Question

In a bank principal increases at the rate of r% per year. Find the value of r if ₹100 double itself in 10 years (log_e 2 = 0.6931).

Answer 1

Let P be the principal at any instant t.
Given:
$$\begin{gathered} \frac{dP}{dt}=\frac{r}{100}P \\ \Rightarrow \frac{dP}{P}=\frac{r}{100}dt \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{dP}{P}=\int \frac{r}{100}dt \\ \Rightarrow \log P=\frac{rt}{100}+\mathrm{C}......\left(1\right) \\ \mathrm{Initially},\mathrm{i}.\mathrm{e}.\mathrm{at}t=0,\mathrm{let}P=P_0. \\ \mathrm{Putting}P=P_0,\mathrm{we}\mathrm{get} \\ \log P_0=C, \\ \mathrm{Putting}C=\log P_0\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \log P=\frac{rt}{100}+\log P_0 \\ \Rightarrow \log \frac{P}{P_0}=\frac{rt}{100} \\ \mathrm{Substituting}{P}_0=100,P=2P_0=200\mathrm{and}t=10\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ \log 2=\frac{r}{10} \\ \therefore r=10\log 2 \\ =10\times 0.6931 \\ =6.931 \end{gathered}$$