According to the meaning of the question, x > y, x < 90, x+y = 92...①,

Then 46 < x < 90, 3 ≤ y ≤ 45;

Then we can know that two schools buy clothes separately, school A carries out the price of 50 yuan/suit, and school B carries out the price of 60 yuan/suit, then we can get: 50x+60y = 5000...②,

Combining the equations ① and ②, we can get X = 52 and Y = 40 by solving the equations.

2. Is it "A pumping 10 student? . "

The best scheme is: buy 9 1 suit, implement the price of 40 yuan/suit, and spend 40× 9 1 = 3640 yuan. Because there were 92 people in two schools, 10 people took it away, and * * had 82 people. If you only buy 82 pieces, you will never buy less than 50× 82 = 4 100 yuan, but if you buy enough 9 1 piece, you can implement the price of 40 yuan/set, so you can buy more clothes.

Reference:

There are X students in A school to participate in the performance, and 92-x students in B school to participate in the performance.

There are more students in school A than in school B, and the number of students in school A is less than 90.

∴x>； 92-x and x

Solution: 46

∴50x+(92-x)*60=5000

Solution: x=52

Then the students who are going to participate in the performance in school B are: 92-52=40 (people).

A: There are 52 students in each of the two schools, and 40 students are going to participate in the performance.

(2)∵ A school has 10 students randomly selected to participate in the painting and calligraphy competition, but they cannot participate in the performance.

The students in A and B schools are: 92- 10=82 (people).

∴ The two schools jointly purchased 82 sets of clothes: 82*50=4 100 yuan.

Two schools, A and B, buy 9 sets more, and the cost is 9 1 set: 9 1*40=3640 (RMB).

∵3640 & lt； 4 100

It is the most economical for school A and school B to jointly buy 9 1 suit.

Answer: It is the most economical for school A and school B to jointly buy 9 1 suit.