An amount of money at compound interest grows | vedclass

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Question

(a) $\quad A=P\left(1+\frac{R}{100}ight)^{T}$

$	herefore \quad 3840=P\left(1+\frac{R}{100}ight)^{4}$

$3936=P\left(1+\frac{R}{100}ight)^{5

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(a) $\quad A=P\left(1+\frac{R}{100}ight)^{T}$

$	herefore \quad 3840=P\left(1+\frac{R}{100}ight)^{4}$

$3936=P\left(1+\frac{R}{100}ight)^{5}$

Dividing equation (2) by equation (1), we have

$\frac{3936}{3840}=1+\frac{R}{100} \Rightarrow \frac{R}{100}=\frac{3936}{3840}-1$

$=\frac{3936-3840}{3840}=\frac{96}{3840}$

$\Rightarrow \quad R=\frac{96}{3840} 	imes 100=2.5 \%$

An amount of money at compound interest grows up to ₹ $3,840$ in $4$ years and up to ₹ $3,936$ in $5$ years. Find the rate of interest (In $\%$)

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