Answer: C) $21,000
Step-by-step explanation:
Given : Pat bought a boat for $37,000 in 2001. In 2006 the boat was worth $27,000.
If the boat depreciation is linear, then the amount by which the value of boat depreciates must be constant.
Let x be the constant depreciation in the value of boat per year.
Then , the value of boat (in dollars) after n years from 2001 is given by :-
[tex]V=37,000-nx[/tex] (1)
For year 2006 , n=5 and V = 27000
Then , [tex]27000=37,000-5x[/tex]
i.e. [tex]5x=37,000-27000[/tex]
i.e. [tex]5x=10,000[/tex]
i.e. [tex]x=2,000[/tex]
Thus , the constant amount of depreciation in the value of boat per year. = $2000
Now for year 2009 , put in n=8 and x= 2000 in (1), we get
[tex]V=37,000-(8)(2000)=37000-16000=21000[/tex]
Hence, the value of boat in 2009 = $21,000
