Question

A car is purchased for Rs 348000. Its value depreciates at 10% per annum during the first year and at 20% per annum during the second year. What will be its value after 2 years?

Answer 1

$$\begin{gathered} \mathrm{Initial}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{car},P=\mathrm{Rs}348000 \\ \mathrm{Rate}\mathrm{of}\mathrm{depreciation}\mathrm{for}\mathrm{the}\mathrm{first}\mathrm{year},\mathit{}p=10\% \\ \mathrm{Rate}\mathrm{of}\mathrm{depreciation}\mathrm{for}\mathrm{the}\mathrm{second}\mathrm{year},\mathit{}q=20\% \\ \mathrm{Time},\mathit{}n=2\mathrm{years}. \\ \mathrm{Then}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{car}\mathrm{after}\mathrm{two}\mathrm{years}\mathrm{is}\mathrm{given}\mathrm{by} \\ \mathrm{Value}=\left\{P\times \left(1-\frac{p}{100}\right)\times \left(1-\frac{q}{100}\right)\right\} \\ =\mathrm{Rs}.\left\{348000\times \left(1-\frac{10}{100}\right)\times \left(1-\frac{20}{100}\right)\right\} \\ =\mathrm{Rs}.\left\{348000\times \left(\frac{100-10}{100}\right)\times \left(\frac{100-20}{100}\right)\right\} \\ =\mathrm{Rs}.\left\{348000\times \left(\frac{90}{100}\right)\times \left(\frac{80}{100}\right)\right\} \\ =\mathrm{Rs}.\left\{348000\times \left(\frac{9}{10}\right)\times \left(\frac{8}{10}\right)\right\} \\ =\mathrm{Rs}.\left(3480\times 9\times 8\right) \\ =\mathrm{Rs}.250560 \\ \therefore \mathrm{The}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{car}\mathrm{after}\mathrm{two}\mathrm{years}\mathrm{is}\mathrm{Rs}250560. \end{gathered}$$