Sonika invested an amount of ₹ 5800 for 2 yea | vedclass

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Question

(a) $594.5=5800\left[\left(1+\frac{r}{100}\right)^{2}-1\right]$

$\frac{594.5}{5800}=\left(1+\frac{r}{100}\right)^{2}-1$

$0.1025+1=\left(1+\frac{

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(a) $594.5=5800\left[\left(1+\frac{r}{100}ight)^{2}-1ight]$

$\frac{594.5}{5800}=\left(1+\frac{r}{100}ight)^{2}-1$

$0.1025+1=\left(1+\frac{r}{100}ight)^{2}$

$1.1025=\frac{(100+r)^{2}}{10000}$

$1.1025 	imes 10000=(100+r)^{2}$

$11025=(100+r)^{2}$

$(105)^{2}=(100+r)^{2}$

$105=100+r$

$r=5 \%$

Sonika invested an amount of ₹ $5800$ for $2$ years. At what rate of compound interest ($\%$ $per$ $annum$) will she get an amount of ₹ $594.5$ at the end of two years?

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