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

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