10a+5b= 1000
5a+3b=550
∴ A = 50 b = 100 for solving equations
∴ It takes 50 yuan to buy a souvenir A and 100 yuan to buy a souvenir B.
(2) Suppose the store buys X pieces of A-type souvenirs and Y pieces of B-type souvenirs.
50x+ 100y= 10000
6y≤x≤8y
The solution is 20≤y≤25.
* * There are six purchase schemes.
(3) Let the total profit be W yuan.
w=20x+30y=20(200-2 y)+30y
=- 10 y+4000
W decreases with the increase of y
When y = 20, w has a maximum.
W Max =- 10× 20+4000 = 3800 (yuan)
∴? Buy 0/60 pieces of Class A souvenirs/KLOC-and 20 pieces of Class B souvenirs, and you can get the maximum profit, with a maximum profit of 3,800 yuan.