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,
\[\frac{dP}{dt} = \frac{6P}{100}\]
\[ \Rightarrow \frac{dP}{dt} = 0 . 06P\]
\[\Rightarrow \frac{dP}{P} = 0 . 06dt \]
Integrating both sides with respect to t, we get
\[\log P = 0 . 06t + C\]
Now,
\[P = P_0\text{ at }t = 0 \]
\[ \therefore \log P_0 = 0 + C\]
\[ \Rightarrow C = \log P_0 \]
Putting the value of C, we get
\[\log P = 0 . 06t + \log P_0 \]
\[ \Rightarrow \log\frac{P}{P_0} = 0 . 06t\]
\[ \Rightarrow e^{0 . 06t} = \frac{P}{P_0}\]
To find the amount after 10 years, we get
\[ \Rightarrow e^{0 . 06 \times 10} = \frac{P}{P_0}\]
\[ \Rightarrow e^{0 . 6} = \frac{P}{P_0}\]
\[ \Rightarrow 1 . 822 = \frac{P}{P_0}\]
\[ \Rightarrow P = 1 . 822 P_0 \]
\[ \Rightarrow P = 1 . 822 \times 1000 =\text{ Rs. }1822\]
To find the time after which the amount will double, we have
\[P = 2 P_0 \]
\[ \therefore \log\frac{2 P_0}{P_0} = 0 . 06t\]
\[ \Rightarrow \log 2 = 0 . 06t\]
\[ \Rightarrow t = \frac{0 . 6931}{0 . 06} = 11 . 55\text{ years }\]