## Depreciation

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This section explores depreciation. By the end you should be able to calculate the depreciated value and the amount of depreciation. You should have prior knowledge of working with percentages, decreasing percentages and good calculator skills.

Depreciation is the opposite of compound interest

...while compound interest increases the initial value each year by a given percentage, depreciation decreases the initial value each year by a given percentage. Depreciation is useful in many mathematical applications. Everyday items/properties when bought lose their value after a certain period of time. Knowing to how calculate depreciation is very useful.
Consider this example; The price of a new car is £25000. The price depreciates by 18% each year (p.a). Suppose we wanted to find its value at the end of 3 years. Here let’s use a table to calculate starting from year 1.

Year 1 | £25000 x 0.82 = 20500 |
---|---|

Year 2 | £20500 x 0.82 = 16810 |

Year 3 | £16810 x 0.82 = 13784 |

£13784.20

It has depreciated by;
£25000 - £13784.20 = £11215.80

...therefore the depreciation is;
£11215.80

## Depreciation formula

Looking at the previous example we saw that;End of 1

^{st}year, = 25000 × 0.82End of 2

^{nd}year = (£25000 × 0.82) × 0.82= £25000 × 0.82²

End of 3

^{rd}year = (£25000 × 0.82²) × 0.82= £25000 × 0.82³

= £13874.20

## Using the depreciation formula

We shall use the formula discovered above to work out some examples.
Example
Using the depreciation formula find the value of a TV after 15 months if it’s price while new is £1400 and it depreciates by 8% per month.

Answer

V = P(1 - r%)

^{n}= 1400(1 - 0.08)

^{15}= 1400 × 0.92

^{15}= £400.82

Explanation
Let’s summarise the known variables and the unknown first then work out. We know that;
We're trying to find the value of V. Using the known values we can conclude that;

P = £1400

r = 8%

n = 15

V = ?

V = 1400(1 - 8%)

We can see that after 15 months the TV is valued at £400.82. It has dropped value by £999.18.
^{15}You have to be careful when working with depreciation. In some cases, depreciation may be calculated monthly, fortnight, weekly and maybe daily. So you have to be careful when working with problems involving depreciation and adjust the rate and number of calculations carefully.

This section explores depreciation. By the end you should be able to calculate the depreciated value and the amount of depreciation. You should have prior knowledge of working with percentages, decreasing percentages and good calculator skills.

Depreciation is the opposite of compound interest

...while compound interest increases the initial value each year by a given percentage, depreciation decreases the initial value each year by a given percentage. Depreciation is useful in many mathematical applications. Everyday items/properties when bought lose their value after a certain period of time. Knowing to how calculate depreciation is very useful.
Consider this example; The price of a new car is £25000. The price depreciates by 18% each year (p.a). Suppose we wanted to find its value at the end of 3 years. Here let’s use a table to calculate starting from year 1.

Year 1 | £25000 x 0.82 = 20500 |
---|---|

Year 2 | £20500 x 0.82 = 16810 |

Year 3 | £16810 x 0.82 = 13784 |

£13784.20

It has depreciated by;
£25000 - £13784.20 = £11215.80

...therefore the depreciation is;
£11215.80

## Depreciation formula

Looking at the previous example we saw that;End of 1

^{st}year, = 25000 × 0.82End of 2

^{nd}year = (£25000 × 0.82) × 0.82= £25000 × 0.82²

End of 3

^{rd}year = (£25000 × 0.82²) × 0.82= £25000 × 0.82³

= £13874.20

## Using the depreciation formula

We shall use the formula discovered above to work out some examples.
Example
Using the depreciation formula find the value of a TV after 15 months if it’s price while new is £1400 and it depreciates by 8% per month.

Answer

V = P(1 - r%)

^{n}= 1400(1 - 0.08)

^{15}= 1400 × 0.92

^{15}= £400.82

Explanation
Let’s summarise the known variables and the unknown first then work out. We know that;
We're trying to find the value of V. Using the known values we can conclude that;

P = £1400

r = 8%

n = 15

V = ?

V = 1400(1 - 8%)

We can see that after 15 months the TV is valued at £400.82. It has dropped value by £999.18.
^{15}You have to be careful when working with depreciation. In some cases, depreciation may be calculated monthly, fortnight, weekly and maybe daily. So you have to be careful when working with problems involving depreciation and adjust the rate and number of calculations carefully.