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Analysis: (1) has two equivalent relationships: tent number+food number =320, tent number-food number =80, directly set the unknown number, list the binary linear equations and solve them;

(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. )