A tour group of 50 people went to a hotel to stay. There are three rooms in the hotel: triple room, double room and single room, among which the triple room is 20 yuan per person per night and the dou

A tour group of 50 people went to a hotel to stay. There are three rooms in the hotel: triple room, double room and single room, among which the triple room is 20 yuan per person per night and the double room. A 50-member tour group stayed in a hotel, which has three rooms: triple room, double room and single room, in which the triple room is in 20 yuan every night, the double room is in 30 yuan every night and the single room is in 50 yuan every night. It is known that the tour group is full of 20 rooms, which makes the total accommodation cost the least, so the most economical accommodation cost is

1 150 1 150

Yuan. Test site: the application of ternary linear equations. Special topic: application problem. Analysis: Let's assume that the tour group lives in a triple room X, a double room Y and a single room Z, and the total accommodation fee is A yuan. List the equations according to the requirements of the topic.

x+y+z = 203 x+2y+z = 5060 x+60y+50z = a？ .

The expressions of y and z expressed by x are obtained respectively, with 0≤x≤20, 0≤y≤20 and 0≤z≤20. According to the expressions of Y and Z expressed by X, the range of X is determined. Substitute the expressions of y and z into 60x+60y+50z=a to get the expression of x represented by a. According to the value of x,

x+y+z = 20 13 x+2y+z = 50260 x+60y+50z = A3？

2x+y=30 from ②-①, that is, y=30-2x ④.

X-z= 10 is obtained from ②-①×2, that is, z=x- 10 ⑤.

∵0≤y≤20, that is, 0≤30-2x≤20, and the solution is 5 ≤ x ≤ 156.

Similarly, 0≤z≤20, that is, 0≤x- 10≤20, and the solution is 10≤x≤30 ⑦.

From ⑥ we know that 10≤x≤ 15.

Substituting ④ ⑤ into ③ gives a = 60x+60 (30-2x)+50 (x-10) =1300-10x? x= 130-a 10

∴ 10≤ 130-

a 10≤ 15？ 1 150≤a≤ 1200

So the answer is 1 150. Comments: This question examines the application of ternary linear equations. The key to solve this problem is to get the expressions of Y and Z represented by X according to the problem equation, and then get the value range of X according to 0≤x≤20, 0≤y≤20 and 0≤z≤20, and then determine the value range of A.