Question

If the interest is compounded continuously at 6% per annum, how much worth Rs 1000 will be after 10 years? How long will it take to double Rs 1000?

Answer 1

Let P₀ be the initial amount and P be the amount at any time t. Then,
$$\begin{gathered} \frac{dP}{dt}=\frac{6P}{100} \\ \Rightarrow \frac{dP}{dt}=0.06P \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{dP}{P}=0.06dt \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}t,\mathrm{we}\mathrm{get} \\ \log P=0.06t+C \\ \mathrm{Now}, \\ P=P_0\mathrm{at}t=0 \\ \therefore \log P_0=0+C \\ \Rightarrow C=\log P_0 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C,\mathrm{we}\mathrm{get} \\ \log P=0.06t+\log P_0 \\ \Rightarrow \log \frac{P}{P_0}=0.06t \\ \Rightarrow e^{0.06t}=\frac{P}{P_0} \\ \mathrm{To}\mathrm{find}\mathrm{the}\mathrm{amount}\mathrm{after}10\mathrm{years},\mathrm{we}\mathrm{get} \\ \Rightarrow e^{0.06\times 10}=\frac{P}{P_0} \\ \Rightarrow e^{0.6}=\frac{P}{P_0} \\ \Rightarrow 1.822=\frac{P}{P_0} \\ \Rightarrow P=1.822P_0 \\ \Rightarrow P=1.822\times 1000=\mathrm{Rs}1822 \\ \mathrm{To}\mathrm{find}\mathrm{the}\mathrm{time}\mathrm{after}\mathrm{which}\mathrm{the}\mathrm{amount}\mathrm{will}\mathrm{double},\mathrm{we}\mathrm{have} \\ P=2P_0 \\ \therefore \log \frac{2P_0}{P_0}=0.06t \\ \Rightarrow \log 2=0.06t \\ \Rightarrow t=\frac{0.6931}{0.06}=11.55\mathrm{years} \end{gathered}$$