Question

Neha borrowed Rs 24000 from the State Bank of India to buy a scooter. If the rate of interest be 10% per annum compounded annually, what payment will she have to make after 2 years 3 months?

Answer 1

$$\begin{gathered} \mathrm{Principal}\mathrm{amount},P\mathit{}=\mathrm{Rs}.24000 \\ \mathrm{Rate}\mathrm{of}\mathrm{interest},\mathit{}R=10\%\mathrm{p}.\mathrm{a}. \\ \mathrm{Time},\mathit{}n=2\mathrm{years}3\mathrm{months}=2\frac{1}{4}\mathrm{years} \\ \mathrm{The}\mathrm{formula}\mathrm{for}\mathrm{the}\mathrm{amount}\mathrm{including}\mathrm{the}\mathrm{compound}\mathrm{interest}\mathrm{is}givenbelow: \\ \mathrm{A}=P\mathit{}\times {\left(1+\frac{R}{100}\right)}^n\times \left(1+\frac{{\displaystyle \frac{1}{4}}R}{100}\right) \\ =\mathrm{Rs}.24000\times {\left(1+\frac{10}{100}\right)}^2\times \left(1+\frac{{\displaystyle \frac{1}{4}}\times 10}{100}\right) \\ =\mathrm{Rs}.24000\times {\left(\frac{100+10}{100}\right)}^2\times \left(\frac{100+2.5}{100}\right) \\ =\mathrm{Rs}.24000\times {\left(\frac{110}{100}\right)}^2\times \left(\frac{100+2.5}{100}\right) \\ =\mathrm{Rs}.24000\times \left(1.1\times 1.1\times 1.025\right) \\ =\mathrm{Rs}.24000\times \left(1.250\right) \\ =\mathrm{Rs}.29766 \\ \mathrm{Therefore},\mathrm{Neha}\mathrm{should}\mathrm{pay}\mathrm{Rs}29766\mathrm{to}\mathrm{the}\mathrm{bank}\mathrm{after}2\mathrm{years}3\mathrm{months}. \end{gathered}$$