Depreciation/ Appreciation

Today’s topic will feature a combination of a variety of topics which includes percentages, depreciation/ appreciation and Geometric Progression. I have not taught Geometric Progression (G.P) but in this very chain of topics I will teach it.

I am going to teach depreciation/ appreciation today, it is the main topic but it involves percentages and (G.P) is a method which depreciation/ appreciation can be done! Let’s look at 2 types of depreciation questions! First we will look at the gradual depreciation type.

E.g. A house of value $500,000 depreciates annually by 2%. What will be the new value of the house 5 years from now, starting January 1^st, 2008? This is the very exact question from 3 posts ago!

Solution: Method 1

Year | value at start | depreciation | end year value

1 | $500,000 | $500,000 x 2% =$10,000 | $490,000

2 | $490,000 | $490,000 x 2% =$ 9,800 | $480,200

3 | $480,200 | $480,200 x 2% =$ 9,064 | $470,596

4 | $470,596 | $470,596 x 2% =$9,411.92| $461,184.08

Therefore the value in the 5^th year is $461,184.08

Note: the end year value for the 4^th year is the value at start on the 5^th year and 5 years from now means after for years. The S.A.T questions are all designed this way, so the trick in the question is the grammar and not the method how to solve it.

Solution: Method 2 (easier method)

Since the depreciation is 2% annually, the value after each year is 98% of the start.

After 1 year the value is $500,000 x .98= $490,000

After 2 years the value is $500,000 x .98 x.98= $480,200

After 3 years the value is $500,000 x .98 x.98 x.98=$470,596
After 4 years the value is $500,000 x (.98)^ 4 =$461,184.08

If you know the original value V, the rate of depreciation d% per annum, and the number of years n, the depreciated value is given by

V x ( (100 – d)/100)^ n
This is known as the depreciation/ appreciation formula.

Note: To do appreciation questions just change the minus sign to a plus sign!

Now let’s look at the other type of depreciation known as fixed appreciation.

E.g. A company vehicle bought for $125,000 in 2004 depreciates by a fixed value by 15% of the vehicle. What would be the value of the vehicle given by the company (book value) after 3 years?

Solution: Method

Value of the vehicle is $125,000.
Depreciation for the first year is $125,000 x 15%=$18,750
Depreciation for 3 years is $18,750 x 3=$56,250
Book value after 3 years is $125,000 - $56,250 =$68,750

This is the simplest for of depreciation. The next post I will show you how to use the (G.P) formula to do these questions.