Question

Martin bought a computer for $1200. At the end of each year the value of the computer is depreciated by 40%.
After how many years will the value of the computer be $491.52? You MUST show your working.

Answer 1

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?

Given:

• Initial price of computer, P = $ 1200
• Rate of depreciation, r = 20 % per annum
• Final price of computer, A = $ 491.52

To Find:

• After how many years will the value of the computer be $491.52 = ?

Formula used:

$\\ \bigstar \: {\underline{\boxed{ \color{red}{\bf \: Amount = P\bigg(1 - \dfrac{R}{100}\bigg)^{n}}}}}$

Where,

• P = Principal
• R = Rate of Interest
• n = Time

Solution:

Let ,

• the number of years be n,

★substituting the values

$\\\sf \dashrightarrow \: \: 491.52 \: = \: 1200 \: {\bigg(1 - \dfrac{20}{100} \bigg)}^{n}$

$\sf \dashrightarrow \: \: \: \dfrac{491.52}{1200} \: = \: \: {\bigg(1 - \dfrac{1}{5} \bigg)}^{n}$

$\sf \dashrightarrow \: \: \: \: \dfrac{49152}{120000} \: = \: \: {\bigg(\dfrac{5 - 1}{5} \bigg)}^{n}$

$\sf \dashrightarrow \: \: \: \: \dfrac{256}{625} \: = \: \: {\bigg(\dfrac{4}{5} \bigg)}^{n}$

$\sf \dashrightarrow \: \: \: \: {\bigg(\dfrac{4}{5} \bigg)}^{4} \: = \: \: {\bigg(\dfrac{4}{5} \bigg)}^{n}$

$\sf \dashrightarrow \: \: \: \: \:n \: = \: 4 \: years$

Hence,

After 4 years, the value of computer 1200 depreciated to1200 depreciated to 491. 52 at the rate of 20 % per annum.

Answer 2

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

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 = 40 % 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

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

So, on substituting the values, we get

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

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

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

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

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

$\begin{gathered}\rm\implies \:n \: = \: 4 \: years \\ \end{gathered}$

$\huge \mathfrak \pink{Himetisha}$