Question

In a bank principal increases at the rate of r% per year. Find the value of r if Rs.100 double itself in 10 yr(log2=0.6391).

Answer 1

Let P, t and r represent the principal, time and rate of interest respectively.
It is given that the principal increases continously at the rate of r% per year.
$\therefore \frac{dP}{dt}$=r% of P $\Rightarrow \frac{dP}{dt}=\frac{rP}{100}\Rightarrow \frac{dp}{p}=\frac{r}{100}dt$
On integrating both sides, we obtain
$\int \frac{dP}{P}=\int \frac{r}{100}dt\Rightarrow log|P|=\frac{r}{100}t+C$ ...(i)
When t=0, let $P=P_0,$ then
$log|P_0|=0+C\Rightarrow C=log|P_0|$
On substituting the value of C in Eq. (i), we get
$log|P|=\frac{r}{100}t+log|P_0|\Rightarrow log|P|-log|P_0|=\frac{r}{100}t\Rightarrow log|\frac{P}{P_0}|=\frac{r}{100}t$ ...(ii)
It is given that when t=10, then $P=2P_0,$ put in Eq. (ii), we get
$log|\frac{2P_0}{P_0}|=log2=\frac{r\left(10\right)}{100}\Rightarrow r=10log2=10\times 0.6931\Rightarrow r=6.931$
Hence, the value of r is 6.93%.