# Mathmatical Problem

Border View 11:29 02 Aug 10
Locked

For those out there who are good at maths this problem will be easy. I cannot figure out how to solve it -

Question Item "A" bought in 2000 cost £18750 by 2010 its value has shrunk to £750. Using the same rate of depreciation Item "B" bought in 2002 cost £23000 what would the value be in 2010

Thank you

Catastrophe 11:45 02 Aug 10

Multiply 23000 by 750 / 18750

Border View 11:52 02 Aug 10

Many thanks Catastrophe

lotvic 12:26 02 Aug 10

What about the 2 years difference?
"A" bought in 2000
"B" bought in 2002

Catastrophe 12:29 02 Aug 10

lotvic

You are correct. Multiply again by 8/10 assuming the annual depreciation rate is constant.

Catastrophe 12:30 02 Aug 10

Ooooops. That's wrong. I guess 10/8 since it is worth more having depreciated over a shorter time.

Border View 12:33 02 Aug 10

Does that mean that "B" will be worth more than £920?

Border View 12:35 02 Aug 10

Working on that basis "B" will be worth £1150?

Catastrophe 12:39 02 Aug 10

Let's do it properly. Annual depreciatorate is

(18750-750)/10 = 1800 per annum.
Hence answer should be
(23000-x)/8 = 1800
so 23000-14400 = 8600

Is that better?

Catastrophe 12:40 02 Aug 10

Really living up to my name today aren't I?

Woolwell 12:48 02 Aug 10

Firstly find the rate of depreciation = Amount lost (£18000) divided by the number of years: in this case 18000/10 - 1800 per year. Change it to a percentage per year = 1800/18000 = 0.096%. This is the rate of depreciation per year.
For the second item the rate of depreciation is over 8 years only. It will have lost 8 x 23000 x 0.096 subtract that from 23000 giving 5336.

This thread is now locked and can not be replied to.

Intel Coffee Lake 8th-gen Core processors release date rumours

1995-2015: How technology has changed the world in 20 years

Apple MacBook Pro with Touch Bar review

Best iPhone games 2017 | Best iPad games 2017: 162 fantastic iOS games that you need to play right…