A car depreciates in value by 15% for the first year and for each later year by 12% if its value at the beginning of that year. Calculate the precent age decrease in the value of the car after 3 years. (Wondering)

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- Apr 25th 2009, 04:35 AMNITARBBY.Depreciation and percentage age decrease
A car depreciates in value by 15% for the first year and for each later year by 12% if its value at the beginning of that year. Calculate the precent age decrease in the value of the car after 3 years. (Wondering)

- Apr 25th 2009, 06:00 AMUnenlightened
Say you buy a car for X on January 1st 2000, say.

After a year it's only worth (100%-15%)X = 85% of X. Let's say 0.85 X.

So January 1st 2001 it's worth 0.85 X.

At the end of that year, it's decreased by 12% again, so is worth only (100% - 12%) = 88% of what it was at the beginning of that year, which was 0.85 X.

So now it's worth (0.88)(0.85)X. (On January 1st 2002).

Then at the end of that year (which is 3 years after you bought it), it is worth 0.88 % of what it was at the start, ie. on January 1st 2002, which is

(0.88)(0.88)(0.85) X

= 0.65824 X,

or 65.824 % of X, X being what you paid in the beginning.

Now that is what it is WORTH, so you're not quite done - you're asked for the percentage decrease, which is of course 100% - 65.824% = 34.176 %

k? - Apr 25th 2009, 10:10 PMpickslides
I'm not sure that is correct

I would say let the starting value be x and the finishing value be y.

Now y depreciates by 15% the first year then 12% for the 2nd and third years.

$\displaystyle y = x-0.15x-0.12(0.15x)- 0.12^2(0.15x) = 0.83x $

$\displaystyle percentage_{change} = \frac{x}{y} -1= \frac{x}{0.83x} -1= \frac{1}{0.83} -1=0.21 $ - Apr 26th 2009, 05:37 AMUnenlightened
- Apr 26th 2009, 04:58 PMpickslides
$\displaystyle y = x-0.85x$

This would suggest that i'm subtracting 85% of the frist year's value.

consider

$\displaystyle y = x-0.15x$

Choose x = 100 and calculate the next year's value

$\displaystyle y = 100-0.15\times100$

$\displaystyle y = 100-15$

$\displaystyle y = 85$ - Apr 27th 2009, 12:07 AMUnenlightened
No, I meant 0.12(0.85x) instead of 0.12(0.15x)

As in, for the second year, you're taking away 12% of the value at the beginning of year 2, as opposed to 12% of the depreciation the first year. - Apr 27th 2009, 02:22 AMSoroban
Hello, NITARBBY!

I agree with Unenlightened.

I'll walk through his reasoning . . .

Quote:

A car depreciates in value by 15% for the first year

and for each later year by 12% if its value at the beginning of that year.

Calculate the percent age decrease in the value of the car after 3 years.

The first year it loses 15% of its present value.

. . It is worth only 85% of its original value.

Its value is: .$\displaystyle 0.85x$

The next year it loses 12% of its present value.

. . It is worth only 88% of its present value.

Its value is: .$\displaystyle 0.88(0.85x)\:=\:0.748x $

The third year is loses 12% of its present value.

. . It is worth only 88% of its present value.

It is value is: .$\displaystyle 0.88(0.748x) \:=\:0.65824x$

The car is worth only 65.824% of its original value.

Therefore, it depreciated 34.176% of its original value.

- Apr 27th 2009, 02:21 PMpickslides