If a piece of real estate purchased for $75,000 in 1998 appreciates at the rate of 6% per year, then its?

Question

If a piece of real estate purchased for $75,000 in 1998 appreciates at the rate of 6% per year, then its?
value t years after the purchase will be f(t)=75,000(1.06^t). According to this model, by how much will the value of this piece of property increase between the years 2005 and 2008?
a.$14,300
b.$21,500
c.$37,800
d.$59,300

Answer 1

b) 21,500

using 75k(1.06^[2008-1998]) - 75k(1.06^[2005-1998]) -> 75k(1.06^10 - 1.06^7) = $21,540

Answer 2

there are 3 years between 2008 and 2005 and 7 years between 2005 and 1998
so price in 2005 is
f(2005) =75000*(1.06^7)=75000*1.5036=112772

in f(2008) = 112772*(1.06^3)=134314and substracting from this value 112772 you find

CORRECT ANSWER b

Answer 3

first figure the value in 2005
f(t) = 75000(1.06^7) = 112772.27
next figure the value in 2008
f(t) = 75000(1.06^10) = 134313.58
next subtract the 2 values to get how much it increased
134313.58 - 112772.27 = 21541.31
so your answer would be letter b.) 21,500

Answer 4

75000 multiplied by (1+6/100 )to the power of (2007 minus 1998)

Answer 5

OK

You need to calculate 1.06 ^3. This is about 1.19

Multiply 1.19 x 75000 = 14,250. So A is the answer.

Hope that helps.