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In a bank principal increases at the rate of $5 \%$ per year. An amount of $₹ 1000$ is deposited with this bank, how much will it worth after 10 years
$$
\left(e^{0.5}=1 \cdot 648\right)
$$
$$
\left(e^{0.5}=1 \cdot 648\right)
$$
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Verified Answer
Let $\mathrm{p}$ be the principal Rate of interest is $5 \%$ $\frac{\mathrm{dp}}{\mathrm{dt}}=\frac{5}{100} \mathrm{p} \quad \therefore \frac{\mathrm{dp}}{\mathrm{p}}=0.05 \mathrm{dt}$
Integrating, $\log \mathrm{p}=0.05 \mathrm{t}+\log \mathrm{c} \Rightarrow \log \frac{\mathrm{p}}{\mathrm{c}}=0.05 \mathrm{t}$
Initially, $\mathrm{p}=₹ 1000, \mathrm{t}=0 \quad 1000=\mathrm{c}, \mathrm{e}^{\circ}=\mathrm{c}$
$\therefore \mathrm{c}=1000 \quad$ Putting this value in (i): $\mathrm{p}=1000 \mathrm{e}^{0.05 \mathrm{t}}$
when $\mathrm{t}=10, \mathrm{p}=1000 \mathrm{e}^{0.05 \times 10}=1000 \mathrm{e}^{0.5}$
$\mathrm{p}=1000 \times 1.648 \Rightarrow \mathrm{p}=1648\left(\because \mathrm{e}^{0.5}=1.648\right)$
After 10 years $₹ 1000$ will amount to $₹ 1648$.
Integrating, $\log \mathrm{p}=0.05 \mathrm{t}+\log \mathrm{c} \Rightarrow \log \frac{\mathrm{p}}{\mathrm{c}}=0.05 \mathrm{t}$
Initially, $\mathrm{p}=₹ 1000, \mathrm{t}=0 \quad 1000=\mathrm{c}, \mathrm{e}^{\circ}=\mathrm{c}$
$\therefore \mathrm{c}=1000 \quad$ Putting this value in (i): $\mathrm{p}=1000 \mathrm{e}^{0.05 \mathrm{t}}$
when $\mathrm{t}=10, \mathrm{p}=1000 \mathrm{e}^{0.05 \times 10}=1000 \mathrm{e}^{0.5}$
$\mathrm{p}=1000 \times 1.648 \Rightarrow \mathrm{p}=1648\left(\because \mathrm{e}^{0.5}=1.648\right)$
After 10 years $₹ 1000$ will amount to $₹ 1648$.
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