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.