A painting, purchased for $10 000 in 1990, increased in value by 8% per year. How do you find the value of the painting in the year 2000?

Question

3067 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-07T18:20:43

You have to compose the interests year by year.

Let's try to find the general law of this.

The initial value of the paint, at zero years, is #V_0# (later we will substitute the $10,000).

After the first year we have a new value that is augmented of the percentage given. We call the percentage P (later we will substitute 8/100).

#V_1=V_0+P*V_0=V_0(P+1)#.

The second year we have the same rule, but applied to #V_1#

#V_2=V_1+P*V_1=V_1(P+1)#

and substituting the previous value of #V_1=V_0(P+1)# we have

#V_2=V_0(P+1)(P+1)=V_0(P+1)^2#.

It seems we found a rule, let's try again for the third year

#V_3=V_2+P*V_2=V_2(P+1)#

and substituting the previous equation for #V_2=V_0(P+1)^2#

#V_3=V_0(P+1)^2(P+1)=V_0(P+1)^3#.

Then, after #N# years, we have

#V_N=V_0(P+1)^N#.

This is the rule that is governing the increases of the value in the years. Now it is enough to substitute

#V_0=10000#

#P=8/100#

#N=2000-1990=10#

#V_N=10000(8/100+1)^10#

#=10000*(108/100)^10#

#=10000*(1.08)^10#

#\approx10000*2.16#

#=21600#.

This paint was a good investment, after #10# years it has a value that is more than double: $21 600.