Question

Martin bought a computer for $1200. At the end of each year the value of the computer is depreciated by 20%. After how many years will the value of the computer be $491.52? You MUST show your working.

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

Martin bought a computer for $1200. At the end of each year the value of the computer is depreciated by 20%.

So, we have

Initial price of computer, P = $ 1200

Rate of depreciation, r = 20 % per annum

Final price of computer, A = $ 491.52

Let assume that the number of years be n, so that the value of computer will be $ 491.52

We know,

If the rate of depreciation is r % per annum and the initial price of the object is P, then depreciated value (A) of the object after n years is given by

$\boxed{\sf{  \: \: \: A \: = \: P \: {\bigg[1 - \dfrac{r}{100} \bigg]}^{n} \: \: \: }}$

So, on substituting the values, we get

$\rm \: 491.52 \: = \: 1200 \: {\bigg[1 - \dfrac{20}{100} \bigg]}^{n}$

$\rm \: \dfrac{491.52}{1200} \: = \: \: {\bigg[1 - \dfrac{1}{5} \bigg]}^{n}$

$\rm \: \dfrac{49152}{120000} \: = \: \: {\bigg[\dfrac{5 - 1}{5} \bigg]}^{n}$

$\rm \: \dfrac{256}{625} \: = \: \: {\bigg[\dfrac{4}{5} \bigg]}^{n}$

$\rm \: {\bigg[\dfrac{4}{5} \bigg]}^{4} \: = \: \: {\bigg[\dfrac{4}{5} \bigg]}^{n}$

$\rm\implies \:n \: = \: 4 \: years$

Hence,

After 4 years, the value of computer $ 1200 depreciated to $ 491. 52 at the rate of 20 % per annum.

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Additional Information :-

1. Amount on a certain sum of money of Rs P invested at the rate of r % per annum compounded annually for n years is given by

$\boxed{\sf{  \: \: \: A \: = \: P \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} \: \: \: }}$

2. Amount on a certain sum of money of Rs P invested at the rate of r % per annum compounded semi - annually for n years is given by

$\boxed{\sf{  \: \: \: A \: = \: P \: {\bigg[1 + \dfrac{r}{200} \bigg]}^{2n} \: \: \: }}$

3. Amount on a certain sum of money of Rs P invested at the rate of r % per annum compounded quarterly for n years is given by

$\boxed{\sf{  \: \: \: A \: = \: P \: {\bigg[1 + \dfrac{r}{400} \bigg]}^{4n} \: \: \: }}$

4. Amount on a certain sum of money of Rs P invested at the rate of r % per annum compounded monthly for n years is given by

$\boxed{\sf{  \: \: \: A \: = \: P \: {\bigg[1 + \dfrac{r}{1200} \bigg]}^{12n} \: \: \: }}$

Answer 2

Step-by-step explanation:

Initial price of computer, P = $ 1200

Rate of depreciation, r = 20 % per annum

Final price of computer, A = $ 491.52

Let assume that the number of years be n, so that the value of computer will be $ 491.52

We know,

If the rate of depreciation is r % per annum and the initial price of the object is P, then depreciated value (A) of the object after n years is given by

$\begin{gathered}\sf{ \: \: \: A \: = \: P \: {\bigg[1 - \dfrac{r}{100} \bigg]}^{n} \: \: \: }\\ \end{gathered}$

So, on substituting the values, we get

$\color{olive} \pmb{ \[ \begin{array}{l} 491.52=1200\left[1-\dfrac{20}{100}\right]^{ n } \\ \\ \tt\dfrac{491.52}{1200}=\left[1-\dfrac{1}{5}\right]^{ n } \\ \\ \tt \dfrac{49152}{120000}=\left[\dfrac{5-1}{5}\right]^{ n } \\ \\ \tt \dfrac{256}{625}=\left[\dfrac{4}{5}\right]^{ n } \\ \\ \\ \tt {\left[\begin{array}{l} 4 \\ 5 \end{array}\right]^{4}=\left[\begin{array}{l} 4 \\ 5 \end{array}\right]^{ n }} \\ \\ \tt \Longrightarrow n =4 \text { years } \end{array} \]}$

Hence,

After 4 years, the value of computer 1200 depreciated to1200depreciatedto 491. 52 at the rate of 20 % per annum.