Question

the value of a machine worth rupees 500000 is depreciating at the rate of 10% every year. in how many years will its value be reduced to rupees 364500?​

Answer 1

Given :

• Value of a machine = Rs. 5,00,000
• Rate of depreciation = 10 %
• Value of machine reduces to = Rs. 3,64,500

To find :

• Number of years in which its value will get reduced

Knowledge required :-

• Formula to calculate amount when the value of a thing depreciates -

⠀⠀⠀$\boldsymbol{Amount = P\Bigg[1 - \dfrac{r}{100}\Bigg]^{n}}$

where,

• P = Principal
• r = The rate of depreciation
• n = Time

Solution :

Substitute the given values in the formula of amount.

We have,

• Amount = Rs. 3,64,500
• Principal = Rs. 5,00,000
• r = 10%
• n = ?

$\\ : \implies\bold{364500 = 500000 \Bigg[1 - \dfrac{10}{100}\Bigg]^{n}}$

$\\ : \implies \bold{ \dfrac{364500}{500000 } = \Bigg[1 - \frac{10}{100} \Bigg]^{n}}$

$\\ : \implies \bold{ \dfrac{3645 \not0 \not0}{5000 \not0 \not0 } = \Bigg[1 - \frac{10}{100} \Bigg]^{n} }$

$\\ : \implies \bold{ \dfrac{3645}{5000} = \Bigg[1 - \frac{10}{100} \Bigg]^{n} }$

$\\ : \implies \bold{ \dfrac{729}{1000} = \Bigg[1 - \frac{10}{100} \Bigg]^{n} }$

$\\ : \implies \bold{ \Bigg(\dfrac{9}{10} \Bigg)^{3} = \Bigg[1 - \frac{10}{100} \Bigg]^{n} }$

$\\ : \implies \bold{ \Bigg(\dfrac{9}{10} \Bigg)^{3} = \Bigg[1 - \frac{1 \not0}{10 \not0} \Bigg]^{n} }$

$\\ : \implies \bold{ \Bigg(\dfrac{9}{10} \Bigg)^{3} = \Bigg[1 - \frac{1}{10} \Bigg]^{n} }$

$\\ : \implies \bold{ \Bigg(\dfrac{9}{10} \Bigg)^{3} = \Bigg[ \frac{10 - 1}{10} \Bigg]^{n} }$

$\\ : \implies \bold{ \Bigg(\dfrac{9}{10} \Bigg)^{3} = \Bigg[ \frac{9}{10} \Bigg]^{n} }$

Bases are equal, powers are equal.

$\\ : \implies \bold{3 = n}$

$\\ \therefore \underline{ \ \sf{ \pmb{Number \: \: of \: \: years \: \: in \: \: which \: its \: \: value \: \: will \: \: get \: \: reduced = 3 \: \: years}}}$

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Some related formulae :-

$\bullet \: \: \bold{Simple~Interest = \dfrac{P \times R \times T}{100}}$

$\bullet \:  \: \bold{C.I. = P \bigg[1 + \dfrac{r}{100} \bigg]^{n}  - P}$

$\bullet\: \: \bold{Compound~Interest = Amount - Principal}$

$\bullet\: \: \bold{Amount = Compound~Interest + Principal}$

$\bullet \: \: \bold{Amount = P \bigg[1 +  \dfrac{r}{100} \bigg]^{n}}$

Answer 2

Given:

• Value of a machine = Rs. 5,00,000
• Rate of depreciation = 10 %
• Value of machine reduces to = Rs. 3,64,500

To find:

• Number of years in which its value will get reduced

Knowledge required :-

Formula to calculate amount when the value of a thing depreciates -

$\boldsymbol{Amount = P\Bigg[1 - \dfrac{r}{100}\Bigg]^{n}}$

where,

• P = Principal
• r = The rate of depreciation
• n = Time

Solution:

Substitute the given values in the formula of amount.

We have,

• Amount = Rs. 3,64,500
• Principal = Rs. 5,00,000
• r = 10%
• n = ?

$\begin{gathered} \\ : \implies\bold{364500 = 500000 \Bigg[1 - \dfrac{10}{100}\Bigg]^{n}}\end{gathered}$

$\begin{gathered} \\ : \implies \bold{ \dfrac{364500}{500000 } = \Bigg[1 - \frac{10}{100} \Bigg]^{n}}\end{gathered}$

$\begin{gathered} \\ : \implies \bold{ \dfrac{3645 \not0 \not0}{5000 \not0 \not0 } = \Bigg[1 - \frac{10}{100} \Bigg]^{n} }\end{gathered}$

$\begin{gathered}\\ : \implies \bold{ \dfrac{729}{1000} = \Bigg[1 - \frac{10}{100} \Bigg]^{n} }\end{gathered}$

$\begin{gathered} \\ : \implies \bold{ \Bigg(\dfrac{9}{10} \Bigg)^{3} = \Bigg[1 - \frac{10}{100} \Bigg]^{n} }\end{gathered}$

$\begin{gathered} \\ : \implies \bold{ \Bigg(\dfrac{9}{10} \Bigg)^{3} = \Bigg[1 - \frac{1 \not0}{10 \not0} \Bigg]^{n} }\end{gathered}$

$\begin{gathered}\\ : \implies \bold{ \Bigg(\dfrac{9}{10} \Bigg)^{3} = \Bigg[1 - \frac{1}{10} \Bigg]^{n} }\end{gathered}$

$\begin{gathered}\\ : \implies \bold{ \Bigg(\dfrac{9}{10} \Bigg)^{3} = \Bigg[ \frac{10 - 1}{10} \Bigg]^{n} }\end{gathered}$

$\begin{gathered}\\ : \implies \bold{ \Bigg(\dfrac{9}{10} \Bigg)^{3} = \Bigg[ \frac{9}{10} \Bigg]^{n} }\end{gathered}$

Bases are equal, powers are equal.

$\begin{gathered}\\ : \implies \bold{3 = n}\end{gathered}$

$\begin{gathered} \\ \therefore \underline{ \ \sf{ \pmb{Number \: \: of \: \: years \: \: in \: \: which \: its \: \: value \: \: will \: \: get \: \: reduced = 3 \: \: years}}}\end{gathered}$

━━━━━━━━━━━━━━━━━

Some related formulae :-

$\bullet \: \: \bold{Simple~Interest = \dfrac{P \times R \times T}{100}}$

$\bullet \:  \: \bold{C.I. = P \bigg[1 + \dfrac{r}{100} \bigg]^{n}  - P}$

$\bullet\: \: \bold{Compound~Interest = Amount - Principal}$

$\bullet\: \: \bold{Amount = Compound~Interest + Principal}$

$\bullet \: \: \bold{Amount = P \bigg[1 +  \dfrac{r}{100} \bigg]^{n}}$

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