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? ​

Answer 1

Step-by-step explanation:

Given ,

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

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

We know that,

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

Substituting the values, we get

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

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

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

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

$\begin{gathered}\sf \: {\bigg(\dfrac{4}{5} \bigg)}^{4} \: = \: \: {\bigg(\dfrac{4}{5} \bigg)}^{n} \\ \\ \end{gathered}$

$\begin{gathered}\sf \red{n \: = \: 4 \: years} \\ \end{gathered}$

Answer 2

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