Question

A new car is purchased for $17,000 and over time its value depreciates by one half every 3 years. What is the value of the car 19 years after it was purchased, to the nearest hundred dollars?

Answer 1

200

Explanation:

Halving Formula:

y=a\left(\frac{1}{2}\right)^{\frac{t}{h}}

y=a(

2

1

​

)

h

t

​

a=17000\hspace{40px}h=3\hspace{40px}t=19

a=17000h=3t=19

h is the halving time

\text{Plug in:}

Plug in:

y=17000\left(\frac{1}{2}\right)^{\frac{19}{3}}

y=17000(

2

1

​

)

3

19

​

y=210.826702215

y=210.826702215

y\approx 200

y≈200

Answer 2

The answer is "$238".

Explanation:

Current worth
`= \$ 17,000`

depreciates by
`(1)/(2)` in 3 years.

time= 19 years

depreciates rate=?

Using formula:

`\to \text{Worth= Current worth}(1- \frac{\text{depreciates rate}}{100})^(time)`

`\to A_t=A_0(1-(r)/(100))^t`

calculates depreciate value in 3 year
`= (1)/(2) * 17,000`

`= 8,500`

so,

`A_t=8,500\\\\A_0=17,000\\\\t=3\ years`

`\to A_t=A_0(1-(r)/(100))^t\\\\\to 8,500= 17,000(1-(r)/(100))^3\\\\\to (8,500)/(17,000)= (1-(r)/(100))^3\\\\\to (1)/(2)= (1-(r)/(100))^3\\\\\to ((1)/(2))^{(1)/(3)}= (1-(r)/(100))\\\\\to 0.793700526 = (1-(r)/(100))\\\\\to (r)/(100) = (1-0.793700526)\\\\\to (r)/(100) = (1-0.8)\\\\\to r= 0.2 * 100 \\\\\to r= 20 \%`

depreciates rate= 20%

`\to \text{Worth= Current worth}(1- \frac{\text{depreciates rate}}{100})^(time)`

`= \$ 17,000 (1- (20)/(100))^(19)\\\\= \$ 17,000 (1-0.2)^(19)\\\\= \$ 17,000 (0.8)^(19)\\\\= \$ 17,000 * 0.014\\\\= \$ 238`