Question

Pat bought a boat for $37,000 in 2001. In 2006 the boat was worth $27,000. If the boat depreciation is linear, how much will the boat be worth in 2009?

A) $27,000
B) $22,000
C) $21,000
D) $20,000

Answer 1

the answer is C) $21,000

hope this helps

Answer 2

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 :-

$V=37,000-nx$                (1)

For year 2006 , n=5 and V = 27000

Then , $27000=37,000-5x$

i.e.  $5x=37,000-27000$

i.e.  $5x=10,000$

i.e.  $x=2,000$

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

$V=37,000-(8)(2000)=37000-16000=21000$      

Hence, the value of boat  in 2009 = $21,000