There were 4000 cars sold in 2005 and 6100 cars sold in 2007, an increase of 2100 cars. Now, what "percentage" increase that is depends on "percentage of what?" Normally, in time problems like this, "percentage change" refers to percent of the initial value. Here that would be \(\displaystyle \frac{2100}{4000}= 0.525\) or 52.5%.
As to what 2004 sales were, obviously there is no "correct" answer to that. There are several different ways to "guess" what the answer should be. What we are given is a 52.5% increase in two years which is an average of 26.25% in one year. IF that held true for the previous year, then we want to find "S" sales such that S+ .2625S= 1.2625S= 4000 so that S= 4000/1.2625.
But we are also given that there were 5200 cars sold in 2006, an increase of \(\displaystyle \frac{1200}{4000}= 0.30\) so perhaps S+ .30S= 1.30S= 4000 and S= 4000/1.3 is more appropriate.
Or, there exist a unique parabola passing through any three points so we could find \(\displaystyle S= at^2+ bt+ c\) such that \(\displaystyle 4000= a(1)^2+ b(1)+ c\), \(\displaystyle 5200= a(2)^2+ b(2)+ c\) and \(\displaystyle 6100= a(3)^2+ b (3)+ c\) where I have used the number of years since 2004 as 't'. You can solve those equations for a, b, and c and the find S when t= 0.
a+ b+ c= 4000, 4a+ 2b+ c= 5200. Subtracting, 3a+ b= 1200.
a+ b+ c= 4000, 9a+ 3b+ c= 6100. Subtracting, 8a+ 2b= 2100.
8a+ 2b= 2100 and 6a+ 2b= 2400. Subtracting 2a= -300 so a= -150. Then 3a+ b= -450+ b= 1200 so b= 1200+ 450= 1650. Finally a+ b+ c= -150+ 1650+ c= 4000. c= 4000- 1650 + 150= 2500.
Of course, evaluating at t= 0 gives just S= c= 2500.
