The difference between compound interest and | vedclass

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Question

C.I. for 2 years $@ 10 \%$ p.a. interest

$\begin{array}{l}=P\left[\left(1+\frac{10}{100}ight)^{2}-1ight] \	ext { S.I. }=\frac{P 	imes 2 	im

Vedclass · Accepted Answer

$31.01$

Vedclass · Answer

C.I. for 2 years $@ 10 \%$ p.a. interest

$\begin{array}{l}=P\left[\left(1+\frac{10}{100}ight)^{2}-1ight] \	ext { S.I. }=\frac{P 	imes 2 	imes 10}{100}=\frac{P}{5}\end{array}$

$(1)-(2) \Rightarrow P\left[\left(\frac{11}{10}ight)^{2}-1-\frac{1}{5}ight]=20$

Or $P\left(\frac{(11)^{2}-(10)^{2}-20}{100}ight)=20$

or $P=\frac{20 	imes 100}{1}=₹ 2000$

$	herefore \quad$ If interest is compounded half yearly,

C.I. $-$ S.I. $=P\left[\left(1+\frac{5}{100}ight)^{4}-1ight]-\frac{P}{5}$

$=P\left[\left(1+\frac{5}{100}ight)^{4}-1-\frac{1}{5}ight]$

$=2000\left[\left(\frac{21}{20}ight)^{4}-1-\frac{1}{5}ight]=2000\left[\frac{(21)^{4}-(20)^{4}}{(20)^{4}}-\frac{1}{5}ight]$

$=2000\left[\frac{\left(21^{2}+20^{2}ight)\left(21^{2}-20^{2}ight)}{20^{4}}-\frac{1}{5}ight]=2000\left[\frac{841 	imes 41 	imes 1}{20^{4}}-\frac{1}{5}ight]$

$=2000\left[\frac{34481}{1,60,000}-\frac{1}{5}ight]=\frac{2000}{1,60,000}[34481-32000]$

$=\frac{1}{80} 	imes 2481=31.0125$

$	herefore \quad$ Difference of interest $=₹ 31.01$

The difference between compound interest and simple interest on a sum for $2$ years at $10 \%$ per annum, when the interest is compounded annually, is ₹ $20$. If the interest were compounded half yearly, the difference in two interests (In ₹) will be

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