A sum of Rs. 1000 after 3 ,years at compound | vedclass

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Question

The amount of Rs. 1000 at $10 \%$ p.a. after 3 years would becomes:

$\Rightarrow P\left(1+\frac{R}{100}\right)^{3}=A \Rightarrow 1000\left(1+\frac{

Vedclass · Accepted Answer

$1.7$

Vedclass · Answer

The amount of Rs. 1000 at $10 \%$ p.a. after 3 years would becomes:

$\Rightarrow P\left(1+\frac{R}{100}\right)^{3}=A \Rightarrow 1000\left(1+\frac{10}{100}\right)^{5}=A$

$\Rightarrow A=R s .1331$

Here, now the principal of Rs. 1728 depreciated at rate of $R \%$ per annum and amounts to Rs. 1331

$\Rightarrow P\left[\left(1-\frac{R}{100}\right)\right]^{3}=1331$

$\Rightarrow 1728\left[\left(1-\frac{ R }{100}\right)\right]^{3}=1331$

$\Rightarrow\left(1-\frac{ R }{100}\right)^{3}=\frac{1331}{1728} \Rightarrow 1-\frac{ R }{100}=\frac{11}{12}$

$\Rightarrow \frac{ R }{100}=\frac{1}{12}$ or $R =8.33 \%$

$\Rightarrow$ Hence rate difference between CI and $\quad$ depreciation is almost $=(10-8.33) \%=1.7$

A sum of $Rs. 1000$ after $3 \,years$ at compound interest becomes a certain amount that is equal to the amount that is the result of a $3 \,year$ depreciation from $Rs. 1728$. Find the difference between the rates (In $\%$) of $C.I.$ and depreciation? (Given $C.I.$ is $10 \%$ $p.a.$)

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