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It is said the cost of the machine depreciates by Rs.4000 during the first year. Depreciation is the value of an asset over time, due to particular wear and tear. It is the decrease in the value of currency.

Now the depreciation by the end of \[{{2}^{nd}}\] year = Rs.3600

Let P be taken as the cost of the machine. Let R be the rate of depreciation in percentage. Let \[{{A}_{1}}\] be the amount. We know the formula,

\[\Rightarrow A=P{{\left[ 1-\dfrac{R}{100} \right]}^{n}}\]

Here, put \[A={{A}_{1}}\] and n = 1 year, we get \[{{A}_{1}}=P{{\left[ 1-\dfrac{R}{100} \right]}^{1}}\].

We know that,

Principal = amount + Depreciation

\[P={{A}_{1}}+D\]

\[\Rightarrow P-{{A}_{1}}=D\]

Let us substitute the values, we get,

\[\begin{align}

& P-P\left[ 1-\dfrac{R}{100} \right]=4000 \\

& \Rightarrow P\left[ 1-1+\dfrac{R}{100} \right]=4000 \\

& \therefore \dfrac{PR}{100}=4000-(1) \\

\end{align}\]

Let \[{{A}_{2}}\] be the amount after \[{{2}^{nd}}\] year. Here n = 1 for time span from \[{{2}^{nd}}\] year stand to end.

\[\therefore {{A}_{2}}={{A}_{1}}{{\left[ 1-\dfrac{R}{100} \right]}^{1}}\]

Here we consider \[{{A}_{1}}\] as the principal after \[{{2}^{nd}}\] year. Put \[{{A}_{1}}=P\left[ 1-\dfrac{R}{100} \right]\].

\[\therefore {{A}_{2}}=P\left[ 1-\dfrac{R}{100} \right]\left[ 1-\dfrac{R}{100} \right]=P{{\left[ 1-\dfrac{R}{100} \right]}^{2}}\]

Hence, \[P-{{A}_{2}}=D\]

i.e. \[{{A}_{1}}-{{A}_{2}}=D\]

Put, \[{{A}_{1}}=P\left[ 1-\dfrac{R}{100} \right]\] and \[{{A}_{2}}=P{{\left[ 1-\dfrac{R}{100} \right]}^{2}}\], D = 3600.

\[\begin{align}

& P\left[ 1-\dfrac{R}{100} \right]-\left( P{{\left[ 1-\dfrac{R}{100} \right]}^{2}} \right)=3600 \\

& P\left[ 1-\dfrac{R}{100} \right]-\left( 1-\left( 1-\dfrac{R}{100} \right) \right)=3600 \\

& \Rightarrow P\left[ 1-\dfrac{R}{100} \right]\times \dfrac{R}{100}=3600 \\

& \left( 1-\dfrac{R}{100} \right)\dfrac{PR}{100}=3600 \\

\end{align}\]

From (1) we got the value of \[\dfrac{PR}{100}=4000\]. Putting this in the above equation we get,

\[\left( 1-\dfrac{R}{100} \right)\times 4000=3600\]

Now let us simplify the above expression and get the value of R.

\[\begin{align}

& 1-\dfrac{R}{100}=\dfrac{3600}{4000} \\

& \Rightarrow 1-\dfrac{R}{100}=\dfrac{9}{10} \\

& \therefore \dfrac{R}{100}=1-\dfrac{9}{10}=\dfrac{10-9}{10} \\

& \therefore R=\dfrac{1}{10}\times 100=10 \\

\end{align}\]

Hence we got the rate of interest as 10%.

Let us put this value of R in (1).

\[\begin{align}

& \dfrac{PR}{100}=4000\Rightarrow \dfrac{P\times 10}{100}=4000 \\

& \therefore P=4000\times 10=40000 \\

\end{align}\]

Thus we got the principal as Rs.40000.

Now let us find the amount after 3 years, put n = 3.

\[\begin{align}

& {{A}_{3}}=P{{\left[ 1-\dfrac{R}{100} \right]}^{3}} \\

& {{A}_{3}}=40000{{\left[ 1-\dfrac{10}{100} \right]}^{3}}=40000{{\left[ 1-0.1 \right]}^{3}} \\

& {{A}_{3}}=40000\times {{0.9}^{3}} \\

& {{A}_{3}}=40000\times 0.729 \\

\end{align}\]

\[{{A}_{3}}\] = Rs.29160

Thus we got the cost at the end of the third year as Rs.29160.

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