A sum of ₹ 3,200 invested at 10 % per cent co | vedclass

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Question

(a) $A=P\left(1+\frac{R}{100}\right)^{T} \Rightarrow \frac{3362}{3200}=\left(1+\frac{10}{400}\right)^{4 t}$

$\Rightarrow \frac{1681}{1600}=\left(\f

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(a) $A=P\left(1+\frac{R}{100}\right)^{T} \Rightarrow \frac{3362}{3200}=\left(1+\frac{10}{400}\right)^{4 t}$

$\Rightarrow \frac{1681}{1600}=\left(\frac{41}{40}\right)^{4 t} \Rightarrow\left(\frac{41}{40}\right)^{2}=\left(\frac{41}{40}\right)^{4 t}$

$\Rightarrow \quad 4 t=2 \Rightarrow t=\frac{1}{2}$ year

A sum of ₹ $3,200$ invested at $10 \%$ per cent compounded quarterly amounts to ₹ $3,362.$ Compute the time period (In $years$)

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