Depreciation

eddy2017

eddy2017

Elite Member
Joined
Oct 27, 2017
Messages
2,525
Otis said:
Hi porterehd. You've posted on the Pre-Algebra board. Has your class been discussing exponential fractions? If so, then how do they define those?

Otherwise, do you know how to calculate (100-13.5)% of 17000 ?

:)
Click to expand...
After a simple calculation ( on paper and with a pen) ;)
86.5 % of 17000
17000/x = 100/86.5 ( simple rule of 3)
you finish portereh, there is no low-hanging fruit on this site. Believe you me!
 
eddy2017 said:
After a simple calculation ( on paper and with a pen) ;)
86.5 % of 17000
17000/x = 100/86.5 ( simple rule of 3)
you finish portereh, there is no low-hanging fruit on this site. Believe you me!
Click to expand...
no

porterehd said:
A new car is purchased for 17000 dollars. The value of the car depreciates at 13.5% per year. What will the value of the car be, to the nearest cent, after 12 years?
Click to expand...

value after 12 years =[imath] 17000(0.865)^{12}[/imath]
 
eddy2017 said:
Then how do you calculate what Otis posted?.
(100-13.5)% of 17000 ?. What is wrong with what I did?.
Click to expand...
The question in OP was the depreciated value after 12 years - time involved. Otis's post in this thread (#5) there is no time involved.
 
Subhotosh Khan said:
[In Otis's post] there is no time involved.
Click to expand...
Of course there is time involved!

My expression gives the depreciated value after one year.

Were the OP to have responded in the affirmative, then I would have explained how one could use an expression to answer the exercise, by hitting the enter button on a calculator twelve times in a row. And that process would have been my precursor to an algebraic explanation.

Of course, I'd never gotten a response.

?
 
100-13.5%, that is: 86.5% from the previous year or 0.865 from the previous year, and so on each year:

17,000 * .865 * .865 ..., twelve times, or: 17,000 * (.865)^12. Then:

17,000 * (. 865) ^ 12 = 17,000 * .1754;

$ 2892.92

Okay. I wrote the above just to remind you that this was the answer to the depreciation value post. But hat was settled.
My question is if the following formula could be applied here as well.
D= C-R / n

D= stands for depreciation
C= stands for cost of the asset in question
R= the so called residual value
N =useful span of life of the asset.

Could I apply the formula here being this a linear depreciation?
 
eddy2017 said:
100-13.5%, that is: 86.5% from the previous year or 0.865 from the previous year, and so on each year:

17,000 * .865 * .865 ..., twelve times, or: 17,000 * (.865)^12. Then:

17,000 * (. 865) ^ 12 = 17,000 * .1754;

$ 2892.92

Okay. I wrote the above just to remind you that this was the answer to the depreciation value post. But hat was settled.
My question is if the following formula could be applied here as well.
D= C-R / n

D= stands for depreciation
C= stands for cost of the asset in question
R= the so called residual value
N =useful span of life of the asset.

Could I apply the formula here being this a linear depreciation?
Click to expand...
Someone told me this is not linear but exponential so the formula does not apply.
What is the diffrence between a linear depreciation and an exponential depreciation?.
 
Are you interested in the mathematical difference or the practical difference? If the latter, are you interested in depreciation as a concept in economics or in accounting.
 
eddy2017 said:
D= C-R / n
Click to expand...
Is that what you wanted write ? or did you want to write:

D = \(\displaystyle \frac{C - R}{n} \) = (C - R)/n .......................... without those parentheses the meaning and the value of the expression changes drastically !!
 
Subhotosh Khan said:
Is that what you wanted write ? or did you want to write:

D = \(\displaystyle \frac{C - R}{n} \) = (C - R)/n .......................... without those parentheses the meaning and the value of the expression changes drastically !!
Click to expand...
without the parentehesis.
 
JeffM said:
Are you interested in the mathematical difference or the practical difference? If the latter, are you interested in depreciation as a concept in economics or in accounting.
Click to expand...
Yes, I am. I am interested in everything I see and I do not understand, mathematically wise. And this forum is a good source of good questions given to students who go to schools and have math teachers. I can't help but when I see that a question was left unanswered because the poster did not follow through and is a good question going to waste I can't help but go ahead and make it my own. Guidelines do not allow me to post anything on the same thread but I was told I can open another and quote this one.
Yes, I am interested. I was watching a tutorial on lineal depreciation and the teacher was explaining this formula bu then again someone told me that does not apply here because this is an exponential appreciation. and this formula is for lineal depreciation.
 
Last edited:
To be honest according to what I have learned here no it doesn't. Looks to me the numerator part ( for lack of a better term) should be enclosed paratheses.
 
eddy2017 said:
To be honest according to what I have learned here no it doesn't. Looks to me the numerator part ( for lack of a better term) should be enclosed paratheses.
Click to expand...
Where did you encounter the equation without parentheses (or written up as \(\displaystyle \frac{numerator}{denominator} \))
 
The formula for annual linear depreciation is [imath]\dfrac{a - s}{y}[/imath], where a is acquisition cost, s is expected salvage value, and y is number of years of expected useful life for items of that type. In many cases, s is assumed to be 0.

The formula for annual exponential depreciation after n years is [math]\left ( \dfrac{y - 1}{y} \right )^n * a.[/math]
 
You must log in or register to reply here.
Top