference}\mathrm{between}\mathrm{C}.\mathrm{I}\&\mathrm{S}.\mathrm{I}=\mathrm{Rs}1298.56-\mathrm{Rs}1200=\mathrm{Rs}98.56$