Question

A car was valued at $39,000 in the year 1995. The value depreciated to $11,000 by the year 2003.

Answer 1

Answer:

14.6328% , $5836.03

Step-by-step explanation:

Here we are going to use the formula

$A_{0}(1-r)^n = A_{n}$

$A_{0}$ = 39000

r=?

$A_{8}$ = 11000

n=8

Hence

$39000(1-r)^8 = 11000$

$(1-r)^8 = \frac{11000}{39000}$

$(1-r)^8 = 0.2820$

$(1-r) = 0.2820^{\frac{1}{8}$

$(1-r) = 0.2820^{0.125}$

$(1-r) = 0.8536$

$(1-0.8536=r$

$r = 0.1463$

Hence r= 0.1463

In percentage form r = 14.63%

Now let us see calculate the value of car in 2003 that is after 12 years

we use the main formula again

$A_{0}(1-r)^n = A_{n}$

$A_{0}$ = 39000

r=0.1463

$A_{12}$ = ?

n=12

$39000(1-0.14634)^{(12} = A_{12}$

$39000(0.8536)^{12} = A_{12}$

$39000*0.1497 = A_{12}$

$A_{12}=5840.34$

Hence the car's value will be depreciated to $5840.34 (approx) by 2003.