(2) First, a set of one-dimensional linear inequalities is obtained from the equivalence relation, then the solution set is obtained, and then the scheme is determined according to the actual meaning;
(3) Calculate the freight of each scheme separately, and then compare the results.
Solution: Solution: (1) If there are x tents packed into pieces, the number of food pieces is (x-80).
Then x+(x-80)=320 (or x-(320-x)=80)(2 points).
X=200, x-80= 120(3 points).
A: There are 200 tents and food packages 120. (3 points)
Method 2: There are X tents, and Y food is packaged into one.
rule
x+y=320x-y=80?
(2 points)
solve
x=200y= 120?
(3 points)
A: There are 200 tents and food packages 120; (3 points)
(Note: Full marks are also given in arithmetic. )
(2) If Z trucks of Class A are rented, then
40z+20(8-z)≥200 10z+20(8-z)≥ 120?
(4 points)
The solution is 2≤z≤4(5 points)
∴z=2 or 3 or 4, the Civil Affairs Bureau has three schemes when arranging A and B trucks.
The design scheme is as follows: ① there are 2 cars a and 6 cars b;
(2) There are 3 vehicles A and 5 vehicles B;
(3) A has 4 vehicles and B has 4 vehicles; (6 points)
(3) The freight rates of the three schemes are as follows:
①2×4000+6×3600=29600 (yuan);
②3×4000+5×3600=30000 yuan;
③4×4000+4×3600=30400 yuan.
∫ Scheme 1 is less than Scheme 2 and less than Scheme 3,
Option ① has the least freight, and the lowest freight is 29,600 yuan.
(Note: The nature of the linear function indicates that Scheme ① will not be deducted at least. )