If the compound interest on a sum for 2 years | vedclass

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Question

(d) Let the principal be ₹P.

$	herefore \quad CI =P\left[\left(1+\frac{R}{100}ight)^{T}-1ight]$

$\Rightarrow \quad 510=P\left[\left(1+\frac

Vedclass · Accepted Answer

$480$

Vedclass · Answer

(d) Let the principal be ₹P.

$	herefore \quad CI =P\left[\left(1+\frac{R}{100}ight)^{T}-1ight]$

$\Rightarrow \quad 510=P\left[\left(1+\frac{\frac{25}{2}}{100}ight)^{2}-1ight]$

$\Rightarrow \quad 510=P\left[\left(1+\frac{1}{8}ight)^{2}-1ight]$

$\Rightarrow \quad 510=P\left[\left(\frac{9}{8}ight)^{2}-1ight]$

$\Rightarrow \quad 510=P\left(\frac{81}{64}-1ight)$

$\Rightarrow \quad 510=P\left(\frac{81-64}{64}ight)=\frac{17 P}{64}$

$\Rightarrow \quad P=\frac{510 	imes 64}{17}=₹ 1920$

$	herefore \quad SI =\frac{	ext { Principal } 	imes Time 	imes Rate }{100}$

$=\frac{1920 	imes 2 	imes 25}{100 	imes 2}=₹ 480$

If the compound interest on a sum for $2$ years at $12 \frac{1}{2} \%$ per cent is ₹ $510$, the simple interest (In ₹) on the same sum at the same rate for same period of time is

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