Search any question & find its solution
Question:
Answered & Verified by Expert
In a bank principal increases at the rate of $r \%$ per year. Find the value of $r$ if $₹ 100$ double itself in 10 years ( $\log _{\mathrm{e}} 2$ $=0.6931)$
Solution:
2114 Upvotes
Verified Answer
Let $\mathrm{P}$ be the principal at any time $t$.
According to the problem
$$
\frac{d \mathrm{P}}{\mathrm{dt}}=\frac{r}{100}, \mathrm{P} \quad \Rightarrow \quad \int \frac{d \mathrm{P}}{\mathrm{P}}=\int \frac{r}{100} d t
$$
$\log \mathrm{P}=\frac{r}{100} t+C_1 \Rightarrow P=e^{r t / 100} \cdot e^{C_1}$
Now $\mathrm{P}=100$, when $t=0 \therefore P=e^{r t / 100} \times 100$
When $\mathrm{P}=200, t=10 \Rightarrow \log 2=\frac{r}{10}$ $\Rightarrow r=6.931 \%$ per annum.
According to the problem
$$
\frac{d \mathrm{P}}{\mathrm{dt}}=\frac{r}{100}, \mathrm{P} \quad \Rightarrow \quad \int \frac{d \mathrm{P}}{\mathrm{P}}=\int \frac{r}{100} d t
$$
$\log \mathrm{P}=\frac{r}{100} t+C_1 \Rightarrow P=e^{r t / 100} \cdot e^{C_1}$
Now $\mathrm{P}=100$, when $t=0 \therefore P=e^{r t / 100} \times 100$
When $\mathrm{P}=200, t=10 \Rightarrow \log 2=\frac{r}{10}$ $\Rightarrow r=6.931 \%$ per annum.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.