Depreciation

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Vv20

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May 14, 2020
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Hello

This is the question, in question:
The furniture in an office was bought 6 years ago. It is being depreciated over 10 years, straight line, to a scrap value of $200. If its current book value is $1,120, how much did it cost?

Okay, i started here:
Depreciation per year
= Asset Cost - Salvage value
Useful life
= 1120 - 200
4 (10 years - 6 years)
Depreciation per year
= $230
= 230 * 4 (10 years - 6 years)
= 920 + 1120
Beginning value = $2040

Ok. Looks good i thought. But then i put details into depreciation calculator, well here i am.
Any advice would be greatly appreciated.
Thank you.
 
Hi Harry_the_cat, thank you for answering
After posting that problem, i did notice that i had only used 8 years. So i did do, what you are suggesting. Unfortunately, with the numbers produced, the depreciation schedule totally passed the $$1120 at 6 years, in fact there wasn't a $1120 figure anywhere on the schedule.
 
Thanks Harry_the_cat.
I looked at the problem differently after your feedback, and i got it.

1120 - 200
5 (The beginning on one year to the end of another year, is a year. So says my teacher, after he viewed a similar question where i was missing a year.)
=$184
=184 x 5 years
= $920
=1120 + 920
Starting Cost
= $2040
 
Vv20 said:
Hello

This is the question, in question:
The furniture in an office was bought 6 years ago. It is being depreciated over 10 years, straight line, to a scrap value of $200. If its current book value is $1,120, how much did it cost?

Okay, i started here:
Depreciation per year
= Asset Cost - Salvage value
Useful life
= 1120 - 200
4 (10 years - 6 years)
Depreciation per year
= $230
= 230 * 4 (10 years - 6 years)
= 920 + 1120
Beginning value = $2040

Ok. Looks good i thought. But then i put details into depreciation calculator, well here i am.
Any advice would be greatly appreciated.
Thank you.
Click to expand...
Straight algebra

[MATH]c = \text {original cost.}[/MATH]
[MATH]\dfrac{c - 200}{10} = \text {annual depreciation.}[/MATH]
[MATH]1120 = \text {book value after 6 years.}[/MATH]
[MATH]\therefore c - \left ( 6 * \dfrac{c - 200}{10} \right ) = 1120 \implies\\ 10c - 6c + 1200 = 11200 \implies\\ 4c = 10000 \implies\\ c = 2500.[/MATH]Let's check.

[MATH]\dfrac{2500 - 200}{10} = 230.\\ 6 * 230 = 1380.\\ 2500 - 1380 = 1120 \ \checkmark[/MATH]Your problem was using 4 years, which is the remaining life. But that is irrelevant to the current book value, which has been depreciated over 6 years. There is the so-called missing year. It was perfectly ok to compute the annual depreciation as you did over the remaining life. The two of us got the same answer as you can see. But you wanted the original cost, and the time difference there is SIX years, not four.
 
Last edited:
Thank you JeffM.
I found that work through totally amazing.
 
Thank you Vv for your kind words. But there was nothing amazing about it, or there would not be if we taught algebra in a sane way.

The main practical use of algebra is to find unknown quantities. The key first step is to "name" things clearly. You wanted to find original cost. Give it a name (I used "c" for obvious reasons). The next step is to figure out what you do know. Here we knew the depreciation method, the years since purchase, and the current book value. The third step is to figure out the quantitative relationship between what we already know and what we want to know. Once we have that, the rest is mechanics that a machine can do better than a human. But we don't teach algebra that way.

I want to reiterate that your way of calculating annual depreciation was perfectly correct. It was not as efficient as my method in getting to the answer, but it would have worked. Efficiency is not a big issue in most cases, and the more work you do in a field the more efficient you will become. The main advantage of my method was psychological: it forced me to think 6 years.
 
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