Question

The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs 291600, for how much was it purchased?

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{initial}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{machine},P\mathrm{be}\mathrm{Rs}x. \\ \mathrm{Rate}\mathrm{of}\mathrm{depreciation},\mathit{}R=10\% \\ \mathrm{Time},n=3\mathrm{years} \\ \mathrm{The}\mathrm{present}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{machine}\mathrm{is}\mathrm{Rs}291600. \\ \mathrm{Then}\mathrm{the}\mathrm{initial}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{machine}\mathrm{is}\mathrm{given}\mathrm{by} \\ \mathrm{Value}=P\times {\left(1-\frac{R}{100}\right)}^n \\ =\mathrm{Rs}.\mathit{}x\times {\left(1-\frac{10}{100}\right)}^3 \\ =\mathrm{Rs}.x\times {\left(\frac{100-10}{100}\right)}^3 \\ =\mathrm{Rs}.x\times {\left(\frac{90}{100}\right)}^3 \\ =\mathrm{Rs}.x\times {\left(\frac{9}{10}\right)}^3 \\ \therefore \mathrm{Present}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{machine}=\mathrm{Rs}291600 \\ \mathrm{Now},\mathrm{Rs}291600=\mathrm{Rs}\mathit{}x\times \left(\frac{9}{10}\right)\times \left(\frac{9}{10}\right)\times \left(\frac{9}{10}\right) \\ \Rightarrow x\mathit{}=\mathrm{Rs}\frac{291600\times 10\times 10\times 10}{9\times 9\times 9} \\ \Rightarrow x=\mathrm{Rs}\frac{291600000}{729} \\ \Rightarrow x\mathit{}=\mathrm{Rs}400000 \\ \therefore \mathrm{The}\mathrm{initial}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{machine}\mathrm{is}\mathrm{Rs}400000. \end{gathered}$$