Upstairs obviously overlooked a point, always need to buy 36 tickets.

There are three combinations: first class and second class, second class and third class, first class and third class.

First of all, the combination of first-class and second-class seats is not feasible, because even if all the second-class seats are bought, 36 tickets will cost more than 7,200 US dollars and 5,025 US dollars.

Therefore, we can only choose two schemes: 1, the combination of first and third seats, and the combination of second and third seats.

Two equations can be obtained.

1, x first-class seats, 36-x third-class seats.

300x+ 125(36-x)=5025

300x+4500- 125x=5025

175x=525

x=3

36 x = 33

There are x second-class seats and 36-x third-class seats.

200x+ 125(36-x)=5025

200x+4500- 125x=5025

75x=525

x=7

36 x = 29

So * * * there are two schemes:

Three first class seats and 33 third class seats.

Seven second-class seats and 29 third-class seats.