Question

A car was valued at $41,000 in the year 1994. The value depreciated to $13,000 by the year 2004

Answer 1

`\begin{gathered} a)\text{ 0.1085} \\ b)\text{ 10.85 \%} \\ c)\text{ \$7,300} \end{gathered}`

Firstly, we have to write the general depreciation fomula as follows;

`V(t)=I(1-r)^t`

where V(t) represents the value of the commodity after a certain t years

I is the initial value of the item

r is the annual percentage change (rate of depreciation)

t is the number of years

a) Here, we want to calculate the annual rate of change

According to the data given in the question;

V(t) = $13,000

I = $41,000

r = ?

t = 2004-1994 = 10 years

Now, we proceed to substitute these values, to find the find of r as follows;

`\begin{gathered} 13,000=41,000(1-r)^(10) \\ (1-r)^{10\text{ }}\text{ = }(13,000)/(41,000) \\ (1-r)^{10\text{ }}=\text{ }(13)/(41) \\ \\ (1-r)\text{ = }\sqrt[10]{(13)/(41)} \\ \\ 1-r\text{ = 0.8915} \\ r\text{ = 1-0.8915} \\ r\text{ = 0.1085} \end{gathered}`

b) To write r in percentage form, we have to multiply the answer in 'a' above by 100

We have this as ;

`0.1085*100\text{ = 10.85 \%}`

c) Here, we want to get the car value by year 2009

In that instance;

V(t) = ?

I = $41,000

r = 0.1085

t = 2009-1994 = 15

Substituting these values, we have;

`\begin{gathered} V(15)=41,000(1-0.1085)^(15) \\ V(15)\text{ = \$7,321.56} \end{gathered}`

To the nearest 50 dollars, this is $7,300