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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

Verified Answer :

Katlynce7

the answer is C) $21,000

hope this helps

JeanaShupp

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 :-

@@MATH_TOKEN_0@@                (1)

For year 2006 , n=5 and V = 27000

Then , @@MATH_TOKEN_1@@

i.e.  @@MATH_TOKEN_2@@

i.e.  @@MATH_TOKEN_3@@

i.e.  @@MATH_TOKEN_4@@

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

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