The difference between the compound interest | vedclass

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Question

Let rate of interest be $R \%$ p.a.

$S.I.  =\frac{18000 	imes 2 	imes R}{100}$

$C.I. =  18000\left[\left(1+\frac{R}{100}ight)^{2}-

Vedclass · Accepted Answer

$15$

Vedclass · Answer

Let rate of interest be $R \%$ p.a.

$S.I.  =\frac{18000 	imes 2 	imes R}{100}$

$C.I. =  18000\left[\left(1+\frac{R}{100}ight)^{2}-1ight]$

$(2)-(1) \Rightarrow 18000\left[\left(1+\frac{R}{100}ight)^{2}-1-\frac{2 R}{100}ight]$

$=405 or 18000\left[\left(\frac{R}{100}ight)^{2}ight]$$=405$

Or   $\left(\frac{R}{100}ight)^{2}=\frac{405}{18000}$

or   $=\frac{45}{2000}$$=\frac{9}{400}$$\frac{R}{100} =\frac{3}{20}$

Or   $R=15 \%$

The difference between the compound interest and the simple interest accrued on an amount of ₹ $18000$ in $2$ years was ₹ $405 .$ What was the rate of interest per cent per annum?

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