A sum of money placed at compound interest do | vedclass

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Question

$2 P=P\left(1+\frac{R}{100}ight)^{5}$

Let the number of years at which amount becomes 8 times be $N$.

$	herefore \quad 8 P=P\left(1+\frac{R}{

Vedclass · Accepted Answer

$15$

Vedclass · Answer

$2 P=P\left(1+\frac{R}{100}ight)^{5}$

Let the number of years at which amount becomes 8 times be $N$.

$	herefore \quad 8 P=P\left(1+\frac{R}{100}ight)^{N}$

$(1) \Rightarrow \quad\left(1+\frac{R}{100}ight)^{5}=2$

$\frac{(2)}{(1)} \Rightarrow \frac{8 P}{2 P}=\frac{P\left(1+\frac{R}{100}ight)^{N}}{P\left(1+\frac{R}{100}ight)^{5}}$

on substitution from $(3),$ we get,

$4=\frac{\left(1+\frac{R}{100}ight)^{N}}{2}$

Or $\left(1+\frac{R}{100}ight)^{N}=8$

(3) $\Rightarrow \quad\left(1+\frac{R}{100}ight)^{5}=2$

$	herefore \quad(4) \Rightarrow \quad\left(1+\frac{R}{100}ight)^{N}=2^{3}=\left[\left(1+\frac{R}{100}ight)^{5}ight]^{3}=\left(1+\frac{R}{100}ight)^{15}$

$	herefore \quad N=15$ years

A sum of money placed at compound interest doubles itself in $5 \,years$. It will amount to eight times itself at the same rate of interest in (In $years$)

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