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Mrs. Rajni deposited Rs. 10,0000 in a bank that pays $4 \%$ interest compounded continuously then the amount she gets after 10 years is Rs. approximately.
(given $e^{(0.4)}=1.49182$ )
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(given $e^{(0.4)}=1.49182$ )
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The correct answer is:
$14918$
$\begin{aligned} & \because A=P\left(1+\frac{R}{100}\right) \\ & \Rightarrow A=10000\left(1+\frac{4}{100}\right)^{10}=10000 \times e^{\frac{4}{100} \times 10} \\ & \Rightarrow A=10000 \times e^{0.4}=14918 \quad\left[\because \lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}=e\right]\end{aligned}$
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