A car depreciates at a rate of 30% per year. How much will a $20,000 car be worth after 5 years?

Printable View

- April 21st 2009, 09:47 PMmath123456exponential functions and depreciation
A car depreciates at a rate of 30% per year. How much will a $20,000 car be worth after 5 years?

- April 21st 2009, 09:57 PMChris L T521
- April 21st 2009, 10:42 PMpickslides
I would use the model

where V = current value, A = initial value, P = percentage change and t = time.

Therefore

then make t = 5

- April 21st 2009, 11:15 PMearboth
I don't want to pick at you but in my opinion the equation you used is only applicable for rates of decrease smaller than 10%.

With the given data you know that the car's value after 1 year is $14.000. If you now calculate the constant k you'll get:

Therefore your result will be too large. - April 21st 2009, 11:26 PMChris L T521
- April 22nd 2009, 04:53 AMpickslides
- April 22nd 2009, 05:27 AMearboth
If you use Chris L T 521's formula:

then where p is the percentage of decrease.

Now compare:

decrease by 2% ====> ln(0.98) = -0.020202...

decrease by 5% ====> ln(0.95) = -0.051293...

decrease by 10% ====> ln(0.9) = -0.10536...

...

decrease by 20% ====> ln(0.8) = -0.22314...

You see that the approximation of k = -p is only sufficient for p < 10%. - April 22nd 2009, 07:30 AMSoroban
Hello, math123456!

Quote:

A car depreciates at a rate of 30% per year.

How much will a $20,000 car be worth after 5 years?

Each year the car is worth only 70% of its value during the previous year.

. . Its value after years is: .

In 5 years, its value is: .

- April 22nd 2009, 09:25 AMSoroban
Seeing the way that Chris approached the problem,

I am obligated to ask: Is someone teaching that the -formula

. . is theway to handle depreciation?*only*

If so, they are doing a great disservice.

While is highly applicable to decay and population problems,

I feel that it requires more work when applied to depreciation problem,

. . especially when they*give us*the depreciation rate.

Consider the original problem . . .

. . Initial value: $20,000

. . Depreciation rate: 30% (annually)

We start with: .

When

. . Hence, the function (so far) is: .

When

. .

Hence, the function is: .

This equation is correct . . . I have no complaint.

But simplify the equation: .

Since , we have: .

We had this equation*way back then*, just "eyeballing" the problem.

- April 22nd 2009, 05:29 PMpickslides
- April 23rd 2009, 02:39 AMearboth
The difficulty lies in the word "always" ...

With the given problem the variable t is a discret number, that means actually you are dealing with a geometric sequence:

If a value is increasing or decreasing**continuously**your method can't be applied, then you have to use the equation which was posted by ChrisLT521