The increase in the principal amount kept at compound interest in a bank is proportional to the product of the principal amount and annual rate of interest.
- Annual rate of interest in a bank is 5%. How many years will it take to double the principal amount?
- At what annual rate of interest, the principal amount will double in 10 years?
Let the principal amount at any instant of time t be $= P$
Let the annual rate of interest $= R\% = \dfrac{R}{100} = r$
Increase in principal amount with time $= \dfrac{dP}{dt}$
$\therefore$ $\;$ As per question, $\;\;$ $\dfrac{dP}{dt} \propto Pr$
i.e. $\dfrac{dP}{dt} = k P r$ $\;\;\;$ [k is the constant of proportionality]
i.e. $\dfrac{dP}{P} = k \; r \; dt$
i.e. $\displaystyle \int \dfrac{dP}{p} = k \; r \int dt$
i.e. $\ln P = k \; r \; t + c$ $\;\;\;$ [c is the constant of integration] $\;\;\; \cdots \; (1)$
At time $t = 0$, initial amount $= P = P_0$
Substituting these values of $t$ and $P$ in equation $(1)$ give
$c = \ln P_0$ $\;\;\; \cdots \; (2)$
Substituting the value of $c$ from equation $(2)$ in equation $(1)$ gives
$\ln P = k \; r \; t + \ln P_0$
i.e. $\ln P - \ln P_0 = k \; r \; t$
i.e. $\ln \left(\dfrac{P}{P_0}\right) = k \; r \; t$
i.e. $\dfrac{P}{P_0} = e^{k\;r \; t}$ $\implies$ $P = P_0 \; e^{k \; r \; t}$ $\;\;\; \cdots \; (3)$
After 1 year, $t = 1$, $P = P_0 + P_0 \; r = P_0 \left(1 + r\right)$
Substituting these values of $t$ and $P$ in equation $(3)$ give
$P_0 \left(1 + r\right) = P_0 \; e^{k \; r}$
i.e. $e^{k \; r} = 1 + r$ $\implies$ $k \; r = \ln \left(1 + r\right)$ $\;\;\; \cdots \; (4)$
$\therefore$ $\;$ In view of equation $(4)$, equation $(3)$ becomes
$P = P_0 \; e^{t \; \ln \left(1 + r\right)}$
i.e. $P = P_0 \; e^{\ln \left(1 + r\right)^t}$
i.e. $P = P_0 \left(1 + r\right)^t$ $\implies$ $P = P_0 \left(1 + \dfrac{R}{100}\right)^t$ $\;\;\; \cdots \; (5)$ $\;\;$ $\left[\because \; r = \dfrac{R}{100}\right]$
- Given: Rate of interest $= R = 5\% = \dfrac{5}{100}$, $P = 2 P_0$
Substituting these values in equation $(5)$ give
$2 P_0 = P_0 \left(1 + \dfrac{5}{100}\right)^t$
i.e. $2 = \left(1.05\right)^t$
i.e. $\log \left(2\right) = t \; \log \left(1.05\right)$ $\implies$ $t = \dfrac{\log \left(2\right)}{\log \left(1.05\right)} = 14.2$
$\therefore$ $\;$ It takes 14.2 years to double the principal when the rate of interest is 5% - Given: Time $= t = 10$ years, Principal amount $= P = 2 P_0$
Substituting these values in equation $(5)$ give
$2 P_0 = P_0 \left(1 + \dfrac{R}{100}\right)^{10}$
i.e. $2 = \left(1 + \dfrac{R}{100}\right)^{10}$
i.e. $\log \left(2\right) = 10 \; \log \left(1 + \dfrac{R}{100}\right)$
i.e. $\dfrac{1}{10} \times \log \left(2\right) = \log \left(1 + \dfrac{R}{100}\right)$
i.e. $\log \left(2^{0.1}\right) = \log \left(1 + \dfrac{R}{100}\right)$
i.e. $2^{0.1} = 1 + \dfrac{R}{100}$ $\implies$ $R = 100 \times \left(2^{0.1} - 1\right) = 7.2$
$\therefore$ $\;$ Annual rate of interest $= 7.2\%$