Question

In how many years will Rs 1800 amount to Rs 2178 at 10% per annum when compounded annually?

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{required}\mathrm{time}\mathrm{be}n\mathrm{years}. \\ \mathrm{Rate}\mathrm{of}\mathrm{interest},R\mathit{}=10\% \\ \mathrm{Principal}\mathrm{amount},P=\mathrm{Rs}.1800 \\ \mathrm{Amount}\mathrm{with}\mathrm{compound}\mathrm{interest},A=\mathrm{Rs}.2178 \\ \mathrm{Now},A\mathit{}=P\times {\left(1+\frac{R}{100}\right)}^n \\ =\mathrm{Rs}.1800\times {\left(1+\frac{10}{100}\right)}^n \\ =\mathrm{Rs}.1800\times {\left(\frac{100+10}{100}\right)}^{\mathrm{n}} \\ =\mathrm{Rs}.1800\times {\left(\frac{110}{100}\right)}^{\mathrm{n}} \\ =\mathrm{Rs}.1800\times {\left(\frac{11}{10}\right)}^{\mathrm{n}} \\ \mathrm{However},\mathrm{amount}=\mathrm{Rs}.2178 \\ Now,\mathrm{Rs}.2178=\mathrm{Rs}.1800\times {\left(\frac{11}{10}\right)}^n \\ \Rightarrow \frac{2178}{1800}={\left(\frac{11}{10}\right)}^n \\ \Rightarrow \frac{121}{100}={\left(\frac{11}{10}\right)}^n \\ \Rightarrow {\left({\displaystyle \frac{11}{10}}\right)}^2={\left(\frac{11}{10}\right)}^n \\ \Rightarrow n=2 \\ \therefore \mathrm{Time},\mathit{}n=2\mathrm{years} \end{gathered}$$